Integrals
The three ideas behind every exact amount:
the rectangles, the area under the curve, and the limit that closes the gap between them.
Start hereWelcome, and how to read this book
This book teaches the integral: the half of calculus that builds a whole back up from infinitely many small pieces.
Picture a car on a straight highway, an hour into a road trip. Two gauges sit side by side on the dashboard. The speedometer keeps no memory: it answers one question, over and over, how fast right now? The odometer could not be more different. It remembers everything. Every mile since the trip began is piled into one number that only grows. It is a running total.
Here is the puzzle. The speedometer needle never held still for a second; it drifted up on the open road, sank in traffic, dipped for every turn. So how does the odometer add up a speed that was different at every instant? You cannot just multiply one speed by the time, because there was no one speed. Slice the hour into tiny stretches instead. Over a stretch of a few seconds the speed barely changes, so distance is very nearly speed times that little bit of time, a thin rectangle of area under the speed graph. Add up all the rectangles. Then slice finer, and finer, and let the rectangles shrink toward zero width. The running total they close in on is the exact distance. That limit of a sum, pushed all the way, is the integral. This book is the story of that sum: what it means, the rules that make it fast, and the shapes and totals it builds.
The speedometer asks the opposite question: how fast is the total changing at this one instant? That is the other half of calculus, and it has its own book. The two questions turn out to be perfect mirrors of each other, which is one of the best surprises in all of math. You meet the mirror by name in Section 1.4, the Fundamental Theorem of Calculus, where the two books shake hands.
You do not need any calculus experience to read this book. You need algebra, a few function facts, and curiosity. Everything else is built here, from the ground up. If a prerequisite is rusty, Appendix A is a friendly refresher you can visit any time: the unit circle, trig, logarithms, exponentials, inverse functions, the geometry of area and volume, and a little physics and probability.
The structure repeats so you always know where you are:
Picture This (a plain-language hook), then Build the Intuition (a picture before the symbols), then Definition (the precise version), then Worked Examples, then Why This Matters in a World That Moves, then Dig In (a rigorous stretch for everyone). An optional Dig Deeper follows for anyone who wants to push further into the beauty of the math. Then Exercises (with solutions you can reveal when ready). Each unit closes with Mixed Practice that pulls the whole unit together, then a Recap and Self-Check to test yourself objective by objective.
Two students can read the same section and both be challenged. If a topic is new, follow the core path; the Dig In box gives everyone a rigorous stretch. The boxes marked Dig Deeper are there whenever you want to push further: they hold the proofs, the connections to later courses, and the farther reaches of each idea, open to anyone curious enough to open them. Nothing in the core path depends on them, so no one is ever left behind and no one is ever held back.
Twelve sections also include a hands-on widget you can drag and manipulate right on the page, no download or internet needed: 1.1, 1.3, 1.4, 1.5, 2.1, 2.2, 3.1, 3.2, 3.4, 3.5, 3.6, and 3.7. Look for the ▶ mark in the Contents to find them.
A few more reading tools, available everywhere in the book: every underlined word shows its meaning on hover, every section has a Listen button to hear it read aloud (every formula included), and every exercise has a Hint you can open before you reveal the full solution. Try one right here:
Hint
This is exactly what a hint looks like inside an exercise: click to open it, click again to close it. A real hint nudges you toward the approach, it never hands you the final answer.
The top bar holds the rest of your reading tools: Aa sets text size and line spacing, Night flips the book to a dark theme that is easy on late-night eyes, and ↓ Print / PDF prints the whole book, one section, or just the problems, with solutions hidden, shown inline, or gathered at the end. To search the whole book, use the Search box in the sidebar, or your browser's own find-on-page (Ctrl+F, or Cmd+F on a Mac).
Three study banks travel with the book: the Glossary and Rules & Theorems in the back matter, and Notation & symbols just below in the front matter. Search any of them, filter by unit, or flip one into flashcards to quiz yourself.
Each unit ends with a Recap & Self-Check where you tick Got it for every objective, and a small nudge brings a finished unit back a few days later, because returning is what makes it stick. Those checkmarks live only in this browser on this device; clearing browsing data or switching devices resets them. Nothing in this book reports to your teacher. Solutions are open on purpose: check yourself after an honest attempt.
Everything in this book works offline: open it once with internet and it keeps working without one, and you can also save this page itself for a copy that never needs a connection. The one exception is , up in the top bar (and, while this page is in view, right here too): a chat you can reach any time you are stuck, not just during class. It gives hints, never the answer. Chats are saved, and your teacher can read them; the TA is fair game for homework. Ask your teacher for the class passphrase.
Quick questions, quick answers
Does the book work offline?
Yes, after one visit with internet; the note just above has the details and the one exception.
Does it work on my phone?
Yes. The whole book, widgets included, runs in your phone's browser, and the layout reshapes itself for a small screen.
Can I use the Ask the TA chat for homework, and who sees my chats?
Yes, you can. The note just above is the whole policy: the TA gives hints, never the answer, chats are saved, and your teacher can read them.
What is the best way to study for a test?
Work the unit's Recap & Self-Check: hit Try it! on every goal cold, check yourself against the answer, and follow the Review link back into any section that wobbled.
Do I need a calculator?
Give the exact value first, the way every answer in this book does, and reach for a calculator only where a problem asks for a decimal. When one does, the book rounds with to three decimal places.
I think I found a mistake in the book.
Tell your teacher. A sharp catch helps every reader after you.
I missed a class. How do I catch up?
Read the matching section, work its Try it, then hit that goal's Recap item to confirm it stuck. The Ask the TA chat is there if you get stuck on the way.
Where do the formula sheet and the method guide live?
Both are in the back matter: the Formula Sheet is the one-page test-legal reference, and Which Tool? Read the Shape walks the how-do-I-choose decision. The Print / PDF dialog can print the formula sheet on its own.
Jump straight to Section 1.1. The legend and notation tables just below are here for whenever you need to look something up.
Front matterThe legend: what every box and style means
This book uses color and shape on purpose, not for decoration. Here is the full key, so nothing is a mystery.
The goals box. One of these opens every unit, section, and appendix page (sometimes labeled "By the end of Unit..." or "Quick reference for") with a short bulleted list of what you are about to learn. Navy bar.
A friendly, jargon-free hook that opens a topic. Read it even if you feel confident. Filled soft sky-blue panel.
The precise meaning of a term. The word being defined is shown in bold navy. Berry bar. These are the sentences worth memorizing.
A fact you can rely on and reuse, often with a short reason. Steel-blue bar on a pale panel.
A fully worked example. The setup is stated, then the solution is worked step by step.
Solution. Steps appear under a dashed line like this one.
The throughline. Every one of these boxes shows where the idea lives in the moving world: speedometers and braking distances, cooling coffee, draining tanks, growing populations, and the instruments that measure change. Berry bar with a navy arrow tag.
A common mistake, stated plainly so you can avoid it. Amber bar (the book's one caution color).
A quick check you can do in your head or on scratch paper before moving on. Dashed berry outline.
A hands-on widget you can manipulate right on the page: drag a slider, switch between methods, and watch the rectangles, the area, or the solid respond. One appears in the section where it teaches the most. No download, no internet, just click and try it. Look for the ▶ mark next to a section in the Contents. Solid berry frame.
A rigorous stretch that every student can follow, no matter what math class you are in now: a short "why is this true," a pattern to generalize, or a proof you can do with algebra. Always shown, because everyone should reach for it. Berry bar.
Dig Deeper Optional higher reach▸
The high-ceiling material, for anyone wanting to challenge themselves even further: proofs, previews of later courses, and the deeper reasons behind the rules. Collapsed until you open it, and nothing in the core path depends on it, so reach for it whenever you are curious.
The three or four things to carry forward. Appears at the end of each section.
Exercise tags (teacher key only)
These tags appear in your Teacher Solutions Key, not in the student book: they tell you what to assign. Every exercise carries one. The first three say how hard it is; the last two say what it connects to. None of them is a warning. The one caution color in the book is amber, and it is used only for the Watch Out box.
Warm up a gentle starter to get moving. Core the standard practice everyone should do. Stretch a harder challenge to push you. Dig In builds on that section's Dig In (rigor for everyone). Dig Deeper builds on that section's Dig Deeper (an optional higher reach).
Typographic conventions
| Style | Meaning |
|---|---|
| Bold navy | A new technical term, at the moment it is defined. |
| Italic | Emphasis, or a term borrowed from everyday language. |
| Berry link | A cross-reference or an outside resource. |
| Math set in a serif | All symbols and equations are typeset, for example . |
Abbreviations, spelled out
| Short | Full form |
|---|---|
| ASTC | All Students Take Calculus (which trig functions are positive in quadrants I, II, III, IV; reviewed in Appendix A.1) |
| iff | If and only if |
| PEMDAS | Parentheses, Exponents, Multiplication/Division, Addition/Subtraction (order of operations) |
A note on dashes: this book uses commas, colons, and parentheses instead of long dashes on purpose, so a dash is never confused with a minus sign.
Front matterNotation & symbols
A single place to look up any symbol. It grows as the book does; here is the core set. The full, searchable version lives in the Glossary. Search it, filter it by unit, or flip it into flashcards to quiz yourself.
"the sum, for i from 1 to n, of a sub i"
sigma notation: run the counter from the bottom value to the top value and add every term along the way. A Riemann sum is written
Unit 1Section 1.1
"delta x, the strip width"
the common width of the strips in a regular partition, for strips across ; with strips there are cut points but only gaps
Unit 1Section 1.1
"the integral of f of x, d x"
the indefinite integral, or general antiderivative, of : the tall S is the integral sign, is the integrand, and the names the variable the derivative is being undone in. It always equals
Unit 1Section 1.2
"plus C"
the constant of integration attached to every indefinite integral: it marks the general antiderivative as a whole family rather than a single function, and an initial value pins it to one number
Unit 1Section 1.2
"the integral from a to b of f of x, d x"
the definite integral: is the integral sign (an elongated S, for sum), and are the limits of integration, is the integrand, and names the variable of integration and marks the vanishing strip width
Unit 1Section 1.3
"F of x, evaluated from a to b"
the evaluation bar: shorthand for , the change in an antiderivative between the two limits, which by FTC Part 2 is the value of
Unit 1Section 1.4
"g of x equals the integral from a to x of f of t, d t"
the accumulation function: a definite integral with a fixed left end and a moving right end , so its value depends on ; the inside variable is a dummy, and
Unit 1Section 1.4
"the integral from a to b of the absolute value of v of t, d t"
the total distance traveled: the integral of the speed, the absolute value of velocity. The absolute value sits inside the integral, on the velocity, so backward stretches add to the distance instead of subtracting; it is not the same as the absolute value of
Unit 1Section 1.5
"f average"
the average value of on an interval : , the height of the rectangle whose area matches the area under the curve
Unit 2Section 2.1
"u equals g of x, and d u equals g prime of x, d x"
the inside function's working name and its differential: naming the inside turns a composition into , while is the piece of the integrand that trades for . is scaffolding, and finished answers rename back in.
Unit 2Section 2.2
"arc sine"
the inverse sine function: is the angle in whose sine is . It is the antiderivative form of , appearing as
Unit 2Section 2.3
"arc tangent"
the inverse tangent function: is the angle in whose tangent is . It is the antiderivative form of , appearing as
Unit 2Section 2.3
"the natural logarithm of the absolute value of x"
the antiderivative of : the absolute value extends the natural logarithm to negative inputs, so is defined wherever is, and its derivative is on each side of zero
Unit 2Section 2.4
"rho"
density: mass per unit length for a rod, , or mass per unit area for a plate, ; a rod's mass is
Unit 3Section 3.5
"x bar"
the center of mass (balance point) of a rod: , the mass-weighted average position
Unit 3Section 3.5
"the work, W"
work, the energy a force delivers as it acts through a distance; , in joules or foot-pounds.
Unit 3Section 3.6
"the spring constant, k"
the stiffness in Hooke's law , the force needed per unit of stretch.
Unit 3Section 3.6
"p of x"
a probability density function: the height of the density curve at , measured in probability per unit , so a probability appears only after integrating across an interval
Unit 3Section 3.7
"mu"
the mean, or expected value, of a continuous random variable: , the balance point of the density
Unit 3Section 3.7
"P of theta"
the point where the angle theta's terminal ray meets the unit circle, with coordinates (cosine theta, sine theta); the whole appendix is a table of this point at the landmark angles
Appendix AAppendix A.1
"sine of theta"
the y-coordinate of the point reached by sweeping angle around the unit circle: bounded between negative 1 and 1, period
Appendix AAppendix A.2
"cosine of theta"
the x-coordinate of the point reached by sweeping angle around the unit circle: bounded between negative 1 and 1, period , a quarter turn ahead of sine
Appendix AAppendix A.2
"tangent of theta"
the ratio , the slope of the radius to that point on the unit circle: unbounded, period , undefined wherever
Appendix AAppendix A.2
"secant of x"
the reciprocal of cosine, ; its square is the antiderivative fact , and clears an in a trigonometric substitution
Appendix AAppendix A.3
"cosecant of x"
the reciprocal of sine, , not the reciprocal of cosine despite the shared "co"; the letters cross over
Appendix AAppendix A.3
"cotangent of x"
the reciprocal of tangent, , equal to ; it integrates to in Section 2.3
Appendix AAppendix A.3
"the natural logarithm of x"
the exponent is raised to in order to give , the inverse of ; defined for , with and
Appendix AAppendix A.4
"log base b of x"
the base- logarithm, the exponent is raised to in order to give ; every base- log is a rescaled natural log, , the change-of-base formula
Appendix AAppendix A.4
"f inverse"
the inverse function of , the rule that reads each output back to its input; an inverse, not the reciprocal , and defined only when is one-to-one
Appendix AAppendix A.6
"r, the radius"
the radius of a circle, cylinder, cone, or sphere: the distance from the center out to the edge, and the input every area and volume formula on this page takes; if a problem gives a diameter, halve it to get r
Appendix AAppendix A.7
"the slant height"
a cone's slant height, the distance straight up its slanted side from the base rim to the apex; it is the long side of the right triangle formed with the radius and the vertical height, so , and the slanted surface uses it, not the height
Appendix AAppendix A.7
"P of A, the probability of event A"
the probability of the event A, a number from 0 to 1; for equally likely outcomes it equals the count of A over the count of S
Appendix AAppendix A.9
"S, the sample space"
the sample space, the set of all outcomes of an experiment; the count of S is how many outcomes it holds
Appendix AAppendix A.9
"E of X, the expected value of X"
the expected value, or mean, of the random variable X: the weighted sum of its values, each times its probability; the continuous version in Section 3.7 is the mean mu, the integral of x times p of x
Appendix AAppendix A.9
Front matter
Full table of contents
One place to jump anywhere. Open a unit to see its sections; every line is a link. Each unit ends with a Mixed Practice set that blends every idea from the unit, and then a Recap & Self-Check (try each goal, reveal an example, or jump back).
Front matterhow to read this book▸
Unit 1 · Foundations of Integralsfrom a stack of rectangles to one exact total▸
- 1.0Unit 1 Opener & Objectives
- 1.1Approximating Area (includes an interactive widget)
- 1.2Antiderivatives
- 1.3The Definite Integral (includes an interactive widget)
- 1.4The Fundamental Theorem of Calculus (includes an interactive widget)
- 1.5The Net Change Theorem (includes an interactive widget)
- 1.PMixed Practice: Unit 1
- 1.RUnit 1 Recap: Check Yourself
Unit 2 · Integration Rulesthe toolkit that reaches every function▸
Unit 3 · Applications of Integralsthe shapes and totals of the moving world▸
- 3.0Unit 3 Opener & Objectives
- 3.1Arc Length (includes an interactive widget)
- 3.2Areas between Curves (includes an interactive widget)
- 3.3Surface Area of a Revolution
- 3.4Volume Methods (includes an interactive widget)
- 3.5Mass and Density (includes an interactive widget)
- 3.6Force and Work (includes an interactive widget)
- 3.7Probability (includes an interactive widget)
- 3.PMixed Practice: Unit 3
- 3.RUnit 3 Recap: Check Yourself
Appendix A · Prerequisite Toolkitreference any time▸
Foundations of Integrals: from a stack of rectangles to one exact total
Your dashboard already knows the two ways to measure motion. The speedometer names this instant and keeps no memory; the odometer remembers the whole trip and pours every mile into one growing total. This unit builds the mathematics that runs the odometer: how to add up a rate that was different at every instant. It starts by estimating an area under a curve with rectangles, sharpens the estimate into an exact limit, and lands on the definite integral. Then the Fundamental Theorem of Calculus shakes hands with the Derivatives volume, and the whole trip's total falls out of a single antiderivative.
- Estimate the area under a curve with left, right, and midpoint rectangles, write the estimate in sigma notation, and sharpen it by using more, thinner rectangles.
- Reverse the derivative rules to find an antiderivative, write the general antiderivative and say why the is there, and pin down from a starting value.
- Read as the limit of those rectangle sums, interpret it as net signed area, and use the basic properties of the definite integral.
- Evaluate a definite integral in one line with the Fundamental Theorem of Calculus, , and differentiate an accumulation function.
- Read the integral of a rate as the net change it produces, and tell net displacement apart from total distance traveled.
More than two thousand years ago, Archimedes wanted the exact area of a circle. He trapped it between two ladders of straight-sided polygons, one drawn just inside the circle and one just outside, and kept doubling their sides. The two areas squeezed toward each other, and the number caught between them was the circle's true area. He had no formula for the limit, but he had the idea that runs this entire unit: chop a curved thing into many straight pieces, add them up, and let the pieces shrink toward nothing. The Derivatives volume ended by handing you a question from the dashboard, how do all those separate instants pile back into one total, and this is the tool that answers it. The surprise, discovered by Newton and Leibniz two thousand years after Archimedes, is that this ancient area problem and the tangent-line problem from the other volume are not two subjects at all. They are exact opposites, inverses of each other, and the sentence that says so is the Fundamental Theorem of Calculus. This unit walks that road in five sections. Section 1.1 starts with Archimedes' rectangles. Section 1.2 learns to run a derivative backward. Section 1.3 pushes the rectangles to their limit and the definite integral appears. Then Section 1.4 ties the two together in one theorem, and Section 1.5 reads any total straight off its rate.
Section 1.1
Approximating Area
- Estimate the area under a curve by lining up rectangles and adding their areas, and say why the estimate is only ever a start.
- Build left, right, and midpoint sums, and read off which one runs high and which runs low for a curve that only rises or only falls.
- Write a long sum compactly in sigma notation, with the strip width and the sample points named.
- Sharpen an estimate by using more, thinner rectangles, and watch the left and right sums close in on one number.
- Read the trapezoid sum as the average of the left and right sums, and see why it, and the midpoint sum, beat both ends.
Back on the dashboard from Welcome. You have been driving for forty minutes, and the speed needle never held still: it climbed on the on-ramp, sagged in traffic, pushed again on the open road. When you park, the odometer shows exactly how far you went, one clean total. Here is the quiet fact this whole book turns on: that total is the area under your speed-versus-time graph. Fast stretches draw tall, slow stretches draw short, and the odometer has been adding up their area the entire drive. The catch is that you cannot watch a graph while you steer. You glanced at the speedometer a few times, and that is all you have: a handful of speeds at a handful of moments. Can you rebuild the odometer's total from those glances alone? Not exactly. But you can box it in, and boxing it in is where this book starts.
Build the intuition
Slice the drive into short time blocks and pretend the speed holds steady across each one. Steady speed times block length is a distance, and a distance under a speed graph is the area of a rectangle. Line the rectangles up under the curve and their areas add to an estimate of the whole trip. The one choice left is which speed to use for each block, and there is no single right answer. Use the speed at the START of each block, and on a stretch where you are speeding up every rectangle comes out a little too short, so the estimate lands low. Use the speed at the END of each block and every rectangle runs a little too tall, so the estimate lands high. The true total is caught between the two. Figure 1.1.1 shows both, not on the drive but on a curve we can check by hand, from 0 to 2.
Figure 1.1.1 Four rectangles under on , drawn to their true heights. Read at the left edges they total , an under-estimate; read at the right edges they total , an over-estimate. The true area, , is trapped between them.
Adding four rectangle areas is easy to write out. Adding four hundred is not, so we need shorthand. The capital Greek letter (sigma) means "add these up": the expression says start a counter at 1, run it up to , and add every term along the way. The small numbers under and over the sigma are where the counter starts and stops. With that one symbol, the estimate we just drew fits on a line.
To estimate the area under from to , cut the interval into strips of equal width. A regular partition chops into equal pieces, each of width
with cut points , , and in general , up to . In each strip pick one input, a sample point , and stand a rectangle there of width and height . Adding all the rectangle areas gives a Riemann sum,
an estimate of the area under the curve that sharpens as the strips get thinner.
Three natural places to sample each strip name three sums. Sample the left edge and you get the left sum; sample the right edge, the right sum; sample the middle, the midpoint sum:
A fourth needs no new picture: the trapezoid sum is just the average of the two end sums,
Every one uses the same strips of the same width ; only the height changes with where you sample. That is the whole reason there is more than one answer.
If only rises across , each strip's shortest edge is on the left and its tallest is on the right, so the left rectangles all fall short and the right rectangles all overshoot:
If only falls, the roles swap: now is the under-estimate and the over-estimate. Reason: on a rising curve you sample low on the left and high on the right, and on a falling curve it is the reverse. A curve that turns around inside the interval gets neither guarantee; split it where it turns and treat each piece on its own.
Estimate the area under from to with four rectangles, first reading heights at the left endpoints () and then at the right endpoints (). The true area is (Section 1.4 will hand it to you in one line); check that your two estimates trap it.
| Strip | Interval | Left height | Right height |
|---|---|---|---|
| 1 | 0.00 | 0.25 | |
| 2 | 0.25 | 1.00 | |
| 3 | 1.00 | 2.25 | |
| 4 | 2.25 | 4.00 |
Solution. The width is , and the cut points are . The left sum adds the left-edge heights (the first four in Table 1.1.1), the right sum the right-edge heights (the last four):
Since only rises on , the left sum falls short and the right sum overshoots, exactly as the rule promised: . The true area sits inside a window of width , and is comfortably in it. That window is still wide; the rest of the section is about shrinking it.
Predict first: if you used only two rectangles instead of four, would the gap between the left and right estimates get wider or narrower? Then compute and for the same on and check your prediction against and .
Show solution
Wider. and , a gap of 4, versus the gap of 2 with four rectangles.
With , and the cut points are . Then and . Fewer, fatter rectangles miss more of the curve, so the estimates spread apart. Thinner strips pull them back together, which is the whole strategy from here.
Merging onto a highway, you note the car's speed every 10 seconds for 40 seconds. Table 1.1.2 holds the readings, in feet per second, and the car only speeds up. (a) Estimate the distance covered with a left sum. (b) Estimate it with a right sum. (c) The odometer's true gain is one number; say what your two estimates tell you about it.
| Time (seconds) | Speed (feet per second) |
|---|---|
| 0 | 44 |
| 10 | 52 |
| 20 | 60 |
| 30 | 64 |
| 40 | 66 |
Solution. Each block is seconds wide, and the heights are the speeds. A speed times a time is a distance, so the sums come out in feet.
(a) The left sum reads speed at the start of each block (the first four readings):
(b) The right sum reads speed at the end of each block (the last four readings):
(c) The car only speeds up, so the left sum runs low and the right sum runs high: the odometer added somewhere between 2200 and 2420 feet. If you want one number, the trapezoid sum splits the difference, feet. You never measured distance directly; you measured speed a few times and let the rectangles rebuild the odometer's total, which is the area under the speed curve.
Back to on , where the true area is . The left and right sums bracket it, but from far away. Two cheap fixes do much better: sample each strip at its midpoint (the midpoint sum ), or just average the two end sums (the trapezoid sum ). (a) Compute and . (b) Watch all four sums as grows.
Solution. (a) With four strips of width , the midpoints are :
Both straddle : the midpoint sum lands just under, the trapezoid just over, and each is off by a small fraction of what the plain left and right sums missed by.
(b) Table 1.1.3 runs all four sums out to sixteen strips.
| Left | Right | Midpoint | Trapezoid | |
|---|---|---|---|---|
| 4 | 1.750 | 3.750 | 2.625 | 2.750 |
| 8 | 2.188 | 3.188 | 2.656 | 2.688 |
| 16 | 2.422 | 2.922 | 2.664 | 2.672 |
Read down the columns. The left and right sums crawl toward one slow step at a time, still wrong in the first decimal at sixteen rectangles. The midpoint and trapezoid sums are already right to two decimals with just four. Sampling the middle cancels most of each strip's overshoot against its undershoot, and averaging the ends does nearly the same thing for free. Every column heads for the same place, ; some columns just get there far faster.
A river does not come with an odometer, but the people who manage a dam need one. A stream gauge on the bank reports the river's flow, in cubic feet per second, every fifteen minutes. Nobody can measure the total water that passed in a day directly; they only have the flow at those sampled moments. So they do exactly what this section does: multiply each flow reading by its fifteen-minute block, add the pieces, and read the running total off a Riemann sum. That total decides how much to hold back and when to warn towns downstream. A storm that spikes the flow gets the gauge sampling faster, because thinner strips track a fast-changing rate more tightly, the same reason four midpoint rectangles beat sixteen crude ones above. The odometer under your dash and the flood gauge on the river run the identical machine: a rate you can only glance at, added up into a total you really need.
A table can hint that the sums close on , but it can never prove a limit. Algebra can pin one down exactly. Take the cleanest case, the area under from 0 to 1, and write the left sum for a general . With and cut points , reading heights at the left edges gives
The term is 0, so the sum runs from 1 to . Here is the fact that cracks it open, worth stating because we will lean on it again: the first perfect squares add up to
which you can check for small and prove by induction. Applied to the sum from 1 to (put in place of ), it gives , so
No rectangle ever hits ; every left sum sits a little below it. But as grows the last two terms melt away, so closes on as tightly as you like, and is the exact area under on . Hold on to this. "The number a Riemann sum closes in on" is about to stop being a hunch and become the definition of the integral in Section 1.3.
Dig Deeper The squeeze that guarantees an answer▸
Why should a Riemann sum close in on anything at all? For a curve that only rises, there is an honest reason, and it needs no new machinery. On with increasing, the left sum is built from each strip's shortest height and the right sum from each strip's tallest, so the true area is trapped between them: . Now measure the gap. Line the two sums up over the same cut points and subtract; every interior height cancels, because the right edge of one strip is the left edge of the next, and only the two outermost heights survive:
The bracket is a fixed number, so the whole gap is a constant divided by . Raise and the gap shrinks toward 0, clamping and , and every sum caught between them, onto a single value. That value is the area. The only limit fact used is that a constant over goes to 0, and that is the point: the estimates have to converge because the room between them runs out. On over , the gap is , which matches the window of width 2 we saw at . Section 1.3 gives this squeeze its formal name and makes the shared value the definition of the definite integral.
Two slips to name. First, direction. For a curve that only rises, the left sum under-estimates and the right sum over-estimates; for a curve that only falls, it is the reverse. Do not memorize "left is small," read the slope: the edge that sits lower builds the shorter rectangles. On the braking car in Exercise 4 the speed falls, so the left sum is the high one. Second, the width. With strips across , each is wide, not . There are cut points but only gaps between them, and it is the gaps you are adding. Count fence posts instead of fence panels and you will build one strip too few.
Let on , a curve that only falls. (a) Find and , and say which one over-estimates the area. (b) Find the trapezoid sum . The exact area is (Section 1.4 will produce it directly); which of your three estimates is closest?
Hint
With two strips on , and the cut points are ; build , , and exactly as in Example 1.1.1, but remember is falling, so the monotone rule from Section 1.1 flips which end runs high.
Show solution
(a) and ; over-estimates. (b) , the closest of the three to .
(a) The heights are , , . Then and . Because falls, the left edges are the tall ones, so is the over-estimate and the under-estimate: .
(b) . Comparing the three to , the trapezoid is off by about 0.333, while and are off by more than 1.6 each. Averaging the ends cancels most of the error, the same payoff Example 1.1.3 showed.
Slide to pack in thinner strips, switch between the Left, Right, Midpoint, and Trapezoid rules, and watch your sum close the gap to the true area under on .
Frozen at midpoint strips under on : the sum already sits just under the true area .
Exercises 1.1
1. Let , a straight line. Using four rectangles on (so ): (a) find the left sum ; (b) find the right sum ; (c) find the trapezoid sum , and compare it with the exact area under the line, a triangle of area . Warm up
Hint 1
The heights are at the cut points ; the left sum uses the first four, the right sum the last four, and the trapezoid sum is their average (Section 1.1).
Show solution
(a) . (b) . (c) , exactly the triangle's area.
The values of at are .
(a) .
(b) .
(c) . For a straight line the trapezoid sum is exact, because a trapezoid's slanted top matches a line's slant perfectly. The left and right sums still miss, one low and one high, but their average lands dead on.
2. You want to estimate the area under a curve on with six rectangles of equal width. (a) Find . (b) List the seven cut points through . (c) Which of those points are the sample points for the right sum ? Warm up
Hint 1
Use with , , , not (the Section 1.1 Watch Out). Step up from by to list the cut points; the right sum samples the right edge of each strip.
Show solution
(a) . (b) . (c) The six right edges: .
(a) .
(b) Starting at 2 and stepping up by : . That is seven points, marking the six strips.
(c) The right sum reads the right end of each strip, which is every cut point except the first: . (The left sum would use every point except the last.)
3. Estimate the area under from to with four rectangles. (a) Find . (b) Find . (c) The exact area is . Confirm your two sums trap it, and find the gap . Core
Hint 1
Here and the cut points are : the same machine as Example 1.1.1 on a new interval.
Hint 2
and . The gap simplifies to .
Show solution
(a) . (b) . (c) They trap ; the gap is .
The heights at are .
(a) .
(b) .
(c) Since rises on , , and is inside. The gap is , which equals , the shortcut from the Dig Deeper box.
4. A car brakes to a stop. Its speed, in feet per second, is logged every 2 seconds: 64 at , then 48, 32, 16, and 0 at . (a) Estimate the stopping distance with a left sum. (b) Estimate it with a right sum. (c) The speed only falls, so one sum runs high and one low; say which is which, and give the trapezoid estimate. Core
Hint 1
The heights are the speeds and the width is seconds; a speed times a time summed is a distance (Example 1.1.2). A falling speed flips the usual over/under, per the Section 1.1 Watch Out.
Hint 2
uses the first four speeds, the last four. Because the speed drops, the left edges are the tall ones, so is the larger estimate.
Show solution
(a) feet. (b) feet. (c) over-estimates and under-estimates; the trapezoid gives feet.
(a) feet.
(b) feet.
(c) The speed falls the whole way, so the left rectangles are too tall and the right ones too short: the true distance is between 192 and 320 feet, with high and low. The trapezoid sum is feet. Because the speed drops in a straight line here, 256 feet is in fact the exact stopping distance.
5. Sigma notation makes a long sum short. For on with four strips (, right sample points ): (a) write the right sum in the form and evaluate it; (b) find the left sum ; (c) the exact area is 4. Do your sums trap it? Core
Hint 1
The right sample points are for , that is ; plug each into and multiply by (the sigma form from Section 1.1).
Hint 2
. For the left sum, use to instead, or subtract from .
Show solution
(a) . (b) . (c) Yes: .
(a)
(b) The left sum drops the tall right-end rectangle and adds a zero-height one at the start: .
(c) Since rises on , : the sums trap the true area 4, though loosely, since four strips is coarse.
6. The midpoint sum often beats both ends. Estimate the area under from 0 to 3 with three rectangles, sampling each at its midpoint (). (a) List the three midpoints. (b) Compute . (c) The exact area is 9. How far off is your estimate? Core
Hint 1
Here , so the strips are , , ; each midpoint sits halfway across its strip (the midpoint sum from Section 1.1).
Hint 2
The midpoints are , so .
Show solution
(a) . (b) . (c) Off by 0.25, on the low side.
(a) The strips , , have midpoints .
(b) The heights there are , so .
(c) The true area is 9, so is off by . Three midpoint rectangles land within a quarter of the exact area, tighter than a three-strip left or right sum could manage.
7. Water flows into a reservoir, and a gauge logs the inflow rate in liters per minute every 5 minutes: 2 at , then 5, 9, 12, and 14 at . The rate only rises. (a) Estimate the total water added over the 20 minutes with a left sum. (b) Estimate it with a right sum. (c) Bracket the true total, and explain in one sentence why a table alone cannot give it exactly. Stretch
Hint 1
An inflow rate times a time block is a volume, summed just like distance in Example 1.1.2; the width is minutes and the heights are the rates. A rising rate means left under, right over (Section 1.1).
Hint 2
and . The true total is caught between them.
Show solution
(a) liters. (b) liters. (c) The true total is between 140 and 200 liters; a table gives no rule for the rate between readings, so the strips cannot be made any thinner.
(a) liters.
(b) liters.
(c) The rate only rises, so liters; the trapezoid sum liters is a fair middle guess. Between the five-minute marks the gauge says nothing, so you cannot refine the estimate past the data. Sometimes all you have is the rectangles, and the bracket is the honest answer.
8. For on , the right and left sums differ by . (a) Show this gap equals . (b) How many rectangles force the gap down to or less? (c) With four rectangles the midpoint sum was already off by only about 0.042. In one sentence, say what that comparison tells you about which method to trust. Stretch
Hint 1
Use and , and multiply (the gap formula comes from the Section 1.1 Dig Deeper). For (b), set the gap at most and solve for .
Hint 2
The gap is . Solving gives .
Show full solution
(a) . (b) . (c) The midpoint sum reaches with 4 rectangles an accuracy the raw left/right gap needs hundreds to match, so the midpoint sum (or the trapezoid) is the one to trust.
(a) and , so the gap is .
(b) Requiring means . At the left and right sums still disagree in the third decimal place.
(c) Four midpoint rectangles already pinned two decimals, while the raw left/right window needs 800 strips to close to the same width. Thinner strips are the brute-force road to accuracy; a smarter sample point is the cheap one.
9. The Section 1.1 Dig In found the left sum for on closes on . Do the same from the other side. (a) Using , show the right sum is and that . (b) Evaluate and the Dig In's exactly, and confirm they trap . (c) Pick your own upper limit and show the right sum for on closes on . Dig In
Hint 1
On , and the right sample points are , so the right sum is (mirror the Section 1.1 Dig In). Pull the constants out of the sum before using the squares formula.
Hint 2
. Simplify, then let grow so the pieces vanish.
Show full solution
(a) . (b) and , trapping . (c) The right sum on is .
(a)
As grows the last two terms vanish, so , meeting the left sum at the same target from above.
(b) , and the Dig In's formula gives . Since , the two sums straddle .
(c) On , and , so , which tends to .
Check your own version. For your chosen , compute as a decimal and confirm it sits just above , then check that sits closer still. If the sums do not head for , make sure you multiplied the height by the width , not by .
10. The Section 1.1 Dig Deeper squeezed a rising curve between its left and right sums, with gap . Make the collapse concrete on over . (a) Explain why, when you subtract from , every height cancels except and , and show the gap is . (b) How large must be to force the gap to or less? (c) Choose your own rising function on your own interval, write its gap formula, and find the that squeezes the gap under a tolerance you pick. Dig Deeper
Hint 1
Write and as sums over the same cut points: samples the right edges ( to ), the left edges ( to ). Line them up and see which heights survive the subtraction (the Section 1.1 Dig Deeper).
Hint 2
The overlap cancels, leaving . Here and .
Show full solution
(a) The gap is . (b) . (c) Design answer: the gap is , which any large enough drives below a chosen tolerance.
(a) Write and . The two sums share every height from to , so subtracting leaves only the ends: .
(b) Requiring gives .
Check your own version. For your function and interval, the gap must be , a fixed number over , so doubling must halve the gap. Test it: compute the gap at your and at and confirm the second is half the first. If it is not, check that your function really only rises on the interval, since a turn inside it breaks the clean cancellation.
The area under a curve is estimated by standing rectangles under it and adding their areas, a Riemann sum with strip width . Where you sample each strip gives it a name: the left sum and right sum bracket a curve that only rises (left low, right high; reversed for a falling curve), while the midpoint sum and the trapezoid sum (the average of the two ends) sit close to the true value with far fewer strips. More, thinner rectangles tighten the estimate: for a rising curve the gap between the end sums is , a constant over that shrinks to 0, so the sums close on one number they never quite land on. That number is the exact area, and it is what Section 1.3 makes the definition of the definite integral. Rectangles get you close but never exact, and two different roads reach the exact answer, one through undoing a derivative and one through this limit. They turn out to be the same road. The next stop, Section 1.2, learns to undo a derivative, no limit required.
Section 1.2
Antiderivatives
- Undo a derivative: reverse the power rule and let constant multiples and sums pass straight through.
- Write the general antiderivative , and explain why every antiderivative carries that .
- Read and write indefinite-integral notation .
- Solve an initial value problem: recover a position from a velocity and one known reading by pinning down .
In Section 1.1 you rebuilt a trip's distance from its speed by stacking rectangles under the speed graph, getting closer the more rectangles you used, but never landing exactly. Here is a completely different road to the same kind of answer, and this one needs no rectangles and no limit at all. Suppose you know your speed at every single instant of a drive. Can you recover your odometer reading, how far you have gone, at every instant too? You are asking for a function whose rate of change is the speed you were handed. In the Derivatives volume you turned a position into a speed by taking its derivative; this section runs that machine backward, from the speed back to the position. The whole job is reverse differentiation, and it is pure algebra.
Build the intuition
Running differentiation backward has a catch, and the catch is the heart of this section. Picture two cars on the same straight road, one starting at mile marker 3 and the other at mile marker 10, that then drive in perfect lockstep: the same speed as each other at every instant, the whole way. Their speedometers agree from start to finish. Their odometers never do: the second car stays exactly 7 miles ahead the entire trip. Speed alone cannot tell the two drives apart, because speed is about change, and a fixed 7 mile head start never changes. So when you rebuild a position from a speed, you recover it only up to an unknown constant, the head start the speed could never see. Pin that constant down and you have the one true trip; leave it free and you have the whole family of trips that share that speed. That free constant gets a plus sign and a name, , and the rest of this section is bookkeeping around it.
A function is an antiderivative of on an interval when for every in that interval. If is one antiderivative of , then adding any constant gives another, and the whole collection , with running over all real numbers, is the general antiderivative: the entire family of functions whose derivative is . The free number is the constant of integration.
If and are both antiderivatives of the same function on an interval, then is a constant. Two functions with the same derivative can differ by nothing more than an added number. The reason is short: their difference has derivative , and a function whose slope is all across an interval never rises and never falls, so it holds one fixed value. This is why a single letter captures the entire freedom in an antiderivative, and it is the fact the Fundamental Theorem leans on in Section 1.4.
Figure 1.2.1 Three antiderivatives of . The curves for are vertical shifts of one another, so at any input they climb at the same rate. The three short segments at are parallel, each with slope 2: the derivative cannot tell the curves apart, and the gap between them is the constant .
The indefinite integral of is its general antiderivative, written
where is any one antiderivative of . Read it as "the integral of of , d x." The tall S is the integral sign, is the integrand, and the names the variable you are undoing the derivative in. The is not decoration: it is the whole family that the left side stands for, so an indefinite integral written without it is only one member of the answer, not the answer.
To undo the power rule, raise the exponent by one and divide by the new exponent:
The one forbidden exponent is , because it would divide by . So is not ; that single case waits for the logarithm, and a refresher lives in Appendix A.4. Two helpers ride along, since differentiation has them and antidifferentiation is differentiation read backward: a constant multiple slides through the integral sign, , and a sum breaks into pieces, . Between them, every polynomial and every sum of powers is now within reach.
Find the general antiderivative of .
Solution. Take one term at a time, raising each exponent by one and dividing by the new exponent, and carry the two helpers along:
The constant term is , whose antiderivative is . One at the end covers the whole expression: you do not write a separate constant on each term, because their sum is still just one unknown constant.
Before you go on, predict the general antiderivative of , then differentiate your prediction to check it.
Show solution
.
Reverse the power rule on each term: and antidifferentiates to . Differentiate back to confirm: , which is .
The power rule only reads an expression that is already a sum of powers of , so a radical or a quotient has to be rewritten before you can undo it. Find each indefinite integral. (a) . (b) .
Solution. (a) Rewrite the radical and the quotient as powers: and . Then the rule applies to each term:
Dividing by is multiplying by , which turns into ; dividing by gives , and reads as .
(b) There is no rule for a quotient, so split it into two powers first: . Now integrate term by term:
An initial value problem gives you a function's rate of change together with one known value, and asks for the function itself. The rate of change fixes the general antiderivative up to ; the known value pins to a single number, selecting one function out of the whole family.
A cart rolls forward along a straight track past a sensor. Its velocity is feet per second, seconds after the sensor triggers, and at that moment the cart is feet past the sensor. (a) Recover the position function . (b) Where is the cart after 3 seconds?
Solution. (a) Velocity is the derivative of position, so position is an antiderivative of velocity:
This is the whole family of trips with that speed record. The reading is the head start that picks out the real one: setting gives , so , and
(b) Read the odometer at :
The velocity alone could never have told you the cart was 5 feet past the sensor to begin with; that is exactly the information the was holding a place for, and is what fills it in.
An airliner crossing an ocean cannot always see a satellite, so it also carries an inertial navigation system: a set of accelerometers that measure the plane's acceleration at every instant. Acceleration is the derivative of velocity, and velocity is the derivative of position, so the computer runs this section in reverse, twice. It antidifferentiates the measured acceleration to rebuild the velocity, then antidifferentiates that to rebuild the position, without a single glance outside the aircraft. Each step hands back a : the velocity is known only up to a constant, and so is the position. Those two constants are exactly what the crew loads before pushback, the plane's starting speed and its starting location, an initial value problem the size of an ocean. Tell the machine where it began, and pure reverse differentiation carries it the rest of the way.
Every antiderivative comes with a free check built in: differentiate your answer and see whether you land back on . If you do, the answer is right; if you do not, the derivative shows you which term slipped. Because the derivative of any constant is , the simply vanishes under the check, which is the very reason different constants can share one derivative. Try it on Example 1.2.1:
and is the we started with, whatever the value of . Now watch two correct answers coexist. Both and are antiderivatives of , since each differentiates to . They are not equal, yet neither is wrong: their difference is , a constant, exactly as the Same Derivative, Same Family theorem promises. Make the differentiate-back check a habit, and you will never have to wonder whether an antiderivative is correct. You can always settle it yourself in one line.
Dig Deeper Why the constant is the whole story▸
The Same Derivative, Same Family theorem said two antiderivatives of one function differ only by a constant. Here is the whole argument in two honest sentences. If and on an interval, then the difference has derivative at every point of that interval; and a function with zero slope all the way across an interval never climbs and never drops, so it sits at one fixed height, which is to say is a constant. That is the entire freedom in an antiderivative: not a stray anywhere, not a coefficient, just one number you are free to choose. The plus sign in is not hiding anything more.
The phrase "on an interval" is doing real work, though, and it is worth seeing why. Consider , which is defined for every except , so its domain is really two separate intervals, and , with a gap between them. An antiderivative is , because . On a split domain like this one, the antiderivative can carry a different constant on each piece: the function equal to for and for still has derivative everywhere it is defined. The two pieces never touch, so no single constant is forced across the gap. The "differ by a constant" rule holds on each interval on its own, not across a break in the domain, and that subtlety comes back to bite in Section 1.4 if you evaluate an integral straight through a point where the function blows up.
Three slips account for most lost points here. First, the missing : an indefinite integral names a whole family, so is not finished; the honest answer is . Second, the trap. The reverse power rule divides by , which is exactly when , so is not ; that lone case is not a power at all, and it waits for the logarithm (Appendix A.4 is the refresher). Third, there is no reverse product rule to guess at. You cannot antidifferentiate by "undoing the product rule"; nothing in this course's toolkit does that. Rewrite or expand into a sum of powers first (here, ), then reverse the power rule term by term.
(a) Find . (b) Rewrite, then integrate: . (c) An object moves with velocity meters per second and starts at meters. Find , then find its position at seconds.
Hint
Every part reverses the power rule from Section 1.2: raise each exponent by one and divide by the new exponent, one term at a time, and keep the . In (b), rewrite and as powers and first. In (c), antidifferentiate to get up to , then use to pin down .
Show solution
(a) . (b) . (c) , and meters.
(a) Term by term: , , and antidifferentiates to , so the integral is .
(b) Rewrite as :
(c) Antidifferentiate the velocity: . The start reading gives , so . Then meters.
Exercises 1.2
1. Find each general antiderivative. (a) . (b) . Warm up
Hint
Reverse the power rule from Section 1.2 on each term: raise the exponent by one, divide by the new exponent, and remember the . A plain is , whose antiderivative is .
Show solution
(a) . (b) .
(a) and , so the integral is . Differentiate back: .
(b) and antidifferentiates to , giving .
2. Let . (a) Is an antiderivative of ? (b) Is an antiderivative of ? (c) If both are, by what constant do they differ? Warm up
Hint
Use the differentiate-back check from the Section 1.2 Dig In: a function is an antiderivative of exactly when its derivative equals . For (c), subtract the two functions.
Show solution
(a) Yes. (b) Yes. (c) They differ by the constant .
(a) , so is an antiderivative.
(b) as well, so is too.
(c) , a constant with no left in it, exactly as the Same Derivative, Same Family theorem requires.
3. Rewrite first, then integrate. (a) . (b) . Core
Hint 1
Neither integrand is a sum of powers yet, so rewrite before you reverse the power rule (Section 1.2). A radical is a fractional power; a quotient by can be split term by term.
Hint 2
In (a), and . In (b), .
Show solution
(a) . (b) .
(a) With :
(b) Split the quotient: , so .
4. A drone rises straight up with velocity meters per second, and at it hovers meter above its launch pad. (a) Find the height function . (b) How high is the drone after 4 seconds? Core
Hint 1
Height is an antiderivative of velocity, so this is an initial value problem (Section 1.2): antidifferentiate to get , then use the known height to pin down .
Hint 2
. Set and use to find before answering (b).
Show solution
(a) . (b) meters.
(a) . Then , so . Check: .
(b) meters.
5. Find the function whose derivative is and which satisfies . Core
Hint 1
This is an initial value problem in disguise (Section 1.2): antidifferentiate to get the family , then use the one known value to select the right member.
Hint 2
. Substitute and set the result equal to to solve for .
Show solution
.
Antidifferentiate: . The condition gives , so and . Thus , and a quick check confirms with .
6. No reverse product rule exists, so expand first. Find each indefinite integral. (a) . (b) . Core
Hint 1
The Watch Out in Section 1.2 says it plainly: you cannot antidifferentiate a product as a product. Multiply each expression out into a sum of powers first.
Hint 2
and . Now the power rule applies term by term.
Show solution
(a) . (b) .
(a) Expand: . Then .
(b) Expand: . Then .
7. A ball is thrown straight up. Its acceleration is the constant feet per second squared (gravity), its velocity at launch is feet per second, and it leaves a hand feet off the ground. (a) Find . (b) Find the height . (c) How high is the ball after 1 second? Stretch
Hint 1
This takes two antidifferentiations, each an initial value problem (Section 1.2): acceleration antidifferentiates to velocity, and velocity antidifferentiates to position. Each step brings its own constant, pinned by its own starting value.
Hint 2
Start with and use to find . Then antidifferentiate that and use .
Show full solution
(a) . (b) . (c) feet.
(a) , and , so .
(b) , and , so .
(c) feet. Two unknown constants, and it took two known readings (the launch speed and the launch height) to nail them both down.
8. Find . Stretch
Hint 1
A quotient by a radical is not a sum of powers yet, so rewrite it (Section 1.2). Divide each term of the top by , remembering that dividing by a power subtracts exponents.
Hint 2
. Now reverse the power rule on each term.
Show solution
.
Rewrite by dividing each term by : . Then
and . Differentiating back returns , the rewritten integrand.
9. The Section 1.2 Dig In built a habit: differentiate an antiderivative to check it, and watch that different constants share one derivative. Put it to work. (a) A classmate claims . Use the differentiate-back check to decide whether the formula is right, and state what, if anything, is missing. (b) Write two different correct antiderivatives of , and confirm by differentiating that both work and that they differ only by a constant. (c) Make up your own , write two of its antiderivatives, and show the check. Dig In
Hint 1
Reread the Section 1.2 Dig In: the test for an antiderivative is that its derivative equals . For (a), differentiate the claimed answer and compare, then ask what an indefinite integral must always carry.
Hint 2
For (b), one antiderivative of is ; build a second by adding any nonzero constant, then differentiate each and subtract to see the constant that separates them.
Show full solution
(a) The formula is right but the is missing; the complete answer is . (b) For example and ; both differentiate to and differ by the constant . (c) Answers vary; see the check below.
(a) , which is the integrand, so the terms are correct. What is missing is the constant of integration: an indefinite integral is a whole family, so the finished answer is .
(b) and , so both are antiderivatives of . Their difference is , a constant, so they belong to the same family.
(c) One sample: for , both and work, since each differentiates to , and they differ by .
Check your own version. Differentiate each of your two antiderivatives: both must return your chosen exactly, and their difference must simplify to a plain number with no in it. If the difference still contains an , one of the two is not really an antiderivative of , and its derivative will show you the offending term.
10. The Section 1.2 Dig Deeper argued that "same derivative" forces "differ by a constant," but only on a single unbroken interval. Explore both halves. (a) Two antiderivatives of are written and . Simplify the second. Are the two the same function, or do they differ by a constant, and which? (b) In two sentences, explain why any two antiderivatives of one on an interval can differ only by a constant. (c) The function is defined for every . Give an antiderivative that equals at and equals at , and explain why that is possible here but impossible for a function defined on a single unbroken interval. Dig Deeper
Hint 1
Reread the Section 1.2 Dig Deeper. For (b), the argument runs through the difference and its derivative. For (c), the domain of splits into two separate intervals, so each piece can carry its own constant.
Hint 2
For (a), multiply out and compare with . For (c), start from the antiderivative and add a constant on the side and a possibly different constant on the side, choosing each to hit the required value.
Show full solution
(a) The second simplifies to ; the two are the same function, differing by . (b) See the two-sentence argument below. (c) for and for ; possible because the domain splits into two separate intervals.
(a) , which is the first form exactly. The two look different but are the identical function, so they differ by . Looking different is not the same as being different.
(b) If and both antidifferentiate on an interval, then there. A function with zero slope all across an interval never rises or falls, so it holds one fixed value, which means is a constant.
(c) Start from , whose derivative is . On the side add a constant to hit at : gives . On the side add a constant to hit at : gives , so . The antiderivative is for and for . This works only because is a gap in the domain, splitting it into two intervals that each carry their own constant; on a single unbroken interval the constant is forced to be the same throughout, so you could not set two separated values independently.
Check your own version. Differentiate your on each piece and confirm you get there. Then test the interval rule: if your is defined on one unbroken interval, any two of its antiderivatives must differ by a single constant everywhere, so subtract two of them and confirm the difference has no . If your has a gap in its domain, check that a different constant on each side still differentiates back to .
An antiderivative of is any function with , and because the derivative cannot see a constant, antiderivatives come in families: the general antiderivative is , written with the indefinite-integral sign as , and the is part of the answer every time. The main tool is the power rule run backward, for , with constant multiples and sums passing straight through, and the standing habit is to differentiate your answer back to check it. The one exponent the rule cannot touch is , the case , which is not a power at all and waits for the logarithm (Appendix A.4). And when a single known value rides along with the rate, you have an initial value problem: antidifferentiate to get the family, then use the known value to pin and recover the one function, the way a starting odometer reading turns a speed record back into a position. Next, Section 1.3 goes back to the area problem from 1.1, which wants a limit, not an undo; the section after that shows the two roads reach the same place.
Section 1.3
The Definite Integral
- Define the definite integral as the single number the Riemann sums close in on as the number of rectangles runs to infinity.
- Read every part of : the integral sign, the limits of integration, the integrand, and the .
- Interpret a definite integral as net signed area, count area below the axis as negative, and use for total area.
- Use the four working properties: zero width, reversing the limits, linearity, and additivity over adjacent intervals.
- Evaluate simple definite integrals by recognizing a familiar shape (a rectangle, a triangle, a semicircle).
In Section 1.1 you covered a stretch of road and estimated the distance with rectangles under the speed graph, then watched the estimate sharpen every time you sliced the trip into more pieces. The estimates never quite landed. The left sums came in low, the right sums came in high, and the true distance sat in the crack between them. Here is the move that closes the crack: do not stop slicing. Let the number of rectangles run off to infinity, and the low estimate and the high estimate get squeezed together until one number is left. That number is the exact area under the curve, and it earns a name and a symbol of its own. The odometer knew it the whole time: the total on the dial is exactly this number, the area under the speed graph, read off with no rectangles at all. This section gives that number its definition, its notation, and the handful of rules that make it easy to handle.
Build the intuition
Picture the region under from to , the same region you boxed in with rectangles in Section 1.1. Because climbs as you move right, a right rectangle reaches up to the tallest point of its strip and pokes out above the curve, so the right sum always overshoots. A left rectangle reaches only to the shortest point and tucks under, so the left sum always undershoots. The exact area is trapped between them. Figure 1.3.1 draws the four right rectangles at : they overshoot with a total of , while the smooth shaded region below them is the true area. Now imagine the width of each strip shrinking toward nothing. The overshoot of every rectangle shrinks with it, the gap between and closes, and both sums settle onto a single value. That value is what we are about to define.
Figure 1.3.1 The region under from 0 to 2, with the four right rectangles from Section 1.1 (each of width 0.5, with heights 0.25, 1, 2.25, and 4). They overshoot: their total sits above the exact area. Slice finer and the rectangles settle onto the shaded region, whose area is .
Let be a function on the closed interval . Cut into strips of equal width , choose a sample point in each strip (the star is a reminder that it can sit anywhere in the strip, the of Section 1.1), and add up the rectangle areas to form a Riemann sum. The definite integral of from to is the number these Riemann sums close in on as runs to infinity:
When this limit exists and gives the same number for every way of choosing the sample points, is integrable on . The numbers and are the limits of integration, and is the integrand.
If is continuous on a closed interval , then is integrable on : the Riemann sums close on one number no matter how you pick the sample points. The reason is the squeeze from Section 1.1: for a continuous the sum built from each strip's shortest height and the sum built from its tallest height trap the true area between them, and their gap shrinks to zero as the strips get thinner, so there is nowhere for the limit to go but the value they both approach. Every function in this book is continuous on the intervals we integrate over, so the integral always exists. We will take this as a fact rather than prove it here.
Reading the notation
The symbol is an elongated letter S, chosen by Leibniz to stand for "sum," because a definite integral is exactly a sum of infinitely many vanishing strips. The whole expression is read aloud as "the integral from to of of , ." Table 1.3.1 names each part. Hold on to the : it does two jobs at once. It tells you the variable you are integrating with respect to, and it is the leftover trace of the strip width after the limit has shrunk it to nothing.
| Part | Name | What it does |
|---|---|---|
| integral sign | an elongated S, for "sum": it stacks up the strip areas | |
| and | limits of integration | start at , stop at ; sits low, sits high |
| integrand | the height of the strip at position | |
| variable of integration | names as the variable, and marks the vanishing strip width |
One warning that saves confusion later: the letter inside is just a placeholder. Whether you write or , you get the same number, because both describe the same area under the same curve between the same two limits. The variable of integration is a dummy variable; renaming it changes nothing.
For integrable functions and and constants and , the definite integral obeys four working rules that come straight from the fact that it is built out of sums:
Zero width and reversing are bookkeeping. Linearity says a constant slides out front and a sum splits into two integrals, exactly as it does for the Riemann sums inside. Additivity says two side-by-side regions glue into one, and it holds for any ordering of the three points once you accept the sign rule.
Use Definition 1.3.1 to compute exactly, taking right endpoints for the sample points. You will need the sum-of-first--squares formula, .
Solution. Split into equal strips. The width is , and the right endpoint of strip is . Build the right sum:
Now feed in the sum-of-squares formula and simplify:
The last form does the heavy lifting. As , the two correction terms and shrink to nothing, so
This is the number the rectangles in Section 1.1 were closing on. Check it against that section: at this formula gives , the right-hand estimate from before, and the left estimate sat just as far below. Both were reaching for , and the limit is where they finally meet.
Example 1.3.1 found for . Predict first, then check. (a) Evaluate at (two strips, right endpoints). (b) Without recomputing, is closer to than is? Say why in one sentence.
Show solution
(a) . (b) Yes, closer: the correction terms shrink as grows, so a larger lands nearer .
(a) At : . You can read this straight off the two strips of width 1 with right endpoints : .
(b) Every extra strip trims the overshoot. At the corrections are only , far smaller than the they were at , so is much closer to the exact .
Computing a definite integral from the limit is honest and exact, but it took most of a page for one clean function. When the region is a shape you already know, there is a shortcut: read the area straight off the geometry.
Evaluate each definite integral by recognizing the region as a familiar shape. (a) . (b) . (c) . (Rusty on the area of a circle? Appendix A.7 is the refresher.)
Solution. (a) The region under from to is a trapezoid standing on the -axis, with parallel vertical sides of height and and width . Its area is the average of the two heights times the width:
(b) The line starts at and drops to , so the region is a right triangle with base and height :
(c) The curve is the top half of the circle , a circle of radius 3 centered at the origin (square both sides to see it). From to the region is a full semicircle, so its area is half the circle's area (Figure 1.3.2):
No limit, no sum-of-squares formula. When the picture is a shape with a known area, the definite integral is that area.
Figure 1.3.2 is the top half of a circle of radius 3 centered at the origin. The shaded region is a semicircle, so equals its area, .
When the curve dips below the axis
Every region so far sat above the -axis, where the integrand was positive and the strip heights were positive numbers. But a strip height can be negative. When is below the axis, is a negative contribution to the sum, so that strip subtracts from the total instead of adding. The definite integral does not measure raw area; it measures net signed area, the area above the axis minus the area below it. If you want the plain geometric area, integrate instead, which flips the negative part back up.
The line crosses the axis at the origin. (a) Evaluate . (b) Find the total area between and the -axis on . Explain why the two answers differ.
Solution. (a) Figure 1.3.3 shows the two pieces. On the line is below the axis, forming a triangle of area , which the integral counts as . On the line is above the axis, an identical triangle of area , counted as . Add the signed pieces:
The two halves cancel exactly. A zero here does not mean there is no region; it means the region below the axis and the region above it are equal and opposite.
(b) Total area ignores the sign of each piece and adds the plain areas:
So the net signed area is and the total area is . They differ because the definite integral lets the lower triangle subtract, while total area counts every piece as positive. Whenever a curve crosses the axis, check which quantity a problem is asking for.
Figure 1.3.3 For , the graph is below the axis on (a triangle of area 2, counted as ) and above it on (a triangle of area 2, counted as ). The two cancel: the definite integral is , while the total area is .
A gauge on a tidal river measures flow rate, positive when the river runs out to sea and negative when the incoming tide shoves the water back upstream. Over a full tide cycle the flow rate traces a curve that spends part of the time above zero and part below. The definite integral of that flow rate over the cycle is the net signed area, and it answers one question exactly: how much water, on balance, left the estuary. The upstream surges subtract, just like the triangle below the axis in Example 1.3.3. If instead you want the total volume that sloshed past the gauge in either direction, you integrate the absolute value of the flow rate, which turns every upstream push back into a positive contribution. Same curve, two honest totals: net change reads the sign, total throughput ignores it. Engineers sizing a culvert care about the second; a hydrologist tracking where the water went cares about the first.
Example 1.3.3 hid a pattern worth naming. The graph of is symmetric through the origin: the piece on the left is the piece on the right flipped upside down and backward, so their signed areas are guaranteed to cancel. That is not special to . It is the payoff of symmetry, and it saves real work.
Recall two kinds of symmetry. A function is even when , so its graph is a mirror image across the -axis (think or ). A function is odd when , so its graph is unchanged by a half-turn through the origin (think , , or ). Split an integral over at zero using additivity:
For an even function the left region is the mirror image of the right region, so it has the same signed area, and the two pieces add:
For an odd function the left region is the right region flipped across the axis, so it carries the opposite sign, and the two pieces cancel:
Test it on what you already know. Since is even and , the shortcut hands you with no new work. Since is odd, , exactly the cancellation Example 1.3.3 computed the long way. When you spot symmetry, use it.
Dig Deeper Building the power pattern from the limit▸
Example 1.3.1 did with specific numbers. Run the same machine with a general upper limit and a pattern appears that the next section turns into a one-line rule. Take first, right endpoints, using the sum-of-first--integers formula . Here and :
So , which you can also read off as the area of a triangle with base and height . Now redo the same way, with the sum-of-squares formula:
Two results, one shape: and . The exponent goes up by one and the new exponent divides. The full pattern is
which you could grind out for each from its own sum formula. Section 1.4 proves it in a single line without any sums at all, by connecting the definite integral to the antiderivatives you built in Section 1.2. That connection is the whole point of what comes next.
The definite integral is net signed area, not total area, and the two come apart the instant a curve crosses the axis. A definite integral can be zero (Example 1.3.3) or negative, and neither means you made a mistake; it means the region below the axis matched or outweighed the region above. If a problem asks for the actual area of a region, hunt for axis crossings first and integrate , splitting at each crossing. Two more slips to avoid. First, a definite integral is a number, so it never carries a ; the belongs only to the indefinite integral (the general antiderivative) from Section 1.2. Second, the variable of integration is a dummy: and are the same number, so do not treat a stray or as if it changed the answer.
(a) Use geometry to evaluate . (b) Evaluate . (c) Suppose and . Find and .
Hint
Part (a) is a piece of the circle from Figure 1.3.2: which fraction of it? Part (b) has equal limits, so look at the zero-width property. Part (c) chains the two given integrals with additivity, then uses the reversing rule for the last one. None of these needs a Riemann sum.
Show solution
(a) . (b) . (c) and .
(a) The region under from to is the left half of the semicircle in Figure 1.3.2, so it is a quarter of the full circle of radius 3: .
(b) The limits are equal, so the interval has zero width and , no matter what the integrand is.
(c) Additivity glues the two adjacent intervals: . Reversing the limits flips the sign: .
Drag the limits and along the axis and watch the area above the axis add while the area below the curve subtracts, then flip to total area to fold the negative piece up.
Frozen on the window , where the curve dips below the axis on the left (subtracting) and rises above it on the right (adding), for a net signed area of .
Exercises 1.3
1. Suppose . Use the properties from this section to evaluate each. (a) . (b) . (c) . Warm up
Hint
Each part is one property from the Section 1.3 Property box applied to the given value 8: reversing the limits, zero width, and linearity (a constant slides out front). No integration required.
Show solution
(a) . (b) . (c) .
(a) Reversing the limits flips the sign: .
(b) Zero width: the interval from 2 to 2 is empty, so .
(c) Linearity pulls the constant out: .
2. Evaluate each definite integral by recognizing the region as a familiar shape, and sketch the region. (a) . (b) . Warm up
Hint
Follow Example 1.3.2 in Section 1.3. In (a) the graph is a horizontal line, so the region is a rectangle. In (b) the graph is a line through the origin, so the region is a triangle: base times height over two.
Show solution
(a) . (b) .
(a) The region under from 0 to 5 is a rectangle of width 5 and height 4: .
(b) The region under from 0 to 3 is a triangle with base 3 and height : .
3. Compute from the limit of the right sum , the way Example 1.3.1 did on . Use . Show that and take the limit. Core
Hint 1
Mirror Example 1.3.1 in Section 1.3 with and : find and first, then build and substitute the sum-of-squares formula.
Hint 2
Here and , so . Simplify and split into a constant plus terms with in the denominator.
Show solution
.
With and right endpoints :
As the last two terms vanish, so .
4. Let on . The graph crosses the axis at . (a) Find as net signed area by splitting into two triangles. (b) Find the total area between the graph and the axis on . Core
Hint 1
This is the net-versus-total contrast from Example 1.3.3 in Section 1.3. Split at the crossing : one piece is below the axis, one is above.
Hint 2
On the graph runs from up to : a triangle of base 2 and height 2, below the axis. On it runs from up to : a triangle of base 3 and height 3, above the axis. Sign the first piece negative for part (a).
Show solution
(a) . (b) .
The below-axis triangle on has area ; the above-axis triangle on has area .
(a) Net signed area counts the first piece negative: .
(b) Total area counts both pieces positive: .
5. The curve is the top half of a circle of radius 2. (a) Evaluate by geometry. (b) Use the even-symmetry shortcut to find . Give each answer exactly and rounded to three decimals. Core
Hint 1
Part (a) is a semicircle, like Example 1.3.2(c) in Section 1.3 but with radius 2. For part (b), notice is even, so the Section 1.3 Dig In shortcut relates the two integrals.
Hint 2
A full circle of radius 2 has area , so the semicircle is half that. For an even function, , so is half of your part (a) answer.
Show solution
(a) . (b) .
(a) The region is a semicircle of radius 2, so its area is .
(b) Because is even, , so . Geometrically, covers a quarter circle, and a quarter of is .
6. Suppose , , and . Evaluate each. (a) . (b) . (c) . Core
Hint 1
Combine the properties from Section 1.3: additivity glues the two intervals, linearity handles the combination in (b), and reversing handles (c).
Hint 2
For (a), . For (b), , using your answer to (a).
Show solution
(a) . (b) . (c) .
(a) Additivity over adjacent intervals: .
(b) Linearity: .
(c) Reversing the limits: .
7. Compute from the limit, and check the bracketing. (a) Show the right sum is and take its limit. (b) Evaluate and the left sum at , and confirm they straddle your limit. Stretch
Hint 1
Same limit machine as Example 1.3.1 in Section 1.3, now on , so and . This is the limit Section 1.1's Dig In pointed toward.
Hint 2
For (a), . For (b), compute both fractions at and compare with .
Show full solution
(a) . (b) and , which straddle .
(a) With and :
(b) At : and . Since , the left sum undershoots and the right sum overshoots the exact value , just as they should for an increasing function.
8. A drone's upward velocity is meters per second for , where a negative value means the drone is descending. (a) Compute as net signed area, and read it as the drone's net change in height. (b) Compute the total distance the drone travels vertically, using . State units. Stretch
Hint 1
The velocity is a straight line crossing zero, so this is the net-versus-total split from Example 1.3.3 in Section 1.3. Find where first, then treat each piece as a triangle.
Hint 2
Set to get . On the graph is below the axis (a triangle of base 2, height 4); on it is above (a triangle of base 3, height 6). Net signs the first piece negative; total counts both positive.
Show full solution
(a) meters (net rise). (b) meters.
The line crosses zero at . Below-axis triangle on : base 2, height , area . Above-axis triangle on : base 3, height , area .
(a) Net signed area signs the descending piece negative: meters. The drone ends 5 meters higher than it started.
(b) Total distance uses , so both pieces add: meters. The drone dropped, then climbed, covering 13 meters of vertical travel to net a 5 meter rise.
9. The Section 1.3 Dig In box gave the even and odd symmetry shortcuts. (a) Evaluate using symmetry alone, with no antiderivative (recall from Appendix A.2 that is odd). (b) Using from Exercise 3, find . (c) Choose your own even function and your own odd function, predict each integral on a symmetric interval , and justify the prediction from symmetry. Dig In
Hint 1
Reread the shortcut in Section 1.3's Dig In: on a symmetric interval an odd function integrates to 0 and an even function integrates to twice its right half. First classify each function as even or odd.
Hint 2
For (a), makes odd, so the interval is symmetric about 0. For (b), is even, so .
Show full solution
(a) . (b) . (c) Any odd function gives 0 and any even function gives twice its right-half integral on .
(a) is odd and is symmetric about 0, so : the hump on the left of the axis cancels the hump on the right.
(b) is even, so .
(c) For example, is odd, so for every ; and is even, so . Each follows because the left region is either the flipped-over (odd) or mirrored (even) copy of the right region.
Check your own version. Confirm your chosen odd function really satisfies and your even function satisfies ; a quick test is to plug in one value like . Then your odd integral over must be exactly 0, and your even integral must equal exactly twice the same function's integral over . If the odd one is not 0, the function was not odd after all.
10. The Section 1.3 Dig Deeper box built the pattern . (a) Use the pattern to write down and with no rectangles. (b) Derive the case, , from the limit, using . (c) Pick your own exponent and upper limit , predict from the pattern, and describe a right-sum test at large that would confirm it. Dig Deeper
Hint 1
Part (a) is direct substitution into the Section 1.3 Dig Deeper pattern. Part (b) reruns the Dig Deeper derivation with the sum-of-cubes formula in place of the sum-of-squares.
Hint 2
For (b), and , so . Simplify the square and take the limit.
Show full solution
(a) and . (b) . (c) The prediction is , confirmed when a right sum with large sits just above it and creeps down toward it.
(a) With the pattern: and .
(b) With and :
(c) For example, and predicts . A right sum evaluated at should land just above and shrink toward it as grows.
Check your own version. Your predicted value must be , and because is increasing on for , a right sum must overshoot it for every finite and settle down onto it as . If your right sum sits below the prediction, recheck whether you used right endpoints, which give the tallest rectangle in each strip here.
The definite integral is the single number the Riemann sums close in on as the number of strips runs to infinity, and any function continuous on is integrable, so that number always exists. Read the notation part by part: the limits of integration and , the integrand , and the that names the variable and marks the vanishing strip width (rename that variable freely, it is a dummy). The integral measures net signed area, area above the axis minus area below, so it can be zero or negative; total area uses . Four properties do the daily work: zero width, reversing the limits flips the sign, linearity, and additivity over adjacent intervals, with even and odd symmetry as a bonus shortcut on . You can already evaluate an integral two ways, from the limit or from a known shape, but the limit is brutal, a page of algebra for a single answer. Section 1.4 ends that work: one theorem turns every definite integral into a quick lookup, and the antiderivatives from Section 1.2 are the table you look up in.
Section 1.4
The Fundamental Theorem of Calculus
- State both parts of the Fundamental Theorem of Calculus and say, in plain words, why the derivative and the integral undo each other.
- Evaluate a definite integral in one line: find any antiderivative , then read off with the evaluation bar .
- Differentiate an accumulation function straight from its integrand, without computing the integral first.
- Pay the chain-rule toll when the upper limit is a function of , not just .
- Check that your antiderivative is continuous across the whole interval before you evaluate, so you never step across a hole.
Back to the dashboard from Welcome. Two dials, one drive. The speedometer names how fast you are moving at this instant; the odometer piles those instants into a total distance. For three sections you have treated them as separate problems, one about slopes and one about areas. They were never separate. The distance on the odometer is nothing but the speed piled up, and the speed is just how fast that pile is growing. This section is the single sentence that says so, out loud, with symbols. It is the handshake between the two halves of calculus, and once you have it, the page of limits you fought through in Section 1.3 collapses into one line of arithmetic.
Build the intuition
Start an integral moving. Fix the left end at , and let the right end slide. What you get is not a single number anymore but a function of where you stopped: the area under from out to . Call it .
For a continuous function and a fixed left endpoint , the accumulation function is
the net signed area under from to the moving right endpoint . The variable inside is written so the upper limit reads cleanly; they are the same graph. At the start there is no width and no area, so .
Now ask the question this whole book has been circling: how fast does grow? Nudge the right end from to , a hair to the right. The area gains one thin strip. That strip is almost a rectangle: its width is the tiny step , and its height is the value of the function right there, . So the area added is about , and the rate grows, its derivative, is the height itself. Figure 1.4.1 draws the strip.
Figure 1.4.1 The accumulation function is the shaded area out to . Push a hair to the right and gains one thin strip, height and width , so grows at the rate . That is the whole of Part 1: .
If is continuous and , then is differentiable and
Reason: the strip picture above. A small step to the right adds an area of about , so is . Differentiating an integral hands back the integrand you started with. One direct consequence is worth saying now: every continuous function has an antiderivative, because is one.
Part 1 says differentiation undoes integration. Turn it around and you get the workhorse of the whole course, the part you will use on nearly every problem from here on. If integrating and then differentiating gets you back where you started, then an antiderivative is exactly the right tool for reading an integral's value.
If is continuous on and is any antiderivative of (any function with ), then
No limit of Riemann sums, no page of algebra: find one antiderivative, subtract its value at the two ends, done. The shorthand for that subtraction is the evaluation bar,
read " of , evaluated from to ." The word "any" is a gift: two antiderivatives of differ only by a constant (the family fact from Section 1.2), and that constant cancels in the subtraction , so you may reach for the simplest antiderivative and drop the entirely. Why it is true is the telescoping story: chop into tiny steps, and the total change in is a chain of small changes that collapses, each end meeting the next until only survives. The Dig In makes that argument exact.
Evaluate .
Solution. An antiderivative of is (check: ; the reverse power rule from Section 1.2). Apply the evaluation bar:
Stop and look at what just happened. That number, , is the same you estimated with rectangles in Section 1.1 and then earned the hard way in Section 1.3, as the limit of a right sum, with the sum-of-squares formula, over most of a page. Here it took one line. The rectangles and the limit built the idea; the Fundamental Theorem turns that idea into a lookup. (And "any" antiderivative really means any: use instead and you get , the 5 gone as promised.)
Before you go on, predict, then check: evaluate with the same antiderivative and the evaluation bar. Should the answer be bigger or smaller than the from Example 1.4.1, and why?
Show solution
. Bigger: the interval is wider, so there is more area to pile up.
Same antiderivative , new right end:
The region now runs out to instead of , over a taller stretch of the parabola, so is exactly what the picture predicts.
The evaluation bar runs the same way for every function whose antiderivative you can name. Here are three at once, one of each flavor you will meet all book long.
Evaluate each definite integral. (a) . (b) . (c) .
Solution. (a) Antidifferentiate term by term: , so .
(b) An antiderivative of is (mind the sign: the derivative of is ). Working in radians,
That is the area of one full hump of the sine curve: a clean 2, no decimals.
(c) Here is the one power the reverse power rule cannot touch, and its antiderivative is (the natural logarithm; Appendix A.4 is the refresher, and Section 2.4 earns this fact in full). On every is positive, so :
The number is precisely the upper limit that makes this area exactly 1. Hold on to that; it is where the logarithm comes from.
Part 1 is not just the reason Part 2 works; it is a tool of its own. When the thing you want is the derivative of an accumulation function, you do not integrate first and differentiate second. You read the answer straight off the integrand.
(a) Find . (b) For a continuous , find , then use it to compute .
Solution. (a) Part 1 says differentiating an accumulation function hands back the integrand, evaluated at the upper limit. The upper limit is , so
You can confirm it the long way: , and . Part 1 skips the middle step.
(b) Now the upper limit is , not , so the accumulation is a composition: an outer accumulation function evaluated at an inner . That calls for the chain rule (from the Derivatives volume). Write , so by Part 1, and the quantity you want is :
The extra factor is the chain-rule toll: a variable upper limit charges you the derivative of that limit. With ,
Look back at the dashboard one more time, because the odometer has been doing Part 2 the whole drive. Your speed is a velocity , and your position is , an antiderivative of it: . The distance added between time and time is the definite integral of speed, and by Part 2 that integral is just , the difference of two odometer readings. The dial does not run a limit of rectangles behind the glass; it reports one subtraction, the same one the theorem names. A car's trip computer resetting "miles this trip," a household water meter totaling gallons from a varying flow, an electricity meter turning fluctuating power into a month of kilowatt-hours: each is a physical machine that integrates a rate by holding an antiderivative and reading the change in it. The theorem is not a classroom trick. It is the arithmetic every meter you own already trusts.
Why does Part 2 follow from Part 1? The whole argument is two honest lines, and it leans on one fact you already proved in Section 1.2: two functions with the same derivative on an interval differ only by a constant (the family fact). Here is the payoff.
Let be continuous, and let be any antiderivative you like, so . Build the particular accumulation function from Part 1, . By Part 1, as well. So and have the same derivative on the interval, which forces them to differ by a constant:
Pin down by feeding in . There , so , which means . Now read the integral you wanted off the top end:
That is Part 2, and notice it never needed a special : the constant carried whatever choice you made and then canceled. The accumulation function was the bridge. Part 1 built it; the family fact walked you across.
Dig Deeper A function you can only see through its integral▸
Part 1 has a quiet superpower: it lets you understand a function completely even when you cannot write down a formula for its integral. Take
None of this section's antiderivatives produce , so you cannot evaluate this with the reverse power rule. It does not matter. Part 1 tells you everything about the shape of straight from the integrand.
Its slope is the integrand: , which is positive for every , so is always increasing, gently, never flat, never falling. Its bending is the integrand's slope: rises up to and falls after, so is concave up for and concave down for , with an inflection at the origin where . You have just sketched a full graph, increasing through the origin and leveling toward horizontal ends, without ever integrating. The integral defined a brand new function, and Part 1 read its personality off the integrand alone. (This particular does have a name from the wider world of functions: it is , the inverse tangent; Appendix A.6 introduces it. The point stands for integrands that have no name at all.)
Part 2's one condition is easy to skip and expensive to skip: the antiderivative must be continuous across the entire interval . If your integrand blows up somewhere inside, the evaluation bar is meaningless there, and it will hand you a confident wrong number. Watch it fail. A student writes . But is never negative, so its net signed area cannot be ; something is broken. The break is at , where shoots to infinity and has a hole: the theorem's continuity condition is violated, so the bar does not apply, and the honest tools of this course simply do not reach across that gap. Before you evaluate, glance at the interval and make sure your integrand has no holes inside it. And on Part 1, do not forget the toll: with a variable upper limit , the derivative is , and the is the piece people drop.
(a) Evaluate . (b) Find for . (c) Find .
Hint
For (a), rewrite and use the reverse power rule for the antiderivative, then the evaluation bar (Section 1.4). For (b), Part 1 hands back the integrand at the upper limit, no integration needed. For (c), the upper limit is , so expect the chain-rule toll from Example 1.4.3.
Show solution
(a) . (b) . (c) .
(a) An antiderivative of is (raise the power to , then divide by ):
(b) By Part 1, differentiating the accumulation returns the integrand at the top end: . No antiderivative required.
(c) The upper limit is , so the integrand is evaluated at and multiplied by the toll :
Drag the upper limit along the axis and watch the signed area trace itself out below, its tangent's slope always equal to the height .
Frozen at : the accumulated area is traced from to , and its tangent rises because is positive, the Fundamental Theorem in one picture.
Exercises 1.4
1. Evaluate with the evaluation bar. (a) . (b) . Warm up
Hint 1
Name an antiderivative with the reverse power rule from Section 1.2 (), then subtract its values at the two ends, . Drop the ; it cancels.
Show solution
(a) . (b) .
(a) .
(b) .
2. Use Part 1 to write directly, without computing the integral. (a) . (b) . Warm up
Hint 1
Part 1 (Section 1.4) says differentiating an accumulation function hands back the integrand, evaluated at the upper limit. Here the upper limit is just , so there is no toll to pay.
Show solution
(a) . (b) .
Part 1 returns the integrand at the top end. The lower limits (0 and ) are fixed constants and do not affect the derivative; they only set where the accumulation starts from zero.
3. Evaluate . Core
Hint 1
Antidifferentiate term by term with the reverse power rule (Section 1.2), then apply the evaluation bar. Mind the signs: the lower limit is negative, so and will appear.
Hint 2
An antiderivative is . Compute and separately, then subtract (which is itself negative, so you are subtracting a negative).
Show solution
.
4. Evaluate, working in radians. (a) . (b) . Core
Hint 1
Read the antiderivatives backward off the derivative table: an antiderivative of is , and of is (mind that sign). A constant multiple slides straight through the integral.
Hint 2
For (b), pull the 2 out front: . Then evaluate the bar and multiply by 2.
Show solution
(a) . (b) .
(a) .
(b)
5. Evaluate . Core
Hint 1
This is the case the reverse power rule skips; its antiderivative is (Example 1.4.2c, and Appendix A.4). On every is positive, so .
Hint 2
Evaluate . Use and .
Show solution
.
6. Differentiate each accumulation function, paying the chain-rule toll where the upper limit is not just . (a) . (b) . Core
Hint 1
Both use Part 1 with a variable upper limit : the derivative is , integrand at the top end times the derivative of that end (Example 1.4.3). You never integrate at all.
Hint 2
In (a), so . In (b), so ; substitute into to get .
Show solution
(a) . (b) .
(a) The integrand at is , and the toll is :
(b) The integrand at is , and the toll is :
7. Evaluate by rewriting the integrand first. Stretch
Hint 1
You cannot antidifferentiate a quotient by inspection, so split it into powers first: divide each term by , the same rewrite-first move from Section 1.2. Then the reverse power rule handles each piece.
Hint 2
. Antidifferentiate to , then apply the bar.
Show full solution
.
Rewrite, then antidifferentiate term by term:
At : and , giving . At : . Subtract:
8. Let . (a) Find . (b) Where is decreasing, and where is it increasing? Find the input where is smallest. (c) Compute and , and say what the signs mean as areas. Stretch
Hint 1
For (a), Part 1 hands back the integrand. For (b), a function decreases where its derivative is negative and increases where it is positive, so read the sign of evaluated at . For (c), evaluate the integral with the bar and read a negative value as net signed area below the axis (Section 1.3).
Hint 2
is negative for and positive for . An antiderivative of is , so .
Show full solution
(a) . (b) Decreasing on , increasing for ; smallest at . (c) , .
(a) By Part 1, .
(b) is negative for and positive for , so falls until and climbs after: its smallest value is at , where the integrand crosses from negative to positive.
(c) Using :
From 0 to 2 the integrand sits below the axis, so accumulates negative signed area, reaching . From 2 to 4 it sits above, adding back exactly of area, so returns to 0: the area below cancels the area above.
9. The Section 1.4 Dig In showed Part 2 by comparing the accumulation function with any antiderivative : they differ by the constant . Make that concrete. (a) Take and . Compute as a formula in , and pick the simple antiderivative . (b) Show is a constant, and check that the constant equals . (c) Confirm that equals both and for your own choice of . Dig In
Hint 1
Reread the Section 1.4 Dig In: the accumulation function is one specific antiderivative, pinned so that . Any other antiderivative sits a constant above or below it, and that constant is .
Hint 2
Evaluate . Then collapses to a single number; compare it with .
Show full solution
(a) . (b) . (c) With , all three equal .
(a) , and .
(b) , a constant with no in it. And , so the constant is exactly , as the Dig In argument promised.
(c) Choose . Then , and , and a direct evaluation . All three agree.
Check your own version. For any continuous , your must simplify to a single constant (no survives), and that constant must equal for your chosen . If an survives, an antiderivative was miscomputed; if the constant is not , recheck that . Then any you pick must give to the last digit.
10. The Section 1.4 Dig Deeper read the shape of off its integrand, with no closed form. Build your own such reading. Let . (a) Is ever decreasing? Justify from the integrand, not from a formula. (b) Find , and say whether is concave up or concave down just to the right of . (c) Explain why stays finite as grows large, in one sentence about how fast the integrand shrinks. Dig Deeper
Hint 1
Reread the Section 1.4 Dig Deeper: Part 1 gives as the integrand at , and the integrand's own slope tells you the concavity of . You never need a formula for itself.
Hint 2
, which is positive for every . For concavity near 0, ask whether is rising or falling just past . For (c), compare with for large .
Show full solution
(a) Never decreasing. (b) ; concave down just right of 0. (c) The integrand shrinks like , fast enough that the piled area approaches a finite ceiling.
(a) By Part 1, . The denominator is always at least 1, so is positive for every : is always increasing and never decreasing.
(b) . Just to the right of 0, the integrand starts falling from its peak value 1 (its denominator grows), so is decreasing there, which makes concave down just right of the origin.
(c) For large , is close to , which drops off steeply, so each new strip of area is tiny and the running total climbs toward a finite ceiling rather than growing without bound.
Check your own version. If you invent your own positive integrand and set : must be positive everywhere is positive (so never decreases), , and is concave up or down exactly where rises or falls. A quick numerical check: a right-endpoint sum of on a fine grid should match your predicted rise, and it should keep increasing if and only if stays positive.
The Fundamental Theorem of Calculus is the sentence where the derivative and the integral shake hands. Part 1 starts an integral moving: the accumulation function grows one thin strip at a time, so its derivative is the height of that strip, . Differentiating an accumulation hands back the integrand, and when the upper limit is a function of you pay the chain-rule toll, multiplying by that limit's derivative. Part 2 spends Part 1: for continuous and any antiderivative , , which turns the page of Riemann-sum algebra from Section 1.3 into a single subtraction (the same , in one line). Two care points carry forward: the cancels, so reach for the simplest antiderivative, and must be continuous across the whole interval, so never evaluate across a hole. With this, any rate integrates to its total. Section 1.5 names that idea and reads its sign: net change.
Section 1.5
The Net Change Theorem
- Read the Net Change Theorem out loud: the integral of a rate of change is the net change in the quantity, so .
- Tell net displacement (, which can be negative) from total distance (), and find total distance by locating where the velocity is zero and splitting there.
- Turn any accumulated rate into a total: water filling a tank, a population growing, a cost rising.
- Estimate a net change from a table of rate measurements with a Riemann sum, and say honestly how good the estimate is.
Picture the dashboard from Welcome once more. The speedometer calls out how fast you are going at each instant; the odometer just piles all of that up into a single number, the miles you have covered. That piling-up is the whole idea of this section, and it turns out to be the plainest sentence in the book: the total change in a quantity is the integral of its rate. The odometer literally sums the speedometer. Unit 1 opened with a needle that would not hold still and asked how the instants stack back into a total. Here is the answer, in one line.
Build the intuition
Slice the trip into tiny stretches of time. Over one stretch of width the speed barely changes, so the distance covered is almost exactly (speed). Add up every thin stretch from the start time to the end time and you have the odometer's total, which is the integral of the speed. Nothing about that argument was special to speed. Replace "speed" with any rate of change and "distance" with the quantity it changes, and the same slicing says: the total change is the integral of the rate.
You have already proved this. A rate of change is a derivative , and adding up from to lands exactly on . That is the Fundamental Theorem of Calculus from Section 1.4, now read as a sentence about rates instead of a rule for evaluating integrals. The odometer reading is the Fundamental Theorem applied to speed.
One warning rides along for free. When the rate can turn negative (a car shifts into reverse, a tank switches from filling to draining), the thin pieces below the axis subtract. Net change lets forward and backward cancel, which is usually what you want. But if you want the total amount of motion, every piece has to count as positive. That single distinction, cancel or count, is the heart of this section.
If is the rate of change of a quantity , then the net change in from to is the integral of that rate (for a rate that is continuous on , as every rate in this book is):
Reason: this is the Fundamental Theorem of Calculus (Section 1.4) with the integrand seen as a rate. The integral adds up the rate over every instant, and that sum is the difference between the ending amount and the starting amount. The integral hands you the change; to get the ending amount you add the change to where you started.
An object moving along a line with velocity on a time interval has two totals worth naming. Its net displacement is
the signed change in position: how far it ended from where it started, forward counted positive and backward negative. Its total distance is
the whole length of the path traveled, with every stretch counted as positive whether the object was moving forward or back.
Net displacement is the plain integral of velocity, , where forward and backward motion cancel. Total distance is the integral of the speed , where nothing cancels. To compute total distance: find every time in where , split the interval at those times, integrate on each piece, and add the magnitudes. Reason: on each piece keeps one sign, so its integral there is the signed area of that piece; taking magnitudes turns every piece into real distance covered.
Figure 1.5.1 The velocity (in m/s) of the car in Example 1.5.1. It is negative on (the car rolls backward) and positive on (forward). The backward region has area , the forward region area . Net displacement subtracts the signed pieces, m; total distance adds the areas, m.
A remote-control car on a straight track has velocity meters per second, with in seconds, over . A positive velocity is forward, a negative velocity is backward. (a) Find the car's net displacement. (b) Find its total distance traveled. (c) Say plainly why the two answers differ. Figure 1.5.1 shows the velocity.
Solution. (a) Net displacement is the plain integral of velocity:
The car ends up 3 meters behind where it started: the backward stretch outweighed the forward one.
(b) Total distance needs , so first find where the car turns around. Set : gives (the only turning time in ). On the velocity is negative, so there; on the velocity is positive, so . Integrate each piece:
Adding the magnitudes gives the total distance:
(c) The car rolled a full meters of track, most of it backward and then a shorter run forward. Its net displacement is only meters because the forward run erased part of the backward trip. Distance measures the odometer's honest total; displacement measures only where the car finished relative to the start.
Before you compute, predict. A drone rises with velocity meters per second on . (a) Will its total distance be larger than, smaller than, or equal to the magnitude of its net displacement? (b) Now find both and confirm your prediction.
Show solution
(a) Equal. (b) Net displacement m and total distance m: they match because the whole time, so the drone never reverses.
Since on , the velocity never changes sign, so and the two integrals are identical:
Displacement and distance only differ when the motion reverses. No reversal here means no gap.
Rates that are not motion
Nothing in the Net Change Theorem mentions cars. Any rate at all, if you can integrate it, adds up to the total change in whatever it is the rate of. Water flows into a tank at some liters per minute, and the integral is the liters added. A colony grows at some cells per hour, and the integral is the cells gained. The recipe never changes; only the units do.
Example 1.5.1's car is on the track below. Drag the time cursor (or press Run) and watch the two meters disagree: net displacement is the signed integral and can FALL, while total distance is the odometer and only climbs. Challenge: park the cursor exactly at , the turnaround. The two meters read and : same size, opposite jobs, and everything after that moment only pushes them apart by double counting the backing up.
Water flows into a tank at a rate of liters per minute, with in minutes, over . The tank starts with 5 liters. (a) How much water is added during the first 10 minutes? (b) How much is added during just the last 4 minutes? (c) How much water is in the tank at ?
Solution. (a) The water added is the integral of the inflow rate:
(b) The last 4 minutes run from to :
The inflow is easing off (the rate falls from 12 down to 2), so the last 4 minutes add much less than the first 4 would.
(c) The amount in the tank is the starting amount plus the net change, exactly the second form of the Net Change Theorem:
A flow meter on a pipe reports the inflow rate into a reservoir every 5 minutes, but gives you no formula. Table 1.5.1 is the readout. Estimate the total water added over the 20 minutes, and be honest about how good the estimate is.
| Time (minutes) | Inflow rate (L/min) |
|---|---|
| 0 | 4.0 |
| 5 | 7.0 |
| 10 | 11.0 |
| 15 | 16.0 |
| 20 | 20.0 |
Solution. The water added is , but with only five readings we cannot evaluate that integral exactly. We can do what Section 1.1 did with a speed table: build rectangles. With minutes, the left sum uses the reading at the start of each 5-minute block and the right sum uses the reading at the end:
The rate only rises across the table, so the left sum sits under the true total and the right sum sits over it: the exact amount added is somewhere between 190 L and 270 L. Averaging the two (the trapezoid estimate from Section 1.1) gives a single best guess:
This is an estimate, not the answer. A finer meter, reading every minute, would tighten the bracket; a formula for would close it exactly. With a table, the rectangles are all you get, and honesty means reporting the bracket along with the estimate.
Your electricity meter never measures energy directly. What it watches is power: the rate you are drawing at each instant, in kilowatts, which jumps every time a heater kicks on or a fridge cycles off. The meter adds that rate up over the whole month, and the kilowatt-hours on your bill are exactly , the net change in energy delivered. Run a 1.5 kilowatt space heater steadily for 4 hours and the arithmetic is easy, kilowatt-hours, because a constant rate is just rate times time. The moment the power stops holding still, that shortcut breaks and the integral takes over. Energy is to power what the odometer is to the speedometer: the running total of a rate that never sits still.
Example 1.5.1's car reversed once. Here is one that reverses twice, so the split has three pieces. A particle moves with velocity meters per second on . Factor it to find the turning times: , so at and . Testing a point in each stretch, the velocity is positive on , negative on , and positive again on : forward, back, forward.
Net displacement is one clean integral, letting the pieces cancel:
Total distance splits at both turning times and adds the magnitudes. Using the same antiderivative on each piece:
The middle piece is negative because the particle was moving backward there. Distance ignores that sign and adds the sizes:
So the particle traveled 4 meters of path but ended only meters ahead. Look back at Figure 1.5.1's shading to see what just happened: net displacement is the signed area under the velocity graph (below the axis subtracts), the same net signed area you met in Section 1.3; total distance is the total area, every piece flipped positive. Two readings of one graph, and the split at each sign change is what separates them.
Dig Deeper The cost of the next hundred units▸
The Net Change Theorem reaches well past the physical world. In economics, a firm's marginal cost is the cost of making one more unit when it is already producing of them: the rate at which total cost rises. Because it is a rate, the cost of increasing production from to is its integral, .
Suppose a workshop's marginal cost is dollars per unit (it falls as the shop finds efficiencies). The cost of raising output from 20 to 60 units is
Those 40 extra units cost $2000 to make. Notice what the integral does and does not tell you. It gives the change in cost, not the total cost of producing 60 units, because the fixed startup cost (rent, machines, the cost that is already there at ) never appears in the rate . To get a total cost you would add this change to a known cost at some production level, exactly the form of the theorem. A rate tells you how much things changed, never where they started.
The cardinal error of this section is treating net displacement as if it were total distance. They are different numbers whenever the motion reverses, and no amount of absolute-value signs on the final answer will fix it. In Example 1.5.1 the net displacement is m, so a student who computes and then writes reports 3 meters of travel. The real distance is meters, more than twice as far. The absolute value belongs inside the integral, on the velocity, not outside on the result. That means the work always goes in this order: find every time , split the interval there, integrate on each piece, then add the magnitudes. Skip the splitting and the backward stretch quietly cancels part of the forward one, and you have computed displacement while calling it distance.
A cart moves along a track with velocity meters per second on . (a) Find its net displacement. (b) Find its total distance traveled.
Hint
Net displacement is straight through. For total distance you need , so first solve to find where the cart reverses, then split there and add the magnitudes (Section 1.5, the Watch out).
Show solution
(a) m. (b) m.
(a) Net displacement is the plain integral:
(b) The cart reverses where , that is . On the velocity is negative, so ; on it is positive. Split and add the magnitudes:
Total distance is m, well past the meters you would get by wrongly absolute-valuing the displacement.
Exercises 1.5
1. A town's population changes at a rate of people per year, where is years from now. Find the net change in the population over the next 5 years. Warm up
Hint
The net change is the integral of the rate (the Net Change Theorem). Integrate from to with the reverse power rule from Section 1.2; the units are people.
Show solution
people.
The town grows by 275 people over the 5 years. The rate itself climbs from 40 to 70 people per year across the interval; the integral bundles all of that into one total.
2. A drone flies straight up with velocity meters per second on . (a) Find its net displacement. (b) Find its total distance. (c) Explain why the two answers are equal for this drone. Warm up
Hint
Net displacement is ; total distance is . Check the sign of on before deciding whether the two can differ (Section 1.5).
Show solution
(a) m. (b) m. (c) The velocity is positive the whole time, so the drone never reverses and no motion cancels.
(a)
(b) Because on all of , the speed equals , so the distance integral is the same one: m.
(c) Displacement and distance split apart only when the object turns around. This drone climbs the entire interval, so there is no backward stretch to subtract, and the honest path length equals the net rise.
3. A cart moves along a track with velocity meters per second on . (a) Find its net displacement. (b) Find its total distance traveled. Core
Hint 1
Net displacement integrates straight through; total distance needs . First find where to see when the cart reverses (Section 1.5, the Watch out).
Hint 2
at . For distance, split at ; on the velocity is negative, so use there.
Show solution
(a) m. (b) m.
(a)
(b) The cart reverses at . Split there and add the magnitudes:
Total distance is m. The cart backed up meters, then rolled meters forward, finishing 8 meters behind the start.
4. A car accelerates down a straight road. Its speed is read off the speedometer every 3 seconds, in Table 1.5.2. (a) Estimate the distance the car travels with a left sum and a right sum. (b) Between which two values must the true distance lie, and what is your best single estimate? Core
| Time (seconds) | Speed (ft/s) |
|---|---|
| 0 | 20.0 |
| 3 | 35.0 |
| 6 | 46.0 |
| 9 | 54.0 |
| 12 | 60.0 |
Hint 1
Distance is the integral of speed, and a table gives you only rectangles (Section 1.1). Build a left sum and a right sum with seconds.
Hint 2
The left sum uses the first four readings, the right sum the last four, each multiplied by 3. Since the speed only rises, one sum is a floor and the other a ceiling.
Show solution
(a) Left sum ft, right sum ft. (b) The true distance lies between ft and ft; best single estimate ft.
With seconds:
The speed increases across the run, so the left sum underestimates and the right sum overestimates: the true distance is between 465 ft and 585 ft. Averaging them gives the trapezoid estimate, ft, the best single guess a table this coarse can offer.
5. A conveyor drops sand onto a pile at a rate of kilograms per minute for the first 15 minutes. The pile starts at 100 kg. (a) How much sand is added over the 15 minutes? (b) How much sand is in the pile at ? Core
Hint 1
Sand added is the integral of the drop rate (the Net Change Theorem). Integrate from 0 to 15; then use .
Hint 2
. Evaluate that, then add the starting 100 kg for part (b).
Show solution
(a) kg. (b) kg.
(a)
(b)
The drop rate slows from 50 kg/min down to 20 kg/min but stays positive the whole time, so the pile only grows.
6. At noon a reservoir holds 5000 cubic meters of water. Over the next 6 hours water flows in at a rate of cubic meters per hour, where is hours after noon. How much water does the reservoir hold at 6 p.m.? Core
Hint 1
The final amount is the start plus the net change, (the Net Change Theorem).
Hint 2
Integrate from 0 to 6 to get the water added, then add it to the starting 5000 cubic meters.
Show solution
cubic meters.
First the water added over the 6 hours:
Then add it to the starting amount:
7. A particle moves with velocity meters per second on . (a) Find the time in this interval when the particle is momentarily at rest. (b) Find its net displacement. (c) Find its total distance traveled. Stretch
Hint 1
The particle is at rest where ; solve . Net displacement is ; total distance is , split at that rest time (Section 1.5).
Hint 2
at . On the velocity is negative, so there; on use unchanged.
Show solution
(a) s. (b) m. (c) m.
(a) Set , so and seconds (the negative root is outside ).
(b)
(c) The particle is at rest at ; it moves backward on and forward on . Split and add the magnitudes:
Total distance is m, against a net displacement of only m.
8. A warehouse's stock changes at a rate of pallets per hour on ; a positive rate means pallets arriving, a negative rate means pallets shipping out. The warehouse starts with 10 pallets. (a) Find the net change in the warehouse's stock over the 6 hours. (b) At what time is the stock highest, and how many pallets are there then? Stretch
Hint 1
The net change in stock is . For the peak moment, remember the stock keeps growing only while (Section 1.5).
Hint 2
at : the rate is positive before it (filling) and negative after (draining), so the stock peaks at . Add to the starting 10 pallets.
Show solution
(a) pallets. (b) Highest at hours, holding pallets.
(a)
The net change is zero, but that does not mean nothing happened: pallets arrived, then shipped back out.
(b) The rate is zero at , positive before it and negative after, so the stock builds until and falls afterward. Its peak is
The stock climbs from 10 to 19 pallets over the first 3 hours, then ships the same 9 back out, ending where it began. Net zero, but a real 9 in and 9 out along the way.
9. The Dig In split a velocity into forward and backward pieces. Design your own version. Find a velocity on , not zero everywhere, whose net displacement over is but whose total distance is positive. (a) Show both integrals for your . (b) Explain, using signed area (Section 1.3), why a net displacement of zero can never force the total distance to be zero. Dig In
Hint 1
You want the forward and backward pieces to cancel in the signed integral but both to count in the distance. Reread the Dig In in Section 1.5 and pick a simple straight-line .
Hint 2
A velocity that is negative on the first half of and positive by the same amount on the second half makes . Then find where , split, and integrate .
Show full solution
One valid choice is : net displacement m, total distance m. (Any that is positive on part of and negative on the rest, with the areas canceling, works.)
(a) With , the net displacement cancels:
For distance, at ; the velocity is negative on and positive on :
Total distance is m.
(b) Net displacement is the signed area under the velocity graph, where the piece below the axis subtracts (Section 1.3). Total distance is the total area, every piece flipped positive. The only way the total area is zero is if is zero the whole time. So the moment is nonzero anywhere, real area exists and counts fully toward distance, even when a canceling piece pulls the net all the way back to zero.
Check your own version. Your must be positive on part of and negative on another part, with the signed areas exactly canceling. Antidifferentiate to confirm ; then find every time , split there, integrate on each piece, and add the magnitudes. That total must be positive, and because the net is zero it will equal exactly twice the area of the forward part.
10. A factory's marginal cost is dollars per unit, where is the number of units produced. (a) Find the added cost of raising production from 100 to 200 units. (b) The factory tells you the first 100 units cost $9000 to make. Find the cost of the first 200 units. (c) Explain why part (a) by itself cannot give you part (b)'s answer. Dig Deeper
Hint 1
Marginal cost is a rate, so the added cost is (the Net Change Theorem, and the Dig Deeper box). For a total, recall you also need a known cost to start from.
Hint 2
For (a), integrate from 100 to 200. For (b), add that added cost to the given cost of the first 100 units.
Show full solution
(a) $2500. (b) $11500. (c) The integral gives only the change in cost; a total needs a known starting cost, which the marginal rate never carries.
(a)
Those 100 extra units add $2500 to the cost.
(b) The Net Change Theorem in its form gives the total at 200 units:
(c) Part (a) is a net change, the cost of the extra units and nothing more. The total cost of the first 200 units includes the fixed startup cost buried in , which never shows up in the rate . Without a known cost at some production level, that startup cost is invisible, so the total is unknowable from the marginal rate alone.
Check your own version. Whatever marginal-cost function and endpoints you choose, integrating from to gives the cost of those extra units, never the total production cost. Add it to a stated cost at to reach the cost at ; if no such anchor is given, the fixed cost is invisible to the rate, so the total genuinely cannot be recovered.
The Net Change Theorem is the Fundamental Theorem read as one plain sentence: the integral of a rate of change is the net change in the quantity, . It closes Unit 1's opening question, since the odometer's total is the integral of the speedometer's rate. In motion it splits into two totals that are easy to confuse: net displacement is the signed change in position and can be negative, while total distance counts every stretch of the path as positive. The one non-negotiable move is to find where first, split the interval there, and add the magnitudes, because you can never fix distance by putting an absolute value on the final answer. The same theorem reads any rate into its total, a filling tank, a growing population, a rising cost, and when all you have is a table of readings, a Riemann sum estimates the change and honestly brackets it. Unit 1 built the whole idea, from a stack of rectangles to the limit, to the Fundamental Theorem, to net change; the Mixed Practice mixes it all, and then Unit 2 builds the toolkit that lets Section 1.2's antiderivatives reach far more functions.
Section 1.P
Mixed Practice: Unit 1
Five sections built one toolbox, and now the labels come off. Below you will find an area to box in with rectangles, a derivative to run backward into an antiderivative, and a definite integral to settle from a shape or from the sign of its region. You will also find a rate to add up into a net change and an accumulation function to differentiate on sight. Nothing tells you which idea a problem wants. Read each one first, decide what kind of question it is, an estimate, an antiderivative, a signed area, a one-line evaluation, or a net change, and only then pick a method. Every problem rewards a real attempt. Hold on to the two habits Unit 1 drilled: every indefinite integral ends in a constant, and every definite integral is a single number.
Mixed Practice 1
1. A surveyor measures the width of an irregular field at five points spaced meters apart along its length. Table 1.P.1 lists the widths. The field's area is the area under this width profile. (a) Estimate the area with a left sum. (b) Estimate it with a right sum. (c) Average the two for a single best estimate. Warm up
| Position (meters) | Width (meters) |
|---|---|
| 0 | 10.0 |
| 4 | 14.0 |
| 8 | 15.0 |
| 12 | 13.0 |
| 16 | 11.0 |
Hint
A width times a length is an area, added up strip by strip just like distance in Section 1.1. The strips are meters wide; the left sum reads the width at the start of each strip, the right sum at the end.
Show solution
(a) square meters. (b) square meters. (c) Best estimate square meters.
Each strip is meters wide, and the heights are the widths.
(a) The left sum reads the width at the start of each strip (the first four readings):
(b) The right sum reads the width at the end of each strip (the last four readings):
(c) The field bulges in the middle and narrows at both ends, so neither sum is a clean floor or ceiling. Averaging them (the trapezoid estimate) gives the best single guess: square meters. A tape measure taken every meter would tighten it; five readings can only bracket.
2. Find each indefinite integral. (a) . (b) . Warm up
Hint
Reverse the power rule from Section 1.2 on each term: raise the exponent by one and divide by the new exponent. An indefinite integral names a whole family, so end with .
Show solution
(a) . (b) .
(a) Term by term: , , and antidifferentiates to , giving . Differentiate back to confirm: .
(b) and , so the integral is .
3. Evaluate each definite integral by recognizing the region as a familiar shape. (a) . (b) . Warm up
Hint
Follow the geometry shortcut from Section 1.3. In (a) the graph is a horizontal line, so the region is a rectangle; in (b) the graph is a line through the origin, so the region is a triangle (base times height over two).
Show solution
(a) . (b) .
(a) The region under from to is a rectangle of width and height : .
(b) The region under from to is a triangle with base and height : .
4. The line crosses the -axis at , as Figure 1.P.1 shows. (a) Evaluate as net signed area. (b) Find the total area between the line and the -axis on . Say why the two answers differ. Core
Figure 1.P.1 The line on drops from , crosses the axis at , and climbs to . It carves a triangle below the axis on and a triangle above the axis on .
Hint 1
This is the net-versus-total contrast from Section 1.3. Split at the crossing : one triangle sits below the axis, one above. Net signed area signs the lower piece negative; total area counts both as positive.
Hint 2
The below-axis triangle on has base and height ; the above-axis triangle on has base and height . You can also evaluate part (a) straight through with the evaluation bar and .
Show solution
(a) . (b) . They differ because the definite integral lets the lower triangle subtract, while total area counts every piece as positive.
The below-axis triangle on has base and height , so its area is . The above-axis triangle on has base and height , so its area is .
(a) Net signed area counts the lower triangle as negative: . The evaluation bar agrees: .
(b) Total area ignores the sign and adds the plain areas: . So the net signed area is while the total area is ; the region below the axis is larger, so it pulls the signed total negative even though square units of region sit between the line and the axis.
5. Evaluate each definite integral with the evaluation bar. (a) . (b) . Core
Hint 1
Name an antiderivative , then read off with the bar from Section 1.4. Drop the ; it cancels in the subtraction.
Hint 2
In (a), . In (b), rewrite first (the rewrite-first move from Section 1.2); its antiderivative is .
Show solution
(a) . (b) .
(a) An antiderivative is :
(b) Rewrite , whose antiderivative is :
6. A savings account's balance grows at a rate of dollars per month, months from now, and today it holds dollars. (a) Find the balance function . (b) What will the balance be after months? Core
Hint 1
The balance is an antiderivative of its growth rate, so this is an initial value problem (Section 1.2): antidifferentiate to get the family , then use the known balance to pin down .
Hint 2
. Set and use to find before answering (b).
Show solution
(a) . (b) dollars.
(a) Antidifferentiate the rate: . The reading gives , so . Check: .
(b) dollars. The starting was exactly the information the held a place for.
7. A cell culture grows at a rate of thousand cells per hour, hours after it is seeded, and it starts with thousand cells. (a) How many cells are added during the first hours? (b) How many cells are in the culture at ? Core
Hint 1
The number of cells added is the integral of the growth rate (the Net Change Theorem, Section 1.5). Integrate from to ; the units are thousands of cells.
Hint 2
. For (b), add that change to the starting count using .
Show solution
(a) thousand cells. (b) thousand cells.
(a) The cells added is the integral of the rate:
(b) The amount at is the start plus the change: thousand cells.
8. Estimate the area under from to with three rectangles of equal width (). (a) Write the right sum in sigma notation and evaluate it. (b) Find the left sum . (c) The exact area is . Confirm your two sums trap it, and say which runs high. Stretch
Hint 1
With on , the cut points are , and the right sample points are for (the sigma form from Section 1.1). Plug each into .
Hint 2
Since and , the right sum is . For the left sum, sample instead. The function only rises on , so one sum is a floor and the other a ceiling.
Show full solution
(a) . (b) . (c) They trap since ; the right sum runs high.
(a) With and right sample points , the right sum is
(b) The left sum reads the left edges , so .
(c) Because only rises on , the left rectangles fall short and the right ones overshoot: . The right sum runs high, the left sum low, and the exact area sits between them.
9. Differentiate each accumulation function, paying the chain-rule toll where the upper limit is not just . (a) ; find . (b) . Stretch
Hint 1
Part 1 of the Fundamental Theorem (Section 1.4) says differentiating an accumulation function hands back the integrand, evaluated at the upper limit. No integration is needed. Check whether the upper limit is plain or a function of .
Hint 2
In (a), the upper limit is , so there is no toll: is just the integrand at . In (b), the upper limit is , so multiply the integrand at by the toll .
Show full solution
(a) . (b) .
(a) By Part 1, differentiating the accumulation returns the integrand at the upper limit. The limit is , so there is no toll: .
(b) The upper limit is , so the integrand is evaluated at and multiplied by the toll :
10. A drone flying along a straight line has velocity meters per second on , where a negative value means it is flying backward. (a) Find the time in this interval when the drone is momentarily at rest. (b) Find its net displacement. (c) Find its total distance traveled. Give exact values and rounded decimals. Stretch
Hint 1
The drone is at rest where ; solve on . Net displacement is straight through; total distance is , which you must split at the rest time (Section 1.5, the Watch out).
Hint 2
at . On the velocity is negative, so there; on it is positive, so . Integrate each piece and add the magnitudes.
Show full solution
(a) seconds. (b) m. (c) m.
(a) Set , so and seconds (the negative root is outside ).
(b) Net displacement is the plain integral of velocity:
(c) The drone flies backward on and forward on . Split at and add the magnitudes:
Total distance is m. The drone covered meters of path but ended only meters behind where it began, because the forward leg erased part of the backward one.
Section 1.R
Unit 1 Recap: Check Yourself
Five sections built Unit 1, and this is the checkpoint. Hit Try it! and work each goal cold, then check yourself and tick Got it. A worked example and the section link sit one click away whenever you want a model or another pass.
Watch the trip meter climb during a two-hour drive. It never asks how fast you are going; it only adds. Yet the total it lands on, say 96 miles, is exactly the area under your speed-versus-time graph, even though that speed drifted up on the open road and sank in every town. Section 1.1 estimated that area with a handful of rectangles; Section 1.3 made it exact by letting the rectangles shrink to nothing; and Section 1.4 revealed the shortcut, that the whole accumulated total falls out of a single antiderivative evaluated at the two endpoints. Section 1.5 gave the idea its plain name: the integral of a rate is the total change it produces. That one sentence runs the trip meter, the water meter, the electric meter, and the savings account. In Unit 2, a handful of techniques will let you find those totals for far more than the tidy curves you have met so far.
Your turn. Find one thing in your week that keeps a running total from a rate that never holds still: miles driven, data used on your phone, water down the drain, dollars spent. What is the rate, what is the total, and how would slicing the time into thin pieces let you rebuild the total from the rate?
Can you estimate the area under a curve with left, right, and midpoint rectangles, write the estimate in sigma notation, and sharpen it by using more, thinner rectangles?
Try it!
Coasting downhill, a cyclist's speed is clocked every 4 seconds, in meters per second: 3, 5, 8, 12, and 15 at the last reading, and the speed only rises. (a) Estimate the distance covered with a left sum and a right sum. (b) Which sum runs high? (c) Give the trapezoid estimate.
How did you do?
(a) meters and meters. (b) The right sum runs high. (c) meters.
Each block is seconds wide, and the heights are the speeds; a speed times a time is a distance. The left sum reads the speed at the start of each block, the right sum at the end:
The speed only rises, so the left rectangles fall short and the right ones overshoot: the true distance sits between 112 and 160 meters, with the right sum on the high side. Averaging the two ends gives the trapezoid estimate, meters, the best single guess a table this coarse can offer.
Missed it? Take another pass at Section 1.1 before moving on.
Reveal a worked example
Estimate the area under from to with three rectangles (). Write the right sum in sigma notation, find the left sum, and check that they trap the exact area, which is 18.
With three strips of width on , the cut points are , and the right sample points are for , the sigma form from Section 1.1. The right sum reads the right edges:
The left sum reads the left edges , so . Since only rises on , the left rectangles fall short and the right ones overshoot, so : the two sums trap the exact area 18, though loosely, since three strips is coarse.
Can you reverse the derivative rules to find an antiderivative, write the general antiderivative and say why the is there, and pin down from a starting value?
Try it!
(a) Find the general antiderivative of , and check it by differentiating. (b) A cart rolls along a track with velocity meters per second, and at it sits meters past a marker. Find the position , then find where the cart is at seconds.
How did you do?
(a) . (b) , and meters.
(a) Reverse the power rule on each term, raising every exponent by one and dividing by the new exponent:
The is part of the answer, since any constant differentiates to 0 and so cannot be seen in . Check by differentiating back: , which is the you started with.
(b) Position is an antiderivative of velocity, so antidifferentiate up to a constant:
The reading pins the constant: setting gives , so . Then meters. The velocity alone could never have known the cart started 6 meters past the marker; that is exactly what the held a place for.
Missed it? Take another pass at Section 1.2 before moving on.
Reveal a worked example
(a) Find . (b) An object moves with velocity meters per second and starts at meters. Find , then its position at seconds.
(a) The power rule only reads a sum of powers, so rewrite the radical first: . Then reverse the power rule term by term, as in Section 1.2:
Dividing by is multiplying by , which turns into .
(b) Position is an antiderivative of velocity: . The start reading gives , so , and meters.
Can you read as the limit of those rectangle sums, interpret it as net signed area, and use the basic properties of the definite integral?
Try it!
(a) The line crosses the -axis at . Evaluate as net signed area by splitting into two triangles, then find the total area between the line and the axis on . (b) Suppose and . Find and .
How did you do?
(a) Net signed area ; total area . (b) and .
(a) The line runs from up through the axis at to . On it is below the axis, a triangle of base 3 and height 3, area , which the integral counts as negative. On it is above the axis, a triangle of base 1 and height 1, area , counted as positive. Add the signed pieces:
Total area ignores the sign and adds the plain areas: . The two differ because the definite integral lets the lower triangle subtract, while total area counts every piece as positive.
(b) Additivity glues the two adjacent intervals: . Reversing the limits flips the sign: .
Missed it? Take another pass at Section 1.3 before moving on.
Reveal a worked example
(a) Evaluate by geometry. (b) Evaluate , then find the total area between and the axis on .
(a) The curve is the top half of the circle , a circle of radius 4 centered at the origin (square both sides to see it). From to the region is a full semicircle, so the integral is half the circle's area, the geometry shortcut from Section 1.3:
(b) The line is odd, so on the symmetric interval the region below the axis on cancels the region above it on :
Total area, though, counts both triangles as positive. Each has base 3 and height 6, so each has area , and the total area is . Net signed area reads the sign; total area ignores it.
Can you evaluate a definite integral in one line with the Fundamental Theorem of Calculus, , and differentiate an accumulation function?
Try it!
(a) Evaluate in one line with the evaluation bar. (b) Find .
How did you do?
(a) . (b) .
(a) An antiderivative of is (check: ). Apply the evaluation bar, minding the negative value at the lower end:
(b) The upper limit is , not , so by Part 1 the integrand is read at the upper limit and multiplied by the chain-rule toll :
No integrating at all: Part 1 reads the answer straight off the integrand.
Missed it? Take another pass at Section 1.4 before moving on.
Reveal a worked example
(a) Evaluate . (b) Find .
(a) Antidifferentiate term by term: , since . Apply the bar from Section 1.4, watching the negative lower limit:
(b) The upper limit is , so the integrand is read at and multiplied by the toll :
Can you read the integral of a rate as the net change it produces, and tell net displacement apart from total distance traveled, splitting the motion wherever the velocity changes sign?
Try it!
(a) A cart on a track has velocity meters per second on , where a negative value means it rolls backward. Find its net displacement and its total distance traveled. (b) Water flows into a tank at liters per minute on , and the tank starts with 8 liters. How much water is added over the 4 minutes, and how much is in the tank at ?
How did you do?
(a) Net displacement m; total distance m. (b) 36 liters added; 44 liters in the tank at .
(a) Net displacement is the plain integral of velocity:
For total distance, first find where the cart turns around: at . On the velocity is negative, so there; on it is positive. Split and add the magnitudes:
Total distance is meters, well past the you would get by wrongly absolute-valuing the displacement. The absolute value belongs inside the integral, on the velocity, not outside on the result.
(b) The water added is the integral of the inflow rate:
The tank holds its start plus the change: liters.
Missed it? Take another pass at Section 1.5 before moving on.
Reveal a worked example
A particle moves with velocity meters per second on . Find its net displacement and its total distance traveled.
Factor to find the turning times, as in Section 1.5: , so at (the root sits outside ). Testing a point in each stretch, the velocity is positive on (forward) and negative on (backward). Net displacement lets the pieces cancel:
For total distance, split at and add the magnitudes. On the velocity is positive, on negative, so there :
Total distance is meters, against a net displacement of only meters, because the forward run erased part of the backward one.
Integration Rules: the toolkit that reaches every function
Unit 1 gave you antiderivatives, but only for functions whose reverse you can already see. The moving world's integrals ask for more: an exact curve length, a tank's exact volume, a wait time's exact odds. This unit builds the reach. Substitution runs the chain rule backward and becomes the workhorse; the trigonometric and the exponential-and-logarithmic families round out the table of antiderivatives; and the average value turns a whole integral into a single honest number, one the curve is guaranteed to hit somewhere.
- Use the properties and comparison bounds of the definite integral fluently, compute a function's average value as the height of the equal-area rectangle, and name the Mean Value Theorem for Integrals.
- Reverse the chain rule with a substitution , and change the limits of integration to finish a definite integral cleanly.
- Integrate the basic trigonometric functions, and use a trigonometric substitution to handle expressions carrying or .
- Integrate , , and , and fill the one gap the power rule left with .
The table of antiderivatives from Section 1.2 is honest but short: it only reaches functions you can un-differentiate at a glance. The 1600s did not have that luxury. To compute the area swept by a planet, the volume of a barrel, or the pull of gravity on a falling weight, mathematicians needed integrals of functions no simple reversal could touch, so they built techniques. The greatest of them is substitution, which is nothing more than the chain rule read backward, and it opens whole families at once. Two of those families, the trigonometric integrals and the exponential-and-logarithmic ones, come up so often they earn their own sections, and the trigonometric section carries one more move, a substitution that turns an awkward square root into a clean trig integral. The unit opens with the properties and the average value, because turning an integral into a single representative number is its own kind of power, then spends three sections widening what you can integrate. Section 2.1 sets the properties and the average value. Section 2.2 builds substitution. Section 2.3 takes on the trigonometric integrals and trigonometric substitution, and Section 2.4 finishes the table with exponentials and logarithms.
Section 2.1
Integral Properties and the Average Value
- Use the full property list fluently: linearity and additivity from Section 1.3, plus the comparison bounds .
- Compute the average value of a function on an interval, and read it as the height of the equal-area rectangle.
- Name and use the Mean Value Theorem for Integrals: a continuous function reaches its own average value somewhere on the interval.
- Bracket an integral you cannot evaluate by hand by trapping it between its smallest and largest possible rectangle.
What was the average temperature yesterday? You could grab a thermometer reading at noon and call it a day, but the temperature moved every minute, so one reading is a guess. You could take the high and the low and split the difference, but that ignores the long warm afternoon and the quick cold dawn. The honest average weighs every instant equally, and there are infinitely many of them. That is an integral in disguise: add up the temperature over the whole day and divide by the length of the day. The same move answers a question the dashboard already knows. Set the cruise control so the needle never moves, and the odometer still lands on the same total miles for the trip, as long as you pick the one steady speed that matches your real average. This section turns that idea into a formula, ties it to the properties of the integral, and names the theorem that says the steady speed you would have to hold is a speed the trip really did hit.
Build the intuition
Start with a picture, before any formula. Draw a curve over that rises and dips, and shade the region under it. Two easy rectangles pin the area down right away. Find the lowest the curve ever gets on the interval, call it , and the highest, call it . A rectangle of height sits entirely under the curve, so its area is too small. A rectangle of height swallows the whole region, so its area is too big. The true area is trapped between them:
Now squeeze from both sides toward the one rectangle that is exactly right. Somewhere between the too-short rectangle and the too-tall one is a rectangle whose area matches the region under the curve exactly. It has the same width , so it is pinned by its height alone. That height is the average value of the function, and Figure 2.1.1 shows all three rectangles at once.
Figure 2.1.1 The area under the curve is trapped between the low rectangle and the high rectangle . The one rectangle whose area is exactly right stands at the middle height : the curve pokes above that line by exactly as much as it dips below it, so the extra and the gap cancel.
Those two ideas, the trapping rectangles and the exact rectangle, are the whole section. The trapping rectangles are the comparison bounds, one more property to add to the four you already have. The exact rectangle is the average value. And the theorem that ends the section says the exact height is a height the curve genuinely reaches.
You already own four properties of the definite integral from Section 1.3: zero width, reversing the limits flips the sign, linearity, and additivity over adjacent intervals. Here are the comparison properties that finish the list. Let and be integrable on with .
The reason is the picture, not new machinery: a taller curve builds taller rectangles, and taller rectangles add to a bigger sum. The last line is the workhorse. It brackets an integral using nothing but the smallest and largest values the function takes, which you can often read off without integrating at all.
The average value of an integrable function on the interval is
Read it as area over width. The definite integral is the (net signed) area under the curve, and dividing by the width flattens that area into a rectangle of the same width. The height of that equal-area rectangle is . It is the single steady value that would fence off the same area as the whole moving curve.
Where does this new height land? The comparison bounds already answer that. Divide through by the positive width , and the middle becomes the average value:
So the average value is never smaller than the function's minimum and never larger than its maximum. That is exactly the room a continuous curve needs to hit it.
If is continuous on , then there is at least one point in where equals its own average value:
In words: the equal-area rectangle's height is a height the curve does reach. Set the cruise control at your average speed and, at some instant of the real drive, the speedometer read exactly that. The bounds above put the average value between the curve's low and high points, and a continuous curve cannot get from its low point to its high point without passing through every height in between, including this one. The Dig In turns that sentence into the proof.
Find the average value of on , draw its equal-area rectangle, and find the point the Mean Value Theorem for Integrals promises.
Solution. The width is , and you already know the area: Section 1.3 Exercise 3 (or one line of the Fundamental Theorem from Section 1.4) gives . Divide the area by the width:
So the wiggly region under , which has area 9, flattens into a rectangle 3 wide and 3 tall, area 9, exactly as it should (Figure 2.1.2). Now the point . The theorem says the curve reaches the height 3 somewhere on , so solve :
(The other root is outside the interval, so it does not count.) At the curve crosses the flat top of the equal-area rectangle. Notice the average value 3 is nowhere near the halfway point between the endpoint heights and , which would be . Averaging the two ends ignores how long the curve spends down low, and spends most of well below 9.
Figure 2.1.2 The region under from 0 to 3 has area 9. Flatten it into a rectangle of the same width 3 and it stands tall. The curve reaches that height at : left of it runs below the line, right of it pokes above, and the poke fills the gap.
Example 2.1.1 found the average value of a curve. Now predict one for a straight line before you compute it. For on , the graph is a plain slanted line, so its average value should sit right at the middle height. Predict that middle height, then confirm it with the formula .
Show solution
, the midpoint of the endpoint heights 0 and 4.
The area under from 0 to 4 is a triangle, . Divide by the width: . That matches the midpoint . For a straight line the average value really is the endpoint average, because a line climbs at a steady rate and spends equal time above and below its midline. Hold on to this: it is the one case where the shortcut works, and the Watch Out shows why it fails for everything that bends.
Average value earns its keep the moment the thing being averaged is a rate that will not hold still. The clearest case is a speed.
A car pulls away from a light, speeds up, then eases back to a stop, so its velocity is meters per second for seconds. (a) Find its average velocity over the 6 seconds. (b) Confirm this is the total distance divided by the time. (c) The Mean Value Theorem for Integrals says the car reached its average speed at some instant. When?
Solution. (a) Average velocity is the average value of . First the area, by the Fundamental Theorem (Section 1.4):
Then divide by the width :
(b) The integral of velocity is displacement (Section 1.5). Here across all of , so the car never reverses and the 36 meters of displacement is also the total distance traveled. Distance over time is meters per second, the same number. Average value and "distance over time" are two names for one idea. Had the car backed up partway, displacement and distance would part company, exactly the split from Section 1.5.
(c) Set equal to the average, , and solve:
Both roots, and , land inside , so the speedometer read exactly 6 meters per second twice: once speeding up, once slowing down. The car's top speed was meters per second, so the average of 6 sits sensibly below the peak.
Bracketing an integral you cannot solve
Some integrals have no elementary antiderivative, so the Fundamental Theorem has nothing to look up and honest exact evaluation is off the table. That is not a dead end. The comparison bounds still fence the answer into a range, which is often all a problem needs. The most famous example runs a bell curve.
The function is the heart of the bell curve, and it has no antiderivative you can write with the tools of this course. You still need to know roughly how big is. Bracket it with the comparison bounds.
Solution. Find the smallest and largest values of on . As runs from 0 to 1, the exponent drops from 0 down to , so falls steadily from to . The maximum is at , and the minimum is at . The width is , so the comparison bounds give
Without integrating anything, you know the answer sits between and . This is the same spirit as the rectangle estimates in Section 1.1: box the area in from below and above. A finer estimate (a midpoint sum, say) lands the true value near , which sits comfortably inside the bracket. When exact is impossible, a bracket you can trust beats a decimal you cannot.
A river's flow rate climbs with the spring melt and thins out by late summer, tracing a curve across the whole year. An engineer sizing a reservoir does not chase every wiggle in that curve, because the reservoir does the integrating for them: it collects water when the river runs high and releases it when the river runs low. What sets the size is the average flow, the height of the equal-area rectangle. Average flow times the seconds in the year is the total volume the reservoir must store and pass along. Over a year, roughly 31.5 million seconds, an average flow of 30 cubic meters per second delivers about 946 million cubic meters of water. Two rivers with completely different flow curves but the same average deliver the same yearly total, so the flat equal-area rectangle, not the jagged real curve, is what the concrete gets poured around. The Mean Value Theorem for Integrals adds a reassurance: that steady design flow is a rate the real river genuinely ran at, on some day of the year, not just a paper number.
Figure 2.1.1 makes the Mean Value Theorem for Integrals look obvious, but a picture is not a proof. Here is the proof, and it rides entirely on the comparison bounds you just met.
Let be continuous on . Because is continuous on a closed interval, it reaches a smallest value and a largest value somewhere on that interval (a continuous curve on a closed span cannot run off to infinity or sneak up on a value it never touches). By construction everywhere on , so the comparison bounds apply:
Divide every piece by the positive width . The middle turns into the average value:
So the average value is some number sitting between the low value and the high value . Now the one step that needs continuity. The function attains at some point and at some point. As travels from the first of those points to the second, moves continuously from to , and a continuous function cannot jump over any height between its start and its end. It must pass through every value from to along the way, and is one of those values. Therefore for some in . Multiply back through by and the theorem reads . The comparison bounds set the stage; continuity walks the curve through the average. Exercise 9 shows what breaks when continuity is missing.
Dig Deeper Integrals that run to infinity▸
Every integral so far has run between two finite numbers. What if the region never ends? Picture the area under starting at and stretching forever to the right. The region is infinitely long, yet its area might still be finite, because the curve dives toward zero fast enough that the new slivers you add out near infinity contribute almost nothing. To handle "forever" honestly, do not pretend infinity is a number. Integrate up to a movable finite wall , then slide the wall out and watch the running total. This is an improper integral:
If that limit is a finite number, the integral converges to it; if the running total grows without bound, the integral diverges. Test the two closest neighbors, and , and watch them split.
The tail vanishes, and the whole infinite region has area exactly 1. Now the same setup one notch shallower:
This one diverges: crawls upward forever, so there is no finite area. The two curves look almost identical near , but decays just fast enough to fence off a finite area while does not. That knife's edge sits exactly at the power 1: steeper than converges, and itself is the borderline that fails. This machinery is what lets a probability spread over an endless range of outcomes (a battery's lifetime, the wait for the next bus) still add up to a total certainty of 1, the you will meet in the probability section of Unit 3.
The average value is area over width, not the average of the two endpoints and not the average of a handful of readings. Averaging the endpoints, , only works when the graph is a straight line, as in the Quick Check. It falls apart the instant the curve bends: for on the endpoint average is , while the true average value is 3, because the curve loiters down low far longer than it spends up high, and area weighs that time. Two more cautions. The Mean Value Theorem for Integrals needs continuous; a function that jumps can skip right over its own average value (Exercise 9). And the comparison bounds only hold when and really are the smallest and largest values on the interval, so find the true minimum and maximum, do not eyeball the endpoints and hope.
(a) Find the average value of on . (b) The Mean Value Theorem for Integrals promises a point in with equal to that average; find it. (c) Is the average value the same as the endpoint average ? Say what that tells you.
Hint
(a) Use ; the area comes from the Fundamental Theorem (Section 1.4). (b) Set equal to your part (a) answer and solve for the root inside , rationalizing the radical. (c) Compare your number with and recall which shape makes the two agree.
Show solution
(a) . (b) . (c) No: the endpoint average is 6, not 4, which confirms the curve is not a straight line.
(a) The area is , so .
(b) Solve , so and , which lies in . (The negative root is outside the interval.)
(c) The endpoint average is , well above the true average value of 4. The gap is the tell: for a bending curve, area over width and the endpoint average are different numbers, and only the first is the honest average.
Drag the endpoints and and watch the average-value line, its equal-area rectangle, and the witness point where the curve meets its own average move together.
Frozen on : the area under is , the equal-area rectangle stands tall to match, and the curve reaches its average at .
Exercises 2.1
1. Compute each average value straight from the definition. (a) on . (b) on . Warm up
Hint
Use . In (a) the region is a rectangle; in (b) it is a triangle, so you can read each area off the geometry as in Section 1.3.
Show solution
(a) . (b) .
(a) A constant averages to itself: , so .
(b) The area under from 0 to 3 is a triangle of base 3 and height 6: , so , the midpoint height of the line.
2. A function satisfies for every in . Use the comparison bounds to trap between two numbers. Warm up
Hint
Apply from the Section 2.1 Comparison Bounds box, with , , and width .
Show solution
.
The width is . The low rectangle has area and the high rectangle has area , so .
3. Let on . (a) Find the average value. (b) Find the point in where the Mean Value Theorem for Integrals guarantees equals that average. Core
Hint 1
Mirror Example 2.1.1 in Section 2.1: get the area from the Fundamental Theorem (Section 1.4), then divide by the width.
Hint 2
. For part (b), set equal to your average value and keep the root that lands in .
Show solution
(a) . (b) .
(a) The area is , and the width is , so .
(b) Solve : the root in is . The curve reaches its average height of 7 at that point.
4. Find the average value of on (radians). Then state, without solving for it, why the Mean Value Theorem for Integrals guarantees a point where the curve equals that average. Core
Hint 1
You already know the area of one sine hump from Section 1.4: . Divide by the width .
Hint 2
The width is , so . For the guarantee, note is continuous, and sits between the minimum 0 and maximum 1 of on .
Show solution
; a continuous curve reaching both 0 and 1 must also reach .
The area under one hump is , and the width is , so . The equal-area rectangle over stands tall. Because is continuous and runs from 0 up to 1 and back to 0 on this interval, it passes through every height between 0 and 1, and is one of them. So the Mean Value Theorem for Integrals guarantees a point in with (in fact two of them, one on each side of the peak).
5. A cyclist's speed along a straight path is meters per second for seconds. (a) Find the average speed over the 2 seconds. (b) Show it equals the total distance divided by the time. State units. Core
Hint 1
Average speed is the average value of , as in Example 2.1.2 of Section 2.1. Get the distance from the Fundamental Theorem, then divide by the width.
Hint 2
. Since throughout, the integral of is both displacement and total distance (Section 1.5).
Show solution
(a) meters per second. (b) Distance 8 m over time 2 s gives meters per second, the same number.
(a) The distance is meters, and the time is seconds, so meters per second.
(b) Because on , the cyclist never reverses, so the 8 meters of displacement is also the total distance. Distance over time is meters per second, matching part (a): average value and "distance over time" are the same computation.
6. Without computing either integral first, use a comparison property to decide which is larger, or . Then evaluate both to confirm your call. Core
Hint 1
Compare the integrands on : for , how do and rank? The comparison property in Section 2.1 says the bigger integrand gives the bigger integral.
Hint 2
On , multiplying by another factor of can only shrink a value, so . Evaluate each with the power rule from Section 1.4 to check.
Show solution
is larger than .
On every satisfies , so multiplying by the extra factor can only shrink it: on the whole interval. By the comparison property, . Confirming by evaluation, and , and indeed .
7. The integral has no antiderivative you can write down. (a) Use the comparison bounds to trap it between two numbers, giving each exactly and rounded to three decimals. (b) A numerical estimate gives . Confirm it falls inside your bracket. Stretch
Hint 1
This is Example 2.1.3 of Section 2.1 with a different integrand. Find the smallest and largest values of on by checking whether it rises or falls.
Hint 2
Since increases on , so does its square root: the minimum is at and the maximum at . Then apply with width 2.
Show full solution
(a) , that is . (b) is between them, so it fits.
(a) On the inside climbs from 4 to 12, and the square root climbs with it, so the minimum is at and the maximum is at . With width , the comparison bounds give
(b) The estimate satisfies , so it lands inside the bracket, as any correct value must. The bracket is wide because the integrand climbs a lot across ; over a shorter interval the minimum and maximum would sit closer, and the bracket would tighten.
8. For what value of (with ) does have average value exactly 12 on ? Stretch
Hint 1
Write the average value of on as a formula in using , then set it equal to 12.
Hint 2
The area is , so the average value is . Solve for the positive .
Show full solution
.
The area under from 0 to is , so the average value is
Set this equal to 12: , so and (taking the positive root). As a check, , and .
9. The Section 2.1 Dig In proved the Mean Value Theorem for Integrals for a continuous function. This exercise probes the continuity hypothesis. (a) For on , find every in where equals the average value 3 from Example 2.1.1. (b) Build a function on that jumps (is not continuous) whose average value is a height the function never reaches, and explain why the theorem's proof breaks for it. Dig In
Hint 1
(a) Solve on and keep only the roots inside the interval. (b) Reread the last step of the Section 2.1 Dig In: a step function that takes only two values can have an average landing strictly between them.
Hint 2
(b) Try on and on . Compute its average value by area, then ask which heights the function ever outputs.
Show full solution
(a) Only . (b) The step function on and on has average value 1, a height it never takes; the proof breaks because a jump lets the curve skip past its average.
(a) Solve on . The roots are , but only lies in , so it is the single point where hits its average value. (A monotone curve meets each height once.)
(b) Take on and on . Its area is , so the average value is . But only ever outputs 0 or 2, never 1, so no gives . The Dig In's final step fails here: the function does not travel continuously from its low value 0 to its high value 2, it jumps between them at , so it skips right over the height 1 in the gap. Continuity was the hypothesis that closed that gap.
Check your own version. Whatever jumping function you build, two audits confirm the counterexample: compute its average value as (area) divided by (width) and confirm that number is strictly between two of the function's output levels with no output equal to it, and confirm the function jumps across that missing height rather than passing through it. If your average value happens to be an output the function does reach, the example does not show the failure, so shift the pieces until the average lands in a gap.
10. The Section 2.1 Dig Deeper defined and showed converges while diverges. (a) Evaluate . (b) Decide whether converges or diverges, and find its value or show it grows without bound. (c) State which powers give a convergent , based on your three examples. Dig Deeper
Hint 1
Follow the Section 2.1 Dig Deeper: replace with a wall , integrate with the power rule, then let and see whether the result settles or runs off.
Hint 2
(a) . (b) , whose antiderivative is ; watch whether settles or grows as .
Show full solution
(a) (converges). (b) diverges. (c) It converges exactly when the power is larger than 1.
(a) Integrate to a wall and take the limit:
The tail vanishes, so the region has area .
(b) With , whose antiderivative is :
Here grows without bound, so the integral diverges.
(c) Line up the three powers: converges, converges, diverges, and diverges. The pattern is that converges exactly when is larger than 1, with the borderline power 1 (the function ) itself diverging.
Check your own version. For any power you try, two audits settle it: integrating to a wall and letting , a convergent case must leave a finite constant with every term vanishing, and a divergent case must leave a term (a positive power of or a ) that grows. Your verdict should match the rule "converges when is larger than 1"; if it does not, recheck the sign of the exponent in the antiderivative.
The property list is now complete: on top of zero width, sign reversal, linearity, and additivity from Section 1.3, the comparison bounds trap an integral between its shortest and tallest rectangle, which brackets even integrals you cannot evaluate, like . The average value is area over width, the height of the equal-area rectangle, and the one steady value that fences off the same total: it is not the endpoint average unless the graph is a straight line. The Mean Value Theorem for Integrals says a continuous function reaches its average value at some point , so the steady design speed is a speed the real trip hit. The optional Dig Deeper stretched the integral to infinite intervals, where converges but diverges, a preview of the endless probabilities in Unit 3. Next, Section 2.2 opens the toolbox that reaches far more antiderivatives than Section 1.2 could: substitution, the chain rule run backward.
Section 2.2
Substitution
- Recognize a composition whose inside function's derivative is already sitting in the integrand, give or take a constant.
- Carry out u-substitution on an indefinite integral, then check the answer by differentiating it back.
- Change the limits on a definite integral, and say why that beats renaming back to .
- Fix a constant mismatch, and recognize when a missing variable factor means substitution simply does not apply.
In Section 1.2 you learned to undo a derivative by reading the power rule backward. That works when the integrand is a plain sum of powers, but the moving world rarely hands you one. It hands you compositions: a function tucked inside another function, the way buries inside a fifth power. In the Derivatives volume, the chain rule told you how to differentiate exactly these, and it did so by stacking rates: a nudge in moves the inside, and that move reaches the outside carrying a factor for each stage. This section runs that machine backward. If the chain rule builds a certain product by differentiating a composition, then spotting that product lets you name the composition it came from. That backward reading is called substitution, and it is the one general technique this whole unit turns on.
Build the intuition
The whole idea fits on one arrow, read in two directions. Suppose is an antiderivative of , and let be some inside function. Differentiate the composition with the chain rule and you get the derivative of the outside, read at the inside, times the derivative of the inside: . Read that left to right and it is differentiation. Read it right to left and it is a fact about antiderivatives: whatever product wears the shape , the function it came from is . Substitution is just the discipline for reading it backward without losing track of the pieces. Figure 2.2.1 draws the single arrow both ways.
Figure 2.2.1 One arrow, read both ways. The chain rule differentiates into the product ; substitution reads that same arrow backward, spots the product, and hands back .
Substitution (also called u-substitution) reverses the chain rule. You give the inside function a working name, , replace by , and integrate in the single new variable . It works exactly when the integrand is a composition multiplied by the inside's derivative , so that trading for leaves nothing behind in .
Let , with continuous, so that . If is an antiderivative of , then
For a definite integral, carry the limits along with the variable: as runs from to , the inside runs from to , and
Reason: this is the chain rule from the Derivatives volume, read right to left. The chain rule builds by differentiating , so is exactly the antiderivative that product came from. The method itself is bookkeeping: rename the inside as , trade for , integrate in , then rename back.
Find , and check the answer by differentiating it back.
Solution. The integrand is a composition, , times an extra factor of . The question worth asking first: is that extra factor the inside's derivative? The inside is , whose derivative is , and there is the , sitting right in front. So the substitution will be clean. Name the inside and record its differential:
Now every piece of the integral has a -costume: is , and is . Trade them in and integrate in :
The last line renames back to , because the question was asked in and the answer must speak too. Now the check, which costs one line. Differentiate with the power-chain rule:
That is the integrand exactly, so the answer is right. Make the differentiate-back check a habit: it settles every substitution in one line, and the drops out under it, right where it belongs.
Before you go on, predict: find . Name the inside , confirm its derivative is present, integrate in , and rename back.
Show solution
.
Set , so , and the is present, no fix needed. Then . Differentiate back to confirm: .
Find . Then look hard at , which is one small step away and yet out of substitution's reach.
Solution. Set , so . This time the integrand offers only , not the full that needs. The gap is a constant factor of , and a constant is allowed to cross the integral sign, so you can supply the missing and pay for it with a . Cleanest is to solve the differential for the piece the integrand does have:
Now every piece has a costume again:
Differentiating back gives , the integrand, so the landed in the right place.
Now the near miss. Drop the and you have . The same inside still wants , but now there is no anywhere to build it from. You cannot manufacture one: an is a variable, and a variable may never cross the integral sign the way that constant just did (the Watch Out below returns to this exact temptation). Substitution does not apply. The only honest route is to expand the fifth power into a polynomial and integrate it term by term, which is precisely the grinding labor substitution saves you whenever the inside's derivative is present. So the real skill of this section is the recognition itself: substitution pays off when the inside's derivative is already in the integrand, give or take a constant, and not otherwise.
Evaluate . Do it once by changing the limits, once by back-substituting, and compare the bookkeeping.
Solution. Borrow one fact for this example: , the exponential read backward. Section 2.4 proves it; here it is just the antiderivative you land on. The inside is , with , so , a constant fix like the last example.
Way one, change the limits. The original limits and are values of . Once you switch the variable to , the limits must switch too: as goes from to , the inside goes from to . Then
Because the limits are now -values, you evaluate in and you are finished: there is no trip back to .
Way two, back-substitute. Set the limits aside, find the indefinite integral in , rename back to , and only then evaluate at the original -limits:
Same number, , and the result is a plain number with no , the way every definite integral must be. Both ways are correct, but count the moves: back-substituting made you rename to and then plug in -values, two extra steps that do no mathematical work. Changing the limits skips that round trip. For a definite integral, change the limits.
Two whole families of integrals open up from this one technique. Substitution turns into a one-liner: with and , it becomes , a constant fix on a linear inside. And it reaches the one antiderivative the power rule from Section 1.2 could never touch, : whenever the top of a fraction is the derivative of the bottom, (for instance ). The trigonometric family is Section 2.3; the exponential and logarithmic family is Section 2.4. Substitution is the key that opens both doors.
Substitution is a change of scenery that leaves the area alone, and below you can watch it happen. The left panel shows territory in (Example 2.2.3's decaying twin, constant fix and all); the right panel shows the same integral after , where the integrand is just . Drag and on the left: the -window follows as , the evenly spaced -strips land as warped -strips, and the two shaded areas never disagree. Challenge: put the window at and look at the guide lines. In they are evenly spaced; in they bunch hard to the left, because near the warp factor is tiny.
A turbine flow meter has no clock. Water pushes a little rotor, the meter counts the rotor's spins, and each spin means a fixed slug of water has gone by. To report a flow in liters per second, the meter has to convert a total measured in spins into a total measured in seconds, and the bridge between the two rulers is the spin rate, spins per second. That is a substitution built in hardware. The volume is an integral of flow against time, but the quantity the rotor really tracks is the spin count , so is the exchange rate that trades one ruler for the other, exactly the factor this section trades for . Put in numbers: a rotor turning a steady spins per second, with each spin passing liters, reports liters per second, and it got there by carrying the spin rate as its conversion factor, never by guessing. The same move shows up wherever an instrument logs a rate against the wrong variable. A pump reads its own strokes, a fuel computer logs grams burned against engine revolutions rather than miles. Integrating back to the total you want is a change of variable, and the derivative of the swap is the gear that makes the units line up.
Substitution is not a lucky trick you memorize; it is the chain rule with the arrow reversed, and here is the whole reason in one pass. Suppose is an antiderivative of , so . Ask the Derivatives volume what happens when you differentiate the composition . The chain rule answers: the derivative of the outside, read at the inside, times the derivative of the inside,
Read that line left to right and it is the chain rule. Read it right to left and it is a statement about antiderivatives: the function whose derivative is is . That is the substitution rule word for word,
with no new machinery invented. Watch it run on Example 2.2.1, where , so , and . Then , and differentiating it by the chain rule returns , the integrand. The toolkit is self-consistent: every derivative rule you already own is an integration rule waiting to be read backward, and the chain rule's backward reading is the busiest one in this unit.
Dig Deeper When there is no inside to substitute: integration by parts▸
Substitution reverses the chain rule. The Derivatives volume had a second product-shaped rule, the product rule, and reversing it gives a second technique for a completely different shape of integrand: a product of two unrelated functions, where neither one is the other's inside derivative. The classic is . There is no composition here and no inside derivative to spot, so substitution has nothing to grab.
Start from the product rule, , and integrate both sides across the interval. The left side antidifferentiates to , and moving one term to the other side gives
the rule called integration by parts. It trades the integral you cannot do for a different one you can, so the whole art is the split: pick to be the part that gets simpler when you differentiate it, and to be a part you know how to integrate. For , choose
because differentiating simplifies it to , while integrates to itself. Then
Differentiate back to confirm: , the integrand. The split is the whole game. Had you chosen and instead, you would trade for , a worse integral with a higher power of . Parts is one of the reaches this book keeps in a Dig Deeper; substitution is the one general technique the main path needs.
Two slips cost the most here. First, the stray : after you substitute, every last must be gone, replaced by and . If an survives, the substitution is not finished, because you cannot integrate an expression that is half and half . The usual cause is a mismatched , so always solve the differential and trade for exactly what the integrand offers. Second, the illegal variable pull. You may move a constant across the integral sign to fix a numeric mismatch, the way the did in Example 2.2.2, but you may never move a variable. Facing , it is tempting to write and call it done, but is not a constant, so it cannot come out front, and that line is simply false. And on a definite integral, if you switch to but forget to switch the limits, you will evaluate a -antiderivative at -values and land on a wrong number. Change the limits, or rename back to before you evaluate: pick one and finish it.
(a) Find . (b) Find . (c) Evaluate , exact value first, then rounded to three decimal places.
Hint
Each part names the inside and trades for (Section 2.2). In (a) the is already exactly . In (b) and (c) the integrand offers only or , so you need a constant fix. For (c), change the limits to -values: runs from to , and you can finish entirely in .
Show solution
(a) . (b) . (c) .
(a) Set , so , already present. Then .
(b) Set , so and . Then
(c) Set , so ; the limits change from to :
Exercises 2.2
1. Find each general antiderivative by substitution. (a) . (b) . Warm up
Hint
In each, name the inside and write (Section 2.2). In (a) the inside is , whose derivative is already there; in (b) the inside is , whose derivative is already there. Both are clean, no constant fix needed.
Show solution
(a) . (b) .
(a) , , so .
(b) , , so .
2. For each integral, name a substitution , write , and say whether the inside's derivative is fully present or needs a constant fix. You do not have to finish the integral. (a) . (b) . (c) . Warm up
Hint
The inside is whatever sits under the power, inside the cosine, or in the denominator (Section 2.2). Once you name it, compute and compare it to the leftover factor multiplying : if they match up to a number, the derivative is present; if the only leftover is with no matching factor, you will need a constant.
Show solution
(a) , ; fully present. (b) , ; needs a constant fix (supply ). (c) , ; fully present.
(a) The inside is , whose derivative is , and the factor is right there in the integrand, so is exactly .
(b) The inside is , whose derivative is . The integrand carries only , not , so the mismatch is the constant : supply it and compensate with outside.
(c) The inside is , whose derivative is , and the factor is exactly the other part of the integrand, so .
3. Find . Core
Hint 1
The inside is , whose derivative is (Section 2.2). The integrand only offers , so this is a constant-fix problem like Example 2.2.2.
Hint 2
From you get , so . Substitute and carry the outside the integral.
Show solution
.
Set , so and . Then
Differentiate back: .
4. Find each integral. (a) . (b) . Core
Hint 1
Each inside is linear, so its derivative is a constant (Section 2.2): both are constant-fix problems. Name and read off before you integrate.
Hint 2
In (a), gives , so . In (b), gives , so ; mind the sign.
Show solution
(a) . (b) .
(a) , , so .
(b) , , so . The negative from survives into the answer.
5. Evaluate the definite integral by changing the limits. Core
Hint 1
Substitute with (Section 2.2), a constant fix since only is present. Because it is a definite integral, change the limits to -values so you can finish without renaming back.
Hint 2
From , and as runs from to the inside runs from to . Evaluate and remember a definite integral is a number, with no .
Show solution
.
Set , so . The limits change from to :
6. Find . Core
Hint 1
The messy factor is a clue, not an obstacle: it is the derivative of something in the integrand (Section 2.2). Ask which inside function has as its derivative.
Hint 2
Set ; then , the whole other factor, so no constant fix is needed. The integral becomes .
Show solution
.
Set , so , exactly the leftover factor. Then . Differentiating back returns , the integrand.
7. Find . The inside's derivative is not present, but a rewrite still lets substitution work. Stretch
Hint 1
Set , so (Section 2.2). The trouble is the loose out front, but you can rewrite it: since , you also know .
Hint 2
Substituting and turns the integrand into . Expand it into and integrate term by term.
Show full solution
.
Set , so and . Rewrite everything in :
Differentiating back gives , the integrand.
8. One of these yields to substitution and one does not. Decide which is which, then evaluate each. (a) . (b) . Stretch
Hint 1
Both have inside , whose derivative is . Ask which integrand carries an to build from (Section 2.2). Example 2.2.2 is the model: the one with no is out of reach.
Hint 2
For (a), , a constant fix. For (b) there is no to make , so substitution fails: expand into a polynomial and integrate term by term.
Show full solution
(a) . (b) .
(a) Substitution works. Set , so , and .
(b) There is no factor, so substitution has nothing to build from. Expand instead: , then integrate term by term:
The contrast is the lesson: the loose in part (a) is exactly what makes substitution possible, and losing it forces the long way.
9. The Section 2.2 Dig In showed that because the chain rule says , the antiderivative of must be . Build your own instance of this. Choose an outside antiderivative and an inside function , form the integrand (where ), integrate it by substitution, and confirm with the chain rule that differentiating your answer returns the integrand. Dig In
Hint 1
Reread the Section 2.2 Dig In. Pick a friendly outside like (so ) and a friendly inside like (so ); then the integrand you build is .
Hint 2
With those choices, . Integrate that by substitution and you should land exactly on , plus .
Show full solution
Worked with and : the integrand is , its integral is , and the chain rule differentiates that back to the integrand. The identity holds.
Here means , and , so the integrand is
Integrate by substitution with , , so :
That is , as promised. The chain-rule check: , the integrand exactly.
Check your own version. Whatever and you chose, two things must line up. Your substitution answer must simplify to , your own outside antiderivative wrapped around your own inside; and differentiating that by the chain rule must return the integrand you built. If your integral does not collapse to , check that the factor multiplying the composition really was , give or take a constant, and not some other function of .
10. The Section 2.2 Dig Deeper introduced integration by parts, , for a product with no inside derivative to substitute. Apply it to a fresh product of your own: pick a product of two unrelated functions (something like or ), name your choice of and , work the formula, and check by differentiating that you recover the integrand. Dig Deeper
Hint 1
Reread the Section 2.2 Dig Deeper and copy its discipline: pick to be the part that gets simpler when differentiated (a plain power of is ideal), and to be the part you can integrate.
Hint 2
For , take (so ) and (so , using from the trig table, previewed for Section 2.3). Then assemble .
Show full solution
Worked with , taking and : the answer is , and differentiating it returns . Parts works.
Choose (so ) and (so , borrowing from the trig table that Section 2.3 develops). Then
Differentiate back: , the integrand.
Check your own version. Differentiate your final answer: it must return the exact product you started with, or a term is misplaced. Then test your split. Swap the roles, letting be the part you first integrated and the part you first differentiated, and set up parts again; if the new integral on the right is worse than the one you started with (a higher power of , say), your original choice of and was the right one.
Substitution is the chain rule read backward: name the inside , trade for , integrate in the single variable , and rename back, carrying a on every indefinite answer. It pays off exactly when the inside's derivative is already in the integrand, give or take a constant you may supply from outside the integral (a variable you may not). For a definite integral, change the limits to -values and finish in , skipping the round trip back to ; the answer is a plain number, with no . The differentiate-back check settles any result in one line, and the Dig In showed why it always must: differentiating by the chain rule rebuilds the very integrand you started from. Two whole families fall to this one key: the trigonometric integrals of Section 2.3, where an inside like is waiting, and the exponential and logarithmic integrals of Section 2.4, including the pattern. Next, the trigonometric family.
Section 2.3
Trigonometric Integrals and Substitution
- Integrate the basic trig functions by reading the derivative table backward: , , , and .
- Use substitution to reach , , and simple products like .
- Integrate a squared trig function by rewriting it with a half-angle identity first.
- Run a trigonometric substitution, or , to clear a or an , producing the and forms.
- Work in radians, and check every answer by differentiating it back.
A tide does not only rise and fall; it moves water. Stand at a narrow channel between a bay and the open sea and watch the current. It rushes in, slows, stops at high water, then reverses and rushes back out, over and over, all day long. The flow rate traces a smooth wave, up and down and up again. Yet the water it carries keeps piling up: so many gallons in on this tide, so many out on the next. The rate repeats, but the total it builds does not reset. Everything that cycles works this way. A pendulum, the current in a wall socket, a spring bouncing a weight: the rate swings back and forth, and the running total is still an area under that swinging rate. This section hands you the antiderivatives of the functions that cycle, sine and cosine and their relatives, so you can total a rate that never holds still and never stops repeating.
Build the intuition
In the Derivatives volume you learned that the slope of the sine curve is the cosine curve, and the slope of cosine is negative sine. An antiderivative just reads that backward. If differentiating sine gives cosine, then an antiderivative of cosine is sine. Every derivative fact you already own is an integral fact, waiting to be read the other way. So the four antiderivatives that open this section are nothing new to memorize; they are the derivative table from your last course, turned around. The one thing to keep an eye on is the sign, which is where the minus in cosine's derivative shows up, and the Watch Out box returns to it. First, a picture of what one of these integrals means: the area under a single arch of the sine curve.
Figure 2.3.1 One arch of rises from 0, crests at 1, and returns to 0 across . The area it traps is exactly , a clean whole number falling out of a curve that never sits still. Example 2.3.1 computes it.
Read straight off the derivative table from the Derivatives volume, each derivative fact reversed:
The reason is one line each: differentiate the right side and you land back on the integrand. Sine's antiderivative carries a minus because cosine's derivative does, the single sign to guard. All four run in radians; a refresher on these functions lives in Appendix A.2, and the identities in Appendix A.3.
(a) Find . (b) Find the area under one arch of the sine curve, .
Solution. (a) Antidifferentiate term by term, each line straight off the rule box, and carry one for the whole expression:
(b) This is a definite integral, so it evaluates to a number and needs no . Use as the antiderivative of , then the evaluation bar:
The shaded arch in Figure 2.3.1 has area exactly 2. Notice the antiderivative is negative cosine, yet the definite integral came out a clean positive 2: the minus sign lives in the antiderivative, not in the area.
Before you go on, predict: what is , the area under the cosine curve from 0 to its first zero? Then evaluate it with the antiderivative .
Show solution
.
The antiderivative of is , so
One square unit of area under the first quarter of the cosine wave.
The four rules cover the functions whose antiderivatives are single named functions. Substitution (Section 2.2) reaches a whole layer more. Find (a) and (b) .
Solution. (a) Tangent is not on the rule box, but rewrite it as a quotient and a substitution appears. Since , let , so , which means :
The absolute value is required, exactly as in ; that logarithm returns as its own topic in Section 2.4. Check by differentiating: .
(b) The integrand is a product, and one factor is the derivative of the inside of the other: is the derivative of . Let , so :
Differentiate back: .
There is no antiderivative of to read off a table, and no substitution helps, since the derivative of the inside, , is nowhere in sight. The move is to rewrite with a half-angle identity (Appendix A.3) before integrating. Find (a) and (b) .
Solution. (a) Replace with its half-angle form , a sum of things you can integrate:
The antiderivative of is ; the is the constant fix from Section 2.2, since the inside differentiates to 2.
(b) Evaluate from 0 to :
Both sine terms vanish, because and are 0. That answer carries a quiet, useful fact: the average height of over the full period is . A repeating square, averaged over its cycle, sits at exactly one half. The connection box below spends that fact.
A trigonometric substitution replaces the variable with a trig function of a new angle , chosen so that a square root or a sum of squares collapses through a Pythagorean identity. For an expression carrying , substitute , with kept in arcsine's range , so is never negative and the root comes off cleanly; for one carrying , substitute . In each case the identity (or its cousin ) clears the awkward part, and the integral left behind is one you can already do.
Figure 2.3.2 The substitution as a right triangle: the side opposite is , the hypotenuse is , so , and the adjacent side is . Reading back out gives .
Two integrals that look impossible give way at once to a trigonometric substitution, and they bring the inverse trig functions into your antiderivative table. Take . (a) Show . (b) Show .
Solution. (a) Let , so . The root is where the substitution earns its keep:
The Pythagorean identity turned into , and because stays in arcsine's range (Definition 2.3.1), is never negative, so the square root came off cleanly as with no absolute value needed. Substitute both pieces:
since means , so (Figure 2.3.2). A concrete case, with : .
(b) Let , so . This time the sum of squares collapses:
Substitute:
since . A definite case, with , from 0 to 2:
Both answers pass the differentiate-back check, the surest confirmation: and .
Example 2.3.4 earned both by substitution, and differentiating the right side confirms each. These are the two inverse-trig forms this course carries. Match a square root of to arcsine and a sum to arctangent, read off the expression, and the answer is one line. Inverse functions themselves get a refresher in Appendix A.6.
The wall socket in your home runs on sine, and on the fact you just proved about . Household current does not flow one way; it alternates, sixty full cycles a second, its voltage tracing a sine wave that swings from a positive peak to an equal negative peak and back. If it swings equally both ways, why does the utility call it 120 volts? Because what heats a filament or turns a motor is not the voltage but the power, and power depends on the voltage squared. Average the square of that sine wave over a cycle and, by the half fact from Example 2.3.3, you get exactly of the peak's square. Take the square root of that average, the root mean square, and you get the peak divided by . Working it the other way, a 120 volt rating means a peak of volts, about 170 in round numbers. The number the meter reports, the one your appliances are built around, is an average of a squared sine, the same integral you just ran. The signed total tells a second story: a buoy bobbing on that kind of wave rises and falls, and over one full period its net rise is zero, the ups and the downs canceling as signed area, even while the distance it covered piles up.
Here is the whole engine behind the rule box, and it is a habit worth making permanent: every derivative fact is an integral fact, read the other way. Take the two rules that look least obvious, the ones with the secants. From the Derivatives volume, . Read that right to left: a function whose derivative is is , so
Likewise , and reading it backward gives
No new work and no new trick, just a derivative table read in the opposite direction, and every line is checkable on the spot by differentiating the answer. Any derivative fact you meet becomes an antiderivative the instant you are willing to read it backward. That single idea is the whole of the basic rule box, and it is why the four antiderivatives were nothing to memorize.
Dig Deeper The other squared cases▸
Example 2.3.3 handled with a half-angle identity. Its partner yields to the matching identity , the plus-sign cousin (Appendix A.3):
The only change from the sine case is a plus where there was a minus, matching the identity. A product of two squares takes one more turn of the same handle. To integrate , first fold the two together with the double-angle identity , then square and reduce again:
Now it integrates term by term:
No reduction formulas and no memorized templates: every squared or product-of-squares trig integral in this course comes apart with one or two of the same identities, applied until nothing is squared.
Three slips cost the most points here. First and worst, the sign: , not . The minus is there because cosine's derivative is minus sine, so cosine is the antiderivative of negative sine, and sine's own antiderivative has to pick up the minus to compensate. When in doubt, differentiate your answer: if you wrote , its derivative is , the wrong sign, and the mistake shows itself in one line. Second, radians, always. Every antiderivative in this section, and every derivative behind it, is true only when is measured in radians; a stray degree mode makes false and drags every integral here off with it. Third, a squared trig function has no shortcut: is not or any pattern like it. There is no inside derivative to substitute, so you must rewrite with an identity first (Example 2.3.3). Guessing a power-rule answer here is the single most common wrong move in the section.
(a) Find . (b) Find by substitution. (c) Find exactly.
Hint
(a) Each term is straight off the basic rule box in Section 2.3; keep one . (b) A product with the derivative of the inside present: let , and mind the sign of (Section 2.2). (c) Match to the arcsine form with .
Show solution
(a) . (b) . (c) .
(a) Antidifferentiate term by term: .
(b) Let , so , which means :
The minus comes from , not from the power rule; differentiate back to confirm.
(c) With , the arcsine form gives , so
The angle whose sine is is (Appendix A.1).
Exercises 2.3
1. Find each general antiderivative. (a) . (b) . Warm up
Hint
Antidifferentiate each term straight off the basic rule box in Section 2.3, and keep one . Watch the sign on the sine terms: .
Show solution
(a) . (b) .
(a) The antiderivative of is and of is , so the integral is . Differentiate back: .
(b) The antiderivative of is , and of is , so the integral is .
2. Evaluate each definite integral, exactly. (a) . (b) . Warm up
Hint
Find an antiderivative, then use the evaluation bar . A definite integral is a number, so there is no . Exact trig values live in Appendix A.1.
Show solution
(a) . (b) .
(a) .
(b) .
3. Use substitution to find each. (a) . (b) . Core
Hint 1
Each is a product where one factor is the derivative of the inside of the other, so substitution fits (Section 2.2), the same move as in Example 2.3.2.
Hint 2
(a) Let , so ; the integral becomes . (b) Let , so ; the integral becomes .
Show solution
(a) . (b) .
(a) With and : . Differentiate back: .
(b) With and : .
4. Find . Core
Hint 1
Cotangent is not on the rule box, but rewrite it as a quotient and a substitution appears, exactly the way tangent did in Example 2.3.2.
Hint 2
Write and let , so . The integral becomes , and remember the absolute value.
Show solution
.
Rewrite and let , so :
The absolute value is required, just as in . Differentiate back: .
5. Match each to an inverse-trig form and integrate. (a) . (b) . Core
Hint 1
Match each to one of the two inverse-trig forms from Section 2.3: a square root of points to arcsine, a sum points to arctangent.
Hint 2
In both, , so . Read the matching form and fill in ; the arctangent form carries an extra out front.
Show solution
(a) . (b) .
(a) With , the arcsine form gives .
(b) With , the arctangent form gives . Differentiate back: .
6. Evaluate each definite integral, exactly, then as a decimal to three places. (a) . (b) . Core
Hint 1
Find the antiderivative with the matching inverse-trig form from Section 2.3, then evaluate with the bar. Keep the exact answer carrying before you round.
Hint 2
(a) , antiderivative ; at the angle has sine . (b) , antiderivative ; at the angle has tangent 1.
Show solution
(a) . (b) .
(a) .
(b) .
7. Find . (An odd power hides a substitution, even though a squared trig function does not.) Stretch
Hint 1
A squared trig function needs an identity, but an odd power hides a substitution. Peel off one sine and convert the rest to cosine with the Pythagorean identity (Appendix A.3).
Hint 2
Write , then let , so . The integral becomes .
Show full solution
.
Split off one sine and rewrite the rest with the Pythagorean identity: . Let , so :
Differentiate back: .
8. Evaluate by running the substitution from scratch: find , clear the radical with the Pythagorean identity, change the limits, integrate in , and convert back. Do not simply quote the arcsine form. Stretch
Hint 1
Match to with , so the substitution is (Section 2.3). Differentiate it to get .
Hint 2
, and by the Pythagorean identity, so the two cancel. The limits and become and .
Show full solution
.
The radical is with , so substitute , which gives . The Pythagorean identity clears the root:
Change the limits: gives , and gives , so . The two factors now cancel:
Running the substitution lands the same the arcsine form would hand you, but now you have built that form instead of quoting it.
9. The Section 2.3 Dig In turned derivative facts into integral facts by reading the table backward. Do the same with two functions the rule box did not list. (a) From the Derivatives volume, . Use it to write . (b) From , write . (c) Choose any derivative fact you know, reverse it into an antiderivative, and confirm it by differentiating. Dig In
Hint 1
Reread the Section 2.3 Dig In: a derivative fact says at once that . For (a) and (b), the derivative fact is handed to you, so read it the other way, minus sign and all.
Hint 2
For (a), the derivative of is , so the derivative of is ; that is your antiderivative. For (b), the derivative of is , so adjust the sign to land on alone.
Show full solution
(a) . (b) . (c) Answers vary; see the check below.
(a) Since , differentiating gives , so .
(b) Since , differentiating gives , so .
(c) One sample: from , reading backward gives , and differentiating returns .
Check your own version. Differentiate whatever antiderivative you wrote: you must land exactly on the integrand, sign and all. If you are off by a sign, the derivative rule you reversed carried a minus you dropped, and differentiating your answer shows it in one line.
10. The Section 2.3 Dig Deeper integrated and with half-angle and double-angle identities. Put those to work. (a) Evaluate . (b) Evaluate . (c) Invent your own squared or product-of-squares trig integral over an interval you choose, evaluate it with an identity, and check it. Dig Deeper
Hint 1
Use the antiderivatives the Section 2.3 Dig Deeper built: and . Then evaluate with the bar.
Hint 2
At , and , so both sine terms drop and only the or piece survives.
Show full solution
(a) . (b) . (c) Answers vary; see the check below.
(a) , since .
(b) , since .
Check your own version. Differentiate your antiderivative back to the squared expression you started with, then confirm your evaluation by checking that every sine term you dropped really is 0 at your chosen limits. If a sine term does not vanish, keep it; a nonzero boundary term is fine, it just means your interval did not land on a clean multiple of the period.
The basic trigonometric antiderivatives are the derivative table read backward: , , , and , with the lone minus sign on sine the one thing to guard. Substitution reaches the next layer: a quotient like becomes once you set , and a product with the inside's derivative present, like , falls to a power of . A squared trig function has no shortcut and no substitution; rewrite it with a half-angle identity first, which also shows that the average of over a full cycle is exactly . And a trigonometric substitution, or , clears a or an through a Pythagorean identity and brings in the two inverse-trig forms, and . One family is left, the one that grows in proportion to itself and fills the hole from Section 1.2: the exponentials and logarithms of Section 2.4.
Section 2.4
Exponential and Logarithmic Integrals
- Integrate , , and for any positive base .
- Fill the gap left open in Section 1.2: , and explain why the absolute value belongs there.
- Use substitution to reach the logarithm pattern .
- Read an exponential accumulation in an application, totaling a proportional rate exactly and then as a decimal.
Two questions have been left open since Unit 1, and this section closes both. Back in Section 1.2 you undid derivatives by reversing the power rule, and it handled every power of but one. The rule raised the exponent by one and divided, so the exponent jammed it: dividing by zero. That single case, the integral of , got a promissory note that read "waits for the logarithm," and it has been waiting ever since. The second loose end is a whole family of real quantities that all share one habit: they change in proportion to how much of them is there. Money in an account earns on the balance it already holds, so a bigger balance grows faster. A population makes new members out of its current members, so a bigger population grows faster. This section hands you the antiderivatives for that self-feeding family and, in the same stroke, pays off the promissory note. When it is done, every function this course meets has an antiderivative you can write down.
Build the intuition
Here is the one move behind the whole section: every derivative fact, read backward, is an antiderivative fact. You built a table of derivatives in the first volume of this pair, and Appendix A.10 keeps it within reach. Antidifferentiating is just reading that table from right to left. Two of its rows carry this section. The first: the natural exponential is the rare function that is its own derivative, so read backward it is its own antiderivative, and . The second: the natural logarithm was built so that , so read backward, the missing antiderivative of is a logarithm. (Rusty on either function? Appendix A.5 refreshes exponentials, Appendix A.4 refreshes logarithms.) Everything else here is those two facts dressed for work: a constant in the exponent, a general base, or an inside function waiting for substitution.
Read straight off the derivative table:
These hold for any nonzero constant and any base with . The reasons are one line each. Since , dividing by gives back the antiderivative, which is why the rides out front. Since , dividing by the constant undoes it. You never have to trust these on faith: differentiate the right side and you land back on the integrand.
Find each indefinite integral. (a) . (b) . (c) .
Solution. (a) A sum, so integrate term by term: is its own antiderivative, and the power rule from Section 1.2 handles :
(b) The exponent is with , so divide by :
(c) A general base with , so divide by :
Every one carries its , and every one can be checked in a single line by differentiating back: , and .
Before you go on, predict each integral, then differentiate your prediction to check it: (a) . (b) . (c) .
Show solution
(a) . (b) . (c) .
Each answers to one row of the rule: with the power rule in (a), the form with in (b), and the general base with in (c). Differentiate (b) back and the chain rule returns , the integrand.
The case the reverse power rule could not touch:
The absolute value is not decoration. The function is defined for negative as well as positive, but itself only exists for . The bars repair the mismatch: is defined everywhere is, and its derivative works out to on each side of zero (the Dig In checks both sides). So one formula antidifferentiates across its whole domain, the negative half included, as long as you keep the bars.
Figure 2.4.1 The shaded area under from to is . Reading the upper limit as a variable turns this picture into a definition of the logarithm.
That picture is the cleanest way to define the natural logarithm in the first place. Let be the running area under starting at :
Then (no width, no area), and by the Fundamental Theorem (Section 1.4) the derivative of this accumulation function is exactly its integrand , which is the derivative fact this section reversed. The logarithm and the antiderivative of are not two ideas that happen to agree; they are one object seen from two sides. Appendix A.4 has the log laws this view makes visual: doubling the upper limit tacks on a fixed strip of area, which is why never changes, no matter what is.
Substitution (Section 2.2) turns the fact into a pattern you will reach for constantly. Whenever the top of a fraction is the derivative of the bottom, the integral is a logarithm:
Reason: let , so , and the integral becomes . The whole skill is spotting a bottom whose derivative is sitting right on top, up to a constant you can supply.
Find each indefinite integral. (a) . (b) , the loose end left in Section 2.3.
Solution. (a) The bottom is , whose derivative is , sitting exactly on top. Let , so :
The bars come off here because is positive for every , so its absolute value is just itself.
(b) Write . Now the bottom is , whose derivative is : the top is the derivative up to the sign. Let , so , which means :
The bars stay this time, since turns negative on part of its range. This is the result promised in Section 2.3, now shown to be one more instance of the same pattern.
A bacteria culture grows so that new cells appear at a rate of cells per hour, hours after noon. (a) How many cells does the culture add during the first 4 hours? Give the exact value, then a decimal. (b) Compare the cells added during the first hour ( to ) with the cells added during the fourth hour ( to ).
Solution. (a) The total added is the integral of the rate, the same move the odometer makes when it sums the speedometer into a distance (Section 1.5). Since has antiderivative , the constant rides along:
(b) Same antiderivative, different limits. Over the first hour,
and over the fourth hour,
The fourth hour adds about four and a half times as many cells as the first, though both stretches are exactly one hour long. That is the whole personality of proportional growth: the rate feeds on the amount, so a later hour, starting from a larger culture, piles on far more.
Figure 2.4.2 The growth rate and the cells it piles up. The whole shaded area from to is the total, cells. Because the rate feeds on the amount, the first-hour slab is a sliver beside the fourth-hour slab, which stacks up more than four times taller.
The same integral turns up wherever a rate is set by an amount. A charged capacitor pushes current through a resistor at a rate that fades as the capacitor empties, and the total charge it delivers over all time is the integral of that current, which collapses to a clean , the whole tank drained. A savings account under continuous compounding grows by the same law, so the interest it stacks up over a stretch of years is one exponential integral. A lump of radioactive material sheds particles at a rate proportional to how much is left, and the count over any window is again this integral. The logarithm sits on the other face of the same coin: because is what antidifferentiates , it is the natural ruler for anything that multiplies rather than adds, which is exactly why earthquakes (Richter), sound (decibels), and acidity (pH) are all read on logarithmic scales. In every one of these the shape is the dashboard's: a rate on one gauge, its running total on another, and a single antiderivative carrying you from the first reading to the second.
The rule rests on one claim worth checking yourself: that on both sides of zero, not just the friendly right side. The absolute value splits into two cases, so take each in turn. For the bars do nothing, since , so and
For the bars flip the sign, since , so . That is a composition with inside , so the chain rule applies:
Both cases land on the same . That is the whole reason the absolute value earns its place: it lets one formula, , serve as an antiderivative of on the entire domain, negative half included. Drop the bars and you keep only the right half, so any integral of that wanders into negative (Exercise 9 has one) falls apart.
Dig Deeper Splitting a fraction into logs▸
Substitution reaches , but plenty of rational functions hide their logarithm until you take them apart. Partial fractions is the technique: rewrite one awkward fraction as a sum of simpler ones, each of which integrates to a logarithm. Take . The bottom factors as , and you can always find constants and so that
Clear the denominators by multiplying through: . Setting kills the term and gives , so ; setting kills the term and gives , so . The decomposition is
Each piece is now a with , so each integrates to a logarithm:
where the last step uses the log law that a difference of logs is the log of a quotient (Appendix A.4). Differentiate that answer and you land back on , the check every antiderivative comes with. Partial fractions is one of three optional reaches Unit 2 tucks into its Dig Deepers, next to integration by parts and improper integrals; the main road stays substitution plus the direct rules. Exercise 10 hands you a denominator of your own.
Three slips cluster here. First, the trap from Section 1.2, now paid off: is not and not any power at all; it is . If you ever catch yourself dividing by a zero exponent, this is the case you have wandered into. Second, do not drop the absolute value. The honest answer is , not , because the bare is undefined for negative and so cannot be the antiderivative there. The one time the bars may come off is when the argument is provably positive, like , where the absolute value would change nothing. Third, carries the . The antiderivative of is , not : differentiating hands back an extra factor of , and the is there to cancel it. It is the same constant-fix that substitution asks for (Section 2.2), just committed to memory.
(a) Find . (b) Find . (c) Find .
Hint
(a) Two rows of the rule box: is its own antiderivative, and integrates to a logarithm with the bars. (b) The exponent is with , so divide by . (c) This is the logarithm pattern from Section 2.4: check whether the top is the derivative of the bottom.
Show solution
(a) . (b) . (c) .
(a) The term is its own antiderivative, and integrates to , so the integral is .
(b) With , divide by : .
(c) The bottom is , whose derivative is , sitting exactly on top. Let , so :
Exercises 2.4
1. Find each indefinite integral. (a) . (b) . Warm up
Hint
Two rows of the Section 2.4 rule box: is its own antiderivative and the power rule handles ; and integrates to a logarithm, so keep the absolute value.
Show solution
(a) . (b) .
(a) The term copies itself, and , so the integral is .
(b) A constant slides through the integral sign: . The bars stay, since lives on both sides of zero.
2. Find each indefinite integral. (a) . (b) . (c) . Warm up
Hint
All three are the form from Section 2.4: read off the exponent (it can be negative, and it can be a fraction), and divide by that .
Show solution
(a) . (b) . (c) .
(a) Here , so .
(b) Here , so .
(c) The exponent is with , so divide by , which multiplies by : .
3. Find each indefinite integral. (a) . (b) . (c) . Core
Hint 1
Parts (a) and (b) are the general-base form from Section 2.4: an antiderivative of divides by . Part (c) is a sum, so integrate each piece on its own.
Hint 2
Write for each base, and remember that is its own antiderivative (its base is , and , so no denominator appears).
Show solution
(a) . (b) . (c) .
(a) With : .
(b) With : .
(c) Integrate term by term: gives and gives , so the sum is . The term needs no denominator because .
4. Find each indefinite integral by recognizing the logarithm pattern. (a) . (b) . Core
Hint 1
Each is from Section 2.4: check that the top is exactly the derivative of the bottom, then the integral is .
Hint 2
In (a), ; in (b), . Both bottoms are positive for every , so the bars are optional.
Show solution
(a) . (b) .
(a) Let , so , and the top is exactly : . The bars drop because always.
(b) Let , so : , again with the bars optional since .
5. These need a constant fix before the logarithm pattern fits. (a) . (b) . Core
Hint 1
The top is not quite the derivative of the bottom, but it is off only by a constant, which you may move across the integral sign (Section 2.2). Set equal to the bottom and write out .
Hint 2
In (a), , so . In (b), gives , so .
Show solution
(a) . (b) .
(a) Let , so and :
(b) Let , so and :
The bars stay in (b), since can be negative.
6. Two trigonometric integrals from Section 2.3, each a logarithm in disguise. (a) . (b) . Stretch
Hint 1
Rewrite each as a quotient of sine and cosine, then look for the logarithm pattern from Section 2.4: is the top the derivative of the bottom, up to a sign?
Hint 2
For (a), gives . For (b), , and gives , which is exactly the top.
Show full solution
(a) . (b) .
(a) Write , let , so and :
(b) Write , let , so , which is exactly the top:
The bars stay in both, since sine and cosine each turn negative on part of their range.
7. Evaluate each definite integral, so the answer is a number with no . (a) . (b) . Core
Hint 1
Find the antiderivative from Section 2.4, then use the evaluation bar from Section 1.4: subtract the value at the lower limit from the value at the upper limit.
Hint 2
In (a), the antiderivative is , and , . In (b), the antiderivative is , and , .
Show solution
(a) . (b) .
(a)
(b)
8. An account's balance grows at a continuous rate of dollars per year, years after it opens. How much does the balance grow over the first 6 years? Give the exact value, then a decimal, with units. Stretch
Hint 1
This is Example 2.4.3 with money instead of cells: the total growth is the integral of the rate over (Section 1.5), and the rate is an form (Section 2.4).
Hint 2
Here , so has antiderivative . Evaluate , and keep exact before rounding.
Show full solution
dollars dollars.
The growth is the integral of the rate over the first 6 years. Since has antiderivative , the constant rides along:
The balance itself keeps climbing across those years, so the interest it throws off grows too, which is why the total is more than the dollars a flat rate would have added.
9. The Section 2.4 Dig In checked on both sides of zero. Put the negative side to work. (a) Starting from for , differentiate with the chain rule to confirm the derivative is . (b) Use to evaluate , and say why the answer should come out negative. (c) Make your own: pick an interval lying entirely on the negative side of zero, evaluate with , and check the sign against the graph of . Dig In
Hint 1
Reread the Section 2.4 Dig In. For (a), the inside function is , so its derivative comes along as a factor. For (b), the absolute value is what lets you evaluate at negative limits at all.
Hint 2
For (a), . For (b), , and is negative all across .
Show full solution
(a) . (b) , negative because on . (c) Answers vary; see the check below.
(a) With inside , the chain rule gives , so the formula really does antidifferentiate on the negative side.
(b)
The sign is right: on the integrand is negative everywhere, so the net signed area sits below the axis and the integral must be negative. Writing instead of here would be nonsense, since and do not exist.
Check your own version. With , your value must be , and because the answer must be negative, matching the fact that is below the axis across the whole interval. If your answer came out positive, you likely swapped the limits or dropped a bar and tried to take the log of a negative number.
10. The Section 2.4 Dig Deeper split into partial fractions. Run the machine on a new denominator. (a) Decompose , find and , and integrate to reach . (b) Make your own: pick two different numbers and , decompose , and integrate it. Check your own version. Dig Deeper
Hint 1
Reread the Section 2.4 Dig Deeper: factor the bottom, write the fraction as a sum of two simpler ones, clear denominators, and read off the constants by plugging in the roots.
Hint 2
Clear denominators to get . Set to find , then to find . Each piece then integrates to a logarithm, and the log law folds the difference into one quotient.
Show full solution
(a) , , and . (b) Answers vary; see the check below.
(a) The bottom factors as . Clear denominators in to get . Setting gives , so ; setting gives , so . The decomposition is , so
Check your own version. Two audits settle it. Differentiate your final answer: it must simplify back to . And recombine your two partial fractions over a common denominator: the numerators must cancel down to , with the terms gone. If an survives in that numerator, one of your constants is wrong, most often from plugging a root into the wrong factor.
You can now integrate the last family the reverse power rule left out. The exponential antiderivatives read straight off the derivative table: , with the out front, and . The gap at is finally filled: , absolute value and all, because is the one formula whose derivative is on both sides of zero. Substitution turns that into the workhorse logarithm pattern : whenever the top is the derivative of the bottom, the integral is a log, which is exactly what makes fall out. With this section the antiderivative table is complete for every function this course meets. Unit 2's Mixed Practice mixes all four techniques, and then Unit 3 spends the whole toolkit on the shapes and totals of the moving world: lengths, areas, volumes, masses, work, and probability.
Section 2.P
Mixed Practice: Unit 2
Unit 2 handed you four tools one at a time, each in its own section with worked examples pointing straight at which tool to use. Out here the labels come off. A definite integral does not tell you whether it wants the average value formula, a substitution, a trig identity, or the logarithm pattern; you have to read its shape first and choose. That reading is the skill these ten problems train. One of them reaches back to Unit 1: it evaluates a definite integral by the Fundamental Theorem and then checks that answer against a rectangle estimate from Section 1.1, so the two methods confirm each other and neither gets the last word alone. Work every problem cold, and name the tool before you pick up the pencil. Stuck on which method? The guide in the back matter walks the decision.
Mixed Practice 2
1. Find , and check your answer by differentiating it back. Warm up
Hint 1
The integrand is a composition times an extra factor, so try substitution (Section 2.2). Name the inside and write ; the is exactly twice what needs, a clean constant.
Show solution
.
Set , so and . Then
Differentiate back to confirm: , the integrand exactly. The rides along because this is an indefinite integral.
2. Evaluate , exactly. Warm up
Hint 1
Each term is straight off the basic trig rule box (Section 2.3): the antiderivative of is , and of is . Then use the evaluation bar, and remember a definite integral is a number with no .
Show solution
.
An antiderivative of is , since . Evaluate from to :
The integral is zero because the area sine loses over this quarter period exactly matches the area cosine gains: the two signed regions cancel.
3. This one reaches back to Unit 1. Let on . (a) Evaluate with the Fundamental Theorem. (b) Confirm the value by trapping it between the left and right Riemann estimates and , five strips of width , the way Section 1.1 boxed in an area. (c) Find the average value of on . Core
Hint 1
Three tools meet here. Part (a) is one line of the Fundamental Theorem (Section 1.4): antidifferentiate and use the evaluation bar. Part (b) is the rectangle bracket from Section 1.1: since only rises, the left sum runs low and the right sum runs high. Part (c) is area over width (Section 2.1).
Hint 2
For (a), . For (b), with the strip heights are at ; adds the first five, the last five. For (c), divide your part (a) answer by the width .
Show solution
(a) . (b) and , so : the estimate confirms the value. (c) .
(a) The Fundamental Theorem turns the whole limit of Section 1.1 into one line:
(b) With five strips of width , the cut points are and the heights there are . The left sum reads the first five, the right sum the last five:
Because only rises on , the left rectangles fall short and the right rectangles overshoot, so the true value is caught between them: . And sits comfortably inside, so the rectangle estimate and the Fundamental Theorem agree. (A midpoint estimate lands even closer: .)
(c) Average value is the integral divided by the width:
4. The integral has no antiderivative you can write with this course's tools. (a) Use the comparison bounds to trap it between two numbers. (b) A numerical estimate gives . Confirm it falls inside your bracket. Warm up
Hint 1
This is the comparison bracket from Section 2.1: find the smallest value and largest value of on , then . Since rises on , so does its square root, so the minimum is at and the maximum at .
Show solution
(a) . (b) is between and , so it fits.
(a) On the inside climbs from to , and the square root climbs with it. So the minimum is at and the maximum is at . With width , the comparison bounds give
(b) The estimate satisfies , so it lands inside the bracket, as any correct value must. When an exact answer is out of reach, a bracket you can trust is still an honest answer.
5. Evaluate , exactly, then as a decimal to three places. Core
Hint 1
Look for the logarithm pattern from Section 2.4: the bottom is , whose derivative is , and the top is , the derivative up to a constant. A constant may cross the integral sign (Section 2.2).
Hint 2
Let , so and . Change the limits: as runs from to , runs from to . Then evaluate .
Show solution
.
Set , so , and change the limits from to :
The absolute value bars drop because is positive for every . The answer is a plain number, with no .
6. Evaluate by changing the limits. Core
Hint 1
Substitution fits (Section 2.2): the inside is , whose derivative is , and the integrand offers , so this is a constant-fix problem. Because it is a definite integral, change the limits to -values and finish in .
Hint 2
From you get , so . As runs from to , runs from to . Evaluate .
Show solution
.
Set , so , and the limits change from to :
Because the limits were switched to -values, there is no trip back to : you evaluate in and you are done.
7. Evaluate , exactly, then as a decimal to three places. Say why the answer should come out negative. Core
Hint 1
The antiderivative is , and the absolute value is what lets you evaluate at negative limits at all (Section 2.4). Without the bars, and would not even exist.
Hint 2
Use the bar: . For the sign, notice that is negative everywhere on , so its net signed area sits below the axis.
Show solution
, negative because all across .
The antiderivative is valid on the negative side because the bars keep the argument positive:
The sign is right: on the integrand runs below the axis the whole way, so the net signed area is negative. Writing instead of here would be nonsense, since neither nor is defined.
8. Evaluate , exactly, then as a decimal to three places. Core
Hint 1
Two tools stack here: a substitution (Section 2.2) tucked inside an exponential (Section 2.4). The inside of the exponential is , and its derivative is present up to a constant.
Hint 2
Let , so and . Change the limits: become . Then use .
Show solution
.
Set , so , and change the limits from to :
The substitution is what makes the exponential reachable: the loose out front is exactly the factor needs.
9. Evaluate , exactly, then as a decimal to three places. Stretch
Hint 1
A squared trig function has no power-rule shortcut and no inside derivative to substitute (Section 2.3). Rewrite it first with the half-angle identity (Appendix A.3), which turns it into a sum you can integrate.
Hint 2
Integrating the half-angle form gives (the inside the sine is the constant fix for the inside ). At the angle is , where , so that term does not vanish; keep it.
Show solution
.
Rewrite with the half-angle identity , then integrate term by term:
Now evaluate from to :
Unlike the full-period case, the boundary sine term does not drop out here: , so the stays in the answer.
10. Find . The inside's derivative is not present, but a rewrite still lets substitution work. Stretch
Hint 1
Set , so (Section 2.2). The trouble is the loose out front, but you can rewrite it: since , you also know .
Hint 2
Substituting and turns the integrand into . Expand it into and integrate term by term.
Show solution
.
Set , so and . Rewrite everything in :
Differentiate back to confirm: , the integrand.
Section 2.R
Unit 2 Recap: Check Yourself
You have just mixed all four of Unit 2's tools in the Mixed Practice; here is one last self-check, one goal at a time, before Unit 3 spends the whole toolkit. Hit Try it! and work each goal cold, then check yourself and tick Got it. A worked example and the section link sit one click away whenever you want a model or another pass.
A dose of medicine enters your blood and then clears out, its concentration decaying like . Your doctor cares about the AVERAGE concentration across the hours between doses, not the spike right after you take it and not the trough right before the next one. That average is Section 2.1's formula, , and finishing the integral is Section 2.4's rule with a constant-fix substitution from Section 2.2. One wild, decaying quantity becomes one honest number, and that number is what tells the doctor whether the dose and its spacing are right. Every technique in this unit exists so that an integral you could once only set up, you can now finish: substitution un-stacks a chain rule, the trigonometric and exponential families fill out the table, and the average value turns a wiggling quantity into a single steady figure.
Your turn. Find an average in your life that is not just the average of a few numbers: your average speed on a drive that sped up and slowed down, the average brightness of a screen that dims, the average balance in an account that rose and fell. What curve would you be averaging, and over what interval?
Can you use the properties and comparison bounds of the definite integral fluently, compute a function's average value as the height of the equal-area rectangle, and name the Mean Value Theorem for Integrals?
Try it!
(a) Find the average value of on , then find the point the Mean Value Theorem for Integrals promises. (b) The integral has no antiderivative you can write with this course's tools. Trap it between two numbers with the comparison bounds.
How did you do?
(a) , reached at . (b) , that is .
(a) Average value is area over width. The area comes from the Fundamental Theorem (Section 1.4), and the width is :
That 8 is the height of the equal-area rectangle over . The Mean Value Theorem for Integrals promises the curve reaches that height somewhere, so solve :
which lands inside .
(b) On the exponent climbs from 0 to 1, so rises steadily from to : the minimum is and the maximum is . With width , the comparison bounds give
A numerical estimate lands the true value near , comfortably inside the bracket, exactly where a correct value has to sit.
Missed it? Take another pass at Section 2.1 before moving on.
Reveal a worked example
Find the average value of on , and the point the Mean Value Theorem for Integrals promises.
Average value is area over width, the area from the Fundamental Theorem and the width :
The equal-area rectangle over stands 14 tall. Set to find where the curve reaches that height:
which sits inside . The average 14 falls just below the endpoint average , because the curve bends and spends more of the interval down low, so area over width is the honest average and the endpoint shortcut is not.
Can you reverse the chain rule with a substitution , spotting the inside function whose derivative is already present, and change the limits of integration to finish a definite integral cleanly?
Try it!
(a) Find , and check by differentiating it back. (b) Evaluate by changing the limits, exact value first, then rounded to three decimal places.
How did you do?
(a) . (b) .
(a) The extra factor is exactly the inside's derivative, so the substitution is clean. Let , so , already present:
Differentiate back to confirm: , the integrand exactly, and the drops out under the derivative.
(b) The integrand offers only , so it is a constant fix, and because it is a definite integral, change the limits. Let , so ; as runs from 2 to 4, the inside runs from 0 to 12:
Because the limits are now -values, there is no trip back to , and a definite integral lands on a plain number with no .
Missed it? Take another pass at Section 2.2 before moving on.
Reveal a worked example
Evaluate by changing the limits.
Let , so and . As runs from 0 to 1, the inside runs from 1 to 3:
Changing the limits kept the whole computation in , so there is nothing to rename back, and the answer is a number with no .
Can you integrate the basic trigonometric functions, and use a trigonometric substitution to handle expressions carrying or , reading the or form off the result?
Try it!
Integrate each in radians. (a) . (b) , exact then rounded to three decimals. (c) , exact then rounded to three decimals.
How did you do?
(a) . (b) . (c) .
(a) Each term comes straight off the basic rule box, and the indefinite answer carries one . The antiderivative of is (the lone minus sign to guard), and of is :
(b) Match to the arctangent form with , the form the trigonometric substitution produced, so the antiderivative is :
since the angle whose tangent is is (Appendix A.1).
(c) Match to the arcsine form with , the form produced, so the antiderivative is :
since and the angle whose sine is is (Appendix A.1).
Missed it? Take another pass at Section 2.3 before moving on.
Reveal a worked example
Use a trigonometric substitution to evaluate .
The sum of squares calls for , so . The Pythagorean identity collapses the denominator:
Substitute both pieces and the integral becomes one you can do at sight:
since means . Now evaluate from 0 to 5:
Can you integrate , , and , and fill the one gap the power rule left with , keeping the absolute value and recognizing the pattern?
Try it!
(a) Find . (b) Find . (c) Evaluate , exact then rounded to three decimals.
How did you do?
(a) . (b) . (c) .
(a) Two rows of the rule box. The form with divides by 4, and integrates to a logarithm, bars kept because lives on both sides of zero:
(b) A general base with , so divide by :
Differentiate back to confirm: .
(c) The top is exactly the derivative of the bottom , the logarithm pattern . Let , so , and change the limits to and :
The bars drop because is positive for every , and a definite integral is a number, with no .
Missed it? Take another pass at Section 2.4 before moving on.
Reveal a worked example
A lab culture's mass grows at a rate of milligrams per hour, hours after the start. How much mass does it add during the first 5 hours? Give the exact value, then a decimal.
The mass added is the integral of the rate, the same move the odometer makes summing the speedometer into a distance (Section 1.5). Since has antiderivative , the constant 100 rides along:
The rate feeds on the amount, so the culture piles on more mass in the later hours than the early ones, and the exact total carries an right up to the last step.
Applications of Integrals: the shapes and totals of the moving world
Unit 1 built the integral and Unit 2 built the toolkit. This unit spends both. Every section here runs the exact same recipe, the one the whole book is named for: slice a quantity into infinitely many small pieces, approximate each piece, add them up, and take the limit. Point that recipe at a curve and you get its length; at a spinning profile and you get a surface or a solid; at a rod, a spring, or a tank and you get mass, work, and force; at a density curve and you get a probability. Seven different questions, one machine.
- Find the true length of a curve by slicing it into tiny straight pieces, .
- Find the area between two curves as , splitting the interval wherever they cross.
- Find the skin of a surface of revolution by summing thin bands, each a circumference times its slant width.
- Find a volume by slicing into cross-sections, , using the disk, washer, and cylindrical-shell methods.
- Find the mass of a non-uniform object from its density, .
- Find the work done by a varying force, , including stretching a spring and pumping a tank.
- Read a probability density as an area, and find interval probabilities and the mean of a continuous quantity.
Here is where the integral stops being a definition and starts being a tool. The rectangle idea from Unit 1 was never really about area; area was just the first thing it measured. The deeper idea is that any quantity you can chop into tiny nearly-uniform pieces can be rebuilt by an integral, and the moving world is full of such quantities. A road's true length is a sum of tiny straight steps. A vase's glaze is a sum of thin bands. A bell's bronze is a sum of thin shells. A spring's work is a sum of tiny pushes, each a little harder than the last. Even a wait time's odds is a sum of slivers of area under a curve. Watch for the same four words under every section, slice, approximate, sum, and limit, because once you see that they are always the same four words, a brand-new application stops being a new formula to memorize and becomes the same old machine pointed somewhere new. That is the whole promise of the book made concrete: build a whole back up from infinite pieces, over and over, until it is second nature.
Section 3.1
Arc Length
- Slice a curve into tiny near-straight pieces and add their lengths, the rectangle-summing move from Section 1.1 aimed at length instead of area.
- Derive and use the arc-length formula .
- Spot a clean curve, where is a perfect square and the length integrates by hand, and tell it from one you can only set up and estimate.
- State the smoothness a curve needs for the formula to hold: continuous.
- Name the slanted piece , the arc-length element that comes back in Section 3.3.
Two towns sit ten miles apart as the crow flies. The road between them is nothing like the crow's path: it bends around a lake, climbs a ridge, doubles back through a pass. Drive it and your odometer reads seventeen miles, not ten. The extra seven miles are real asphalt you paid for in gas and time, and the map ruler never saw them. That gap between the straight-line distance and the true length of a curve is the whole question of this section. Here the book's subtitle stops being a slogan and turns literal: to measure the road, you rebuild its length from infinitely many tiny straight pieces, each so short the curve barely bends across it. Sum the pieces and you have the odometer's number. This is the first stop in Unit 3, and every stop after it runs the same play.
Build the intuition
Every application in Unit 3 follows one recipe: slice the quantity into thin pieces, approximate each piece, add the pieces, and take the limit as they get thin, which turns the sum into an integral. For length, the thin pieces are short segments of the curve. Zoom in on any curve far enough and a small stretch looks straight, like a tiny ramp. That ramp has a horizontal run and a vertical rise, so its length is the hypotenuse of a right triangle: the Pythagorean theorem, one slice at a time.
Take a strip of width . Across it the curve climbs by some amount . The straight segment joining the two ends (a chord) has length
where the second form comes from pulling a out of the root (factor from under it and it leaves as ). String these chords end to end and you get an inscribed polygon whose total length is a little short of the curve's. As the strips shrink, closes on the slope , the polygon hugs the curve tighter, and the sum of chord lengths closes on the true length. That is the Riemann-sum picture from Section 1.1, now measuring length.
Figure 3.1.1 Left: chords strung between points on the curve form an inscribed polygon, its total length a little short of the curve's; more, shorter chords close the gap. Right: one slice is the hypotenuse of a right triangle, so its length is .
The arc length of a curve is the distance measured along the curve itself, from one end to the other, the number an odometer would read if it rolled along the curve. The formula below needs the curve to be a smooth curve: one where is continuous on , so the slope never jumps and the curve has no corners. A corner would break a single piece into two directions at once, and the Pythagorean slice would not know which way to point.
For a smooth curve from to , the arc length is
Reason: one slice of width contributes a chord of length ; as the slices shrink, , the chord lengths sum to (a Riemann sum), and the limit is the integral. The piece under the root, , is the length of one tiny slanted step along the curve; call it the arc-length element. Hold on to it: Section 3.3 spins a curve around an axis and multiplies this same slanted step by a circumference to measure the skin of a solid.
Find the exact length of from to .
Solution. Everything rides on the slope. Differentiate with the power rule and the chain rule (the inside is ):
Now build , and watch it fold into a perfect square:
That is the whole trick of a clean arc length: the root of a perfect square is a plain expression, no radical left behind. Since is never negative, , and the integral is one you already own:
The curve is exactly units long, no rounding needed. It climbs steeply over a run of , so a length well past is exactly what a bent curve should give.
A curve is a straight line with on . Predict first: is its arc length more than, less than, or equal to the straight-line distance between its two endpoints? Then run the formula and check.
Show solution
Equal: , exactly the straight-line distance, because a straight line is its own shortest path.
With constant, the integrand is constant too: . Over a run of the line rises , so its endpoints are apart, the same number. A line never bends, so its arc length and its chord agree. Every other curve is longer than its chord.
Clean curves like that one are rare and engineered. Change the function a little and the perfect square vanishes, the root refuses to simplify, and no antiderivative in this book's toolkit will touch it. That is the common case, and it is not a failure. Setting the integral up correctly is the mathematics; getting a number is then a job for estimation, exactly the way Section 1.1 estimated an area it could not yet compute.
Estimate the length of the parabola from to .
Solution. Set it up honestly. Here , so
Now try the clean-curve move: is a perfect square? It is not, and no substitution in this course clears the root, so the exact value is out of reach by hand. Estimate instead. Use a midpoint sum with four strips, , sampling each strip at its middle (the midpoint sum from Section 1.1). The integrand is .
| Strip | Midpoint | Height |
|---|---|---|
| 1 | 0.125 | 1.0308 |
| 2 | 0.375 | 1.2500 |
| 3 | 0.625 | 1.6008 |
| 4 | 0.875 | 2.0156 |
Add the heights and multiply by the width:
Want more digits? Double the strips: , and the estimates settle near . The setup is exact and the number is an honest estimate, which is the truth about most arc lengths. Figure 3.1.2 shows the curve with four chords doing the same job by hand.
Figure 3.1.2 Four chords inscribed under on . Their total length falls just short of the curve's, and shrinking them closes the gap, the same limit that turned rectangles into an integral in Section 1.1.
A fabricator bends a steel guide rail to the profile (in meters) over a horizontal run from to . How long a rail should be cut?
Solution. The slope is a clean power:
No perfect square this time, but no trouble either: the inside of the root, , is linear, so the shifted power rule from Section 2.2 integrates directly. The antiderivative is , and
Cut the rail meters long. The profile spans a -meter run, and the bent rail is longer than the run, as any curved length must be.
Arc length is what you get whenever something real follows a curve instead of the straight shot between its endpoints. A runner in an outer lane of a track covers more ground than one on the inside, which is why the starts are staggered by exactly the arc-length difference of the lanes. A factory feeding wire, pipe, or trim onto a curved form has to cut the stock to the curve's length, not the gap it spans, or the piece comes up short. Even a coastline's length is an arc-length question, and a famously slippery one, since the answer keeps growing as you measure with a finer ruler, the same shrinking strips that sharpen every estimate in this book.
The formula is worth building once from the ground up, because every step is the Section 1.1 recipe wearing new clothes. Cut into strips of width . On the -th strip the curve runs from to , climbing by . The straight chord across that strip has length, by the Pythagorean theorem,
Here is the one honest fact that closes the gap. On a smooth curve, the average slope across the strip, , is matched by the curve's own slope at some point inside the strip: somewhere in every little rise, the curve is climbing at exactly its average rate for that rise. So the chord length is , and stringing all the chords together gives
That sum is a Riemann sum, precisely the object Section 1.1 built, with integrand . Let the strips shrink, , and the sum becomes the integral Section 1.3 named the definite integral:
No new machine, just the rectangles of Unit 1 pointed at a curve's length instead of the area beneath it.
Dig Deeper A curve on the clock, and the circle's ▸
Some curves are not the graph of any : a circle fails the test, since one can carry two heights. The fix is to let a clock drive the curve. Give a point a position that depends on time ; as runs, the point traces the curve. A slice now takes a short time , and in that time the point moves across and up. The chord length is the same Pythagorean hypotenuse, with a factored out this time:
so summing and taking the limit gives the parametric arc length
Test it on the shape geometry already handed you. A circle of radius is traced by and as runs from to (radians). Then and , so
and the root collapses to the constant :
The circumference from Appendix A.7, rebuilt from nothing but slicing. A formula you have known since grade school turns out to be an arc length in disguise.
Figure 3.1.3 A circle traced by , . The speed is the constant , so one full turn has length .
The root does not come apart. is not , and more generally : the root of a sum is never the sum of the roots. Check it on numbers, , not , and never split it in a setup. The second slip is quieter: dropping the . If you write you get , and integrating that measures only how much the curve climbed, the vertical change, not the slanted distance it traveled. The under the root is the horizontal run; without it there is no triangle and no length. Keep the , and keep it inside the root.
Find the exact length of from to , then give the rounded decimal.
Hint
Try the clean-curve move from Example 3.1.1: compute , then check whether is a perfect square. The two middle terms are designed to cancel a into a , leaving a square you can read off.
Show solution
.
Differentiate: . Squaring, the cross term is , so
The root is (positive on ), so
Slide to double the inscribed polygon's straight segments and watch its length climb toward the true arc length , always a little short.
Frozen at : the eight-segment inscribed polygon hugs the curve, its length just short of the true arc length .
Exercises 3.1
1. Set up, do not evaluate. Write the arc-length integral for each curve, simplifying the integrand. (a) on . (b) on . Warm up
Hint 1
The integrand is always (the Section 3.1 formula). Find , square it, add under the root; do not try to integrate, since neither of these has a clean antiderivative.
Show solution
(a) . (b) .
(a) , so and the integrand is .
(b) , so and the integrand is . The sign of disappears in the square, which is why the formula never cares whether the curve is rising or falling.
2. A straight line has on . (a) Find its arc length exactly, and confirm it equals the straight-line distance between the endpoints. (b) A classmate writes . In one sentence, say what is wrong. Warm up
Hint 1
For a line is a constant, so the integrand is a constant and the integral is just that constant times the width of (the Section 3.1 Quick check). For part (b), recall the Watch Out: a root of a sum is not the sum of roots.
Show solution
(a) , equal to the endpoint distance . (b) , not ; you cannot split a root across a sum.
(a) , so the integrand is and . The endpoints are and , a run of and a rise of , so their distance is : a line's arc length is its chord.
(b) , while . The root of a sum is not the sum of the roots.
3. Find the exact length of from to , then give the rounded decimal. Core
Hint 1
This is a clean curve like Example 3.1.1 in Section 3.1: compute and check whether is a perfect square.
Hint 2
. The cross term in the square is , so . Integrate the root.
Show solution
.
Differentiate: . Squaring gives a cross term , so
Then
4. Find the exact length of from to , then give the rounded decimal. Core
Hint 1
Another clean curve: compute and look for a perfect square. The antiderivative of is from Section 2.4.
Hint 2
, and . Integrate term by term.
Show solution
.
Here , and the cross term in the square is , so
On the base is positive, so
5. Estimate the length of from to with a midpoint sum of four strips. Round to three decimals. Core
Hint 1
Set up the integrand as in Example 3.1.2 of Section 3.1. It has no clean antiderivative, so estimate with the midpoint sum from Section 1.1.
Hint 2
Here and the four midpoints are . Evaluate at each, add, and multiply by .
Show solution
(a finer sum gives about ).
With , the integrand is , and , which has no antiderivative in this course. Use and midpoints :
More strips settle near . The setup is exact; the number is an estimate.
6. A curved handrail is bent to the profile (in feet) from to . Find the exact length of the rail, then the rounded decimal, with units. Core
Hint 1
Compute and ; the inside of the root comes out linear, so the shifted power rule from Section 2.2 handles it, just like the rail in Example 3.1.3 of Section 3.1.
Hint 2
, so and the integrand is . Its antiderivative is .
Show solution
feet.
Differentiate: , so and . Writing the integrand as ,
7. The center line of a curved road follows (in meters) from to . (a) Set up the arc-length integral and estimate it with a midpoint sum of four strips. (b) Find the straight-line distance between the endpoints, and explain why it is a floor your estimate must clear. Stretch
Hint 1
The integrand is , which has no clean antiderivative, so estimate as in Example 3.1.2 of Section 3.1. The straight-line distance uses the endpoints and .
Hint 2
With , the four midpoints are . For part (b), a curve is never shorter than the straight segment between its ends, so the chord is a lower bound (the Section 3.1 Quick check).
Show full solution
(a) , and meters. (b) The chord is meters, a floor the estimate clears because no curve is shorter than its chord.
(a) With , the integrand is . Using and midpoints ,
(b) The endpoints and are meters apart. The road bends, so it must be longer than that straight shot, and indeed . If a midpoint estimate ever came out below the chord, that would be a signal to recheck the setup.
8. A curve is described by for from to . It is a mess to solve for in terms of , but easy the other way. Find the exact length, then the rounded decimal. Stretch
Hint 1
When is the clean function of , integrate in : the same formula with the roles swapped, . This choose-the-easier-variable move returns in Section 3.2.
Hint 2
, and the cross term makes . Integrate the root over .
Show full solution
.
Since is given cleanly as a function of , integrate in . Differentiate: , with cross term , so
Then
9. The Section 3.1 Dig In built the arc-length formula from one slice. Make it yours: pick your own smooth curve on your own interval , run the one-slice argument (chord length, then the slope, then the sum, then the limit), and say out loud where the Section 1.1 Riemann sum shows up. Finish by evaluating your integral if the curve is clean, or estimating it if it is not. Dig In
Hint 1
Reread the Dig In in Section 3.1: one slice gives a chord , and the average slope becomes . The sum of chords is the Riemann sum.
Hint 2
To keep the algebra clean, pick a curve of the perfect-square family (like Example 3.1.1's ) on a short interval, so the final integral is one you can finish by hand.
Show full solution
Worked with on : one slice gives , the sum is a Riemann sum, and its limit is .
One slice of width has chord length . The average slope is matched by at some point in the strip, and , so each chord is . Summing gives , a Riemann sum for . Its limit is
Check your own version. Your one-slice length must come out , your sum must be a Riemann sum for , and your final number must be at least the straight-line distance between your two endpoints. If it comes out smaller than that chord, a sign or a square slipped: no bent curve is shorter than the straight path between its ends.
10. The Section 3.1 Dig Deeper rebuilt a circle's circumference from the parametric arc-length formula. Make it yours: choose your own radius , write the circle as , , and use to recover both the full circumference and the length of a quarter circle. Dig Deeper
Hint 1
From the Section 3.1 Dig Deeper, and . Square and add before you take the root.
Hint 2
The Pythagorean identity collapses the root to the constant . Integrate over from to for the full circle and from to for a quarter (radians throughout).
Show full solution
Worked with : the full circle ( from to ) gives , matching ; a quarter circle ( from to ) gives .
With and , and , so
and the root is the constant . The full circle and the quarter:
The full length is , exactly the circumference.
Check your own version. Whatever radius you pick, the speed must collapse to the constant , the full turn from to must give exactly , and any arc from to must give . If your full circle does not land on , check that you squared and added the two derivatives before taking the root, not after.
Arc length is the book's subtitle made literal: chop a curve into tiny near-straight pieces, and each piece's length is by the Pythagorean theorem. Add the pieces and take the limit, the Section 1.1 recipe, and the sum becomes , valid whenever is continuous (a smooth curve). A curve is clean when folds into a perfect square and the root disappears, so the length integrates by hand; most curves are not clean, and then you set the integral up exactly and estimate it with a midpoint sum, which is as honest as the answer gets. Keep the under the root, and remember . The slanted piece , the arc-length element, is not finished with you: Section 3.2 runs the same slice-and-sum on the area trapped between two curves, and Section 3.3 wraps this very element around an axis to measure the skin of a solid.
Section 3.2
Areas between Curves
- Find the area between two curves by integrating the top curve minus the bottom curve in .
- Locate the crossings first, and split the region wherever the top curve changes identity.
- Add the pieces of a region whose curves cross, keeping the sign of each piece right.
- Integrate in (right curve minus left curve) when the region would take two pieces in .
- Set the integral up straight from a sketch.
Two cars pull away from the same light at the same instant. You have both speedometers on video, so you can plot each car's speed against time. For the first stretch the red car reads higher at every moment, and the gap between the two speed graphs opens into a sliver of space. That sliver is not just a shape. Its area is the distance the red car pulls ahead, foot by foot, because at each instant the extra speed times a sliver of time is a sliver of extra distance. Stack the slivers over the whole stretch and you have the lead in feet, read straight off the region between two curves. This section takes any two curves and measures the region caught between them, and it turns out to be the same slicing you already know, aimed at a new target.
Build the intuition
Every section in this unit runs one recipe, named in the unit opener: slice the quantity into thin pieces, approximate each piece, sum, and pass to the limit to get an integral. Section 3.1 aimed it at the length of a curve. Here the target is the area of the region between a top curve and a bottom curve over an interval .
Slice the region into thin vertical strips, exactly the rectangles from Section 1.1. A strip standing at position reaches from the bottom curve up to the top curve, so it is a rectangle of width and height . Its area is height times width:
Add the strips across the interval and let their width shrink to nothing, and the sum becomes the integral. The height of each strip is a difference, so the region's own position does not matter: whether both curves ride high above the axis or dip below it, only the gap between them counts. Figure 3.2.1 shows one strip standing in the region.
Figure 3.2.1 One vertical strip in the region between two curves. Its area is . Sum the strips and take the limit, and the region's area is the integral of that height.
The area between curves and over , where the top curve stays at or above the bottom curve on that interval, is the region caught between the two graphs from to . Its area is the integral of the strip height .
If on , the area of the region between them is
When the region is easier to slice sideways, cut it into horizontal strips instead. A strip at height runs from a left curve to a right curve , so with on ,
Reason: a vertical strip is a rectangle of width and height , so its area is . Summing the strips and shrinking their width is Section 1.1's recipe, and the sum becomes the integral. Slicing sideways swaps the roles of and .
Find the area of the region between and on .
Solution. First decide which curve is on top. Test a point inside the interval, say : the line gives and the parabola gives . The line sits higher, so and all the way across . The two curves meet exactly at the endpoints, since gives , so and . Set up the integral of top minus bottom and evaluate it with an antiderivative, the way Section 1.4 makes routine:
The region between the line and the parabola on has area (Figure 3.2.2). Notice the height is zero at both ends, where the curves touch, and widest near the middle, which is exactly what the picture shows.
Figure 3.2.2 The region between (top) and (bottom) on . The curves meet at and , and the shaded area is .
Example 3.2.1 found area between and on . Predict first, then check: is the area between and on more than , less, or exactly the same?
Show solution
Exactly the same, .
On the curve is on top (at , beats ), so
It is no accident. The graph of is the mirror image of across the line , so this region is Example 3.2.1's region reflected. A reflection keeps area, so the two match.
In Example 3.2.1 one curve stayed on top the whole way. That is the friendly case. When two curves cross inside the interval, the roles swap partway through, and a single top-minus-bottom integral would let the two parts cancel. The fix is to find the crossings first and handle each piece on its own.
Find the total area of the region enclosed by and .
Solution. Find the crossings before anything else. Set the curves equal:
so the curves meet at , , and . Two crossings sit inside, so the region comes in two pieces, and the top curve is different on each. Test a point in each piece:
- On , try : beats , so is on top.
- On , try : beats , so is on top.
Integrate top minus bottom on each piece, using the right top curve for each:
Add the pieces: the total area is (Figure 3.2.3).
Here is the trap the split avoids. If you had ignored the crossing and integrated the single height straight across , you would get
because on the left piece is negative (there is the bottom, not the top), and that negative exactly cancels the positive right piece. Zero is the right answer to "net signed height," but it is not the area. Find the crossings, split, and use the true top curve on each piece.
Figure 3.2.3 The two lobes enclosed by and . The cubic tops the line on , the line tops the cubic on , and each lobe has area , for a total of .
Example 3.2.2's two curves are live below. Drag the endpoints and along the axis and watch the enclosed region reshade, lobe by lobe. Slide the window out to : the plain unsplit integral of reads while the true area reads , because the two lobes cancel unless you split at the crossings. That is the whole Watch Out, played out in front of you.
Sometimes the region fights back in : the top or bottom boundary changes identity even though the curves never cross, because a sideways curve has two branches. When that happens, turn the picture on its side and slice in instead.
Find the area of the region bounded by the sideways parabola and the line .
Solution. The parabola opens to the right, so a vertical line can hit it twice. Slice in and the trouble disappears: each horizontal strip runs from a single left curve to a single right curve. Find where the two boundaries meet by setting the -values equal:
so they cross at and . Test a height between them, : the line gives and the parabola gives , so the line is the right curve and the parabola is the left curve. Integrate right minus left in :
The region has area (Figure 3.2.4). Slicing sideways took one clean integral. Slicing in would have taken two: for the region runs between the two branches of the parabola, and for it runs between the parabola and the line, so you would compute
the same answer the hard way. When a boundary splits into branches, the variable that keeps each strip whole is the one to integrate in.
Figure 3.2.4 The region between (left) and (right), sliced by a horizontal strip. The boundaries meet at and , and the area is .
The two cars from Picture This are the honest version of this idea. Plot each car's speed against time, and the region between the two speed curves has an area measured in distance, because speed times time is distance. While the red car reads faster, the gap fills in above the blue car's curve and the lead grows. If the blue car later reads faster, the curves cross, blue's curve climbs on top, and the new sliver counts against the lead instead of for it: this is the same crossing that made Example 3.2.2 split into pieces. Keep the sign straight and the integral hands you the net lead; take the total area between the curves and you get how far apart the two trips ever drifted, coming and going. A simple case makes it concrete: hold 60 miles per hour against a rival's 50 for one hour and the gap is a flat rectangle, area miles ahead. Two runners, two pumps filling one tank, a company's revenue against its costs: the same picture prices them all, and the area is the answer.
The rule says area is the integral of , with no mention of where the axis is. That is worth trusting even when both curves fall below the -axis, where "area under a curve" turns negative. Watch it work.
Take Example 3.2.1's region between and on , and slide the whole picture straight down by 2. The top curve becomes and the bottom becomes . On both new curves live below the axis, so each one, measured against the axis, has a negative signed area, the net signed area idea from Section 1.3. Compute the two signed areas:
Now subtract, top signed area minus bottom signed area:
the same as before. The reason is short. Sliding both curves down by the same amount leaves the height unchanged, since , so the strips are identical and the area cannot move. When you write the area as , the top curve's signed area carries the negative from the axis, the bottom curve's carries a bigger negative, and the two negatives cancel to leave exactly the gap. The axis is only a reference line; the difference of the curves does not care where you draw it.
Dig Deeper The area between sine and cosine▸
Curves that repeat still trap regions, and a clean one lives between and . To find one full patch of it, find where the two curves cross. They are equal when , which in radians happens at and again at (rusty on these exact angles? Appendix A.1 is the refresher). Between those two crossings, test the midpoint : there and , so sine is on top across the whole span.
Integrate top minus bottom, using the trig antiderivatives from Section 2.3 (the antiderivative of is ):
So one crossing-to-crossing patch has area (Figure 3.2.5). Because sine and cosine repeat every , every patch between consecutive crossings is a copy of this one, each with area . The exercises ask you to confirm that on the next span.
Figure 3.2.5 One patch between and , from to , where sine is on top. Its area is .
Three slips cost the most points here, and all three are about setup, not arithmetic. First, it is top minus bottom, never the reverse. If your area comes out negative, you swapped the two curves; a real area is never negative, so flip the subtraction and move on. Second, find the crossings before you integrate, not after. The crossings are where the top and bottom curves change places, and they set the limits and the split points; guessing them from a rough sketch is how a piece ends up with the wrong top curve. Third, split when the top curve changes identity. If two curves cross inside your interval, one integral of a single height lets the pieces cancel, exactly the zero that fooled the shortcut in Example 3.2.2. Slice into pieces at every crossing, use the true top curve on each piece, and add the pieces. And if slicing in forces a boundary to split into branches, slice in instead.
(a) Find the area of the region enclosed by and . (b) Find the total area of the region enclosed by and .
Hint
For each part, set the two curves equal to find the crossings, then test a point to see which curve is on top. Part (a) has one span with a single top curve; part (b) crosses at three points, so it splits into two pieces with different top curves, the way Example 3.2.2 did. Integrate top minus bottom on each piece and add.
Show full solution
(a) . (b) .
(a) Setting gives , so and . At the curve reads 8 and reads 0, so is on top all the way across:
(b) Setting gives , so . Two crossings sit inside, so the region is two pieces. On , test : beats , so is on top. On , test : beats , so is on top.
so the total area is . Skipping the split and integrating straight across would have given 0, which is not the area.
Exercises 3.2
1. Find the area of the region enclosed by and . Warm up
Hint
Set to find where the horizontal line meets the parabola; those are your limits. The line is the top curve. Integrate top minus bottom, following the Rule box in Section 3.2.
Show solution
.
The curves meet where , so and . The line is on top:
2. Find the area of the region between and on . Warm up
Hint
Test a point such as to see which curve is on top. The two meet at and , so no split is needed; just integrate top minus bottom, as in Example 3.2.1 of Section 3.2.
Show solution
.
At the line beats , so the line is on top:
3. Find the area of the region enclosed by and . Core
Hint 1
The two curves enclose a single region, so find both crossings first by setting . They become the limits of integration.
Hint 2
The crossings are and . Test a point between them to confirm the line is the top curve, then integrate from to .
Show full solution
.
Setting gives , so and . At the line reads 3 and the parabola reads 0, so the line is on top:
4. Find the total area of the region enclosed by and . Core
Hint 1
This is the crossing case from Example 3.2.2 of Section 3.2. Set and factor; you will get three crossings, so the region is two pieces.
Hint 2
The crossings are . On the cubic is on top; on the line is on top. Integrate top minus bottom on each piece and add.
Show full solution
.
Setting gives , so . Test a point in each piece: on , gives , so the cubic is on top; on , gives , so the line is on top.
so the total area is .
5. Find the area of the region bounded by and by integrating in . Core
Hint 1
Both boundaries are given as in terms of , so slice horizontally and integrate right minus left in , the way Example 3.2.3 of Section 3.2 did.
Hint 2
Set to get the crossings and . Test : the line is to the right of the parabola , so integrate .
Show full solution
.
Setting gives , so and . At the line is the right curve, so
6. Find the area of the region between and on . Give the answer exactly, then rounded to three decimals. Core
Hint 1
Decide which curve is on top on : compare and 1 at a point like . The antiderivative of comes from Section 2.4.
Hint 2
Since on , the curve is on top. Integrate , using and .
Show full solution
.
On , , so is the top curve:
7. Two cyclists start together. Cyclist A's speed is feet per second and cyclist B's is feet per second, for seconds. The distance A leads by is the area between the two speed curves. Find how far A is ahead at seconds, with units. Stretch
Hint 1
This is the two-cars picture from the "Why this matters" box in Section 3.2. The lead is over the interval, provided A stays faster the whole time.
Hint 2
Check that A is faster on by setting ; they are equal only at and , and A leads in between. Integrate from 0 to 2.
Show full solution
feet.
The speeds are equal when , that is , so and . Between them A is faster (at , ), so A leads the whole way:
At the two speeds match again, so the lead stops growing at 4 feet.
8. Find the area of the region bounded above by the two lines and and below by the -axis. Stretch
Hint 1
Sketch it: the two lines meet at a peak, and the region is a triangle sitting on the -axis. The top boundary changes identity at the peak, so this is a split, as the Watch Out in Section 3.2 warns.
Hint 2
The lines meet where , at . To the left the top is ; to the right the top is . The bottom is throughout, so integrate each top curve over its own piece.
Show full solution
.
The lines cross at (where ), hits the axis at , and hits it at . The top curve changes at , so split there and take the bottom as :
As a check, the region is a triangle with base 6 and height 3, so its area is .
9. The Section 3.2 Dig In showed that sliding both curves down by a constant leaves the area unchanged, because the height is a difference. (a) Take the region from Exercise 3 (between and ) and slide both curves down by 5, giving and . Show the area is still . (b) Pick any region from this exercise set, shift both of its curves by a constant of your choice, and confirm the area does not change. Dig In
Hint 1
Reread the Section 3.2 Dig In: shifting both curves by the same constant subtracts from top and from bottom, so the height is unchanged.
Hint 2
For (a), the new height is , exactly the old height. The crossings are the same, so the integral is the same.
Show full solution
(a) The shifted height equals the original height, so the area is still . (b) Any constant shift leaves the area unchanged.
(a) The new top minus new bottom is , the same integrand as Exercise 3. The curves still cross where , that is , giving and again. So
(b) Shifting both curves by any constant turns the height into , which is the same height, and it leaves the crossings where they were. The integral, and so the area, cannot change.
Check your own version. After shifting your chosen region by your constant, recompute both the crossing points and the integral. The crossings must land at the same -values as before, and the area must match your original to the last digit. If either moved, recheck that you shifted both curves by the same amount and in the same direction.
10. The Section 3.2 Dig Deeper found area between and from to . (a) Find the area of the next patch, from to the following crossing , being careful about which curve is on top there. (b) Explain in one or two sentences why every crossing-to-crossing patch has the same area. Dig Deeper
Hint 1
Reread the Section 3.2 Dig Deeper. The crossings of and are spaced apart, and the top curve switches at each crossing, so test a midpoint to see who is on top this time.
Hint 2
The midpoint of is ; test , where beats . So cosine is on top: integrate , using the trig antiderivatives from Section 2.3.
Show full solution
(a) . (b) The curves repeat every and swap places at each crossing, so each patch is a shifted copy of the same shape.
(a) On , cosine is on top (at , ). The antiderivative of is , so
(b) Sine and cosine both repeat with period , and their crossings are evenly spaced. From one crossing to the next the roles of top and bottom simply switch, so each patch is a congruent copy of the one before, and congruent regions have equal area.
Check your own version. Pick any crossing-to-crossing span you like, say , decide who is on top by testing the midpoint, and integrate. The answer must again be . If it is not, check that you used the correct top curve for that span; the top curve alternates every crossing.
The area between two curves is the same slice-sum-integrate recipe with a new height: cut the region into thin strips and integrate top minus bottom, . Because the height is a difference, the region's position does not matter, so the rule holds even when both curves sit below the axis (the axis cancels, connecting straight back to Section 1.3's signed area). When the curves cross inside the interval, find the crossings first, split the region at each one, use the true top curve on each piece, and add the pieces; a negative result just means you subtracted in the wrong order. When a boundary in splits into branches, turn the picture on its side and integrate right minus left in , . Next, revolve a curve around an axis and the same slicing measures the skin it sweeps out: Section 3.3 finds the surface area of a revolution.
Section 3.3
Surface Area of a Revolution
- Set up the surface area of a solid of revolution by slicing its skin into thin circular bands.
- Use , reading each band as its circumference times the slant width from Section 3.1.
- Evaluate the clean cases (a cone, a sphere, a parabolic dish) and cross-check them against the geometry formulas in Appendix A.7.
- Say why the slant width, not the flat , measures a sloping band, and where the radius is measured from when the axis changes.
A potter drops a lump of clay on the wheel, and as it spins her hands pull a flat disk up into a curved bowl. The bowl is one curve, the profile under her fingertips, spun all the way around. When the bowl is fired she has to glaze it, and glaze is sold by how much surface it has to cover. So how much skin does that spinning curve make? A tall narrow vase and a wide shallow dish can start from the same rim, yet one drinks far more glaze than the other, because the steeper the wall climbs, the more surface hides in the slant. This section measures that skin exactly. The same question follows the curve wherever it spins: the chrome on a car's curved bumper, the reflective coating inside a headlight, the bronze poured into a bell. Each is a curve carried once around an axis, and each has a surface area you can read straight off the profile.
Build the intuition
Every section of this unit runs the one recipe: slice the thing into thin pieces, approximate each piece, sum the pieces, and let the slices go to zero so the sum becomes an integral. Section 3.1 sliced a curve into short near-straight segments and added their lengths. Section 3.2 sliced a region into thin strips and added their areas. Now slice a surface into thin bands and add their areas.
Picture the profile curve sitting above the x-axis, then spin it once around that axis. A single short piece of the curve, sitting over a tiny width , sweeps out a thin band, like one ring cut from the wall of the bowl (Figure 3.3.1). Two numbers fix that band's area. The first is how far around it goes. The piece is at height above the axis, so when it spins it traces a circle of radius , and that circle has circumference
The second is how wide the band is. Here is the one trap of the whole section: the band is not wide. The curve is sloping, so the little piece is longer than its shadow on the axis. Its true length is the slant you already met in Section 3.1, one leg across and one leg up, joined by the Pythagorean theorem:
A band is a thin strip, and a thin strip's area is its length around times its width across. Multiply the circumference by the slant width and you have the area of one band. Add every band from to and shrink the slices, and the sum becomes an integral. That is the whole formula, assembled from parts you already own.
Figure 3.3.1 One band of the spun surface. The curve piece at height traces a circle of circumference as it revolves, and it is wide because it slopes. Area of the band is the one times the other.
Take a curve over with , and spin it once around the x-axis. The skin it sweeps out is a surface of revolution. Its area counts only the curved skin, not the flat disk caps at the ends. The radius of the surface at position is the distance from the axis to the curve, which is itself when the axis is the x-axis.
If has a continuous derivative on and , the area of the surface swept out by revolving about the x-axis is
Reason: one band is a circle of circumference times a slant width , so is the area of a single band, and the integral adds them all. The is exactly the arc-length integrand from Section 3.1: a surface area is an arc length carried around a circle.
Spin the line from to around the x-axis. You get a cone (picture a paper megaphone) whose open end has radius . Find the area of its curved skin, then check it against the cone formula in Appendix A.7.
Solution. The slope is constant, , so the slant factor is a single number for the whole cone:
Drop that and the radius into the formula, then integrate:
Now the geometry check. The cone has base radius and height (the run along the axis), so its slant height is (Figure 3.3.2). The lateral surface area of a cone from Appendix A.7 is :
Same number, two roads. The integral did not need to know it was a cone; it added bands. The geometry formula knew the shape but nothing else. When a shape is simple enough to have its own formula, the integral has to agree with it, and here it does, exactly.
Figure 3.3.2 The cone from on : base radius 3, height 4, slant 5, a 3-4-5 right triangle. Its skin has area whether you slice it into bands or read off the triangle.
Predict before you compute. Spin the flat line from to around the x-axis and you get a tube (a cylinder with no end caps) of radius 3 and length 5. Because the line is flat, and the slant width is just . What surface area does the formula give, and does it match the cylinder's side area you know from Appendix A.7?
Show solution
, which is exactly .
With , the slant factor is , so the band width really is . Then . The formula collapses to the familiar cylinder side area whenever the curve is flat, because a flat curve has no slant to stretch the band. The moment the curve tilts, the slant factor climbs above 1 and the band grows.
The cone came out clean because its slope never changed. Curves that bend make the slant factor move with , and most of them give an integral no elementary antiderivative can finish. A few bend in exactly the right way to collapse anyway, and the most beautiful of them is the sphere.
The top half of a circle of radius is on . Spin it around the x-axis and you sweep out a whole sphere. Find its surface area.
Solution. Differentiate the profile with the power-chain move:
Now build the slant factor, and watch what happens when you add 1. Put the two pieces over the common denominator :
Here is the reveal. The band area multiplies the radius by that slant factor, and the two square roots cancel:
Every band has the same area per unit of axis length, no matter where you cut it. (One honest footnote: at the poles the slant factor blows up, because the profile turns vertical there. But it blows up exactly as fast as the radius shrinks to zero, so their product, the band area, holds steady at and the integral is untroubled.) The messy profile is gone, and the integral is a constant:
The surface area of a sphere is , matching Appendix A.7 exactly. A globe of radius 5 has skin square units. The cancellation is not a coincidence; it is Archimedes' old discovery that a sphere's skin equals the side of the cylinder that just wraps it (radius , height ), whose area is . Every band of the sphere carries the same area as the matching band of that cylinder.
Figure 3.3.3 Slicing the sphere into bands, here two cuts at mid-latitudes. Wherever the cut lands, the radius and the slant trade off exactly: a smaller radius comes with a steeper, longer slant. The two effects cancel to a flat per band, so the whole surface is .
A small reflector is made by spinning the curve from to around the x-axis (the depth of the dish is 2, and its rim has radius ). Find the area of the reflecting surface.
Solution. Write , so and . Build the slant factor over a common denominator:
Now the radius cancels the in the denominator, which is what makes this one workable:
So . This is a substitution from Section 2.2: let , so and . The limits change from to :
The reflecting surface is square units, about . No geometry formula covers a paraboloid, so the integral is the only route here, and substitution is what finished it.
The mirror inside a headlight, or at the bottom of a reflecting telescope, is a curve spun around an axis, and its shine is a whisker-thin layer of aluminum laid down atom by atom in a vacuum chamber. The cost, the reflectivity, and the coating time all scale with the surface area of that curved skin, which is exactly the integral in this section. The paraboloid dish from Example 3.3.3 has skin square feet, so if the process deposits, say, 2 grams of aluminum per square foot, the mirror drinks about grams of it. Change the curve and the number moves, because a deeper dish hides more surface in its steeper walls even when the rim stays the same size. Metal spinning works the same way: a craftsperson presses a flat disk against a spinning wooden form and it flows into a bowl or a satellite dish, and the anodizing bath that finishes it charges by the spun surface, not by the flat disk it started as. The whole trade lives on the fact that spinning a curve makes measurably more skin than the flat blank you began with.
The formula rests on one claim from the intuition section: a thin band's area is its circumference times its slant width. That deserves a real derivation, and it comes from the shape a real band is, a frustum, which is a cone with its tip sliced off (Figure 3.3.4).
Take a straight piece of the profile running from a point at radius to a point at radius , and spin it. It sweeps out a frustum: a short cone-shaped collar with a small circle of radius at one end and a bigger circle of radius at the other, joined by a slant of length . The lateral area of such a frustum is
(You can read this straight off the cone formula by subtracting the small cone that got sliced away; Exercise 9 walks that through.) Rewrite it to expose the circumference. Pull out a 2 and pair it with the average radius:
Read that in words: the band area is the circumference of the average circle, , times the slant length . Now shrink the band down to a slice over a tiny width . The two radii and squeeze together onto the single value , so the average radius . And the slant length is the near-straight curve piece from Section 3.1, . Put those limits in:
Sum the bands and let , the same slice-approximate-sum recipe from Section 1.1, and the sum becomes . The whole formula is nothing but Section 3.1's arc length carried once around a circle: take the length of a curve piece and multiply it by how far it travels around the axis.
Figure 3.3.4 One real band is a frustum: two circles of radii and joined by a slant . Its area is the average circumference times the slant, and as the band thins it becomes .
Dig Deeper Pappus spins a doughnut▸
The Dig In said a surface area is a curve length carried around a circle. An old theorem makes that literal for the whole curve at once. Pappus of Alexandria noticed that when you revolve a curve about an axis it does not cross, its centroid (its balance point) traces a circle, and the surface area is simply the length of the curve times how far that centroid travels:
where is the length of the curve and is the distance from the curve's centroid to the axis. It is the band formula freed of calculus: instead of tracking the radius band by band, you use the one radius of the whole curve's balance point.
The clean case is a doughnut, or torus. Take a full circle of radius whose center sits a distance from the axis, with so the circle never touches the axis, and spin it. The curve is the circle, so its length is , and its centroid is its own center, at distance . Pappus gives the surface in one line:
A doughnut with tube radius and center distance has skin square units. Setting up that same surface as a direct integral would be a genuine fight; Pappus reads it off the balance point. The theorem also handles a different axis for free, which answers the last worry of this section. Revolve around the y-axis instead and the radius of each band becomes the horizontal distance to that axis, namely , not , so the surface integral becomes . The radius is always the distance to whichever axis you spin around.
The band is not a flat ring of width ; it is a sloping collar of width . Dropping the slant factor treats the cone as a stack of flat rings and undercounts its skin. For the cone in Example 3.3.1, the flat-ring mistake would give , short of the true by exactly the slant factor , since . A cone really does have more skin than a stack of rings, and the slant is where that extra hides. Two more cautions. The radius is the distance from the axis to the curve, so it is only when you revolve about the x-axis; spin about the y-axis and the radius becomes (see the Dig Deeper). And this surface area counts the curved skin alone, never the flat end caps, so a spun bowl's rim disk is not part of unless the problem asks for it on purpose.
Spin the line from to around the x-axis. (a) Find the area of the curved surface with the integral. (b) Confirm it against the cone formula from Appendix A.7.
Hint
The slope is constant, so and the slant factor comes out of the integral as one number, just like the cone in Example 3.3.1 of Section 3.3. For part (b), the cone's base radius is , its height is the run , and its slant height is .
Show solution
(a) . (b) , the same number.
(a) With , the slant factor is , so
(b) The cone has base radius and height , so its slant height is . Then , matching part (a). The integral and the geometry agree, as they must for a shape this simple.
Exercises 3.3
1. Spin from to around the x-axis, making a cone. (a) Find the area of its curved skin with the integral. (b) Check it against . Warm up
Hint
The slope is 1, so and the slant factor is one constant for the whole cone, as in Example 3.3.1 of Section 3.3. The base radius is , the height is , and .
Show solution
(a) . (b) , the same.
(a) With , the slant factor is , so .
(b) The base radius is , the height is , and , so . The two routes agree.
2. Write the surface-area integral (do not evaluate it) for revolving from to around the x-axis. Warm up
Hint
Find , drop and into from Section 3.3, and stop. This integrand has no antiderivative you can write with this course's tools, so setting it up correctly is the whole task.
Show solution
.
Since , we have , so and the slant factor is . The radius is . Assembling the formula gives . Unlike the sphere and the dish, nothing cancels here, so this one is left set up for a numerical method rather than evaluated by hand.
3. A megaphone is the surface from spinning from to (in centimeters) around the x-axis. Find the area of its skin, exact and to three decimals, with units, and confirm it with . Core
Hint 1
The slope is , a constant, so the slant factor is one number for the whole cone. Mirror Example 3.3.1 of Section 3.3.
Hint 2
Compute . The base radius is and the height is , a 5-12-13 triangle.
Show full solution
square centimeters, and confirms it.
The slant factor is . Then
The geometry check: the base radius is , the height is , and the slant height is , so . Same skin, two ways.
4. Find the surface area from revolving from to around the x-axis. Give the exact value and a three-decimal decimal. Core
Hint 1
Follow Example 3.3.3 of Section 3.3: find , build over a common denominator, and watch the radius cancel the leftover .
Hint 2
With , you get , and the integrand collapses to . Then substitute as in Section 2.2.
Show full solution
.
Here , so and . Then , so . The radius cancels the :
Substitute , , with running from 1 to 4:
5. A band is cut from a sphere of radius 5 by revolving over just around the x-axis (a cap near the pole). (a) Find its area. (b) Explain why the answer depends only on the width of the slab, not on where along the sphere it sits. Core
Hint 1
This is the sphere of Example 3.3.2 in Section 3.3 with , integrated over a shorter interval. Rebuild the slant factor and watch the same cancellation happen.
Hint 2
The integrand collapses to the constant , so . A constant integrand means the answer is times the width of the interval.
Show full solution
(a) . (b) The integrand is the constant , so the area is times the slab width, no matter where the slab sits.
(a) With , the same cancellation as Example 3.3.2 gives . Then .
(b) Because every band of a sphere carries the same area per unit of axis length, the total for a slab depends only on how wide the slab is along the axis, here , not on whether it is cut near the equator or near the pole. This is Archimedes' band result: parallel slabs of equal width on a sphere have equal skin.
Check your own version. Cut any other width-2 slab of the same sphere, say or , and its area must again come out to ; if a slab of a different width gives anything other than , recheck that the slant factor really did cancel to the constant .
6. Revolve the line from to around the y-axis (not the x-axis) and find the surface area. Core
Hint 1
Spinning about the y-axis changes what the radius is. Reread the Watch Out and Dig Deeper in Section 3.3: the radius is the distance from the y-axis to the curve, which is , so the band is .
Hint 2
With , the slant factor is , a constant. Integrate from 0 to 2.
Show full solution
.
Revolving about the y-axis makes the radius the horizontal distance to that axis, which is . The slope is 3, so the slant factor is , and
As a check, this is a cone about the y-axis with base radius 2 (the largest ) and height , so its slant height is , and .
7. A lampshade is the band from revolving from to (in inches) around the x-axis, so its two rims have radii 1 and 4. Find its surface area, exact and to three decimals with units, and confirm it with the frustum formula . Stretch
Hint 1
The slope is 1, so and the slant factor is , constant. Set up the integral of as in Example 3.3.1 of Section 3.3, but with a radius that starts at 1 rather than 0.
Hint 2
For the frustum check, the two rim radii are and , and the slant of the whole band is , one straight run.
Show full solution
square inches, and confirms it.
With , the slant factor is , so
The frustum check: the rims have radii and , and the slant length of the band is . Then , matching the integral.
8. The curve from to is revolved about the x-axis, sweeping out a smoothly flaring horn. Find the area of its curved surface, exactly, then as a decimal to three places. (Unlike a cone, this profile is genuinely curved.) Stretch
Hint 1
Use the surface-area formula (Section 3.3). Here , so ; build and watch the factors meet.
Hint 2
, so . Then : the radius cancels the slant's denominator and the integral is clean.
Show full solution
.
With , , so and . The slant factor is , and the radius cancels it down:
So the surface integral closes in one step:
A curved wall, not a straight one, and yet the slant integral still comes out exactly.
9. The Dig In built each band as circumference times slant. Use the surface integral to derive the cone formula itself. (a) Revolve the general cone profile from to around the x-axis, and show the skin comes out to with . (b) Explain why replacing the slant width with a flat would undercount the cone's skin. Dig In
Hint 1
Treat and as fixed numbers and run the same steps as Example 3.3.1 of Section 3.3. The slope is , a constant, so the slant factor comes out of the integral.
Hint 2
The slant factor is . After integrating , simplify and recognize as the slant height . For (b), compare the slant factor with 1.
Show full solution
(a) The integral gives , the cone's lateral area. (b) The slant factor is larger than 1 whenever the cone slopes, so replacing it with shrinks every band and undercounts the skin.
(a) The slope is , so the slant factor is , a constant. Then
Since is the slant height, this is , the Appendix A.7 formula, now derived by calculus.
(b) The slant factor is at least 1, and it is strictly larger than 1 the moment , which is true everywhere on a sloping cone. Using a flat drops that factor and treats each band as a vertical ring rather than a slanted collar, so every band is measured too narrow and the total skin comes out too small. The extra area lives entirely in the slant.
Check your own version. Pick any specific and (for instance , ) and confirm the integral returns ; it must equal the value the geometry formula gives. If the flat- version ever matches the true area, check whether you accidentally set , which would mean a cylinder, not a cone.
10. The Dig Deeper gave Pappus's theorem and used it to spin a torus. Revolve a full circle of radius whose center sits a distance from the axis (so ) around that axis. (a) Find the doughnut's surface area with Pappus. (b) Say what and are for this curve and why they are what they are. Dig Deeper
Hint 1
Use from the Dig Deeper in Section 3.3. The curve being revolved is the whole circle, so is its circumference and is the distance from its center to the axis.
Hint 2
A circle of radius has length , and its centroid is its own center, at distance . Multiply.
Show full solution
(a) . (b) (the circle's circumference) and (its center is its centroid, and that center is 5 from the axis).
(a) By Pappus, . In the general form, , the same.
(b) The curve revolved is the full circle of radius , whose length is its circumference . A circle's centroid (balance point) is its center, and here that center sits from the axis, so . Pappus then carries the whole curve length around a circle of radius , turning both facts into one surface.
Check your own version. For any and with , Pappus must give ; confirm your is the circle's full circumference and your is the center distance . If you picked the axis would cut the circle and the theorem no longer applies, so keep .
Revolving about the x-axis sweeps out a surface of revolution, and slicing its skin into thin bands gives . Each band is a circle of circumference times the slant width , which is Section 3.1's arc-length integrand carried once around the axis: a surface area is a curve length spun in a circle. The clean cases are worth remembering. A cone returns , a sphere collapses (the two square roots cancel) to , and a parabolic dish needs a substitution from Section 2.2 to finish. The one habit to keep: use the slant width, never the flat (a cone has more skin than a stack of rings), and measure the radius from whichever axis you spin around, so it is , not , about the y-axis. Next, the skin done, comes the solid it wraps: the same slice-sum-integrate recipe reads a volume three different ways.
Section 3.4
Volume Methods
- Slice a solid into cross-sections and find its volume with .
- Use the disk method and the washer method , always naming the axis of revolution.
- Use cylindrical shells, , to revolve a region about the y-axis.
- Pick the method that fits the solid: disks and washers cut perpendicular to the axis, shells wrap parallel to it.
- Match the variable of integration to the method, and read every volume off the same slice, sum, integrate recipe.
Slice a loaf of bread and each slice is nearly a flat slab. Its volume is just the area of the cut face times how thick you cut it, and the whole loaf is the stack of those slabs. That is one way to build a volume: know the area of every cross-section and add up the thin slices. The other way starts with a spin. Take a curve, spin it around a line, and it sweeps out a smooth solid. Spin a straight slant and you get a cone. Spin a half circle and you get a ball. Spin the outline of a bowl and you get the bowl, glass and all. A bead is a ball with the hole from the string drilled out. Everything in this section is one of those two pictures: a solid you slice into faces, or a solid you spin from a curve. Both come down to adding up thin pieces, which is exactly what an integral does.
Build the intuition
Every application in this unit runs the same recipe from Unit 3: slice the thing into tiny pieces, approximate each piece, then sum and take the limit to land on an integral. Volume is the cleanest case yet. Set the solid along an x-axis and cut it into thin slabs with planes perpendicular to that axis, spaced apart. The slab at position has a flat face whose area is , and since the slab is thin its shape barely changes across its width, so its volume is close to
Add up every slab and you have a Riemann sum from Section 1.1, . Let the slabs get infinitely thin and the sum becomes the integral that Section 1.3 built.
Figure 3.4.1 A solid cut into thin slabs. The shaded slab has a cross-section of area and thickness , so it holds about . Sum every slab and pass to the limit: the volume is the integral of the cross-sectional area.
A cross-section of a solid is the flat face you see when you cut it with a plane. When those cuts are perpendicular to the x-axis, the cross-section at position has some area . A solid of revolution is the solid you get by spinning a flat region all the way around a fixed line, called the axis of revolution. Its cross-sections perpendicular to that axis are always circles (or rings), which is what makes revolution solids so friendly to slice.
If a solid runs from to and its cross-section perpendicular to the x-axis at position has area , then its volume is
This is the whole theorem. Every method below is just a different formula for . The recipe never changes: find the area of one slice as a function of where it sits, then integrate that area across the solid.
Revolve the region under about the x-axis and each slice is a solid circle, a disk of radius . Its area is a circle's area, so the disk method reads
Revolve the region between two curves and each slice is a circle with a circle punched out of its middle, a flat ring. Call the far edge (the outer radius) and the near edge (the inner radius) , both measured from the axis. The ring's area is the big circle minus the hole, so the washer method reads
State the axis before you write either one: the radius is always the distance from the axis to the curve, and a different axis gives a different radius.
Revolve the region under from to about the x-axis. Find the volume of the solid it sweeps out.
Solution. The axis is the x-axis, and the region touches it, so there is no hole: this is a disk problem. Each slice is a disk of radius , so its area is
The square and the square root cancel, which is why disk problems love a square root. Integrate the area from 0 to 4:
The volume is cubic units. Notice the shape: spinning gives a solid that flares open like a trumpet bell, wide at and pinched to a point at the origin (Figure 3.4.2).
Figure 3.4.2 Spinning about the x-axis. The vertical radius sweeps a disk of area , and stacking the disks from 0 to 4 gives volume .
Before you go on, predict this one, then check it. Revolve the region under the straight line from to about the x-axis. You already know the shape it makes: a cone with radius 3 and height 3. Use the geometry formula to predict the volume, then confirm it with the disk method.
Show solution
cubic units, from both the cone formula and the disk method.
The cone has and , so . The disk method agrees: each slice has radius , so and
The geometry you learned in Appendix A.7 is a special case of the disk method, and it is a good habit to cross-check calculus against a formula you trust whenever the solid has a name.
Add a hole and a disk becomes a washer. The setup is the same, but now two curves bound the region, so the slice is a ring instead of a full circle. The region between two curves is exactly the kind you met in Section 3.2.
Revolve the region between and , from to , about the x-axis. Find the volume.
Solution. First name which curve is on top. On , the line sits above the parabola (at , the line gives and the parabola gives ). Spun about the x-axis, the line is farther from the axis, so it draws the outer edge and the parabola draws the hole:
Each slice is a washer of area . Integrate:
The volume is cubic units. The one move that separates a washer from a disk is subtracting the hole: square each radius on its own, then subtract, never square the difference of the radii (Figure 3.4.3 shows why the two radii are separate distances).
Figure 3.4.3 The region between and . Spun about the x-axis, the outer radius reaches the line and the inner radius reaches the parabola, so each slice is a ring of area .
The shell method: slice the other way
Disks and washers cut perpendicular to the axis. Sometimes that cut is a nightmare. Try to revolve the region under about the y-axis with washers and you would have to cut horizontally, in , which means solving for and juggling curves that change identity partway up. There is a cleaner cut. Instead of flat slices, peel the solid into thin nested tubes, like the rings of an onion or a set of nested paper-towel rolls. Each tube is a cylindrical shell. Slit one open and unroll it, and it flattens into a thin rectangular sheet: its width is the circumference of the tube, its height is the curve value , and its thickness is (Figure 3.4.4). So one shell holds about , and summing the shells gives the volume.
Figure 3.4.4 A cylindrical shell unrolled. The tube at radius becomes a flat sheet of width (its circumference), height , and thickness , so its volume is about .
Revolve the region under , from to with and , about the y-axis. Peeling the solid into thin shells gives
Read the integrand as circumference times height times thickness: is the distance around a tube of radius , is how tall that tube stands, and is how thick its wall is. The radius here is the distance from the axis to the shell, which is when the axis is the y-axis.
Revolve the region under , from to , about the y-axis. Find the volume, and say why shells beat washers here.
Solution. The axis is the y-axis, so a shell at radius has height . The shell method gives
The volume is cubic units. Now the comparison. To use washers you would cut horizontally in , which forces you to solve for and set up an integral in with a different-looking region. Shells let you keep the original and integrate in , no inverting. That is the general lesson worth memorizing: disks and washers cut perpendicular to the axis; shells run parallel to it. When revolving about the y-axis leaves you solving for , reach for shells.
Anything made on a lathe is a solid of revolution. A block of wood or metal spins on the axis while a blade shapes its profile, and the finished chair leg, baseball bat, or engine piston is exactly the curve you cut, spun all the way around. The volume integral tells the shop how much material each part uses and how much it will weigh before a single one is turned. A glassblower spinning a bowl, a potter pulling a vase on the wheel, a machinist boring a bead: all of them are running the disk, washer, and shell setups by hand. Fuel and water tanks get the same treatment. A tank truck's cylinder lying on its side, a spherical propane tank, a bottle with a curved shoulder, each one is sliced into cross-sections so a gauge can turn a depth reading into a volume of liters left. And the cross-section idea reaches past round things. An architect finding the concrete in a ramp, or a biologist estimating a lung's volume from a stack of scan slices, adds up slice areas exactly the way you just did, whether or not anything ever spun.
Disks and washers can feel like two separate rules to memorize. They are not: a disk is just a washer with no hole. Look at the areas side by side. A washer slice has area . Set the inner radius , meaning nothing is punched out, and the formula collapses to , which is the disk. The hole simply closed up.
Take Example 3.4.1. As a disk, the slice area was . As a washer with outer radius and inner radius , it is , the exact same thing, and it integrates to the same . So there are not really two methods, there is one idea, with a circular slice, and the washer formula is the general one that already contains the disk. Whenever a revolved region touches the axis, its inner radius is 0 and the washer politely becomes a disk on its own. Exercise 9 asks you to run this collapse on a solid of your choosing.
Dig Deeper Solids that never spun▸
Revolution is a great way to make a solid, but it is a special case, not the main event. The real theorem is , and can be the area of any shape at all, not just a circle. When the slices are not circles, no shows up.
Here is a solid that never spun. Start with a flat base: the region under , from to , lying in the plane. Now stand a square up on each slice, so that at position the cross-section perpendicular to the x-axis is a square whose side equals the height of the base there, . Picture a row of square tiles standing on edge, each taller than the last. The area of that square slice is
a plain square area, no circle, no . Integrate it just like every other volume:
The volume is cubic units exactly, a clean rational number with no in sight, which is the tell that nothing round was involved. Compare it with Example 3.4.3, which spun that same base region under about the y-axis and got : same base, totally different solid, because the cross-section shape changed from a circle to a square. That is the point. The disk, washer, and shell formulas are three handy shortcuts for the case when the slice is a circle or a ring, but is the actual rule, and it happily builds solids with square, triangular, or semicircular cross-sections too. Exercise 10 lets you design one.
Three slips cost the most volume points. First, the missing hole. When a revolved region does not touch the axis, the slice is a washer, and forgetting the inner radius overfills the solid. Ask every time whether the region reaches the axis; if it does not, you owe a . And square each radius separately: is not . Second, the shell radius. In a shell integral the radius is the distance from the axis to the shell, not the height of the region. When you revolve about the y-axis, the radius is and the height is , and swapping them silently computes the wrong solid. Third, matching the variable to the method. Disks and washers about the x-axis integrate in ; shells about the y-axis also integrate in , but washers about the y-axis would integrate in . Pick the method whose slice you can describe without solving the curve backward, and the variable comes along for free. Always name the axis first, and the rest follows.
Take the region under from to . (a) Revolve it about the x-axis and find the volume with the disk method. (b) Revolve the same region about the y-axis and find the volume with shells. (c) Say in one sentence why the two answers differ.
Hint
(a) About the x-axis the radius is , so ; integrate over . (b) About the y-axis a shell at radius has height , so use , the rule from Section 3.4. (c) The region is the same, but a different axis makes a different solid.
Show solution
(a) cubic units. (b) cubic units. (c) Same region, different axis, so a different solid and a different volume.
(a) Disk about the x-axis: the radius is , so and
(b) Shells about the y-axis: a shell at radius has height , so
(c) The flat region is identical both times, but the axis you spin it around is not, so the two solids have different shapes and different volumes. The axis is part of the problem, which is why you name it first.
Toggle Disk, Washer, and Shell, then slide the representative slice and watch the swept solid fill in. Disk and shell both climb to the same ; Washer now carves a real hole, the inner cone , so its running volume climbs to instead, the full solid minus the core. Challenge: park the slice at in Washer and check the hole radius reads exactly 1, half the slice position, while the outer radius reads .
Frozen on the disk method with a representative disk at , straddling the axis of revolution, its running volume climbing toward the true .
Exercises 3.4
1. The region under from to is revolved about the x-axis, sweeping out a cone. (a) Set up and evaluate the disk-method integral for its volume. (b) Confirm your answer with the cone formula . Warm up
Hint
Each slice is a disk of radius , so ; integrate over as in Example 3.4.1 of Section 3.4. For part (b), read the cone's radius and height off the endpoints: at the height of the region is 2.
Show solution
(a) cubic units. (b) The cone formula gives the same .
(a) The radius is , so and
(b) The cone has radius and height , so , matching part (a).
2. No computing, just planning. For each solid, name the method (disk, washer, or shell), the variable you would integrate in, and the axis of revolution. (a) The region under from 0 to 3, revolved about the x-axis. (b) The region between and from 0 to 2, revolved about the x-axis. (c) The region under from 0 to 3, revolved about the y-axis. Warm up
Hint
Two questions from Section 3.4 settle it: does the region touch the axis (disk if yes, washer if a gap), and does revolving about that axis force you to solve the curve for the other variable (then shells)? Disks and washers about the x-axis integrate in ; shells about the y-axis also integrate in .
Show solution
(a) Disk, integrate in , about the x-axis. (b) Washer, integrate in , about the x-axis. (c) Shell, integrate in , about the y-axis.
(a) The region under a single curve touches the x-axis, so there is no hole: disk method, , integrated in .
(b) Two curves bound the region, with on the outside and on the inside (on , never exceeds 4, so it stays below 6): washer method, , integrated in .
(c) Revolving about the y-axis would force washers to solve for , so shells are cleaner: a shell at radius has height , integrated in .
3. Revolve the region under from to about the x-axis. Find the volume. Core
Hint 1
The region sits under a single curve and touches the axis, so it is a disk problem, like Example 3.4.1 in Section 3.4.
Hint 2
The radius is , so . Integrate from 0 to 1 with the power rule from Section 1.4.
Show solution
cubic units.
Each slice is a disk of radius , so and
4. Revolve the region between and , from to , about the x-axis. Find the volume. Core
Hint 1
Two curves bound the region, so this is a washer, like Example 3.4.2 in Section 3.4. Decide which curve is farther from the axis on : test a point such as .
Hint 2
On , , so (outer) and (inner). Then ; integrate over .
Show solution
cubic units.
On the square root sits above the line (at , beats ), so and . The slice area is
and integrating,
5. Revolve the region under from to about the y-axis. Use cylindrical shells to find the volume. Core
Hint 1
About the y-axis, reach for shells (Example 3.4.3 of Section 3.4). A shell at radius has height .
Hint 2
Use with , so the integrand is . Integrate from 0 to 3.
Show solution
cubic units.
A shell at radius has height , so
6. Revolve the first-quadrant region under , from to , about the y-axis. Use shells to find the volume. Core
Hint 1
About the y-axis, shells keep you in and avoid solving for . A shell at radius has height .
Hint 2
The integrand is . Integrate from 0 to 2; the region meets the x-axis at , which sets the upper limit.
Show solution
cubic units.
A shell at radius has height , so
7. Revolve the region between and about the x-axis. Find the volume. Stretch
Hint 1
First find where the curves cross to get the limits: solve . Then, since a gap opens between them, this is a washer as in Example 3.4.2 of Section 3.4.
Hint 2
The curves meet at and , and on the line is on top, so and . Integrate from 0 to 2.
Show full solution
cubic units.
Setting gives , so the curves cross at and . On the line lies above the parabola (at , ), so and :
8. Revolve the region under , from to , about the y-axis this time. (a) Use shells to find the volume. (b) Example 3.4.1 revolved the same region about the x-axis and got ; explain in a sentence why this axis gives a different answer. Stretch
Hint 1
About the y-axis, shells keep the curve as with no inverting. A shell at radius has height , so use .
Hint 2
Combine the powers: , so the integrand is . Its antiderivative is ; evaluate from 0 to 4, and recall .
Show full solution
(a) cubic units. (b) It is a different solid, because spinning the same region about a different axis carves a different shape.
(a) A shell at radius has height , so the integrand is :
(b) The flat region is the same as in Example 3.4.1, but spinning it about the y-axis instead of the x-axis produces a different solid, so the volumes have no reason to match. The axis is part of the problem.
9. The Section 3.4 Dig In showed that a disk is a washer with inner radius 0. Make it yours: pick your own curve and interval (new numbers, not the section's), revolve the region about the x-axis, compute the volume once by the disk method, then recompute it as a washer with inner radius , and confirm the two agree. Dig In
Hint 1
Reread the Section 3.4 Dig In: the washer area with becomes , the disk. Choose any whose square is easy to integrate.
Hint 2
Write both integrands and check they are identical before integrating: for the disk, and for the washer. They must match term for term.
Show full solution
Worked with on : both the disk method and the washer-with- give cubic units.
Revolve the region under , from 0 to 2, about the x-axis. The disk method:
Now as a washer with outer radius and inner radius :
The two integrands are literally the same once erases the hole, so the volumes match.
Check your own version. Whatever curve and interval you chose, your disk integrand and your washer integrand must be identical before you integrate, so the two volumes must come out exactly equal. If they differ, an inner radius other than 0 sneaked in, or a radius got squared after subtracting instead of before.
10. The Section 3.4 Dig Deeper built a solid from square cross-sections, no revolution and no . Design your own known-cross-section solid: pick a base region and a cross-section shape (a square, or an equilateral triangle, whose area for side is ), set up , and evaluate it. Dig Deeper
Hint 1
Reread the Section 3.4 Dig Deeper: the side of each cross-section is the height of the base region at , and is that shape's area formula applied to that side.
Hint 2
Pick a simple base such as the region under a line, so the side at is easy to write. For equilateral-triangle slices of side , use , then integrate across the base.
Show full solution
Worked with base under on and equilateral-triangle cross-sections of side : cubic units, and no appears.
Let the base be the region under from 0 to 3, so the height of the base at position is . Stand an equilateral triangle of side on each slice. Its area is
and integrating across the base,
Because the slices are triangles, not circles, the answer carries a and no , exactly as the Dig Deeper predicted.
Check your own version. Whatever base and shape you chose, two audits confirm it: the side you plug in must equal the base's height at , and the answer should contain no unless your cross-section is a circle or half circle. If a stray shows up from a non-circular slice, you used a circle's area by mistake; recompute from your shape's own area formula.
Every volume in this section is one integral, : find the area of a single slice as a function of where it sits, then add the slices up. When you spin a region about an axis, the slice is a circle. The disk method handles a region that touches the axis, and the washer method handles a region with a gap, with both radii measured from the axis. The shell method peels the solid into tubes instead, and it wins when revolving about the y-axis would otherwise force you to solve the curve for . Two habits carry it all: name the axis before you write anything, since the radius is always a distance to the axis, and remember disks and washers cut perpendicular to the axis while shells run parallel to it. And the deepest lesson is that revolution is optional: builds any solid whose cross-sections you can measure, circle or square or triangle. Next, that same slice, sum, integrate move reads the mass of a rod whose density changes along its length.
Section 3.5
Mass and Density
- Find a rod's mass from a variable linear density with , keeping the units straight.
- Find a circular plate's mass from a radial density using a thin ring slice, .
- Set up the "slice, mass of a slice, sum" recipe for any object whose density will not hold still.
- Estimate a total mass from a density table with a Riemann sum when there is no formula.
Pick up a baseball bat by the thin handle and it feels light. Slide your grip toward the fat barrel and it feels heavy in a hurry. The bat is one solid piece of wood, but its mass is not spread evenly: there is far more of it packed into the barrel than into the handle. So what is the bat's total mass? You cannot just multiply one density by the length, because there is no single density to multiply. The wood is dense in some places and thin in others, and the answer has to account for every part on its own terms. That is the whole problem of this section, and it yields to the same move every application in this unit uses. Cut the object into slices so thin that each one is nearly uniform, find the mass of one slice, and add them all up. Slice, sum, integrate.
Build the intuition
Start with the simplest heavy object, a straight rod, and lay it along the x-axis from to . Its linear density is a function , the mass packed into each unit of length at position , measured in something like kilograms per meter. Where the rod is thick, is large; where it thins out, is small.
Now take a slice. Cut a tiny piece of the rod at position , only wide. Because the slice is so short, the density barely changes across it, so treat the whole slice as if it had the one density . A piece with a single density is the easy case: its mass is density times length,
That is the payoff of slicing thin. The hard fact (density keeps changing) is true across the whole rod, but almost false across one sliver. Adding up the slivers and letting their width shrink to zero turns the sum into an integral, exactly the limit of Riemann sums from Section 1.1. Figure 3.5.1 draws it: the height of the density curve at is , one thin slice is a rectangle of height and width , and the total mass is the whole area underneath.
Figure 3.5.1 The mass of a rod is the area under its density curve. Each thin slice at position is a rectangle of height and width , with mass . Add every slice and shrink the width to zero: the sum becomes .
The linear density of a rod lying along is a function giving the mass per unit length at each position . Its units are mass over length, such as kilograms per meter or grams per centimeter. A rod is uniform when is a constant, and non-uniform when it varies. (The symbol is the Greek letter rho; a refresher on density and its units is in Appendix A.8.)
For a rod along with linear density , the total mass is the integral of the density:
For a flat circular plate of radius whose density depends only on the distance from the center, slice it into thin rings instead of straight strips. The ring at radius , of thickness , has area , so its mass is , and the total mass is
The ring area is the key. Unroll a thin ring and it straightens into a strip as long as the ring's circumference, , and as wide as its thickness, , so its area is . This is the same circumference-times-width slicing behind the surface bands of Section 3.3 and the cylindrical shells you meet in Section 3.4. Both mass rules are the one recipe from the start of the unit: slice, find one slice's mass, sum, integrate.
Figure 3.5.2 A thin ring at radius and thickness . Unrolled, it is a strip of length (the circumference) and width , so its area is , not (which is the whole disk out to radius ).
A metal rod 3 meters long lies along and thickens toward its far end, so its linear density is kilograms per meter (with in meters). Find the rod's mass.
Solution. Slice the rod, give one slice at position the mass , and integrate over the whole length:
The rod weighs kilograms. Watch the units carry through: density in kilograms per meter, times a length in meters, gives a mass in kilograms, and the integral only adds masses together, so the answer lands in kilograms too. As a sanity check, the density runs from up to kilograms per meter. A uniform rod at the low density would weigh only kg, and one at the high density would weigh kg, so kg sitting between them is exactly right.
Before you compute, predict. A rod along meters has linear density kilograms per meter, so its density climbs steadily from at one end to at the other. Because that climb is a straight line, the rod behaves as if it had one steady density sitting right in the middle of and . Predict the mass from that idea, then confirm it with .
Show solution
kg.
The density averages to kilograms per meter over the -meter rod, so the prediction is kg. The integral agrees:
The prediction worked because the density is a straight line, so its average sits at the midpoint. Hold on to this idea of an average density; the Dig In makes it exact for any density, straight or curved.
The straight rod slices into flat strips. A round plate is denser in some rings than others, so it slices into thin rings instead, and the ring area does the rest.
Example 3.5.1's rod is below, thickening as you go. Drag the cut along it and watch the mass pile up as the area under the density line, while the shortcut "density times length" runs wrong in both directions: the start density undercharges, the end density overcharges. Challenge: park the cut at . The honest mass reads 33 kg, and that is exactly the AVERAGE density times the 3 m length: for a steadily thickening rod, the integral is what finds that average for you.
A flat circular plate of radius meters is denser at its center and thins toward the rim, with area density kilograms per square meter (with in meters, measured out from the center). Find its mass.
Solution. Slice the plate into thin rings. The ring at radius has area and density , so its mass is . Integrate from the center out to the rim at :
The plate weighs kilograms, about kg. Notice the inside the integral: it is not a constant you can pull out and forget, because a ring far from the center holds more material than a ring near it, even at the same density. A ring at is a longer loop than a ring at , so it counts for more. Forgetting that factor is the mistake the Watch Out returns to.
Sometimes there is no density formula at all, only measurements. Then the mass is an estimate, and the tool is the Riemann sum from Section 1.1.
A crew measures the linear density of a -meter cable at every meters and records the readings in Table 3.5.1. There is no formula, just data. Estimate the cable's total mass with a left sum and a right sum, and say what those two numbers tell you.
| Position (m) | Density (kg/m) |
|---|---|
| 0 | 5.0 |
| 2 | 5.8 |
| 4 | 6.4 |
| 6 | 6.9 |
| 8 | 7.3 |
| 10 | 7.6 |
Solution. The mass is , but with only sampled values you estimate that integral. Each strip is meters wide. A left sum uses the density at the left edge of each strip, a right sum the density at the right edge:
The density readings climb steadily, so the cable is denser toward one end. For a rising density the left sum uses the lower reading in each strip and under-estimates, while the right sum uses the higher reading and over-estimates, the same left-under, right-over pattern from Section 1.1. So the true mass is trapped:
Splitting the difference gives the trapezoid estimate kg, a better single guess than either edge. With only six readings you cannot get the exact mass, but you can honestly bracket it, and that is often all a design needs.
A wind turbine blade is not a uniform plank. It is packed with structural spar near the hub, where the forces are largest, and tapers to a thin skin at the tip. Its density changes all along its length, so its total mass is the integral of that density, never one density times the length. Two numbers that both fall straight out of this section decide the machine: the blade's mass, since the tower and bearings have to carry three of them spinning, and its balance point, since that is where the weight loads the hub as it turns. A rocket is the same problem in motion. As it burns fuel, the density along its body drops second by second, and the integral of that changing density is the mass the engines still have to push. Anywhere a thing is heavier in some places than others, from a bridge cable that thickens under load to a chain piling onto a deck, the honest weight is a sum of thin slices, and a sum of thin slices is an integral.
Why is the mass instead of the grade-school "density times length"? Because density times length only tells the truth when the density never changes. Watch what the integral does when is a constant :
So the old formula is not wrong, it is the special uniform case hiding inside the integral. The moment the density varies, no single number is the right one to multiply by, and the slicing takes over.
But there is still one honest density to multiply by, and finding it ties this section straight back to Section 2.1. Divide the true mass by the length:
That is the average value of the density, the one steady density that would give the same total mass. For the rod in Example 3.5.1, kilograms per meter, and kg reproduces the mass exactly. And the rod really is that dense somewhere: solving gives , the midpoint, so at the middle of the rod the density hits its own average. Mass over length is the average density, and the average density is a density the rod genuinely reaches. You cannot know it in advance, though; you have to integrate to find it first.
Dig Deeper Where does the rod balance?▸
Rest a bat across one finger and slide it until it holds level. That balance point is the center of mass, and it sits nearer the heavy end, not the geometric middle. Slicing gives it too. Each slice at position has mass , and its tendency to tip the rod is its mass times its distance from the pivot. Balancing means those tipping tendencies cancel, and working that out gives a clean formula:
The bottom is just the mass. The top weights each position by how much mass sits there, so a heavy far end pulls the balance point toward itself. Take the rod from Example 3.5.1, on , whose mass is kg. The top integral is
so the balance point is
The rod is meters long, so its geometric center is at m, but it balances at about m, shifted toward the heavy far end exactly as your finger would find. A uniform rod would balance dead center; the extra mass out near is what drags the balance point past the middle.
Two slips cost the most points here, and both come from grabbing a formula without checking it. First, the units. Density times length is a mass only when the units reduce to a mass: kilograms per meter times meters gives kilograms, but grams per centimeter times meters gives nonsense until you match the units first (see Appendix A.8). If your final units are not a mass, the setup is wrong, no matter how clean the algebra looks. Second, the ring. The area of a thin ring at radius is , not . The expression is the area of the entire disk out to radius , a solid circle, while the thin ring is only its outer edge: a loop of circumference with a sliver of width . Reach for inside a plate's mass integral and you count every ring's worth of material many times over.
(a) A rod along meters has linear density kilograms per meter. Find its mass. (b) A flat circular plate of radius meters has a uniform area density of kilograms per square meter. Find its mass with a ring slice, then check your answer against area times density.
Hint
(a) Integrate the density over the rod: using the power rule from Section 1.4. (b) Use with constant and ; the whole-disk area is in Appendix A.7 for the check.
Show solution
(a) kg. (b) kg kg.
(a) The mass is
(b) With a constant density , the ring slice gives
The check: a uniform plate is just its area times its density. The area is square meters, so the mass is kg, matching the ring-slice answer. When the density is constant, the ring machinery reduces to the plain area-times-density formula, exactly as it should.
Exercises 3.5
1. A rod along meters has linear density kilograms per meter. Find its mass. Warm up
Hint
Integrate the density over the rod, , using the power rule from Section 1.4. State the units.
Show solution
kg.
Kilograms per meter times meters gives kilograms, so the mass is kg.
2. A flat washer (a ring) has inner radius cm and outer radius cm. (a) Find its exact area from the two circles. (b) Compare that with the thin-ring estimate using cm and cm. Warm up
Hint
(a) The area between two circles is . (b) The thin-ring formula from Section 3.5 is circumference times thickness ; use the middle radius .
Show solution
(a) cm cm. (b) cm, the same value.
(a) The exact area is
(b) The thin-ring estimate is cm, an exact match. That is no accident: the difference of squares equals the thickness times twice the middle radius, which is exactly . The whole-disk area cm would be wildly too big: it counts the solid circle, not the thin ring.
3. A bat is modeled as a rod along meters with a thin handle and a thick barrel, so its linear density is kilograms per meter. Find its mass, exact and then rounded to three decimals. Core
Hint 1
Set up as in Example 3.5.1 of Section 3.5: integrate the density over the length.
Hint 2
Integrate term by term with the power rule: and . Evaluate from to , then convert the exact fraction to a decimal.
Show solution
kg kg.
Most of that mass lives in the barrel end, where the term grows fastest.
4. A flat circular plate of radius meters is denser toward its rim, with radial density kilograms per square meter. Find its mass with a ring slice, exact (carrying ) and then rounded to three decimals. Core
Hint 1
Mirror Example 3.5.2 in Section 3.5: the ring at radius has mass , so integrate .
Hint 2
Distribute first: . Pull the out front, integrate with the power rule, and evaluate from to .
Show full solution
kg kg.
The factor makes the outer rings count for more, and since the density also grows outward, most of this plate's mass sits near the rim.
5. A crew measures a -meter beam's linear density every meters: at the readings are kilograms per meter. (a) Estimate the mass with a left sum and a right sum (). (b) Which is the over-estimate and which the under-estimate, and why? (c) Give the trapezoid estimate. Core
Hint 1
This is Example 3.5.3 of Section 3.5 with a density that falls instead of rising. There are four strips of width ; a left sum uses the reading at each strip's left edge, a right sum the reading at its right edge.
Hint 2
For a decreasing density, think about which edge of each strip holds the higher reading. Recall the left-under, right-over rule from Section 1.1 and how it flips when the function decreases. The trapezoid estimate is .
Show full solution
(a) kg, kg. (b) The left sum over-estimates and the right sum under-estimates, because the density is decreasing. (c) kg.
(a) With four strips of width :
(b) The density falls from left to right, so the left edge of each strip holds the higher reading. The left sum therefore uses the larger value in every strip and over-estimates, while the right sum uses the smaller value and under-estimates. That is the Section 1.1 rule with the sign flipped, since the readings decrease. So .
(c) Averaging the two brackets, kg, the best single estimate from these readings.
6. A rod along meters has linear density kilograms per meter, so it is much denser near . Find its mass. Core
Hint 1
Integrate . First rewrite the integrand as a power of , the way Section 1.2 rewrites before integrating.
Hint 2
Write , whose antiderivative is . Evaluate from to .
Show solution
kg.
The rod is heavy near and thins out fast, so even though it is meters long its mass is a modest kg.
7. A flat circular plate of radius meters has radial density kilograms per square meter for some positive constant . Its total mass is kilograms. Find . Stretch
Hint 1
Set up the mass as a ring-slice integral in terms of , following Section 3.5: . Then set it equal to and solve.
Hint 2
The constant pulls out front: . Evaluate the integral, giving a multiple of , then divide both sides by that multiple of .
Show full solution
(in kilograms per cubic meter).
The ring-slice mass is
Set this equal to the given mass and solve:
Since has units of kilograms per square meter and is in meters, carries units of kilograms per cubic meter.
8. A rod along meters has linear density kilograms per meter, where is a positive constant, so the rod is twice as dense at the far end as at the near end. (a) Find the mass in terms of and . (b) Find the average density, and compare it with the densities at the two ends. Stretch
Hint 1
Integrate the density over the rod, , treating and as constants. The average density is , the average value idea from the Section 3.5 Dig In and Section 2.1.
Hint 2
Split the integral: and . Add them, then divide by for the average density.
Show full solution
(a) kg. (b) The average density is , the midpoint of the end densities and .
(a) Integrate:
(b) The average density is . The end densities are and , and is their exact midpoint, because a density that rises in a straight line averages to its middle value (the Quick Check and Dig In again). A uniform rod of that one density, , over length would weigh the same .
9. The Section 3.5 Dig In showed that a rod's mass divided by its length is the average value of its density from Section 2.1. Make it yours. Pick your own increasing linear density (choose new positive numbers for and ) on an interval , find the mass, find the average density , and confirm it equals at the midpoint . Then explain in one sentence why the midpoint works for your straight-line density but would fail for a curved density like . Dig In
Hint 1
Reread the Section 3.5 Dig In: the average density is . Your job is to run one concrete straight-line case and check the midpoint claim.
Hint 2
Integrate from to , divide by , and compare with . For the curved case, compare the average of on against .
Show full solution
Worked with on : mass kg, average density , and matches the midpoint. The midpoint works for a line but not for , whose average beats its midpoint value .
Take , , , so . The mass is
The average density is kilograms per meter, and the midpoint density is . They agree. For a straight-line density the graph is symmetric about its midpoint, so every unit below the average on the low side is matched by a unit above it on the high side. A curved density has no such symmetry: for the average value on is , while the midpoint value is , and is larger because the curve loiters low and then shoots up.
Check your own version. For any straight-line you pick, two audits confirm it: the average density must come out to , and that must equal exactly. If the two disagree, recheck the integral, since a linear density and its midpoint can never come apart. If you swap in a curved density, expect the average and the midpoint to part ways, with the gap telling you which way the curve leans.
10. The Section 3.5 Dig Deeper found a rod's balance point (center of mass) as . Design your own. Pick an increasing linear density on (new positive numbers), find the balance point, and confirm it lands past the geometric center , toward the heavy far end. Then state where a uniform rod would balance and explain the difference. Dig Deeper
Hint 1
Reread the Section 3.5 Dig Deeper: the bottom integral is the mass, and the top integral weights each position by the mass sitting there.
Hint 2
Compute both integrals for your density, divide, and compare the result with . For a uniform rod, symmetry puts the balance point exactly at ; an increasing density must push it beyond that.
Show full solution
Worked with on : mass , balance point m, past the geometric center m. A uniform rod would balance at m; the extra mass toward shifts it outward.
Take on . The mass is
The top integral weights position by mass:
So the balance point is
The rod is meters long, so its geometric center is at m, but it balances at about m, shifted toward the heavy far end. A uniform rod balances dead center at m by symmetry; here the density grows toward , so more mass sits out there and drags the balance point past the middle.
Check your own version. Whatever increasing density you pick, two audits confirm it: the balance point must come out strictly greater than , and setting the density constant must send back exactly to . If your lands at or below with a density that rises toward the far end, recheck the top integral , since more mass on the far side can only pull the balance outward.
Mass is the first physics that this unit's recipe reads off an object: slice it thin, find one slice's mass, sum, and integrate. For a rod, a slice at position has mass , so the total is , the area under the density curve. For a circular plate with radial density, slice into thin rings whose area is (the circumference times the thickness, not ), so , the same shell-style slicing you will spin into volumes in Section 3.4. Density times length is only the constant-density shortcut; in general, mass over length is the average density, the average value from Section 2.1, a density the object genuinely reaches somewhere. When there is no formula, a density table still yields an honest bracket through a left and right Riemann sum, straight from Section 1.1. Next, the same slice-sum-integrate move turns a force acting across a distance into work.
Section 3.6
Force and Work
- Compute work as an integral, , whenever the force changes with position.
- Use Hooke's law : find a spring's constant from one measurement, then the work to stretch it between any two lengths.
- Set up a pumping problem by slicing the liquid, where each thin slice has weight (density times area times ) and rises its own distance, so .
- Find the work to lift a heavy rope or chain, and keep the units straight (joules, foot-pounds).
Grab a spring and pull. The first centimeter comes easily. The next one fights back a little harder, and the one after that harder still, because a spring pushes against you more the farther it is already stretched. So the force in your hand is never one number. It grows the whole way. If you learned in a physics class that work is force times distance, this is the moment that rule runs out of road, because there is no single force to multiply. The honest answer is the same move this whole unit has been making: slice the pull into tiny steps, over each step the force barely changes, add up the work of every step. What comes out is an integral, and it has a clean picture, the area under the force curve.
Build the intuition
Here is the same recipe you have run all unit (slice, approximate, sum, take the limit), now aimed at work. Push an object along a line from to against a force that depends on where the object is. Cut the trip into thin steps of width . Over one step the position hardly moves, so the force is nearly the constant value , and the little bit of work done on that step is
force times the short distance, which is the one case where "force times distance" is safe. Add up every thin step from to and the sum becomes an integral, exactly as in Section 1.1's rectangles and the shared recipe of this unit's opener. If you graph the force against position, each term is the area of a thin rectangle under the curve, so the total work is the area under the force graph. When the force happens to be constant, that area is just a plain rectangle, and you recover force times distance as the special case it always was.
The work done by a constant force pushing an object a distance along the direction of the force is . When the force varies with position, the work done moving the object from to is the integral of the force over the distance:
In SI units, force is in newtons and distance in meters, so work is in newton-meters, called joules (J): one joule is one newton times one meter. In US customary units, force is in pounds and distance in feet, so work is in foot-pounds (ft-lb). Rusty on these units? Appendix A.8 is the two-minute refresher.
If a force acts along the line of motion, the work done moving from to is
Reason: over a slice of width the force is nearly constant, so the slice's work is ; summing the slices and taking the limit is the definite integral. Read as an area, is the area under the force-versus-position graph. Constant force is the flat case, where the area is a rectangle and .
A spring stretched (or compressed) a distance from its natural length pulls back with a force proportional to :
This is Hooke's law, and the constant is the spring constant, a stiffness measured in newtons per meter (or pounds per foot). A single measurement pins it down: if a known force holds the spring stretched a known distance , then . Because the force grows with , the work to stretch a spring is a textbook case of an integral, never a plain product.
A force of 20 N holds a spring stretched 0.1 m beyond its natural length. (a) Find the spring constant . (b) Find the work done stretching the spring from its natural length to 0.2 m beyond. (c) Find the work done stretching it from 0.2 m to 0.3 m beyond natural length, and compare with part (b). Figure 3.6.1 shows the force.
Solution. (a) Hooke's law says . The measurement is N at m, so
The force as a function of stretch is .
(b) Work is the integral of force over distance. Stretching from the natural length means runs from 0 to 0.2:
(c) The same integrand, now from to :
Look at what that comparison says. The first 0.2 m of stretch cost 4 J. The next 0.1 m alone cost 5 J, more than the entire first stretch, even though it covers half the distance. Slice the stretch into three equal steps of 0.1 m and the works come out 1 J, then 3 J, then 5 J: the same distance, more work each time, because the spring pulls back harder the farther it is already open. This is precisely why work is an integral and not force times distance. The force is a moving target, and the integral is what tracks it.
Figure 3.6.1 The spring force (in newtons) against stretch (in meters). Work is the area under the line. The three equal-width strips have areas , , and J: the same m of extra stretch costs more work each time, because the force keeps climbing.
Before you compute, predict. A spring has constant N/m. Does the work to stretch it from its natural length to 0.4 m equal the work to stretch it from 0.4 m to 0.8 m, the same 0.4 m of extra stretch? Then find both.
Show solution
No. The first stretch takes J and the second takes J, three times as much, because the spring is already open and pulling back harder.
With , the two integrals are
Equal distances, unequal work. The area under the line is skinnier near the start and taller near the end.
Example 3.6.1's spring is live below, with the Dig In's collision built in. Drag the stretch and watch three numbers race: the honest work (the shaded area under ), the end-force shortcut (the dashed rectangle), and the start-force shortcut . Challenge: park anywhere and check that the shaded triangle is exactly HALF the dashed rectangle. That is the average-force story: for a force that climbs steadily from 0, work is the average force times the distance, and the integral is what computes that average for you.
Pumping a liquid, one slice at a time
Lifting water out of a tank is the same recipe wearing new clothes. You cannot lift the tankful all at once with a single force times a single distance, because the water at the bottom has to travel farther than the water near the top. So slice the water into thin horizontal layers. One layer is small enough that every drop in it rises the same distance, so you can find its little bit of work honestly, and then you add the layers up. The work of one slice is its weight times how far it rises, and the weight of a slice is the liquid's weight density times the slice's volume, which is its cross-sectional area times its thickness .
To find the work to pump a liquid, put a depth coordinate on the tank, then look at a thin horizontal slice at depth , of thickness :
- Weigh it. Its weight is (weight density)(cross-sectional area at depth ).
- Lift it. Read off how far that slice must rise to reach the outlet; this rise depends on .
- Add. The work of one slice is (slice weight)(rise), and the total is the integral over the range of the liquid occupies:
Reason: within one thin slice every drop rises the same distance, so its work is an honest product; summing the slices and taking the limit is the integral. The rise lives inside the integral because it changes from slice to slice.
A rectangular tank is 2 m long and 1 m wide and is filled to a depth of 3 m with water. Water weighs 9800 newtons per cubic meter (that is, each cubic meter of water weighs 9800 N). Find the work done pumping all the water up and out over the top rim. Figure 3.6.2 shows one slice.
Solution. Measure depth with , taking at the rim and at the floor. A thin horizontal slice at depth is a flat sheet of water. Its cross-section is the tank's footprint, an area of square meters, and its thickness is , so its volume is and its weight is
That slice sits meters below the rim, so to leave the tank it must rise exactly meters. The work to lift this one slice is its weight times its rise:
Add up every slice from the rim down to the floor:
Pumping the tank dry takes 88200 joules. Notice what varied and what did not. Every slice weighs the same, because the tank has straight walls, so the cross-section never changes. What changes is the rise: the slices near the floor pay for a longer lift, and that growing distance is the sitting inside the integral. Re-derive the coefficient 19600 yourself from the density and the area every time; it is one line, and it is the line a misquoted number would hide in.
Figure 3.6.2 A slice of water at depth , thickness . Its weight is newtons, and it must rise meters to clear the rim, so . The weight is the same for every slice; the rise is what changes.
A uniform chain 20 ft long, weighing 5 lb per foot, hangs straight down from a winch at the top of a building. (a) Find the work done winding the whole chain up to the top. (b) Now suppose a 50 lb toolbox hangs from the bottom of the chain. Find the total work to wind up the chain and the toolbox together.
Solution. (a) Slice the chain by position, not the liquid this time, but the same idea. Let be the distance below the winch, from 0 at the top to 20 at the bottom. A short piece of chain at distance , of length , weighs pounds, and to reach the top it must rise feet. Its work is , and the total is
Winding up the chain takes 1000 foot-pounds. Each link near the bottom is lifted the full 20 feet, while a link near the top barely moves, and the integral bundles all those different lifts into one total.
(b) The toolbox is different in kind. It is a single 50 lb weight, and the whole box rises the same 20 feet, so its force never varies with position. That is the constant-force case, plain :
The two jobs add, so the total work is
One object needed an integral and the other needed a product, and the difference was not the size of the load. It was whether the force changed as the thing moved. The chain unspools link by link, so its lift is an integral; the toolbox goes up as one piece, so its lift is a rectangle.
A pumped-storage hydroelectric plant is a battery made of water and gravity. When the grid has power to spare at night, the plant runs its pumps in reverse and lifts water from a low reservoir to a high one, and every parcel of water rises its own height against gravity, so the energy stored is exactly the pumping integral of this section. Come the afternoon peak, that water falls back through turbines and pays the energy back as electricity. The scale is enormous, millions of cubic meters moved, but the accounting is the one line you just learned: lift a mass a height and you store joules, so one tonne of water raised 300 m banks joules, and a full reservoir is that sum taken over the whole lake. The same "each piece travels its own distance" bookkeeping runs a mine hoist whose steel cable is so long its own weight rivals the car it carries, and a rocket that grows lighter as it burns fuel it no longer has to lift. Wherever a force changes as the load moves, the total is an integral, not a product.
Why is work really and not ? Watch the two answers collide on the spring from Example 3.6.1. Suppose you insist on force times distance. Which force? If you grab the ending force, N, and multiply by the 0.3 m stretched, you get 18 J, far too much, because the force was only 60 N at the very end and smaller the whole way there. If you grab the starting force, 0 N, you get 0 J, absurd. The force was never one number, so no single product can be right.
The integral fixes this by never asking for one force. It slices the stretch into steps of width , and over a single step the position barely changes, so the force is nearly constant and the step's work is honestly . Summing the steps is a Riemann sum from Section 1.1, and its limit is the integral, the area under the force curve, the same "signed area under a graph" you met in Section 1.3 and the same slice-and-sum move as Section 3.1. For a spring the force graph is a straight line, so that area is a triangle:
There is a clean way to read that. The average height of the line over is , the average force, and the work is that average force times the distance , since . So for a force that climbs steadily, "force times distance" is rescued by using the average force, and the integral is what computes the average for you. Check the arithmetic against the numbers directly. For the Example 3.6.1 spring (, ) the exact work is J. A right-endpoint Riemann sum with three slices gives J, a left sum gives J, and the true 9 J sits between them, right where a Riemann sum promises the area will be.
Dig Deeper Pumping a tank whose slices are all different sizes▸
Example 3.6.2's tank had straight walls, so every slice weighed the same and only the rise varied. A cone makes both vary at once, and it is the classic place a printed coefficient goes wrong, so trust nothing you have not re-derived. A tank shaped like a cone with its point at the bottom is 3 m deep and 2 m across the radius at the top, and it is full of water (9800 newtons per cubic meter). How much work pumps it all out over the top rim?
Set as the depth below the rim, from 0 to 3. The trouble is that a slice at depth is a disk whose radius shrinks as you go down (Figure 3.6.3). Similar triangles fix the radius: at the rim () the radius is 2, at the point () it is 0, and it falls in a straight line between, so
That is the geometry to re-derive yourself, not to copy. Now the slice at depth is a disk of that radius and thickness , so its area is , its weight is
and it rises meters. Both the weight and the rise now depend on , so both go inside the integral:
You can check the coefficient a second way without the integral. The whole cone of water weighs (weight density)(volume) N, and the balance point of a point-down cone sits a quarter of the height from its wide base, that is m below the rim. Lifting the whole weight that average distance gives J, the same number the slicing integral found. Two roads, one answer, and neither one borrowed a coefficient it did not earn. Rusty on a cone's volume? Appendix A.7 has it.
Figure 3.6.3 A point-down conical tank. A slice at depth is a disk of radius , from similar triangles, and rises meters. Here both the slice's size and its lift change with depth.
The trap is reaching for out of habit. That formula is only the constant-force special case, the flat rectangle under the force graph. The moment the force changes as the object moves, a spring stiffening, a chain unspooling, a rocket lightening, there is no single to multiply, and the honest total is the integral . In a pumping problem the same trap wears a second disguise: students find one slice's rise, say the depth of the deepest water, and multiply the whole weight by it, as if every drop rose that far. It did not. Each slice rises a DIFFERENT distance, so that rise must stay INSIDE the integral, not out front as a constant. If you ever see a lone number multiplying an integral where a variable rise belongs, stop and re-slice. And keep the units honest: a newton times a meter is a joule, a pound times a foot is a foot-pound, and an answer in "newtons" or "pounds" alone is a force that forgot to travel.
A cylindrical tank of radius 1 m and height 4 m is full of water, which weighs 9800 newtons per cubic meter. Find the work done pumping all the water up and out over the top rim.
Hint
Use the pumping recipe from Section 3.6: measure depth from the rim, take a slice of thickness , and find its weight (weight density times cross-sectional area times ) and its rise. A cylinder's cross-section is a circle of radius 1, so its area is , the same for every slice; only the rise changes.
Show full solution
J.
Let be depth below the rim, from 0 to 4. A slice at depth is a disk of radius 1, so its area is square meters, its thickness is , and its weight is
The slice must rise meters to clear the rim, so , and
The cross-section was constant, like the rectangular tank, so only the rise varied. Carrying exactly to the last line and rounding once at the end keeps the answer honest.
Exercises 3.6
1. A force of 45 N stretches a spring 0.3 m beyond its natural length. Find the spring constant . Warm up
Hint
Hooke's law is , so (Section 3.6). Put the measured force over the measured stretch; the units are newtons per meter.
Show solution
N/m.
The spring pushes back with 150 newtons for every meter it is stretched.
2. A crane lifts a 200 kg steel beam straight up 15 m at a steady speed. Taking meters per second squared, find the work done against gravity. Warm up
Hint
The beam's weight is a constant force (mass times ), and it rises the same 15 m the whole way, so this is the constant-force case , not an integral (Section 3.6). Find the weight in newtons first.
Show solution
J.
The beam's weight is N, a force that does not change as it rises. The lift is a constant force over a fixed distance:
No integral needed, because nothing about the force varied with height.
3. A spring has spring constant N/m. Find the work done stretching it from 0.1 m to 0.4 m beyond its natural length. Core
Hint 1
Work is the integral of the force, and the force is (Section 3.6). The stretch runs from to , so those are your limits.
Hint 2
Evaluate . Do not start from 0; the spring is already stretched 0.1 m when the interval begins.
Show solution
J.
Starting the interval at 0.1 m instead of 0 matters: the first bit of stretch is already spent before this job begins.
4. A force of 25 N holds a spring stretched 0.05 m beyond its natural length. (a) Find the spring constant. (b) Find the work done stretching the spring from its natural length to 0.2 m beyond. Core
Hint 1
Part (a) is one measurement in . Part (b) is an integral of that force from 0 to 0.2 (Section 3.6).
Hint 2
For (a), . For (b), integrate with the you just found.
Show solution
(a) N/m. (b) J.
(a)
(b) With :
5. A rectangular tank 2 m long and 3 m wide is filled to a depth of 2 m with water (weight density 9800 newtons per cubic meter). The water is pumped out through a spout 1 m above the top rim. Find the work done. Core
Hint 1
Follow the pumping recipe (Section 3.6): a slice at depth below the rim has weight (weight density)(area) with a constant area of . The twist is the rise.
Hint 2
The spout is 1 m above the rim, so a slice at depth must rise , not just . Integrate .
Show solution
J.
Measure depth from the rim, 0 to 2. The cross-section is square meters, so a slice weighs newtons. To reach a spout 1 m above the rim, a slice at depth rises meters:
The extra 1 m of lift rode along inside the integral, added to each slice's own depth.
6. A uniform rope 20 ft long weighs 0.8 lb per foot and hangs straight down from the top of a building. Find the work done pulling the whole rope up to the top. Core
Hint 1
This is Example 3.6.3's chain without the toolbox (Section 3.6). Let be the distance below the top; a piece of rope at , of length , weighs and rises .
Hint 2
Integrate . Each piece rises a different distance, which is exactly why this is an integral and not weight times a single height.
Show solution
ft-lb.
A piece of rope at distance below the top weighs pounds and must rise feet, so
The whole 16 lb rope was lifted, but not all of it 20 feet: only the very bottom traveled that far, and the integral averaged the rest in.
7. A rectangular aquarium has a base 1 m by 2 m and stands 1.5 m tall. It is filled with water (9800 newtons per cubic meter) to a depth of 1 m, so the water surface sits 0.5 m below the rim. Find the work done pumping all the water out over the top rim. Stretch
Hint 1
The tank is not full, so the water does not start at the rim. Set as depth below the rim; the water occupies only part of that range (Section 3.6).
Hint 2
The surface is 0.5 m below the rim and the floor is 1.5 m below it, so runs from 0.5 to 1.5, not from 0. Each slice still rises its own depth , and the area is . Integrate .
Show full solution
J.
Measure depth from the rim. The water starts 0.5 m down (its surface) and ends 1.5 m down (the floor), so those are the limits. Each slice has area , weight , and rise :
The empty top of the tank never enters the integral, because there is no water there to lift. The limits carry that fact.
8. The work done stretching a certain spring from its natural length to 0.2 m beyond is 6 J. (a) Find the spring constant. (b) Find the work done stretching the same spring from 0.2 m to 0.4 m beyond natural length. Stretch
Hint 1
Run the spring work formula backward. The work from 0 to a stretch is (Section 3.6, the Dig In), and you are told this equals 6 J at . Solve for .
Hint 2
From , find . Then integrate with that .
Show full solution
(a) N/m. (b) J.
(a) The work from 0 to 0.2 is , and this equals 6 J:
(b) With :
The second stretch costs three times the first, the same odd-number climb you saw in Example 3.6.1.
9. The Dig In showed that for a force climbing steadily along a line, the work equals the AVERAGE force times the distance, because the integral is the area under the force graph. Build your own example to test this. Pick a linear force (choose your own positive slope ) and an interval . (a) Compute the work two ways: as , and as (average force)(distance), and show they match. (b) Explain why the two ways can only ever agree when the force graph is a straight line. Dig In
Hint 1
Reread the Dig In in Section 3.6. The average of a straight line over is the average of its two endpoint values, . Compute that, multiply by , and compare with the integral.
Hint 2
For part (b), think about what "average value equals the average of the endpoints" requires of a graph. A curve that bends spends more of its run on one side of the straight-line average, so the midpoint of the endpoints is no longer the true average.
Show full solution
Worked with on : the integral gives J and (average force)(distance) gives J. They match because the graph is straight; a curved force graph breaks the shortcut.
(a) Take and , so . By the integral,
The average force is the average of the endpoint values and , which is N, and times the distance 4 m that is J. Same number.
(b) The trick "average force = average of the two endpoints" is exactly the statement that the graph's average height is the midpoint of its endpoint heights, and that is true precisely for a straight line. A line spends as much run above its midline as below, so the midpoint of the ends is the true average. Bend the graph and it lingers longer on one side, pulling the true average off the endpoint midpoint, and the shortcut fails; only the integral still gives the right area.
Check your own version. Whatever slope and endpoint you chose, your integral must give , and your (average force)(distance) must give as well. If the two disagree, one common slip is using alone as the "average force" instead of ; the average is halfway between the ends, not the end.
10. The Dig Deeper pumped a point-down cone by re-deriving every coefficient from the geometry. Design your own conical tank and do the same. Pick your own top radius and depth for a point-down cone full of water (9800 newtons per cubic meter). (a) Use similar triangles to write the slice radius at depth below the rim, then set up and evaluate the work to pump it all over the rim. (b) Re-derive your answer a second way from the cone's weight and its balance point (a quarter of the height from its wide base), and confirm the two agree. Dig Deeper
Hint 1
Reread the Dig Deeper in Section 3.6. Similar triangles give : radius at the rim, 0 at the point. The slice weight is and the rise is .
Hint 2
The setup collapses to , and , which gives the tidy closed form . For part (b), the cone's water weighs and its balance point is below the rim.
Show full solution
Worked with m, m: J, and the weight-times-balance-point route gives the same J.
(a) Take a point-down cone with top radius and depth . Similar triangles give . A slice at depth is a disk of area , weight , and rise :
(b) The cone of water weighs N, and its balance point is m below the rim, so lifting the whole weight that far gives J, matching part (a).
Check your own version. Whatever and you chose, your integral should collapse to , and your weight-times-balance-point route, , must give the identical expression. If the two disagree, the usual culprit is the similar-triangles step: check that is at the rim and at the point, not the reverse.
When a force changes as the object moves, work is an integral, not a product: , the area under the force-versus-position graph, with surviving only as the flat, constant-force special case. For a spring, Hooke's law gives that force: read off a single measurement, then integrate to get the work to stretch between any two lengths, which grows faster the farther the spring is already open. A pumping problem runs the unit's slice-sum-integrate recipe on a liquid: a thin slice at depth has weight (density)(area) and rises its own distance, so , and the rise stays inside the integral because it changes from slice to slice; lifting a heavy rope or chain works the same way, piece by piece. Keep the units on the answer, joules or foot-pounds, and re-derive every coefficient from the geometry rather than trusting a printed one. The last application in this unit turns the integral loose on chance itself, where an area under a curve becomes a probability. Next.
Section 3.7
Probability
- Read a probability density function: a curve whose total area is exactly 1.
- Find the probability that a continuous outcome lands in an interval as the area .
- Compute the mean of a distribution, , and read it as the balance point of the area.
- Work the three densities you meet most: the uniform, the linear or triangular, and the exponential wait-time model .
- See how the counting-and-summing probability from your Probability and Matrices course becomes an integral once the outcomes turn continuous.
Ask how long you will wait for the next bus, and there is no clean list of answers. The wait could be 3 minutes, or 3.5, or 3.41 minutes, any real number in a range. In your Probability and Matrices course a probability was a count: favorable outcomes over total outcomes, or the height of a bar in a histogram, and you added those heights to answer a question (a two-minute refresher is in Appendix A.9). That works when you can list the outcomes and stack them into bars. It stops working the instant the outcome can be any real number, because now there are infinitely many of them and no single one has a bar of its own. So the picture changes. Instead of bars you add, you get a smooth curve you integrate, and a probability becomes an area under that curve. This is the last idea in the book, and it is the whole book in miniature: a total built out of infinitely many pieces too small to count one at a time.
Build the intuition
Start from the histogram you already know. Divide a range of outcomes into narrow bins, and let each bar's height be the probability of landing in that bin. Add every bar and you get 1, because the outcome is certain to fall somewhere. Now let the bins shrink. The staircase of bars smooths into a single curve, and the sum of bar heights becomes the area under that curve. That curve is the density, and its total area is still 1.
Figure 3.7.1 The bars you summed in Probability and Matrices, shrunk until the sum becomes an area. Both totals are 1: the outcome is certain to happen.
Every application in this unit has run one recipe: slice the quantity into thin pieces, approximate each piece, add them, and let the pieces shrink to a limit, which lands on a definite integral (Section 1.3). Probability is the same recipe pointed at chance. Slice the interval of outcomes into strips of width ; the strip at holds about of probability; add the strips and take the limit, and the probability of the whole interval is an integral. The one honest question left is what the curve has to be for this to work, and there are just two conditions.
A continuous random variable is a quantity whose outcome can be any real number in some range, like a wait time, a height, or a battery's life. A probability density function (pdf) for is a function with two properties: it is never negative, for every , and the total area under it is exactly 1, . The probability that lands between and is the area over that interval,
A function is a probability density exactly when both hold:
Then every probability is an area under :
Reason: the total area is the probability of the outcome landing anywhere at all, which is certainty, so it must equal 1; and the probability of a smaller interval is just the slice of that total area sitting above it. Most densities live on a finite interval or on , and is simply 0 outside their range, so the total integral only has to be taken where is nonzero.
The mean (or expected value) of a continuous random variable with density is
Reason: in Probability and Matrices the expected value was a weighted sum, each outcome times its probability, . Here the strip at carries probability , so its contribution to the average is the value times that probability, . Adding the strips and taking the limit turns the weighted sum into a weighted integral. The mean is the balance point of the density: the place where the area under the curve would sit level if you set it on a knife edge.
A train is equally likely to arrive at any moment in the next 15 minutes, so your wait is spread evenly across . Its density is constant, on and 0 elsewhere. (a) Check that is a valid density. (b) Find the probability you wait at most 5 minutes. (c) Find the mean wait.
Solution. (a) The height is positive, and the total area is a single rectangle:
Both conditions hold, so is a density. The general pattern is the same for any interval: a uniform density on has constant height , the one height that makes the rectangle's area 1.
(b) A probability is an area, so integrate the density from 0 to 5:
A one-third chance, which matches plain sense: 5 minutes is one third of the 15-minute window, and the density is flat.
(c) The mean is the balance-point integral:
The mean sits at the midpoint of , which is where a flat, symmetric slab balances. For any uniform density the mean is the midpoint .
Figure 3.7.2 The uniform density on . The shaded slab is , one third of the width. The mean sits at the midpoint.
Before you go on, predict, then verify by integrating. A parking meter's leftover time is uniform on minutes, so on that interval. (a) What is ? (b) What is the mean?
Show solution
(a) . (b) minutes.
(a) The area from 0 to 4 is , which is 4 out of the 20 minutes. (b) A uniform density balances at its midpoint, so , the same value gives.
A part is held under a steady load for one minute of stress testing, and the longer it is held the likelier it is to crack, so the instant (in minutes) at which it cracks has a density that rises straight from zero: on . (a) Find the constant that makes this a density. (b) Find the probability the part lasts past the halfway mark. (c) Find the mean cracking time.
Solution. (a) The constant is whatever forces the total area to 1. Integrate and set it equal to 1:
So . Notice the height at the right end is , well above 1. That is allowed: a density is a height, not a probability, and only its area has to stay at 1.
(b) "Past the halfway mark" means between and 1, so integrate there:
Three quarters of the cracks happen in the second half of the minute, even though it is only half the time, because the density is heavier there.
(c) The mean weights each instant by its density:
The balance point is of a minute, pulled toward the heavy right end, not the midpoint . A lopsided density balances off-center.
Calls arrive at a help desk at random, one every 5 minutes on average. The time (in minutes) until the next call has the exponential density for , with , so . (a) Confirm the total area is 1. (b) Find the probability the next call comes within 5 minutes. (c) Find the mean wait.
Solution. Heads up: this example reaches a little past the section. Part (a) uses an improper integral (the optional Section 2.1 Dig Deeper) and part (c) uses integration by parts (the optional Section 2.2 Dig Deeper). If you worked those Dig Deepers, the moves will look familiar; if you skipped them, they are worth a peek now, or you can take the boxed answers as given and check the integrals with Desmos or another tool of your choice. (a) The outcomes run out to infinity, so the total area is an improper integral, and the antiderivative of is the form from Section 2.4 with :
As runs to infinity the term fades to 0, and at it is , so the area closes on exactly 1. It is a density.
(b) Integrate the density from 0 to 5:
About a 63 percent chance the next call lands inside 5 minutes. Push the window out and the tail keeps some weight: , still more than 1 chance in 8 that you wait past 10 minutes.
(c) The mean is . This is a product with no inside-derivative to substitute, so it calls for the reverse product rule, integration by parts, from the Section 2.2 Dig Deeper. Taking and , so :
The boundary term is 0 at both ends: at the factor kills it, and as the decaying exponential beats the growing , the exponentials-outrun-polynomials race from Appendix A.5, so . The mean wait is 5 minutes, exactly . That clean result is the exponential's signature: cut the arrival rate in half and the average wait doubles, every time.
Figure 3.7.3 The exponential density . The shaded area is . The whole curve, out to infinity, encloses area 1, and the mean is .
Probability is an area you can grab. Drag the endpoints and under the help desk's wait-time density from Example 3.7.3 and watch fill in; then flip to a flat uniform density with the SAME mean of 5 and drag the same window. In your Probability and Matrices course you summed dots; here the dots have melted into a curve, and the sum has become this area. Challenge: put the window at on the exponential and notice it already holds half the probability, even though the mean sits out at 5. The tail drags the mean right; the median stays put.
Reliability engineers live in this integral. A bearing spinning inside a machine fails at random, so the time until it goes follows the same exponential density as the wait for a call: . Say a run of bearings lasts 8000 hours on average, so . The fraction that fail inside a 2000-hour warranty is the area from 0 to 2000, which is , about 22 of every 100. Stretch the warranty and that area grows; the same integral prices the guarantee before a single part ships. The identical density sets the gap between random breakdowns on an assembly line, the interval between clicks of a Geiger counter, and the time a queue keeps you standing, and in every one of them a probability is nothing more than the area under the curve across the stretch you asked about.
The rule does real work: it is often what pins down an unknown constant, and the move is worth doing slowly once. Suppose someone hands you on and asks whether it can be a probability density. Two things have to be true. First, on the interval, and since for , any positive keeps the curve above the axis. Second, the total area is 1, and that single condition fixes . Integrate the whole thing:
Set that equal to 1 and solve: , so . The finished density is . The number is called the normalizing constant, and finding it is never a technicality. The condition says the variable is certain to land somewhere, so the total area under the curve is the whole of certainty, one. Normalizing is nothing but scaling the curve up or down until the area it fences off is exactly that one whole certainty.
Dig Deeper Measuring the spread with variance▸
The mean names the center of a distribution, but two distributions can share a mean and still look nothing alike: one bunched tight, the other spread wide. Variance measures that spread. It is the average squared distance from the mean, with each distance weighted by the density:
Squaring keeps the distances from canceling and punishes far-out values harder than near ones. Take the rising density on from Example 3.7.2, whose mean is . A little expansion makes it cleaner to compute the spread through the shortcut , which is the same quantity rearranged. First the piece :
Then subtract the square of the mean:
The standard deviation, the square root, is , a typical distance from the mean in the original minutes. Small means the density hugs its center; large means it sprawls. Exercise 10 hands you a uniform density, where the variance comes out to a formula worth remembering.
The one slip that trips everyone here is reading a density as a probability. The value is not the chance that equals . For a continuous variable that chance is always 0: a single point has no width, so , which means for every single . Only an interval, with real width, carries probability, and that probability is an area. This is also exactly why a density is allowed to top 1. In Example 3.7.2 the density reached , and a uniform variable on has density 2 across its whole range (Exercise 2), yet neither breaks a rule, because a height is not a probability. What must stay at or below 1 is the area, never the curve itself. Keep the two apart: is probability per unit , a rate, and you get an actual probability only after you integrate it across a stretch of real width.
A quantity has the linear density on . (a) Find the constant . (b) Find . (c) Find the mean.
Hint
For (a), force the total area to 1: integrate over , set the result equal to 1, and solve for (this is the normalizing move from the Section 3.7 Dig In). Then (b) is the area , and (c) is .
Show solution
(a) . (b) . (c) .
(a) Set the total area to 1:
(b) The probability is the area from 0 to 2:
(c) The mean weights each value by the density:
The balance point sits two thirds of the way out, pulled toward the heavier right side, just like Example 3.7.2.
Exercises 3.7
1. A stopwatch is stopped at a random moment, so its final digit is uniform on seconds. (a) Write down . (b) Find . (c) Find the mean. Warm up
Hint
A uniform density on has constant height (Section 3.7, Example 3.7.1). A probability is the area of a slab, and the mean of a uniform density is the midpoint.
Show solution
(a) on . (b) . (c) seconds.
(a) The width is 8, so the height is on and 0 elsewhere.
(b) .
(c) A uniform density balances at its midpoint, so seconds.
2. A machine cuts a wire to a random length that is uniform on meters. (a) Find , and notice its value. (b) Find . Warm up
Hint
Use for a uniform density (Section 3.7). The height here comes out bigger than 1, which the Section 3.7 Watch Out says is fine: a density is a height, not a probability.
Show solution
(a) on , a density above 1. (b) .
(a) The width is 0.5, so on . The height 2 is above 1, and that is allowed: only the area has to equal 1, and here the area is .
(b) .
3. A random variable has density on . (a) Find . (b) Find . (c) Find the mean. Core
Hint 1
The constant is set by (Section 3.7, Dig In): integrate over , set it equal to 1, and solve. Then a probability is an area and the mean is .
Hint 2
You will need for part (a), and for the mean . Keep every fraction exact.
Show solution
(a) . (b) . (c) .
(a) Set the total area to 1: , so .
(b) .
(c)
4. Customers arrive at a counter with an exponential wait time averaging 4 minutes, so for . (a) Find the probability the next customer arrives within 4 minutes. (b) Find . (c) State the mean wait. Core
Hint 1
Each probability is an area under (Section 3.7, Example 3.7.3). The antiderivative of is the form from Section 2.4 with .
Hint 2
The antiderivative of is . For (b), integrate from 8 to ; the exponential's mean is , so you can read (c) straight off.
Show solution
(a) . (b) . (c) minutes.
(a)
(b)
(c) The exponential's mean is , and here , so minutes.
5. A bus is equally likely to arrive at any time in the next 12 minutes, so your wait is uniform on . (a) Find . (b) Find . (c) Find the mean. Core
Hint 1
The density is on (Section 3.7). Each probability is the area of a slab: its width times the height .
Hint 2
For (a) the slab runs from 3 to 7, a width of 4. For (b) it runs from 10 to 12. The mean of a uniform density is the midpoint of the interval.
Show solution
(a) . (b) . (c) minutes.
(a) .
(b) .
(c) The midpoint of is minutes.
6. A random variable has the linear density on . (a) Find . (b) Find . (c) Find the mean. Core
Hint 1
Set to find the normalizing constant (Section 3.7, Example 3.7.2). Then (b) is the area from 4 to 6, and (c) is .
Hint 2
The area condition gives . For the mean you will integrate , so does the heavy lifting.
Show full solution
(a) . (b) . (c) .
(a) , so and .
(b)
(c)
The mean sits at , two thirds of the way out, pulled toward the heavy right end.
7. An LED bulb's lifetime (in years) is exponential with a mean of 10 years, so . (a) Find the probability it fails within its first 5 years. (b) Find the median lifetime: the time with . Compare it to the mean. Stretch
Hint 1
The lifetime has density , and by integrating from 0 to (Section 3.7, Example 3.7.3). For the median, set that expression equal to .
Hint 2
Solving gives ; take the natural logarithm of both sides (Section 2.4) to free .
Show full solution
(a) . (b) years, less than the mean of 10.
(a)
(b) Set the accumulated probability to :
The median 6.931 is below the mean 10. The exponential leans right: a few very long lifetimes drag the mean past the halfway point, so more than half of the bulbs fail before the average.
8. A random variable has density on , a smooth arch. (a) Find . (b) Find . (c) Find the mean, and explain the answer from the shape. Stretch
Hint 1
Expand so you can integrate term by term, then set the total area to 1 (Section 3.7, Dig In) to find .
Hint 2
The arch is symmetric about , which tells you the mean before you compute it. To confirm, evaluate ; the symmetry also makes exactly half.
Show full solution
(a) . (b) . (c) , the center of the symmetric arch.
(a) With , , so .
(b)
(c)
The arch is a mirror image about , so it balances there and splits its area evenly at the center. Both answers follow from the symmetry alone.
9. The Section 3.7 Dig In normalized on . (a) Do the same for on : find the that makes it a density. (b) Design your own: pick a function that stays nonnegative on some interval , find the normalizing constant that turns it into a density, and then use it to find one probability. Dig In
Hint 1
Reread the Section 3.7 Dig In: check the function is nonnegative, then set its total integral equal to 1 and solve for the constant.
Hint 2
For (a), ; the symmetry about 0 lets you double the integral from 0 to 3 if you like. Set times that total equal to 1.
Show full solution
(a) . (b) Answers vary; see the check below.
(a) The function on , so any positive keeps it a valid shape. Fix with the area condition:
so and .
Check your own version. Whatever function you chose, run two audits. First, confirm it is at or above 0 across your whole interval; a density that dips below the axis is invalid no matter what constant you attach. Second, integrate your finished over the full interval and confirm you get exactly 1; if you do not, your constant is off. Then any probability you compute must land between 0 and 1, since it is a slice of that total area of 1.
10. The Section 3.7 Dig Deeper measured spread with variance. (a) Show that the uniform density on has mean and variance , using the shortcut . (b) Design your own density on a finite interval, then compute its variance two ways: straight from , and again through the shortcut. Confirm they agree. Dig Deeper
Hint 1
The uniform density on is . Reread the Section 3.7 Dig Deeper for the shortcut, then compute and separately.
Hint 2
You should find and . Then ; simplify over a common denominator.
Show full solution
(a) , . (b) Answers vary; the two methods must match.
(a) The uniform density is . Its mean is the midpoint,
For the variance, first , then subtract :
As a check, the train in Example 3.7.1 had , giving .
Check your own version. Compute your variance both ways and the two numbers must be identical, since expands algebraically into . If they disagree, the usual culprit is a mean that is slightly off, since enters the shortcut squared and any error there is amplified. Your must also come out positive: a squared distance, averaged, can never be negative.
Four ideas to carry out the door. A probability density function is any curve that stays at or above the axis and fences off a total area of exactly 1; that second condition, , is often what pins down an unknown constant. A probability is an area: , so a single point carries none, and a density may climb above 1 as long as its area does not. The mean is the balance point of that area, the continuous echo of the weighted-sum expected value you built in Probability and Matrices (Appendix A.9). And three densities cover most of what you meet: the uniform, flat with mean at the midpoint; the linear or triangular, a straight slope that balances off-center; and the exponential for wait times, whose mean is .
One last look at the dashboard, because this is where the book set out. The odometer never measured distance on its own; it summed a rate, the speedometer's reading, into a total. Probability does the very same thing with a different gauge: it sums a density into a certainty, and the total area of 1 is the whole of what is sure to happen. Same machine, new cargo. And that machine has now run in both directions across these two volumes. The first book took a whole and broke it into instants, naming the rate at each one; this book took the instants and piled them back into a whole, whether the whole is a distance, an area, a volume, a mass, a work, or a certainty. Breaking a whole down to its instants, and building a whole back up from its pieces: two halves of one trick, and you have now seen both halves close. Mixed Practice shuffles this unit first; the recap after it closes the loop, and then the odometer rolls over one more digit and comes to rest.
Section 3.P
Mixed Practice: Unit 3
Unit 3 ran one recipe seven times: slice the thing into thin pieces, approximate each piece, sum, and take the limit to land on an integral. What changed each section was only the target, a length, an area, a skin, a volume, a mass, a work, a certainty. Inside a section the heading already told you which target you were aiming at. Out here nothing does, so the first move in every problem is a decision, not a computation: what quantity is this, what does one thin slice of it look like, and which variable keeps that slice whole? The problems below are shuffled on purpose, with no labels to give the tool away. A few reach back for a step from earlier units, a net change read off a rate or a substitution to finish an integral, and one asks two different questions of a single object. Try each one cold before you open the hints, and keep the units on every applied answer. Stuck on which method? The guide in the back matter walks the decision.
Mixed Practice 3
1. Find the area of the region enclosed by and . Warm up
Hint 1
Set the curves equal to find where they meet, and those two crossings become your limits. Test a point between them to see which curve is on top, then integrate top minus bottom, the rule from Section 3.2.
Show solution
.
The curves meet where , that is , so and . At the line reads and the parabola reads , so the line is on top all the way across:
2. A metal rod lies along meters and thickens toward one end, so its linear density is kilograms per meter. Find its mass. Warm up
Hint 1
Integrate the density over the rod, , the mass rule from Section 3.5. Keep the units straight: kilograms per meter times meters gives kilograms.
Show solution
kg.
As a sanity check, the density runs from up to kilograms per meter, so its average is , and a uniform rod at that average density would weigh kg.
3. A force of N holds a spring stretched m beyond its natural length. (a) Find the spring constant . (b) Find the work done stretching the spring from its natural length to m beyond. Warm up
Hint 1
Hooke's law is , so . Then work is the integral of that force, (Section 3.6), never a plain product, because the force keeps climbing as the spring opens.
Show solution
(a) N/m. (b) J.
(a) From the single measurement, N/m, so .
(b) Work is the integral of the force over the stretch:
4. A curved metal trim strip is bent to the profile (in meters) from to . Find the exact length of the strip, then the rounded decimal, with units. Core
Hint 1
The arc-length integrand is (Section 3.1). Find , then check whether folds into a perfect square, the clean-curve move from Example 3.1.1.
Hint 2
Here , and the cross term in the square is , so . The base is positive on , so integrate it directly.
Show full solution
meters.
Differentiate: . Squaring gives a cross term , so
The base is positive on , so the root is that base, and
5. A metal funnel is the surface swept out by spinning from to (in centimeters) around the x-axis. (a) Find the area of its skin with the surface integral. (b) Confirm the answer with the cone formula from Appendix A.7. Core
Hint 1
The slope is constant, so the slant factor is one number that pulls out of the integral, exactly as in Example 3.3.1 of Section 3.3.
Hint 2
Compute . For the check, the base radius is , the height is , and , a 6-8-10 right triangle.
Show full solution
(a) square centimeters. (b) , the same number.
(a) The slope is , so the slant factor is . Then
(b) The base radius is , the height is , and the slant height is , so . Same skin, two roads.
6. A thickening rod lies along meters with linear density kilograms per meter. Find its mass, exact then rounded to three decimals. Core
Hint 1
Integrate the density over the rod, (Section 3.5). The out front is the derivative of the inside of the root up to a constant, so a substitution finishes it, the move from Section 2.2.
Hint 2
Let , so and . The antiderivative works out to ; evaluate it from to , using and .
Show full solution
kg.
The in front is almost the derivative of the inside , so substitute , , which turns into . The limits run from at to at :
7. Cars arrive at a toll booth at random, one every minutes on average, so the wait (in minutes) until the next car has the exponential density for . (a) Find the probability the next car comes within minutes. (b) Find . (c) State the mean wait. Core
Hint 1
Each probability is an area under (Section 3.7, Example 3.7.3). The antiderivative of is , the form from Section 2.4 with .
Hint 2
For (a) integrate from to . For (b) integrate from to , where the upper limit contributes because fades to nothing. The exponential's mean is always , so read (c) straight off .
Show full solution
(a) . (b) . (c) minutes.
(a)
(b)
(c) The exponential density has mean , and here , so the mean wait is minutes.
8. Two cyclists start together and ride for seconds. Cyclist A's speed is feet per second, and cyclist B's is feet per second. (a) The distance A leads by is the area between the two speed curves; find it, with units. (b) Using the net change idea from Section 1.5, find how far cyclist A travels in those seconds. Stretch
Hint 1
For (a), this is the two-riders picture from Section 3.2: the lead is over the interval, provided A stays faster the whole way. For (b), the distance A travels is the net change of its position, (the Net Change Theorem, Section 1.5).
Hint 2
Confirm A is faster on by setting : the two speeds are equal only at and , with A ahead in between. Integrate from to for the lead, and integrate from to for A's own distance.
Show full solution
(a) A leads by feet. (b) A travels feet.
(a) The speeds are equal when , that is , so at and . Between them A is faster (at , ), so A leads the whole way, and the lead is the area between the curves:
(b) A's speed never turns negative, so the distance it travels is the net change of its position, the integral of its speed:
As a cross-check, B travels feet, and A's feet minus B's feet is the -foot lead from part (a).
9. A cylindrical water tank has radius m and height m and is full of water, which weighs newtons per cubic meter. This one object asks two questions. (a) Find the volume of water it holds, setting it up as over cross-sections. (b) Find the work done pumping all the water up and out over the top rim. Stretch
Hint 1
For (a), every cross-section perpendicular to the tank's axis is a disk of radius , so is the same all the way up (Section 3.4); integrate it over the m height. For (b), switch to the pumping recipe (Section 3.6): a slice at depth has weight (weight density)(area) and must rise its own distance .
Hint 2
In (b) the cross-section never changes, so only the rise varies. The slice weight is newtons, and it rises meters, so integrate from to . Re-derive the coefficient yourself from the density and the area rather than copying it.
Show full solution
(a) cubic meters. (b) joules.
(a) Each cross-section is a disk of radius , so square meters, the same at every height. The volume is
(b) Measure depth from the rim, from at the top to at the floor. A slice at depth has weight (weight density)(area) newtons, and it must rise meters to clear the rim, so its work is :
Both answers are accumulations off the same tank, but of different things. The volume sums equal disks; the work sums slices that all weigh the same yet each rise a different distance, which is why the rise stays inside the integral instead of sitting out front as one number.
10. A random variable has density on and elsewhere. (a) Find the constant that makes a valid density. (b) Find . (c) Find the mean. Stretch
Hint 1
The constant is fixed by the total area, (Section 3.7, Dig In): integrate over , set the result equal to , and solve. Rewrite as before integrating, the way Section 1.2 rewrites first.
Hint 2
The antiderivative of is . For the mean, , which brings in from Section 2.4.
Show full solution
(a) . (b) . (c) .
(a) The curve is positive on , so is fixed by forcing the total area to :
(b) A probability is an area, so integrate the density from to :
(c) The mean weights each value by the density, and , whose antiderivative is :
Section 3.R
Unit 3 Recap: Check Yourself
You have just mixed all seven of Unit 3's ideas in the Mixed Practice; here is one last self-check before you move on. Hit Try it! and work each goal cold, then check yourself and tick Got it. A worked example and the section link sit one click away whenever you want a model or another pass.
Stand a water tower on its curved profile and one shape answers a dozen questions. Spin that profile around its axis and Section 3.4 gives the volume it can hold; the same curve, run through Section 3.3, gives the surface area the painters have to coat; and Section 3.6 gives the work the pumps must do to lift every slice of water to its own height. Trace the curve itself and Section 3.1 measures how much steel the seam takes. Not one of these is a formula to memorize. Each is the same four-word recipe, slice the thing into thin pieces, approximate each piece, add them, and take the limit, pointed at a different quantity. That is the whole book in one habit: whenever something is too curved, too uneven, or too continuous to measure in one shot, cut it into infinitely many pieces small enough to be simple, and let the integral build the whole back up.
Your turn. Pick one object near you that is round, curved, or uneven: a mug, a lightbulb, a spiral staircase, a hill. Name one quantity you would want to know about it (its volume, its surface, its weight, or its length) and describe the thin slice you would cut to measure it.
Can you find the true length of a curve by slicing it into tiny straight pieces, read it off , and spot the clean curves where folds into a perfect square so the length comes out exactly?
Try it!
Find the exact length of from to , then give the rounded decimal.
How did you do?
.
Everything rides on the slope. Differentiate, then build and watch it fold into a perfect square:
The cross term in the square is , which is what sets up the perfect square. The base is positive on , so the root is that base, and
Quick check: the run is only units wide, so a length of about says the curve climbs steeply, which a rising parabola-minus-log does.
Missed it? Take another pass at Section 3.1 before moving on.
Reveal a worked example
Find the exact length of from to .
Differentiate and assemble the slant integrand:
The base is positive on , so
Can you find the area between two curves as , finding the crossings first and splitting the region wherever the top curve changes identity?
Try it!
Find the total area of the region enclosed by and .
How did you do?
.
Find the crossings first. Set the curves equal:
Two crossings sit inside, so the region is two lobes with a different top curve on each. At , beats , so the cubic is on top there; at , beats , so the line is on top. Integrate top minus bottom on each lobe:
The total area is . Integrating the single height straight across would give , the net signed area, because the left lobe cancels the right; that is why you split at the crossing.
Missed it? Take another pass at Section 3.2 before moving on.
Reveal a worked example
Find the total area of the region enclosed by and .
Set them equal: gives , so . On the cubic is on top; on the line is on top. Integrate each lobe:
The total is .
Can you find the skin of a surface of revolution by summing thin bands, each a circumference times its slant width , and cross-check a cone against ?
Try it!
Spin the line from to around the x-axis, making a cone. Find the area of its curved skin, then confirm it with the cone formula .
How did you do?
, and confirms it.
The slope is constant, so the slant factor is one number for the whole cone:
Drop the radius and that slant factor into the surface formula:
Geometry check: the base radius is , the height is , and the slant height is , a 5-12-13 triangle, so . Same skin, two roads.
Missed it? Take another pass at Section 3.3 before moving on.
Reveal a worked example
Spin the line from to around the x-axis and find the area of its curved skin, then confirm it with .
The slope is constant, so the slant factor is . Then
Geometry check: , , , an 8-15-17 triangle, so , the same.
Can you find a volume by slicing into cross-sections, , using the disk, washer, and cylindrical-shell methods, and squaring each radius on its own before you subtract the hole?
Try it!
Revolve the region between and about the x-axis. Find the volume.
How did you do?
cubic units.
Find the crossings for the limits: gives , so and . On the line is farther from the axis (at , beats ), so it is the outer radius and the parabola is the inner:
Each slice is a washer of area . Square each radius on its own, then subtract:
The one move that separates a washer from a disk is subtracting the hole, and squaring and separately, never .
Missed it? Take another pass at Section 3.4 before moving on.
Reveal a worked example
Revolve the region between and about the x-axis, and find the volume.
The curves meet where , so and . On the line is outside (at , beats ), so and :
Can you find the mass of a non-uniform object from its density, , keeping the units straight?
Try it!
A rod along meters has linear density kilograms per meter. Find its mass.
How did you do?
kg.
The mass is the integral of the density over the rod:
The units carry through: kilograms per meter, times a length in meters, gives kilograms. As a sanity check, the density runs from to , and kg sits between the low estimate kg and the high estimate kg.
Missed it? Take another pass at Section 3.5 before moving on.
Reveal a worked example
A rod along meters has linear density kilograms per meter. Find its mass.
Integrate the density over the length:
Most of that mass sits toward the far end, where the term grows fastest.
Can you find the work done by a varying force, , whether it is a spring you read off Hooke's law or a tank you pump slice by slice, integrating rather than multiplying?
Try it!
A force of 48 N holds a spring stretched 0.15 m beyond its natural length. (a) Find the spring constant . (b) Find the work done stretching the spring from its natural length to 0.5 m beyond.
How did you do?
(a) N/m. (b) J.
(a) Hooke's law says , so the one measurement pins down the constant:
and the force as a function of stretch is .
(b) Work is the integral of the force over the stretch, from to :
The force is a moving target, from N at the start to N at m, so multiplying one force by the distance would miss; the integral tracks it.
Missed it? Take another pass at Section 3.6 before moving on.
Reveal a worked example
A force of 54 N holds a spring stretched 0.15 m beyond its natural length. Find the spring constant, then the work to stretch it from its natural length to 0.5 m beyond.
From the single measurement, N/m, so . Then
Can you read a probability density as an area, use to pin down an unknown constant, then find an interval probability and the mean ?
Try it!
A quantity has the linear density on and elsewhere. (a) Find the constant . (b) Find . (c) Find the mean.
How did you do?
(a) . (b) . (c) .
(a) The constant is whatever forces the total area to :
(b) A probability is an area, so integrate the density from to :
(c) The mean weights each value by the density:
The balance point sits two thirds of the way out, pulled toward the heavier right side, exactly as a rising density should.
Missed it? Take another pass at Section 3.7 before moving on.
Reveal a worked example
A quantity has the linear density on and elsewhere. Find , then , then the mean.
Force the total area to : , so . Then the interval probability and the mean:
Prerequisite Toolkit
Integration leans on a handful of skills from earlier courses: the unit circle and the trig functions, logarithms and exponentials and how they undo each other, and the plain geometry and physics that the applications in Unit 3 assume. This appendix gathers those refreshers in one place, so a rusty skill never stops you mid-integral.
Each section here stands on its own and points back at the calculus sections that use it, so you can reach for exactly the one you need and skip the rest. Nothing in the main text waits on the whole appendix; treat it as a shelf of quick references, assigned as needed.
- A.1 The Unit Circle: special angles in radians and their exact coordinates.
- A.2 Trigonometric Functions: sine, cosine, and tangent as coordinates and as graphs.
- A.3 Trigonometric Identities: the Pythagorean and half-angle identities the trig integrals need.
- A.4 Logarithmic Functions: the natural logarithm, the log laws, and .
- A.5 Exponential Functions: growth and decay, and the number .
- A.6 Inverse Functions: undoing a function, with and as the key pair.
- A.7 Area, Surface Area, and Volume: the geometry formulas the applications cross-check against.
- A.8 Basic Physics: density, force, work, and their units.
- A.9 Basic Probability: sample spaces and expected value, the discrete side of Section 3.7.
- A.10 Tables of Derivatives and Integrals: the derivative facts and their reversed antiderivative facts, side by side.
Appendix A.1
The Unit Circle
- Reading an angle in degrees and radians and converting fluently between the two, since every integral in this book measures angles in radians.
- Reciting the exact sine, cosine, and tangent of the five landmark angles from to , always rationalized.
- Finding the reference angle for any angle and reflecting the landmark values into all four quadrants, with the sign read off the quadrant by ASTC (All Students Take Calculus).
- Reading a landmark value at the ends of a definite integral, the instant lookup Section 2.3 leans on when it evaluates something like .
Spin a bike wheel and keep your eye on the valve stem. It rides the same circle every turn, the radius never changes, yet its height above the ground climbs and falls in a smooth, repeating rhythm. Your trig course turned that rhythm into two functions, sine and cosine, read straight off a circle of radius 1. You built their exact values at the friendly angles, the ones that drop out of the two special right triangles from geometry, and you charted where each one runs positive and where it turns negative. None of that is new here. This section is the reference card, the one you flip open the moment a definite integral of a repeating rate lands on the page. Totaling such a rate means evaluating an antiderivative at the two ends of an interval, and those ends are almost always landmark angles like and . If you cannot read their exact values on sight, the integral stalls right at the finish line.
Build the intuition
Two triangles from geometry carry this whole appendix. Cut a unit square along its diagonal and you get a 45-45-90 triangle. Cut an equilateral triangle of side 1 through one vertex and you get a 30-60-90 triangle. Stand each one up with its right angle on the x-axis and its hypotenuse (length 1) running from the origin out to a point on the unit circle. That point's coordinates are the cosine and the sine, read directly off the triangle's two legs, no calculator anywhere. If the shapes of the sine and cosine curves themselves feel rusty, Appendix A.2 is right next door.
Figure A.1.1 Three landmark points on the unit circle. Each dashed pair of legs is a right triangle with hypotenuse 1: the vertical leg is the sine, the horizontal leg is the cosine.
The unit circle is the circle of radius 1 centered at the origin. For an angle measured counterclockwise from the positive x-axis, let be the point where the angle's terminal ray meets the circle. Three functions are read straight off that point: is its x-coordinate, is its y-coordinate, and is the ratio of the two, defined wherever . The radius being 1 is what keeps the reading this clean: on a circle of any other radius, you would have to divide each coordinate by that radius first.
| Angle | Radians | |||
|---|---|---|---|---|
| undefined |
Table A.1.1 is the whole appendix in miniature. Reason: the five points come from the two special right triangles in Figure A.1.1, the 30-60-90 (half an equilateral triangle) and the 45-45-90 (half a square), and the Dig In below rebuilds both from scratch. Every other angle in this section is one of these five landmarks, reflected into a different quadrant.
On the unit circle, an angle of 1 radian is the angle whose terminal ray cuts off an arc of length 1, measured counterclockwise from the positive x-axis. In general, an angle of radians cuts off an arc of length . Since the full circumference is (the radius is 1), one complete trip around measures radians, the same trip degree measure calls . That single equivalence, radians equal , is the whole conversion factor.
, and the reverse, . Reason: radians and measure exactly the same half turn, so the ratio (or its flip) changes only the units, the same way 5280 turns miles into feet. Calculus runs on radians from here forward: a bare angle number with no degree symbol means radians, and the definite integrals in Section 2.3 read their limits, like and , as radian angles.
(a) Convert to radians. (b) Convert to degrees. (c) Convert to radians.
Solution. (a) Multiply by and reduce the fraction of a full circle:
(b) Multiply by :
(c)
Every conversion is the same one multiplication; only the direction of the fraction changes.
Before you go on, predict: (a) convert to radians, and (b) convert to degrees. Each is the same one multiplication, only the direction of the fraction flips. Do it in your head, then check.
Show solution
(a) . (b) .
(a) ; (b) .
The reference angle of is the acute angle between the terminal ray of and the x-axis, always taken as a positive angle between and . Every angle in this course reduces to one of the five landmarks in Table A.1.1: , , , , or . Finding it is a matter of distance from the nearest axis, never from the starting ray.
Figure A.1.2 ASTC, read counterclockwise from Quadrant I: All three positive, Sine only, Tangent only, Cosine only.
, and the same for cosine and tangent: reflecting into a new quadrant changes signs, never magnitudes. The sign comes from which coordinates are positive in that quadrant, remembered by ASTC, All Students Take Calculus: Quadrant I, All three positive; Quadrant II, only Sine; Quadrant III, only Tangent; Quadrant IV, only Cosine. Reason: the axes are mirrors. Reflecting a point across the y-axis flips the sign of x (the cosine) only; across the x-axis flips y (the sine) only; through the origin flips both. The Dig Deeper below writes out exactly which reflection produces each quadrant.
The angle is a landmark in Quadrant I. Use its reference angle and ASTC to find the exact sine, cosine, and tangent at the three angles that share that reference angle in the other three quadrants: , , and .
Solution. Each of the three angles has reference angle , so every value has the same size as Table A.1.1's row, , , ; only the signs change, and ASTC supplies those.
is in Quadrant II (only Sine positive):
is in Quadrant III (only Tangent positive):
is in Quadrant IV (only Cosine positive):
Run the same reflection on the and families, and on the quadrantal angles where the point lands right on an axis, and the whole unit circle falls out of five numbers and one mnemonic. Table A.1.2 collects the result.
| Angle | Radians | |||
|---|---|---|---|---|
| undefined | ||||
| undefined | ||||
A definite integral of a repeating rate is finished by evaluating an antiderivative at the two limits, and those limits are landmark angles. Find each exact value on sight, then use them to close out one integral. (a) . (b) . (c) . (d) . (e) Using (a) and (b), evaluate , given that an antiderivative of is .
Solution. (a) The rightmost point of the circle: .
(b) The leftmost point: .
(c) The point at sits on the x-axis, so its height is : . This is the same reason , the clean zero an integral like starts and ends from.
(d) The top of the circle: .
(e) Evaluate the antiderivative at the top limit minus the bottom limit:
No calculator, no scratch work: the whole evaluation rides on knowing and instantly. Section 2.3 does exactly this, over and over.
Ride a Ferris wheel and your height above the center is , where is the angle the car has swept from the loading platform. At a quarter turn you are at the top, above center; at a half turn you are back level; at three quarters you are below. Over one full turn you climb and fall and end exactly where you started, so the net rise is zero, which is the signed area under across a whole period: . Buoys on a swell, pistons in an engine, and the current in a wall socket all trace the same repeating rate, and totaling any of them is a definite integral that reads its endpoints off this exact-value table. Section 2.3 takes the next step and asks for the running total of a rate that never stops repeating, which is precisely where these landmark values go to work.
Table A.1.1's five values are not handed down; two right triangles force every one of them, using nothing past the Pythagorean theorem.
At : cut a unit square along its diagonal to get a right isosceles triangle with hypotenuse 1 and two equal legs, call each one . The Pythagorean theorem gives , so and once the denominator is rationalized. Both legs measure , and since has both coordinates equal to that leg length, , and .
At and : cut an equilateral triangle of side 1 in half through one vertex. The half-base measures , the hypotenuse is still 1 (an untouched side of the original triangle), and the Pythagorean theorem finds the remaining leg : , so and . That triangle carries a angle and a angle at its two acute corners. Read it from the corner: (adjacent over hypotenuse) and (opposite over hypotenuse). Read the same triangle from the corner and the two legs swap roles: and . Tangent finishes each: , and after rationalizing (multiply top and bottom by ).
Check every pair against the circle's own defining equation, : at , . At , . The unit circle enforces this sum at every angle, not only these five; Appendix A.3 states that fact as a standing identity.
Dig Deeper The Reflection Principle Behind ASTC▸
ASTC works because the coordinate axes are mirrors, and mirrors are easy to write down. Start at an angle in Quadrant I, with point .
Reflect across the y-axis (flip the sign of x only) and the image lands at the angle , in Quadrant II:
Cosine flips, sine does not, which is exactly why Quadrant II keeps sine alone positive.
Reflect through the origin (flip both signs) and the image lands at , in Quadrant III:
Both flip, so their ratio does not: , positive whenever was positive in Quadrant I. That is exactly why Quadrant III keeps tangent alone. The third mirror, the reflection across the x-axis that produces Quadrant IV, is yours to derive in Exercise 10.
Two mirrors, two quadrants, one pattern each time: the reflection that flips only x kills cosine's sign, and the one that flips both leaves the ratio untouched. ASTC is that pattern, memorized instead of re-derived every time it is needed.
Three traps live in this section. First, an exact value is never left unrationalized: is correct, is not, even though the two are equal. A square root does not belong in a denominator in this course's answers; multiply top and bottom by the root to clear it, because a later integration step matches your answer against the rationalized form and stops recognizing it otherwise. Second, is undefined at and , not zero and not some huge number: the point's x-coordinate is there, and dividing by has no answer, full stop. Third, ASTC gives a sign, never a value: it tells you is negative, but you still need the reference angle to know it is and not some other negative number. Stopping at a sign and calling it an answer is the single most common way this section's homework goes wrong.
(a) Convert to degrees. (b) Find the exact sine, cosine, and tangent of . (c) Find the reference angle of , then state exactly.
Hint
(a) Multiply by . (b) The reference angle of is , and the angle lies in Quadrant II. (c) Distance from the nearest axis gives the reference angle; ASTC gives the sign in Quadrant III.
Show solution
(a) . (b) , , . (c) Reference angle ; .
(a) .
(b) has reference angle and sits in Quadrant II, where only sine keeps its sign: the magnitudes match Table A.1.1's row, and cosine and tangent both flip negative.
(c) , so the reference angle is . Quadrant III keeps only tangent positive, so cosine is negative: .
Exercises A.1
1. Convert each angle to radians, or to degrees if it is given in radians. (a) (b) (c) Warm up
Hint 1
Degrees to radians multiplies by ; radians to degrees multiplies by , exactly as in Example A.1.1.
Show solution
(a) . (b) . (c) .
(a) .
(b) .
(c) .
2. Evaluate each exactly, straight from Table A.1.1. (a) (b) (c) Warm up
Hint 1
Each is a direct lookup, no reflection needed: all three angles already sit in Quadrant I. Part (c) is a tangent, so keep the answer rationalized.
Show solution
(a) . (b) . (c) .
(a) . (b) . (c) , the rationalized form of .
3. Find the reference angle for each, and name the quadrant. (a) (b) (c) Core
Hint 1
The reference angle is the gap between the angle and the nearest x-axis crossing (, , or ), always reported as a positive angle under .
Hint 2
(a) is short of ; subtract it from . (b) is past ; subtract from it. (c) is short of ; subtract it from .
Show solution
(a) , Quadrant II. (b) , Quadrant III. (c) , Quadrant IV.
(a) , and sits between and , Quadrant II.
(b) , and sits between and , Quadrant III.
(c) , and sits between and , Quadrant IV.
4. Use the reference angle and ASTC to find the exact sine, cosine, and tangent of . Core
Hint 1
This is Example A.1.2's method on a new angle: find the reference angle, look up its magnitudes in Table A.1.1, then attach signs from ASTC (Appendix A.1).
Hint 2
The reference angle is , and lies in Quadrant III, where tangent alone keeps its sign.
Show solution
, , .
The reference angle is , whose magnitudes are , , and . Quadrant III keeps only tangent positive, so sine and cosine both flip negative: , , . This is the value a problem needs when it asks where has a Quadrant III solution.
5. Every angle below shares reference angle . State , , and exactly at each. (a) (b) (c) Core
Hint 1
Every magnitude is or , straight from Table A.1.1's row. Only the quadrant, and so the sign pattern from ASTC, changes between parts.
Hint 2
(a) is Quadrant II, (b) is Quadrant III, (c) is Quadrant IV. Match each to the one letter of ASTC that stays positive there.
Show solution
(a) . (b) . (c) .
(a) Quadrant II, sine alone positive: , , .
(b) Quadrant III, tangent alone positive: , , .
(c) Quadrant IV, cosine alone positive: , , .
6. Section 2.3 evaluates using the antiderivative . Supply the landmark values that finish it, in one line each. (a) (b) (c) (d) . (e) Combine (c) and (d) to state , the value of that integral. Core
Hint 1
Parts (a) to (d) are direct lookups at the two points where the circle crosses the x-axis, and (Appendix A.1, Table A.1.2).
Hint 2
The evaluation bar means bottom subtracted from top: . Plug in your (c) and (d).
Show solution
(a) . (b) . (c) . (d) . (e) .
(a) . (b) : both endpoints of the arch sit on the x-axis. (c) . (d) .
(e) , so , the area of one arch of the sine curve.
7. A paddle on a waterwheel of radius meters sits at height above the axle's center, where is measured counterclockwise from the horizontal rest position. Find the exact height, then a three-decimal approximation where the value is not already a clean decimal, at (a) (b) (c) , and say what a negative height means. Stretch
Hint 1
Look up for each angle from Table A.1.1 or A.1.2 (Appendix A.1), then multiply by . A negative height just means below the height where the paddle started.
Hint 2
(b) needs , the Quadrant II reflection of ; its exact height carries a .
Show solution
(a) m. (b) m. (c) m. A negative height means the paddle is below the axle's center.
(a) , so the height is m.
(b) , so the height is m.
(c) , so the height is m: the paddle has swung past the rest line, 0.25 meters below the axle.
8. A point sits at on the unit circle. (a) Find its reference angle. (b) Before computing anything, use ASTC to predict the sign of , , and . (c) Compute the exact values and confirm they match your predictions. Stretch
Hint 1
Locate between the quarter marks and to name its quadrant before touching a formula.
Hint 2
The reference angle is . That quadrant keeps exactly one function positive; the other two are negative by ASTC.
Show solution
(a) . (b) Quadrant IV: sine negative, cosine positive, tangent negative. (c) , , , matching the predictions.
(a) .
(b) lies between and , Quadrant IV, where ASTC keeps only cosine positive: sine and tangent should come out negative.
(c) The reference angle's magnitudes are , , and . Attaching the predicted signs: , , . Every sign lands where part (b) said it would.
9. The Appendix A.1 Dig In built the - triangle from an equilateral triangle of side 1. Rebuild it starting from a different side length: (a) cut an equilateral triangle of side 2 through one vertex, find the half-base and the height, then divide every length by the hypotenuse to rescale the triangle onto the unit circle. (b) Confirm you land on and , the same values the Dig In found. Dig In
Hint 1
Reread the Dig In in Appendix A.1: cutting through a vertex halves the base and leaves a full side as the hypotenuse. Every leg-to-hypotenuse ratio (the actual coordinate) does not care about the starting size.
Hint 2
A side-2 triangle has half-base 1 and hypotenuse 2, so the height is . Divide each length by the hypotenuse 2 to land on a radius-1 circle.
Show full solution
The side-2 triangle gives half-base 1, hypotenuse 2, and height ; dividing by the hypotenuse gives and , matching the Dig In exactly.
(a) An equilateral triangle of side 2, cut through one vertex to the midpoint of the opposite side, splits into a right triangle with half-base 1 and hypotenuse 2 (a full, untouched side). The Pythagorean theorem gives the height : , so and . To sit this triangle on the unit circle its hypotenuse must be 1, so divide every length by 2: the half-base becomes , the height becomes , and the hypotenuse becomes .
(b) Read from the corner, the adjacent leg is the rescaled half-base and the opposite leg is the rescaled height: and , the same coordinates the Dig In found from a side-1 triangle. The starting size never mattered; only the ratio of leg to hypotenuse survives the rescaling, which is exactly what a unit-circle coordinate is.
Check your own version. Whatever side length you started with, two tests must pass: your rescaled hypotenuse equals exactly 1 (divide every side by it), and your two rescaled legs come out to and , with no leftover square root in any denominator. A height that is not the shorter side times means the Pythagorean step slipped.
10. The Appendix A.1 Dig Deeper derived the y-axis and origin reflections. Derive the third: reflecting across the x-axis lands at . (a) State the formulas for and in terms of and . (b) Test your formulas at against Table A.1.2's entry for , since and name the same point on the circle. Dig Deeper
Hint 1
Reread the Dig Deeper in Appendix A.1: reflecting across the x-axis flips the sign of y only, leaving x untouched. Translate that geometric fact into a statement about cosine and sine.
Hint 2
and are the same angle, one measured clockwise and one counterclockwise a full turn later; your formula's output at must match Table A.1.2's row exactly.
Show full solution
and ; at these give and , matching Table A.1.2's row for exactly.
(a) The x-axis reflection sends to , and , so the reflected point is . That reflected point is also , so (x is untouched) and (y flips).
(b) At : , and . Table A.1.2's row reads , : an exact match, since and are the same point on the circle, one full turn apart.
Check your own version. Whichever landmark angle you tested, two things must hold: your formula never introduces a sign change, and your formula always flips one. Then convert your to its positive equivalent by adding and look up that row in Table A.1.2; a mismatch means a sign was dropped in part (a).
Five landmark angles carry the whole unit circle: , , , , and , with exact sine, cosine, and tangent built from two right triangles and always rationalized (, never ). Every other angle reduces to one of these five through its reference angle, with the sign supplied by ASTC, All Students Take Calculus, read counterclockwise from Quadrant I as All positive, Sine alone, Tangent alone, Cosine alone. Calculus measures every angle in radians, converted from degrees by the single factor . The payoff is speed at the ends of an integral: Section 2.3 evaluates definite trig integrals like by reading landmark values such as and straight off this table, so the whole appendix earns its keep the moment a repeating rate needs totaling.
Appendix A.2
Trigonometric Functions
- Reading and as the y- and x-coordinates of a point moving around the unit circle, and as the slope of the radius to it.
- Sketching and reading the graphs of , , and : period, amplitude, and the vertical asymptotes where tangent breaks.
- Stating the domain and range of each function, including the exact angles where the tangent graph is undefined.
- Recalling the three derivative facts, , , and , so the antiderivatives in Section 2.3 read as their reverse.
- Working in radians (Appendix A.1 is the unit-circle and radian refresher), this course's standard angle unit.
Ride a Ferris wheel through one full turn and watch your height above the ground. It climbs to the top, eases back down past the loading platform, sinks to the bottom, and climbs again, the same rise and fall on every rotation, for as long as the wheel spins. That repeating shape is a sine wave, and you met it before in your trig course as sine, cosine, and their restless cousin tangent. This appendix does not reteach them from the ground up. It hands the tools back sharpened, radians and all, and recalls the three derivative facts that Section 2.3 runs in reverse. If a problem sent you here mid-integral, you are in the right place, and this section stands on its own.
Build the intuition
Start with the unit circle: a circle of radius 1, centered at the origin, with a point that starts at and travels counterclockwise. This course measures how far it has gone in radians, the arc-length angle unit; a full trip is radians, since the circumference is times the radius. Every calculus statement in this book runs in radians, never degrees. If the special-angle values or that unit are rusty, Appendix A.1 is the two-minute refresher.
After sweeping an angle , the point sits at some location on the circle, and that location is the whole definition. is the point's x-coordinate, and is its y-coordinate. The Ferris wheel is this picture set spinning: as grows, the height rises and falls in the exact repeating wave the seat traces. The ratio is rise over run from the origin out to the point, the slope of that radius. A flat radius has slope 0, and a radius standing straight up has no slope at all, which is the whole reason tangent misbehaves where sine and cosine stay calm.
Figure A.2.1 A point on the unit circle after sweeping angle . Its coordinates are the definitions: is the horizontal coordinate, is the vertical one, and is the slope of the radius . Here , the point Example A.2.1 reads.
On the unit circle (radius 1, centered at the origin), let be the point reached by sweeping angle counterclockwise from the positive x-axis. Then , , and , defined only where .
A function's period is the smallest positive length after which its graph repeats exactly. For a wave that swings between a highest and a lowest value, its amplitude is half the vertical distance between crest and trough, the reach above (and below) its middle line. Amplitude only describes a bounded wave; a graph with no highest or lowest point, like tangent's, has none.
For every real , and , while takes every real value, racing toward positive or negative infinity near the angles where it is undefined. The reason: is a point on a circle of radius 1, so neither coordinate can stray past 1 or below . Tangent has no such leash, because it is a slope, and a radius tipping closer to vertical has a slope racing off to infinity.
| Function | Domain | Range | Period | Amplitude |
|---|---|---|---|---|
| all real numbers | 1 | |||
| all real numbers | 1 | |||
| all real numbers | not defined (unbounded) |
A point on the unit circle sits at , reached by sweeping some angle . Find , , and , and confirm lies on the unit circle.
Solution. The coordinates hand over two of the three answers on sight: (the x-coordinate) and (the y-coordinate). The third divides them:
For the check, every point claimed to be on the unit circle must satisfy :
It checks out. This is the 3, 4, 5 right triangle from geometry class, scaled down so its hypotenuse is exactly 1.
Before you go on, predict: a point on the unit circle sits at . Give , , and for the angle that reaches it, then confirm lies on the unit circle.
Show solution
, , , and , so is on the circle.
The x-coordinate is and the y-coordinate is , so , and . This is the 5, 12, 13 triangle scaled to the unit circle.
The graphs, unwrapped from the circle
Let the swept angle become the horizontal axis instead of a rotation, and plot the height against it. The Ferris wheel's rise and fall unwinds into a wave: up to 1, back through 0, down to , back through 0, repeating every . Cosine draws the same wave a quarter turn early, since it tracks the horizontal coordinate instead of the vertical one. Tangent, the slope of the radius, behaves nothing like them: it climbs, races off to infinity as the radius nears vertical, then reappears climbing from negative infinity on the far side. Figure A.2.2 draws all three on one radian axis, so cosine hitting 0 above lines up exactly with tangent breaking apart below.
Figure A.2.2 Top: and over one full period, both amplitude 1, on a radian axis. Bottom: on the same axis, breaking into vertical asymptotes exactly under the two angles (dashed) where cosine crosses 0 above; its period is , half of sine and cosine's.
Using Figure A.2.2's top panel, state (a) the period and amplitude of and of , (b) every x-intercept of on , and (c) where each curve reaches its maximum and minimum on that interval.
Solution. (a) Both curves finish one full wave and start over at , so both have period . Both reach 1 unit above their middle line and 1 below, so both have amplitude 1.
(b) Sine crosses the axis at , , and ; Table A.2.1's landmark angles confirm .
(c) Sine crests at (value 1) and bottoms out at (value ). Cosine starts at its crest, (value 1), bottoms out at (value ), and climbs back to its crest at . The two waves share every landmark, staggered by a quarter turn.
A vertical asymptote is a vertical line that a graph races toward without ever crossing, as the output climbs or falls without bound near . Tangent has one at every angle where the radius stands vertical, since a vertical radius has no slope to report.
(a) Find every in where is undefined. (b) The slope of at any point equals the height of there, which is the derivative fact . Use it to state the slope of the sine graph at and at , and name the antiderivative fact Section 2.3 gets by running it backward.
Solution. (a) breaks only where the denominator vanishes, so solve on . The x-coordinate of the unit-circle point is 0 exactly at the top and bottom of the circle, so and , the two dashed asymptotes in Figure A.2.2.
(b) The slope of at is : the sine wave is climbing at its steepest as it leaves the origin. At the slope is : the wave is flat at its crest, for one instant neither rising nor falling. Read the fact backward and it becomes an antiderivative fact:
the first line of the trig-integral table Section 2.3 builds. Every derivative fact in this box turns into an integral fact the same way.
In radians, sine, cosine, and tangent differentiate to
where . Section 2.3 reads each line right to left to get an antiderivative: , , and . The lone minus lives on cosine's derivative, and it is the one sign to guard when you reverse the table, since it lands on sine's antiderivative.
Watch the water in a tidal bay. In a stripped-down model, its height above the mid-tide line is meters, with the hours since the water last passed that line on the way up. The height is a sine wave: 0 at mid-tide, up to 3 meters at high tide a quarter cycle later, back down through the line, and on. The water's vertical speed is the rate of change of its height, the derivative , a cosine wave. So the current runs fastest, 3 meters per hour, right at mid-tide, where and . It stalls to a standstill at high and low tide, where and , the pause before the water turns. That is made of real water. Section 2.3 runs the machine the other way: hand it the current and it rebuilds the height, , the same reversal that turns every derivative fact in this appendix into an integral fact.
The Rule above said sine and cosine never leave and named exactly where tangent breaks. Both facts fall out of the unit circle's own equation, , with no graph needed.
Bounded. Since and no square is negative, , which forces ; the same argument with the roles traded forces . In trig terms, and , for every angle , no exceptions. (This is the same equation Appendix A.3 turns into the Pythagorean identity and puts to work simplifying expressions.)
Where tangent breaks. is undefined only when . On the unit circle, at exactly two points: the top, , and the bottom, , reached at and within one lap, and every after that, alternating top and bottom. Nowhere else on the circle does the x-coordinate vanish, so nowhere else does tangent break.
Dig Deeper Why Tangent's Period Is Half of Sine and Cosine's▸
Example A.2.3 pinned tangent's undefined angles a half lap apart. Here is why it repeats on that half-lap rhythm, always. Sweep an extra radians from any point on the unit circle and you land on the exact opposite point, : a half turn about the center reverses both coordinates, since the new point sits the same distance from the origin in the reverse direction. Track each function across that flip:
Sine and cosine each flip sign, so a half lap is not enough to return either one; they need the full before the graph genuinely repeats. Tangent is a ratio, and the two minus signs in cancel every time, for any point at all. That cancellation, not a smaller fact about tangent, is the whole reason its period is rather than .
Try a fresh point: lies on the circle since , and gives . Its opposite point gives again, while , the reverse of . Same story, new numbers. Exercise 10 asks you to run it with a triple of your own.
Three mix-ups travel together, plus one sign to bank for later. First, lending tangent sine and cosine's period: tangent repeats every , not , half as far apart. Second, hunting tangent's breaks at sine's zeros: , a perfectly defined value, even though ; the breaks live at cosine's zeros, since cosine is the denominator. Third, asking for tangent's amplitude: it has none, because amplitude needs a highest and a lowest point, and tangent's graph has neither. Fourth, the sign to guard: cosine's derivative carries a minus, , while sine's does not, which is why Section 2.3 writes and not . Reverse the facts carefully.
(a) A point on the unit circle sits at . Find , , and for the angle that reaches , and name the quadrant. (b) State and , and say which one carries a minus sign. (c) List the two values of in where is undefined, and say why in terms of the unit circle.
Hint
(a) The coordinates are directly; divide to get tangent, as in Example A.2.1, and the quadrant is set by which coordinates are negative. (b) is the Rule box above. (c) is the Dig In: which coordinate has to hit 0?
Show solution
(a) , , ; quadrant II. (b) and ; cosine's carries the minus. (c) and , where .
(a) The x-coordinate is cosine and the y-coordinate is sine, from Definition A.2.1: , . Dividing, . The x-coordinate is negative and the y-coordinate positive, which places the point in quadrant II.
(b) From the Rule box, sine differentiates to cosine with no sign change, while cosine differentiates to . That single minus is the one to reverse with care in Section 2.3.
(c) Tangent breaks exactly where its denominator, , is 0, which on happens only at the top and bottom of the unit circle, and .
Exercises A.2
1. A point on the unit circle sits at . Find , , and , and name the quadrant. Warm up
Hint
The coordinates are directly, as in Example A.2.1 of Appendix A.2; the quadrant is set by which coordinates are negative.
Show solution
, , ; quadrant III.
Read the coordinates directly: , . Dividing, , positive, since two negatives cancel. Both coordinates are negative, which places the point in quadrant III.
2. State the domain, range, and period of . Then do the same for , and explain in one sentence why tangent's domain has gaps that cosine's does not. Warm up
Hint
Table A.2.1 in Appendix A.2 lists all of this directly; the one-sentence reason is the Dig In box's second claim, about which coordinate has to hit 0.
Show solution
: domain all real numbers, range , period . : domain , range all real numbers, period . Tangent has gaps because it divides by cosine, and cosine hits 0 twice per lap.
is defined at every real input, its outputs never leave , and it repeats every . is undefined wherever , at for every integer ; away from those gaps its outputs cover every real number, and it repeats every . Cosine never divides by anything, so it never breaks; tangent is a quotient, and a quotient breaks wherever its denominator does.
3. A point on the unit circle sits at . (a) Confirm the point lies on the unit circle. (b) Find , , and . Core
Hint 1
Part (a) is the check from Example A.2.1 in Appendix A.2: square both coordinates and add, aiming for exactly 1. Part (b) reads off the coordinates once (a) is settled.
Hint 2
8, 15, and 17 are a Pythagorean triple: check before you square and add the fractions, and the sum-to-1 will follow.
Show solution
(a) Confirmed: . (b) , , .
(a) , since .
(b) The coordinates give and directly. Dividing, .
4. State the period and amplitude of , and give its maximum and minimum values. Then do the same for . Core
Hint 1
A constant out front stretches the wave vertically without changing how often it repeats. The amplitude is the size of that stretch, and the extremes ride sine and cosine's own to reach (Appendix A.2, Table A.2.1).
Hint 2
For , track the sign: the extreme of that would normally give the maximum now gives the minimum, since the multiplier is negative.
Show solution
: period , amplitude 4, maximum 4, minimum . : period , amplitude 2, maximum 2, minimum .
A constant multiplier does not change how often the wave repeats, so both keep period . The amplitude is the size of the multiplier: 4 and . For , the extremes ride sine's to reach: maximum , minimum . For , the negative multiplier swaps which extreme is which: at , , so , the maximum; at , , so , the minimum.
5. Find every in where is undefined, and justify each one using the unit-circle coordinates: name the coordinate that hits 0 and where on the circle that happens. Core
Hint 1
Tangent's denominator is ; solve on the interval, the same move as the Dig In box in Appendix A.2.
Hint 2
On the unit circle, the x-coordinate is 0 at exactly two places: the very top and the very bottom.
Show solution
and .
Tangent is undefined exactly where , since it is the denominator of . On the unit circle, the x-coordinate vanishes at the top, , reached at , and at the bottom, , reached at . At every other point of the x-coordinate is nonzero, so tangent stays defined.
6. (a) State the derivatives of , , and . (b) The slope of at a point equals there; find the slope at by computing , and say whether the sine graph is rising or falling. (c) In one sentence, name the antiderivative fact Section 2.3 gets by reversing . Core
Hint 1
Parts (a) and (c) are the Rule box "The Derivative Facts, Ready to Reverse" in Appendix A.2. For (b), just evaluate and read the sign.
Hint 2
A negative slope means the graph is falling; check that against Figure A.2.2, where sine crosses downward through 0 at .
Show solution
(a) , , . (b) Slope ; the sine graph is falling. (c) Reversing it gives .
(a) These are the three facts from the Rule box, in radians.
(b) The slope at is . A negative slope means the curve is descending, which matches Figure A.2.2: sine crosses downward through 0 at , on its way to the trough.
(c) Reading from right to left says is an antiderivative of , so .
7. Explain, from the unit-circle definition, why and for every angle , with no exceptions. Stretch
Hint 1
Think about what sweeping an extra full does to the traveling point on the unit circle, described in Build the intuition in Appendix A.2.
Hint 2
A full lap returns the point to exactly where it started; sine and cosine depend only on the point's location, not on how many laps it took to get there.
Show solution
Both hold because sweeping an extra returns the traveling point to exactly the same location on the unit circle.
Sweeping angle lands the point at some location on the circle. Sweeping means going one extra full lap and arriving back at that identical location, since a full lap is exactly radians. Sine and cosine care only about the point's final resting place, and , so a point landing in the same spot must give the same coordinates: and . This holds for every starting angle, because a full lap always closes back on itself.
8. A float bobs on tidal water, its height above the mid-tide line modeled by meters, with in hours. Its vertical speed is the derivative . (a) Find and . (b) Find every in where the float is momentarily at rest, . (c) Say what the float is doing at those instants, and connect it to the derivative fact . Stretch
Hint 1
Part (a) is direct substitution with sine's landmark values from Table A.2.1. For (b), the speed is 0 exactly where , the same angles tangent breaks (Appendix A.2, Example A.2.3).
Hint 2
Compare where the speed is 0 with where the height reaches its crest and trough; a wave is momentarily flat at its extremes.
Show solution
(a) m, m. (b) and . (c) The float is at its crest ( m) and its trough ( m), momentarily flat, since the height's rate of change is , which is 0 exactly there.
(a) and .
(b) The speed is 0 where , which on is and .
(c) At the height is , the crest; at it is , the trough. Both are moments the float pauses before reversing, which is exactly the fact : the height's slope is the cosine wave, and it passes through 0 at each extreme.
9. The Appendix A.2 Dig In showed forces and . Make it yours: pick a Pythagorean triple of your own, different from , , and , and build a point on the unit circle from it in a quadrant of your choosing. Confirm the point lies on the circle, and state , , and with signs matched to your quadrant. Dig In
Hint 1
Any Pythagorean triple (with ) gives a unit-circle point ; choose the signs to land in whichever quadrant you like.
Hint 2
The quadrant sets the signs before you compute: quadrant II wants a negative x-coordinate and a positive y-coordinate, and so on. Check that your final tangent sign matches sine divided by cosine.
Show full solution
Worked with the triple 12, 35, 37 in quadrant II: point , confirmed on the circle since ; , , .
The triple 12, 35, 37 satisfies . Placing it in quadrant II (negative x, positive y) gives the point , and
confirming it lies on the unit circle. Then , , and : cosine negative, sine positive, tangent negative, exactly the sign pattern quadrant II demands.
Check your own version. Whatever triple and quadrant you picked, two tests must pass: your point's coordinates must square and add to exactly 1 with no rounding, and your three signs must obey (matching signs on top and bottom give a positive tangent, opposite signs a negative one). If your tangent's sign does not match that rule, a coordinate sign was copied wrong.
10. The Appendix A.2 Dig Deeper proved always, because sweeping an extra sends to and the two minus signs cancel in the ratio. Run the argument on a triple the box did not use: pick a Pythagorean triple of your own (not , , , or ), build a first-quadrant point from it, find its opposite point, and verify numerically that tangent matches while sine does not. Dig Deeper
Hint 1
Reread the Dig Deeper box in Appendix A.2: your opposite point is both coordinates negated, and you are checking that while (unless ).
Hint 2
Compute from your first-quadrant point, then compute from the opposite point separately, and compare the two fractions directly rather than assuming they match.
Show full solution
Worked with the triple 33, 56, 65: point gives ; the opposite point gives again, while .
The triple 33, 56, 65 satisfies . The first-quadrant point is , giving , , and .
The opposite point, reached after sweeping an extra , is , so , identical to . But , the opposite of : sine did not return after only half a lap.
Check your own version. Whatever triple you chose, two tests must pass: your two tangent values, computed as separate fractions from the two points, must come out identical after reducing, and your two sine values must be exact opposites, not equal. If your tangents disagree, the opposite point was likely built by flipping only one coordinate instead of both.
Sine, cosine, and tangent all read off one picture: a point sweeping angle around the unit circle. is that point's x-coordinate, its y-coordinate, and the slope of the radius, which is why a vertical radius breaks it. Their graphs carry that geometry outward: sine and cosine repeat every , stay between and 1, and share one wave shape a quarter turn apart, while tangent repeats every and is unbounded, racing to a vertical asymptote at every angle where cosine is 0. Keep the three derivative facts close, , , and , because Section 2.3 reverses each into an antiderivative, and the minus on cosine's derivative is the one sign to guard. This course works in radians throughout; if the special-angle values or the identities that simplify these expressions are shaky, Appendix A.1 and Appendix A.3 are one click away.
Appendix A.3
Trigonometric Identities
- The Pythagorean identity and its two divided forms.
- Tangent as a ratio , and secant, cosecant, and cotangent as reciprocals.
- The half-angle (power-reduction) forms and , the exact rewrite Section 2.3 uses to integrate a squared trig function.
- Simplifying a squared or product trig expression down to something you can integrate, the same first move Section 2.3 makes.
Walk once around a circle of radius 1, and at every step your position is two numbers: how far right of center you are, and how far up. Those two numbers are cosine and sine, and they cannot move independently. Geometry chains them together, and every identity in this section is one link in that chain, not a separate fact to memorize. One link does real work in the calculus ahead: Section 2.3 cannot integrate until you rewrite it with the half-angle identity from this page. If a problem sent you here mid-integral, you are in the right place. This section stands on its own, and you can take just the piece you need.
Build the intuition
Stand at angle on the unit circle and drop a straight line down to the horizontal axis. That one move builds a right triangle: the hypotenuse is the radius, length 1; the bottom leg is , how far right of center you are; the side leg is , how far up. The Pythagorean theorem does not care that the legs carry trig names. It squares them, adds, and sets the sum equal to the hypotenuse squared. Figure A.3.1 draws it, and a refresher on the functions themselves lives in Appendix A.2.
Figure A.3.1 A point at angle x on the unit circle, with its right triangle. Hypotenuse 1, legs and : the Pythagorean theorem on this triangle is the Pythagorean identity.
An identity is an equation that holds for every value of the variable where both sides are defined, not only at a few special values. The equation is a puzzle with one answer, . The equation is true at every angle, no solving needed: it is a fact about sine and cosine themselves. Every boxed rule below is an identity in that sense.
For every angle , . Reason: the right triangle in Figure A.3.1 has legs and and hypotenuse 1, and this identity is just the Pythagorean theorem on that triangle, whatever angle drew it.
Divide the Pythagorean identity through by or by , and two more identities appear, true wherever the division is legal: (for ) and (for ). The Dig In runs both divisions in full, and the first form is the one Section 2.3 leans on to clear an in a trigonometric substitution.
For , . Reason: on the unit circle the point at angle sits at height over run , and rise over run from the origin to that point is exactly the ratio tangent measures. Turned upside down, the same idea gives .
Wherever the denominator is not zero: , , and . Each name flips one basic ratio. Watch the crossover: cosecant flips sine, not cosine, and secant flips cosine, not sine, so the "co" in a name points at the function it does not share a root with.
Two more identities rewrite a squared sine or cosine as a plain first power of a doubled angle:
Each is a half-angle identity (also called a power-reduction identity). Where they come from, in one line: the double-angle identity (a fact from your trig course), solved for and for . Example A.3.2 runs both. This is the exact rewrite Section 2.3 needs, because a squared trig function has no antiderivative you can read off a table.
Simplify each to a single trig function. (a) . (b) .
Solution. (a) Rewrite tangent as a ratio with the quotient identity, then cancel:
(b) The numerator is the Pythagorean identity in disguise; collapse it to 1 first, then read off the reciprocal identity:
Two moves, both from the rule boxes above: name a ratio, and collapse a Pythagorean sum to 1. Those are the whole toolkit for simplifying a trig expression.
Predict, then verify: simplify to a single trig function. Rewrite with the quotient identity first, then look for the cancellation.
Show solution
.
Since , the sine cancels: .
Derive (a) and (b) from the double-angle identity .
Solution. (a) Start from the first face, , and solve for :
(b) Start from the other face, , and solve for :
Both trade a square for a doubled angle. That trade is exactly what turns a squared trig function into a sum you can integrate, one step before Section 2.3 does the integrating.
A squared trig function has no antiderivative to read off a table, so Section 2.3 rewrites it with a half-angle identity first. Rehearse that move. (a) Rewrite . (b) Rewrite , the same form with the angle in place of .
Solution. (a) Straight from the rule box:
(b) The identity holds for any angle. Put everywhere was, so the doubled angle becomes :
Each right side is a constant plus a single cosine, and both of those integrate in one step. That is the whole reason the rewrite is worth making.
Set a mass bouncing on a spring, or a pendulum swinging through a small arc, and its position traces a cosine while its speed traces a sine. Watch where the energy goes over one full swing. The energy stored in the stretch, the potential energy, rises and falls with of the phase; the energy of the motion, the kinetic energy, rises and falls with . At the turning points all of it is stored and none is motion; halfway between, all of it is motion and none is stored. Add the two at any instant and the Pythagorean identity does the bookkeeping: , so the total never changes. That constant sum is conservation of energy, written in trigonometry. The same balance shows up wherever something turns or rocks: a point on a spinning bowl's rim, a wheel rolling down a road, a wave on the water, a quantity split between two perpendicular parts that trade back and forth while their squares always total the whole. When Section 2.3 swaps for 1 in the middle of an integral, it is spending exactly this fact.
Two of the identities above were not read off a triangle; they were divided out of the Pythagorean identity, and here is the full division so nothing is taken on faith. Start from . Divide every term by , legal wherever :
Two ratios turned into named functions: is , and is . Divide the original identity instead by , legal wherever :
Same identity, a different denominator each time, a new pair of named ratios each time. Neither division created a new fact; each one relabeled the one fact you already had. The first form is what lets clear an in Section 2.3, since there , and the awkward sum is gone.
Dig Deeper The Double-Angle Family, and the Product That Powers Section 2.3▸
Example A.3.2 pulled the half-angle forms out of the double-angle identity . That identity has a family. Setting the two angles equal in the angle-addition identities from your trig course (worth a look if it is rusty) gives
The cosine version wears three faces, each one step from the next through the Pythagorean identity: substitute to get , or to get . Those are the two faces Example A.3.2 solved for and .
| Form | Best when you know |
|---|---|
| both and | |
| only | |
| only |
The sine version earns its keep in Section 2.3's Dig Deeper. Halve it and read it as a rule for the product :
That is what folds into a single square, , which a half-angle identity then reduces one more time. The whole toolkit for squared and product trig integrals is these few identities, applied until nothing is squared.
Two mistakes cost the most here. First, the missing half: is , not . The 2 in the denominator is not optional, and dropping it doubles every answer that flows out of it in Section 2.3. When you are unsure, test the form at one angle: at , , and checks, while does not. Second, the co- crossover: cosecant is the reciprocal of sine, not cosine, and secant is the reciprocal of cosine, not sine, however the syllables rhyme. When a reciprocal is in play, write it out longhand, , rather than trusting the name; a written fraction cannot swap sine for cosine the way a half-remembered pairing can.
(a) Simplify to a single trig function. (b) Simplify . (c) Rewrite with a half-angle identity.
Hint
(a) Rewrite secant as a reciprocal first, then look for the ratio. (b) The numerator is a divided form of the Pythagorean identity in disguise. (c) Use with the angle in place of .
Show solution
(a) . (b) . (c) .
(a) .
(b) The numerator is a divided form itself, , so .
(c) With in place of , the doubled angle is : .
Exercises A.3
1. Simplify each expression using the identities of this section. (a) (b) (c) Warm up
Hint 1
Parts (a) and (b) are reciprocal or quotient cancellations. Part (c) is the divided form , rearranged.
Show solution
(a) . (b) . (c) .
(a) .
(b) .
(c) From , moving the 1 across gives .
2. Given with in Quadrant I, use the Pythagorean identity to find , then . Warm up
Hint 1
Solve for , then pick the sign from the quadrant: every ratio is positive in Quadrant I.
Show solution
; .
, so . Quadrant I keeps sine positive, so . Then .
3. Simplify to a single trig function. Core
Hint 1
Rewrite both cosecant and cotangent as ratios built from and before doing anything else.
Hint 2
Once both are fractions, dividing means multiplying by the reciprocal, and the factors cancel, leaving one reciprocal identity.
Show solution
.
4. Simplify . Core
Hint 1
Distribute first, then use a reciprocal identity on one of the two resulting terms.
Hint 2
collapses to 1 by the reciprocal identities. What remains is a rearranged Pythagorean identity.
Show solution
.
the Pythagorean identity, solved for .
5. Rewrite each squared trig function with a half-angle identity, the first step toward integrating it (Section 2.3). (a) (b) Core
Hint 1
Use for (a). For (b), the same sine form holds with the angle in place of .
Hint 2
In (b), doubling the angle gives inside the cosine: .
Show solution
(a) . (b) .
(a) Straight from the rule box: .
(b) The sine form with in place of : .
6. Verify numerically, at and , that , using exact values from Appendix A.1. Core
Hint 1
Compute the left side as and the right side from at each angle, and check that the two match.
Hint 2
At , and ; at , and .
Show solution
Both sides equal at and both equal at .
At : , so . And . They match.
At : , so . And . They match again.
7. Starting from the double-angle identity , derive the half-angle form . Then use it to rewrite . Stretch
Hint 1
Solve the double-angle equation for : add 1 to both sides, then divide by 2.
Hint 2
Once you have the form, put everywhere was; the doubled angle inside the cosine becomes .
Show full solution
; then .
With in place of , the doubled angle is : .
8. Show that , using the product identity and a half-angle identity. Stretch
Hint 1
Write as , then replace with the product identity.
Hint 2
You now have . Apply the half-angle form to , where the angle is , so the doubled angle is .
Show full solution
.
The half-angle form was applied to , doubling into . This is the exact reduction Section 2.3's Dig Deeper uses to integrate .
9. The Appendix A.3 Dig In divided the Pythagorean identity by and by to reach its two divided forms. Make the method yours. (a) Pick an angle of your own and confirm numerically. (b) Divide the Pythagorean identity by a quantity of your own choosing other than or , simplify each term with the reciprocal identities, and say what identity you land on. (c) State which divided form Section 2.3 uses to clear an . Dig In
Hint 1
Reread the Appendix A.3 Dig In: dividing every term of by the same nonzero quantity keeps a true equation, and the reciprocal identities rename the pieces.
Hint 2
A clean divisor to try is : each of the three terms then becomes one of , , or .
Show full solution
(a) At : and . (b) Dividing by gives . (c) .
(a) At , and , so and : they agree.
(b) Divide by , term by term: , which the reciprocal identities rename .
(c) The trigonometric substitution uses to turn into .
Check your own version. Whatever divisor you chose in (b), each term of your result must simplify to a single named reciprocal or ratio, and the equation must check numerically at one test angle to at least three decimal places. If a term will not reduce to a named function, your divisor mixed sine and cosine in a way the reciprocal identities do not name; try a pure power of , , or their product.
10. The Appendix A.3 Dig Deeper collected the double-angle family and pulled the product identity out of it. Put the family to work. (a) Pick an angle of your own and confirm numerically. (b) Use the product identity to rewrite as a single reduced expression, the first step Section 2.3's Dig Deeper takes to integrate it. (c) Confirm all three faces of agree at your angle. Dig Deeper
Hint 1
Any angle works, exact or decimal; a calculator in radian mode handles the direct values of and .
Hint 2
For (b), start from and reduce with a half-angle identity, as in Exercise 8. The three faces in (c) are , , and .
Show full solution
Worked with : (a) both sides give ; (b) ; (c) all three faces give .
(a) At : , and . They agree.
(b) .
(c) At : ; ; . All three match .
Check your own version. At your angle, the two sides of the double-angle sine identity must agree to at least three decimal places, and all three faces of must land on the same number. If one face disagrees, a sign slipped in the Pythagorean substitution that produced it, and checking at your angle will show which term is off.
Every identity here is one geometric fact seen from a new side. The Pythagorean identity is the Pythagorean theorem on the unit circle's own right triangle, and dividing it by or gives and . The quotient identity and the reciprocal identities for secant, cosecant, and cotangent are unit-circle ratios wearing names, with the co- crossover the one thing to keep straight. The half-angle identities and trade a square for a doubled angle, and they are the exact rewrite Section 2.3 makes before it integrates or . None of this is new mathematics invented for calculus; it is the trigonometry you already knew, filed where the integrals can reach it.
Appendix A.4
Logarithmic Functions
- Reading : the exponent must be raised to, the domain , and the two anchor values and .
- The three log laws, product, quotient, and power, run in both directions: expanding one log and combining several.
- Why reads a negative input, the piece Section 2.4 needs for .
- Seeing as the inverse of (Appendix A.6), the mirror of the exponential across the line .
- Solving a simple equation by trading an exponential for a logarithm, or a logarithm for an exponential.
A dish of bacteria doubles on a schedule: one cell, then two, four, eight, and on. The exponential from Appendix A.5 answers the forward question, start at 1 and double for a while, how big is the colony? A logarithm asks the reverse: you walked in and counted the colony, so how many doublings got it here? That reverse question is the whole idea of a logarithm, a machine that takes a size and hands back the exponent that built it. The natural logarithm uses the special base instead of 2, but the job is the same. Its real gift is that it turns multiplying into adding: multiply two amounts and their logarithms add. That single trick is why quantities with runaway range (earthquake energy, sound power, acidity) are quoted on logarithmic scales, so a span of factors in the millions becomes a short, walkable number line. If a problem sent you here mid-course, you are in the right place; this section stands on its own.
Build the intuition
Everything in this section rests on one relationship: the natural logarithm is the inverse of the exponential (Appendix A.6). To undo is to answer "what exponent produced this output," and that is exactly what reports. So and : each function walks back the other's step. Two anchor points fall right out. Since , the exponent that makes 1 is 0, so . Since , the exponent that makes is 1, so .
The domain comes free from the same idea. The outputs of are always positive (Appendix A.5), so those positive numbers are the only inputs a logarithm can undo: is defined for and nowhere else. And the log laws are the exponent laws seen through the mirror. Multiplying two powers of adds their exponents, , so multiplying two numbers adds their logarithms. Keep that one sentence; it drives the rest.
The natural logarithm of a positive number , written , is the exponent that must be raised to in order to give . In symbols, means exactly . Because is positive for every , only positive numbers have a logarithm, so the domain of is . Two values anchor all the rest: , since , and , since .
The graph of : passes through , since ; passes through , since ; is defined only for ; is negative for , zero at , and positive for ; climbs as grows, but ever more slowly; and drops toward as closes in on 0 from the right, so the y-axis is a vertical asymptote. Reason: is the inverse of (Appendix A.6), the mirror of the exponential across the line . Every fact about reflects into a fact about : the exponential's y-intercept becomes the logarithm's x-intercept , and the exponential's floor at becomes the logarithm's cliff at the left edge.
Figure A.4.1 The natural logarithm is the mirror of across the dashed line . Where passes through , passes through ; where reaches , reaches . So and , and because is always positive, only accepts positive inputs.
For positive and and any power :
Reason: a logarithm is an exponent, and these are the exponent rules read back through the inverse (Appendix A.5). Multiplying two numbers adds their exponents, so it adds their logs; dividing subtracts them; raising to a power multiplies. Each law trades a harder operation for an easier one: a product becomes a sum, a quotient a difference, a power a plain multiple. A useful special case is , the quotient law with .
One extension carries into calculus. The logarithm refuses negative inputs, but the function does not: it is defined on both sides of zero. To antidifferentiate across its whole domain, Section 2.4 reads the input by its size, its distance from zero, and writes . For a negative , the size is positive, so is defined, and . The bars are the whole reason one formula covers the negative half too.
Evaluate each exactly, using only that means . (a) . (b) . (c) . (d) . (e) .
Solution. (a) Since , the exponent that makes is 1, so .
(b) Since , the exponent that makes 1 is 0, so .
(c) The exponent is already on display: .
(d) Rewrite the reciprocal with a negative exponent: , so .
(e) A square root is the power : , so . Every answer is just the exponent on , which is the entire meaning of the logarithm.
Before you go on, predict each value, then check. Evaluate exactly: (a) , (b) , and (c) .
Show solution
(a) . (b) . (c) .
Each answer is the exponent on . In (a) it is showing already. In (b), . In (c), , so the logarithm is ; a reciprocal always hands back a negative exponent.
(a) Expand into logs of , , and (all positive). (b) Combine into a single logarithm and evaluate it. (c) Write in terms of and , then give a decimal to three places.
Solution. (a) A quotient is a difference, the top is a product so it opens to a sum, and each power comes out front as a multiple:
(b) The half-power is a square root: . Now fold the three logs into one, a sum for the plus and a quotient for the minus:
(c) Factor 24 into powers of 2 and 3: . The product opens to a sum, and the power law pulls the exponent out:
Evaluate each exactly, using : (a) . (b) . (c) . Then say why this is the form Section 2.4 needs.
Solution. (a) The bars read the size first: , so .
(b) The size of is , so .
(c) The size of is , so .
Section 2.4 antidifferentiates , which lives on both sides of zero, so it needs a logarithm that accepts negative inputs. Reading each input by its size does exactly that, which is why , bars and all.
Logarithms turn multiplying into adding, and that is what a measuring scale needs when a quantity ranges over factors of thousands or millions. Sound power spans a factor of a trillion from a whisper to a jet, so loudness is read in decibels, a logarithm, where every tenfold jump in power adds the same fixed number. Earthquake energy runs over a similar range, so magnitude is logarithmic: a quake one step higher on the scale releases roughly the same fixed multiple more energy. Acidity, the pH of a solution, is the logarithm of a concentration for the same reason. In each case a runaway multiplying quantity is folded back into a short additive one. The natural logarithm carries a second job in this book: because is the function whose rate of change is , it is the antiderivative Section 2.4 reaches for whenever a rate falls off in proportion to the amount still left, from a draining tank to a cooling cup.
The anatomy rule leaned on one fact: is the inverse of . That same fact is where the log laws come from, and the product law is short enough to derive in three lines. Let and . By the definition of the logarithm, that means and . Now multiply:
using the one exponent rule that says equal steps add (Appendix A.5). Read that last line backward through the definition: the exponent you raise to in order to reach is . That is precisely
Try it with numbers to feel it work: , because and the exponents add. The quotient and power laws ride the same rails: gives the difference law, and gives the power law. Every log law is an exponent law seen in the mirror, and Exercise 9 hands you the other two to derive.
Dig Deeper Every logarithm is the natural one in disguise▸
You have met base- logs here, but a calculator has a base-10 key, and a problem may ask for or any other base. You never need a separate table: every logarithm is a scaled copy of . Write , the base- logarithm, which by definition means . Take of both sides and use the power law:
So , the change-of-base formula. The base only sets the constant out front: , so every log curve is stretched by the fixed factor . For base 10 that factor is . This same turns up downstairs in Section 2.4's rule : the base's own natural log is the exchange rate between base and base , running in both directions. Exercise 10 hands you a base of your own.
Three log slips are worth naming. First, the product law is about products, not sums: is not , and there is no rule that opens the log of a sum at all. The true statement is . Second, moving a power is not the same as squaring the output: the power law gives , but squares the whole logarithm and is a different number, so watch where the exponent sits. Third, is undefined for ; if an integral of ever runs through negative values, you need , not (Section 2.4). If your work ever asks for the logarithm of a negative number, or opens the log of a sum into a sum, a law got stretched past what it says.
(a) Evaluate exactly: , , and . (b) Write each as a single logarithm and give a decimal to three places: and . (c) Evaluate .
Hint
(a) Use , and rewrite . (b) A difference of logs folds into one quotient, and a multiple folds into one power (Appendix A.4). (c) Take the size first: .
Show solution
(a) , , . (b) and . (c) .
(a) Each is the exponent on : ; gives ; and .
(b) The difference is a quotient, . The multiple is a power, .
(c) The size of is , so .
Exercises A.4
1. Evaluate each exactly, without a calculator: (a) . (b) . (c) . (d) . (e) . Then say what (a) and the pattern across (c) to (e) show about . Warm up
Hint 1
Each answer is the exponent is raised to. Use and rewrite (Appendix A.4).
Show solution
(a) . (b) . (c) . (d) . (e) ; and shows undoes , with as the anchor.
Each value is the exponent on : gives ; gives ; gives ; gives ; and gives .
Part (a) is the anchor . Parts (c) to (e) all read , which is just the statement that the logarithm undoes the exponential.
2. Without computing, say whether each is positive, negative, or zero: (a) . (b) . (c) . (d) . (e) . Warm up
Hint 1
Compare each input with 1 (Appendix A.4): is positive when , zero at , and negative when .
Show solution
(a) positive. (b) negative. (c) zero. (d) negative. (e) positive.
The sign is set by where the input sits relative to 1. Both and are above 1, so their logs are positive. Both and are between 0 and 1, so their logs are negative. And exactly, the crossing point.
3. Use the log laws to write each as a single logarithm, then give its exact value: (a) . (b) . (c) . Core
Hint 1
Run the laws from the sum, difference, and multiple side back to one log (Appendix A.4): a sum is a product, a difference is a quotient, a multiple is a power.
Hint 2
(a) . (b) . (c) .
Show solution
(a) . (b) . (c) .
(a) The sum is a product: .
(b) The difference is a quotient: .
(c) The multiple is a power: .
4. Expand each into logs of the single letters (all letters positive): (a) . (b) . (c) . Core
Hint 1
Run the laws in the opposite direction from Exercise 3: a product opens to a sum, a quotient to a difference, a power to a multiple (Appendix A.4).
Hint 2
A square root is the power : , so .
Show solution
(a) . (b) . (c) .
(a) The product opens to a sum, and the square pulls out front: .
(b) The quotient is a difference, and the cube pulls out: .
(c) The root is the power , and the product is a sum: .
5. Using and , find each to three decimal places with the laws: (a) . (b) . (c) . Core
Hint 1
Break each number into factors of 2 and 3, then the laws turn the product into a sum of the logs you were handed (Appendix A.4).
Hint 2
, , and ; the power law pulls each exponent out front.
Show solution
(a) . (b) . (c) .
(a) .
(b) .
(c) .
6. The size reading . Evaluate exactly: (a) . (b) . (c) . Then explain why is defined for every while needs , and name the Section 2.4 fact that uses it. Core
Hint 1
The bars take the size first, so and ; then read the ordinary log of that positive number (Appendix A.4).
Hint 2
works because is positive whenever . The Section 2.4 fact is the antiderivative of .
Show solution
(a) . (b) . (c) . And is defined for every because is then positive; the Section 2.4 fact is .
(a) , so .
(b) , so .
(c) , so the log is .
The bars turn any nonzero input into a positive size, which is the only thing can accept. That is exactly why the antiderivative of , a function living on both sides of zero, must be written .
7. Solve for , exact value first and then a decimal to three places: (a) . (b) . (c) . Stretch
Hint 1
A logarithm and an exponential undo each other (Appendix A.6). To free from a log, raise to both sides; to free from an exponent, take of both sides.
Hint 2
(a) . (b) . (c) .
Show solution
(a) . (b) . (c) .
(a) Raise to both sides of : .
(b) Take of both sides of : .
(c) Raise to both sides of : , a positive number, as every output of must be.
8. A colony grows as , with in hours. (a) Find the doubling time by solving ; give the exact value, then a decimal to three places. (b) Find the tripling time the same way. Stretch
Hint 1
Doubling means the colony reaches twice its start, so , and the cancels. Free with a logarithm (Appendix A.6).
Hint 2
Take of both sides: , so . Tripling replaces the 2 with a 3.
Show full solution
(a) hours. (b) hours.
(a) The starting size cancels, leaving . Take of both sides:
(b) Same steps with 3 in place of 2:
Tripling takes longer than doubling but not three halves as long, because growth compounds: the colony is already larger when it starts the climb from double to triple.
9. The Appendix A.4 Dig In derived the product law from the inverse relationship. Make it your own. (a) Run the same argument to derive the quotient law . (b) Verify your law numerically with two positive numbers of your own choosing. (c) Derive the power law the same way. Dig In
Hint 1
Reread the Appendix A.4 Dig In: set and , so and , then combine and read the exponent backward (uses the exponent rules of Appendix A.5).
Hint 2
For the quotient, . For the power, .
Show full solution
(a) gives . (b) Answers vary; with , : . (c) gives .
(a) With and , so and , dividing subtracts the exponents: . Read backward, the exponent that reaches is , so .
(b) With and : , and . The two sides agree.
(c) Raising to the power multiplies the exponent: , so .
Check your own version. Whatever two numbers you chose in (b), your left side and right side must agree to full calculator precision, since the exponents subtract exactly. If they miss, you likely swapped which number is and which is , which flips the sign of the difference.
10. The Appendix A.4 Dig Deeper gave the change-of-base formula . (a) Use it to evaluate and exactly. (b) Show that every base- log is times one fixed constant, and give that constant for to three places. (c) Pick a base and an argument that is a power of , evaluate it with change of base, and connect your constant to why carries a (Section 2.4). Dig Deeper
Hint 1
Reread the Appendix A.4 Dig Deeper: . When is a power of , the two natural logs cancel to a whole number.
Hint 2
, and makes . The constant out front is ; for it is .
Show full solution
(a) and . (b) ; for the constant is . (c) Answers vary; see the check below.
(a) With : . With : .
(b) The formula is multiplied by the fixed number , which depends only on the base. For , , so a base-10 log is always about times the natural log of the same number.
Check your own version. With your base and an argument , change of base must return exactly the whole number , since the on top and bottom cancel. And your constant is the same that sits under in Section 2.4's : both come from the fact that , so if your evaluated power misses , recheck that your argument really is a power of your base.
The natural logarithm is the exponent that turns into , so it is the inverse of (Appendix A.6): defined only for , anchored by and , and drawn as the mirror of the exponential across . The three log laws are exponent rules seen through that mirror: , , and , each trading a harder operation for an easier one. Read a negative input by its size: is defined for every , which is exactly the form Section 2.4 needs for . And every logarithm in any base is just rescaled, , with the same that Section 2.4 places under .
Appendix A.5
Exponential Functions
- Reading the parts of : the initial value , the base , and the variable riding up in the exponent.
- Telling growth () from decay () at a glance, and sketching either shape.
- The three graph facts every base shares: a y-intercept at the initial value (at 1 when ), outputs always positive and never zero, and the horizontal asymptote .
- Evaluating at zero, negative, and fractional inputs by hand, and getting a numeric feel for the base and change.
- Why an exponential eventually outruns every power of , and why is the base whose exponential is its own rate of change.
Lay one grain of rice on the first square of a chessboard, two on the second, four on the third, and keep doubling square by square. Ten squares in, the pile on that square is grains, a small handful. By square 20 it is grains, a sackful. Keep the rule going to the last square, number 64, and it alone holds grains, more than nine quintillion, more rice than the planet's farms grow in centuries. Nobody could ever lay that board out, but the arithmetic is honest, and it is the signature of one family of functions: the ones that grow by multiplying instead of adding. You met this family in algebra. This section re-arms it for calculus, and if a problem sent you here mid-course, you are in the right place; this section stands on its own.
Build the intuition
A parking meter that eats one dollar every hour drains by subtraction: each hour takes off the same amount, and the graph is a straight line. A pond of algae that doubles every day grows by multiplication: each day the whole current amount gets multiplied by 2, so a fuller pond gains more than an emptier one did. That is the split that matters. Linear functions add or subtract a fixed amount per step. Exponential functions multiply by a fixed factor per step, and everything surprising about them follows from that one habit.
The function built for multiplying is , because exponents count factors: moving from to means one more factor of , so . Equal steps in multiply the output by equal factors. That sentence is the fingerprint of the whole family, and it is the property Section 2.4 builds its exponential integrals on, so hold onto it.
An exponential function has the form , where the base is a positive constant with and the multiplier is a positive constant. Since , the multiplier is the function's initial value: . When , each step right multiplies by more than 1, so the outputs climb: exponential growth. When , each step right multiplies by less than 1, so the outputs shrink: exponential decay.
A horizontal asymptote is a horizontal line that a graph closes in on as runs far out to one side: the gap between graph and line shrinks toward zero, though the graph never needs to touch the line. Every function has the x-axis, the line , as its horizontal asymptote. On the graph's low side the outputs get as close to 0 as you like while staying positive.
For any base with , the graph of passes through , since ; passes through , so the base is the height one step right of the axis; stays positive at every and never reaches zero; climbs for and falls for ; and hugs the asymptote on its low side. Reason: equal steps multiply. Each step right multiplies the output by the fixed positive factor , and multiplying a positive number by a positive factor can make it larger or smaller but never zero or negative. Fine print: a multiplier scales every height, so crosses the y-axis at instead of 1; everything else on this list survives untouched.
Figure A.5.1 Growth and decay are mirror images. Both curves cross the y-axis at , both stay above the x-axis everywhere, and each hugs the asymptote on its low side without ever landing on it.
Let . Evaluate (a) , (b) , (c) , and (d) , exact values first.
Solution. (a) Any base to the zero power is 1:
(b) Three factors of 2: .
(c) A negative exponent means reciprocal, not negative:
(d) A one-half power is a square root: . Notice all four outputs are positive, the ones from negative and fractional inputs included: the anatomy rule in action.
Before you go on, predict each value, then check. Let . Find (a) , (b) , and (c) , exact values.
Show solution
(a) . (b) . (c) .
The zero power is 1; a negative exponent means reciprocal, so ; and the one-half power is a square root, . All three outputs stay positive.
Classify each function as growth or decay, give its y-intercept, and describe where its graph heads far to the right. (a) . (b) . (c) .
Solution. (a) The base sits below 1, so is decay. Its y-intercept is . Far to the right the outputs slide toward the asymptote , always positive, never arriving.
(b) The base , so is growth. The multiplier moves the intercept: . Far to the right the outputs climb with no ceiling.
(c) The disguise comes off with one negative-exponent rewrite:
So is decay with base , intercept , sliding toward on the right. A minus sign on the flips growth to decay: it mirrors the graph across the y-axis, exactly the mirror pair in Figure A.5.1.
Between the bases 2 and 3 sits one that calculus prizes above the rest: the number , an unending nonrepeating decimal that starts . One way to meet it is compounding. Grow one dollar at 100 percent annual interest, but split the year into equal pieces and compound each piece, and the year-end balance is dollars. Squeeze larger and the balance closes in on , as Table A.5.1 shows.
| 1 | 2.000000 |
| 2 | 2.250000 |
| 10 | 2.593742 |
| 100 | 2.704814 |
| 1,000 | 2.716924 |
| 10,000 | 2.718146 |
What earns its nameplate is a calculus fact: it is the one base whose exponential is its own rate of change. The steepness of at any point equals its height there, and at the intercept that steepness is exactly 1. Because its rate of change matches itself, is also its own running total, which is why Section 2.4 reaches for when it adds these curves up; there, the integral of comes out as clean as the function itself. For now, carry one fact: is a fixed number, so is an ordinary exponential with everything on the anatomy list, and since it is growth.
Because an exponential multiplies by a fixed factor per step, the time to change by a fixed ratio is constant. For a decaying quantity, the half-life is the time it takes to fall to half of what it was; the next half-life halves it again, and again. For a growing quantity, the doubling time is the time it takes to double. A population with a doubling time of 3 years is four times its start after 6 years and eight times after 9.
The race: doubling against powering
Powers of grow fast. has reached 1,000 by , and has reached 100,000. So here is a question with a surprising answer: can plain doubling, , keep pace with a power like ? Start small, where you can score every round by hand.
Race against at the integers through . Who leads when, and who wins?
Solution. Table A.5.2 scores every round.
| Leader | |||
|---|---|---|---|
| 1 | 1 | 2 | the doubler |
| 2 | 8 | 4 | the power |
| 4 | 64 | 16 | the power |
| 6 | 216 | 64 | the power |
| 8 | 512 | 256 | the power |
| 9 | 729 | 512 | the power |
| 10 | 1,000 | 1,024 | the doubler |
The doubler leads at , falls behind for eight straight rounds, and retakes the lead at : edges out . And the lead is permanent. Each step right multiplies by 2, while it multiplies by , a factor that shrinks toward 1 as grows: from the power's step factor is at most , no match for 2. The doubler gains ground every round from here on.
Now a heavier power. Against the doubler loses early and loses big: at it trails by nearly a hundredfold. Watch what happens anyway.
| Leader | |||
|---|---|---|---|
| 2 | 32 | 4 | the power |
| 5 | 3,125 | 32 | the power |
| 10 | 100,000 | 1,024 | the power |
| 22 | 5,153,632 | 4,194,304 | the power, barely |
| 23 | 6,436,343 | 8,388,608 | the doubler, for good |
| 30 | 24,300,000 | 1,073,741,824 | the doubler, forty-four times over |
The lead flips between and , and seven rounds later the doubler is ahead by a factor of more than forty. The reason is the same step-factor argument as the small race, and the Dig Deeper box below runs it in full. The headline deserves a box of its own.
For any base and any fixed power , the exponential eventually passes and stays ahead forever. Reason: each step right multiplies by the same factor , while it multiplies by , a factor that slides toward 1 as grows. Once the power's step factor drops below , the exponential gains ground on every step after that, and no head start survives a rival who wins every remaining round.
Multiply-per-step is how the moving world compounds. A bacteria colony doubles on a fixed schedule while its food lasts, so its census follows with and a steady doubling time. A dose of medicine leaves the bloodstream by halves, a fixed fraction gone each hour, so what remains follows a decay curve with a constant half-life and a asymptote: the level never snaps to zero, it fades, which is exactly why doses repeat on a schedule. A savings account compounds the same way with a gentler base. And the race is the arithmetic behind the dread in "growing exponentially": a quantity that multiplies per step will pass any quantity that merely powers along, whatever polynomial head start it enjoys. When calculus adds all the tiny changes in these curves back into a running total, one base, , makes that total astonishingly clean; that is the discovery waiting in Section 2.4.
The anatomy rule made two absolute claims: never zero, never negative. "The graph looks like it" is not a reason, so here is the algebra, and it is short. Both claims run on the exponent rules from algebra.
Never zero. Suppose some input gave . Pair it with its opposite input and multiply, using the exponent product rule:
But if , the left side is 0 times something, which is 0. That forces , which is absurd, so no such input exists: an exponential output is never zero. Try the pairing with real numbers to feel it work: .
Never negative. Every output is a perfect square. Halve the exponent, then square, using the power-of-a-power rule:
A square of a real number is never negative, and we just showed the output is never zero, so every output is strictly positive.
So what does the graph do on its low side? It gets close to zero without arriving. For and negative inputs, , and , less than one millionth. Name any tiny tolerance and the curve eventually slips inside it while staying positive the whole way: that is the asymptote earning its definition. Exercise 9 hands you the pairing argument with a base of your own.
Dig Deeper Why the doubler always wins▸
Table A.5.3 showed the flip; here is the machine that makes it inevitable. Compare step factors. One step right multiplies by exactly 2, every time. The same step multiplies by
and this factor shrinks as grows, sliding toward . Ask when it drops below 2. Taking fifth roots, that happens when , which solves to . So from onward the power's step factor stays below 2; at it is , and it only falls from there.
Now the endgame is bookkeeping. From on, every single round multiplies the doubler by more than it multiplies the power, so the ratio of power to doubler shrinks every round, relentlessly. However large the power's lead at , a lead that shrinks by a factor bounded below 1 per round must eventually fall under 1: the tables locate the flip between and , and after it the gap explodes in the doubler's favor.
Nothing here is special to base 2 or power 5. For any base and power , the power's step factor still slides toward 1, so it eventually drops below , and the same endgame runs. That is the full proof of the rule above, and Exercise 10 hands you the machine with a matchup of your own.
Three wrong moves each force an exponential output below its floor, and all three are impossible. First: a negative input does not make a negative output. The slip is the classic; the truth is , positive, because the minus sign in an exponent means reciprocal, never negative. Second: is 1, not 0 and not ; every exponential crosses the y-axis at its multiplier, never at the origin. Third: the graph does not "finally hit zero somewhere off the page." It cannot; the Dig In proved that hitting zero would force . If your work on an exponential ever produces a zero or negative output, do not shrug: something upstream broke, and the negative-exponent slip is the usual suspect.
(a) Let . Evaluate , , and . (b) Classify and as growth or decay, and name each y-intercept and the asymptote both graphs share. (c) At , which is bigger, or , and by what factor?
Hint
(a) Remember , and a negative input flips the base: . (b) Compare each base with 1; the multiplier is the intercept. (c) Compute both values; can be built by doubling fifteen more times.
Show solution
(a) , , . (b) Growth; decay; intercepts 1 and 1; shared asymptote . (c) beats , by a factor of exactly .
(a) , the initial value. . For the negative input, flip and square:
(b) The base exceeds 1: growth. The base sits below 1: decay. Each has multiplier 1, so each crosses the y-axis at 1, and both graphs hug the same asymptote on their low sides.
(c) while . The doubler leads by , a lead it took near and never gave back.
Exercises A.5
1. Evaluate at , , , , and . Then say what your five outputs illustrate about the y-intercept and the sign of every exponential output. Warm up
Hint 1
Work left to right: each step down in the exponent divides by the base once. A zero exponent gives 1, and a negative exponent means one over. The anatomy rule in Appendix A.5 predicts both patterns you should see.
Show solution
, , , , ; the middle output shows the y-intercept at , and all five outputs are positive.
Each step down divides by 4:
At the output is 1, which is the y-intercept fact. And the negative inputs produced reciprocals, not negatives: the outputs shrank toward 0 but stayed positive, exactly the never-zero, never-negative anatomy.
2. Classify each function as exponential growth or exponential decay, and give its y-intercept: (a) . (b) . (c) . (d) . Warm up
Hint 1
One comparison decides each verdict: is the base bigger or smaller than 1 (Appendix A.5)? For the intercept, evaluate at and remember the multiplier is the initial value.
Show solution
(a) Growth, intercept 1. (b) Decay, intercept 1. (c) Growth, intercept 1. (d) Decay, intercept 4.
(a) Base : growth, and .
(b) Base : decay, intercept 1.
(c) Base is just barely past 1, but past is past: growth, intercept 1. A base near 1 grows slowly, not "sort of."
(d) Base : decay. The multiplier 4 is the initial value, so the graph crosses the y-axis at 4, not 1.
3. Let . (a) Give the y-intercept and classify growth or decay. (b) Evaluate . (c) Describe where the graph heads as runs far to the right, and name the line it approaches. Core
Hint 1
Read the anatomy rule's fine print in Appendix A.5: the multiplier scales the intercept, and the base alone decides growth or decay. Part (c) is asking for the horizontal asymptote by name.
Hint 2
For (b), a negative exponent flips the base: .
Show solution
(a) Intercept 8; decay. (b) . (c) The outputs slide toward 0, staying positive; the graph approaches the horizontal asymptote .
(a) , and the base sits below 1, so each step right quarters the output: decay.
(b) Flip the base to undo the negative exponent:
(c) Quartering over and over drives the output toward 0 without ever reaching it, so far to the right the graph hugs the line , the horizontal asymptote. Decay curves are tall on the left and low on the right; against shows the tilt.
4. Evaluate at (a) , (b) , and (c) , exact values. Core
Hint 1
A denominator of 2 in the exponent is a square root, and 16 is a perfect square, so no radical should survive.
Hint 2
Write as : root first keeps the numbers small. The negative sign in (c) means reciprocal, applied after the root.
Show solution
(a) . (b) . (c) .
(a) .
(b) Root first, then cube:
(c) Reciprocal of the root: . All three outputs are positive, fractional inputs included: the anatomy holds at every input, not just the whole numbers.
5. A culture starts at 50 cells and doubles every hour, so after hours it holds cells. (a) Find , , and . (b) After how many whole hours does the count first top 1,000? (c) Does the growth ever level off? Name the reason, and state the doubling time. Core
Hint 1
The model doubles per hour, so build the counts hour by hour (Appendix A.5). For (b), keep doubling past part (a) until you clear 1,000; for (c), reread the anatomy rule.
Hint 2
The hour-4 count lands just under 1,000; take one more doubling.
Show solution
(a) , , cells. (b) After 5 hours, when . (c) No: base , so it climbs with no ceiling; the doubling time is 1 hour.
(a) Each hour doubles the count:
(b) At the count is , still under 1,000. One more doubling: , past 1,000 for the first time. So the answer is 5 whole hours.
(c) It never levels off. A base above 1 multiplies the count by 2 forever, so climbs without bound; there is no asymptote on the high side. Since the count doubles once per hour, the doubling time is 1 hour.
6. A function satisfies and , with . (a) Find and . (b) Compute and classify the function as growth or decay. Core
Hint 1
The value at zero hands you directly, since (Appendix A.5). Then the second condition becomes one equation in alone.
Hint 2
From , divide off the known to get , then take the real cube root.
Show solution
(a) , . (b) ; growth.
(a) The initial value is the multiplier: . Then
(b) , and base makes this growth: each step right doubles the output, and checks it.
7. Race against at the integers through . (a) Find the integers where the tripler leads and where the power leads. (b) Explain, with step factors, why once the tripler retakes the lead at it can never lose it again. Stretch
Hint 1
Score all nine rounds by hand, as in Example A.5.3 of Appendix A.5. For (b), compare what one step right does to each racer: the tripler's factor is always 3, and the power's factor is .
Hint 2
For , the power's step factor is at most . Compute that number and compare it with 3.
Show solution
(a) The tripler leads at and ; the power leads at through ; the tripler retakes the lead at (6561 against 4096) and keeps it. (b) From on, the power's step factor is at most , so the tripler gains every round.
(a) The nine rounds: runs 0, 1, 16, 81, 256, 625, 1296, 2401, 4096 while runs 1, 3, 9, 27, 81, 243, 729, 2187, 6561. The tripler is ahead at and , the power takes over from through , and the tripler retakes the lead at , where .
(b) One step right always triples . The same step multiplies by , and for that factor is at most
and it keeps shrinking toward 1 afterward. So from onward every round multiplies the tripler by more than the power, and a rival who wins every remaining round never falls behind again.
8. Return to the chessboard: 1 grain on the first square, doubling each square, so square holds grains. (a) How many grains sit on square 20? (b) Which is the first square whose pile tops 1,000,000 grains? (c) Show that square 64 alone, with grains, holds more than squares 1 through 63 put together. Stretch
Hint 1
Rewrite "square " as the power , then hunt the first power of 2 past each target: doubling a running product beats computing powers from scratch (Appendix A.5).
Hint 2
For (c), squares 1 through 63 hold . That sum of powers of 2 is exactly .
Show full solution
(a) grains. (b) Square 21, holding grains. (c) Squares 1 through 63 total grains, one less than the on square 64, so the last square beats all the rest combined by a single grain.
(a) Square 20 holds grains.
(b) Hunt the first power of 2 past 1,000,000: falls short and clears it. Since square holds , the pile sits on square 21.
(c) The earlier squares hold
the standard fact that a run of doublings sums to one less than the next power. That total is one grain short of the on square 64, so the last square holds more than every earlier square combined. That single square carries over half the whole board.
9. The Appendix A.5 Dig In proved no exponential output is ever zero, by pairing with , and never negative, by writing as a square. Make both arguments your own: pick a base the box did not use, write out the pairing argument with your base, verify the pairing numerically at one input, and then verify the square trick numerically at one odd power. Dig In
Hint 1
Reread the Dig In in Appendix A.5: both arguments run on the exponent rules, the product rule for the pairing and power-of-a-power for the square. Your base changes the arithmetic, not the moves.
Hint 2
For the numeric pairing, compute and separately and multiply. For the square trick on , write it as and check that the two sides agree.
Show full solution
Worked with : if then , absurd, so never zero; numerically ; and checks as , a square, so never negative.
The pairing argument with base 5: suppose some input gave . Then
but a product with a zero factor is 0, forcing . No input can do that, so is never zero. Numerically at : , as promised.
The square trick at the odd power 3: by power-of-a-power, and , whose square is on the nose. Every output of is a real number squared, so none is negative.
Check your own version. Whatever base you chose, two audits settle it: your paired product must come out exactly 1 with no rounding, since the exponents cancel to zero, and your square-trick check must reproduce from squaring . A paired product that misses 1 usually means one exponent's sign got dropped; a square that misses usually means the fractional exponent was halved twice.
10. The Appendix A.5 Dig Deeper machine compares step factors: the exponential's is always , the power's is , sliding toward 1. Run the machine on a matchup of your own, a base and a power the box did not use: find the beyond which your exponential wins every step, then locate by direct computation the integer where it takes the overall lead for good. Dig Deeper
Hint 1
Reread the Dig Deeper in Appendix A.5: the threshold comes from solving , which an n-th root turns into . A modest matchup like against keeps the numbers printable.
Hint 2
The step threshold only says the exponential gains from there on; the overall lead can flip later. Tabulate both racers at whole numbers past your threshold until the exponential's value passes the power's, and confirm the round before still had the power ahead.
Show full solution
Worked with against : the threshold is , so the quadrupler out-steps the power from on; the overall lead flips at , where passes , while at the power still led, to .
Step factors first. The quadrupler's is always 4. The power's is , and it drops below 4 when
So from onward the quadrupler multiplies by more every round; at the power's factor is , already beaten. The lead itself flips later: at the power still holds it, against , and at the quadrupler takes it for good, against . Winning every step from made that flip a matter of time.
Check your own version. Whatever matchup you chose, three audits must pass: your threshold must satisfy its own inequality (plug it back into and land almost exactly on ); at every integer past the threshold your power's one-step factor must compute to less than ; and at your flip integer the exponential must exceed the power while the integer before shows the reverse. A flip that lands before your threshold means the root in the threshold formula was taken with the wrong .
An exponential function grows by multiplying: equal steps in multiply the output by equal factors, the fingerprint the whole family shares. Read its anatomy at a glance: the initial value is the y-intercept, base means growth and means decay, every output is positive and never zero, and the x-axis is the horizontal asymptote on the low side. A minus sign in the exponent means reciprocal, never a negative output. In any long race against a power of , the exponential eventually takes the lead and keeps it, because its step factor holds steady at while the power's slides toward 1. And one member of the family, base and change, is the one calculus builds on: it is its own rate of change, so it is its own running total, and Section 2.4 spends that fact on the cleanest integral in the book.
Appendix A.6
Inverse Functions
- Reading an inverse function: the rule that runs the original backward, returning the input that produced a given output.
- The reflect-across- picture, and building an inverse's graph by swapping each point's coordinates.
- The one-to-one test (the horizontal line test) that decides whether a function has an inverse at all.
- The pair and (Appendix A.4, Appendix A.5) as the inverse pair Section 2.4 leans on.
- and , the inverse trig functions the trigonometric substitution in Section 2.3 lands on.
Think of a function as a machine: drop in a number, it hands one back. An inverse is the same machine run backward, so you hand it the output and it returns the input that produced it. Some machines reverse cleanly. Add 5, and subtracting 5 walks it back; multiply by 3, and dividing by 3 walks it back. Others jam. Square the number 4 and you get 16, but square and you also get 16, so "take the number that was squared" has no single answer. The machines that reverse cleanly are exactly the ones that never send two different inputs to the same output, and this section names that idea, draws the reverse machine as a mirror image, and meets the pair the rest of the book leans on: and . If a problem sent you here mid-course, you are in the right place; this section stands on its own.
Build the intuition
An inverse swaps the roles of input and output, and that swap is the whole story. If sends 2 to 7, then its inverse sends 7 back to 2: the pair on the graph of becomes the pair on the graph of the inverse. Swapping the two coordinates of every point is a reflection across the line , the diagonal where the two coordinates are equal, so the graph of an inverse is the mirror image of the original across that diagonal.
The same swap explains when an inverse can exist. To run backward you must read each output back to one input, so no output may come from two different inputs. A function with that property is called one-to-one, and only a one-to-one function has an inverse. You can spot the property on a graph: a horizontal line meets the graph of at every input whose output is , so if any horizontal line crosses the graph twice, two inputs share an output. That is the horizontal line test, and reflecting across turns it into the ordinary vertical line test on the mirror image.
The inverse of a function , written , reverses : exactly when . Reversing then applying, or applying then reversing, returns the original input: and . Because the two functions trade inputs for outputs, the domain of is the range of and the range of is the domain of . The symbol is an inverse, never the reciprocal .
A function is one-to-one when different inputs always give different outputs: forces . This is the exact condition for an inverse to exist, because reversing requires reading each output back to a single input. A one-to-one function passes the horizontal line test; a function that fails it, like on all real numbers, has no inverse until its domain is trimmed.
A function has an inverse exactly when it is one-to-one, and you can test that on its graph: is one-to-one when no horizontal line crosses its graph more than once. Reason: a horizontal line meets the graph wherever , so a second crossing is a second input with the same output, the one thing a one-to-one function forbids. A steadily rising or steadily falling graph always passes; a graph that turns around, like a parabola, fails at every height it hits twice.
The graph of is the mirror image of the graph of across the line . Reason: swaps each point's coordinates, turning into , and those two points are reflections of each other across . Reading it off the graph: wherever passes through , its inverse passes through , and the domain and range trade places. The composition confirms it from the algebra side: and , each function undoing the other's step.
Figure A.6.1 An inverse is a reflection across . The steep line and its shallower inverse are mirror images across the dashed diagonal; the point on reflects to on , and the segment joining them crosses at its midpoint at a right angle (slope ).
Let . (a) Find . (b) Verify it undoes . (c) Check that the point on reflects to a point on .
Solution. (a) Set and solve for , since the inverse reads an output back to its input :
Now rename the input : .
(b) Feed into :
The other order lands the same way: .
(c) Since , the point is on ; the inverse reads it back, , so is on . The two points swap coordinates, mirror images across .
Before you go on, predict, then check. Let . (a) Find . (b) Use it to compute .
Show solution
(a) . (b) .
Solve for : subtract 2, divide by 5, giving , so . Then , the input that makes .
The most useful inverse pair in this book is and . The natural logarithm is defined to reverse the exponential (Appendix A.4): is the exponent is raised to in order to give . (a) Write the two compositions that say they undo each other. (b) Use the undo to solve and , exact then decimal. (c) Describe how their graphs sit relative to .
Solution. (a) Each walks the other back:
the first for every , the second for , where is defined.
(b) To free from an exponent, apply the inverse ; to free it from a logarithm, apply the inverse :
(c) Because they are inverses, their graphs are mirror images across : through reflects to through , the picture drawn in Appendix A.4. This is the pair Section 2.4 integrates, where fills the one gap the power rule leaves open.
Does have an inverse? (a) Apply the horizontal line test. (b) If it fails, trim the domain so it passes, and give the inverse there.
Solution. (a) The line meets the graph at and at , since and . Two inputs, one output: fails the test on all real numbers, so it has no inverse there. Every positive height is hit twice, once on each side of the axis.
(b) Trim the domain to , keeping only the right half. Now the graph rises steadily, no height is repeated, and the test passes. On that half, solving with gives , so , with domain . The square root is the inverse of squaring only after the domain is cut, a move Section 2.3 makes again to turn sine and tangent into and .
A function that tells you where a process is at a given time has an inverse that tells you when it reaches a given place, and the moving world asks both questions constantly. A cup of coffee cools on a known curve, so the inverse answers when it will hit drinking temperature. A savings balance or a bacterial colony grows on an exponential, and the inverse, a logarithm, answers how long until it doubles: that is why solving for runs through (Appendix A.4). Whole measuring scales are built from this reversal: decibels and pH read a runaway multiplying quantity through its inverse logarithm, folding factors of millions into a short additive number. And inverses are not only a lookup trick in this book, they are answers the integral hands back. When the trigonometric substitution in Section 2.3 unwinds a square root, the antiderivative it produces is an inverse function, or ; reading an accumulated total back to the angle that made it is the same reverse question, now answered by calculus.
The reflection rule deserves a reason, not just a picture, and the reason is short. Claim: if is a point on , then is a point on , and the two points are mirror images across the line .
First half, the swap. Saying is on means . By the definition of the inverse, is the very same statement as , which says is on . So every point of has its coordinate-swapped twin on .
Second half, the mirror. Two points and reflect across when that line is the perpendicular bisector of the segment joining them. Check both parts. The midpoint is
whose two coordinates are equal, so it sits on . And the segment's slope is
the negative reciprocal of the line's slope 1, so the segment meets at a right angle. Perpendicular and bisected: a reflection. Try it on : the point and its twin have midpoint on the diagonal, and the segment between them has slope . Exercise 9 hands you the argument to run on a function of your own.
Dig Deeper Turning sine and tangent into inverses▸
Sine and tangent repeat, so on their full domains they fail the horizontal line test badly: the line meets at infinitely many angles. The rescue is the one from Example A.6.3, domain restriction, done with care so the trimmed piece still reaches every output.
Trim sine to the single rising stretch . There it climbs steadily from to , hitting each height once, so it is one-to-one and has an inverse. That inverse is : it takes a value between and and returns the angle in whose sine is that value. For instance , because . Trim tangent to the same way and its inverse is , which accepts any real number and returns an angle in ; for instance .
Here is why an integrals book cares. The trigonometric substitution in Section 2.3 meets a square root like or a sum and clears it with or ; when the dust settles, the antiderivative is one of these inverse functions:
The integral asks a reverse question, which angle produces this accumulated amount, and an inverse function is exactly the tool that answers it. Exercise 10 hands you one of these to explore.
The most expensive slip on this page is reading as a reciprocal. The exponent here marks an inverse function, not a power: is the input that sends to , while is one divided by an output, a completely different number. The same trap wears a trig costume: means , the inverse, not , so an inverse-trig answer from Section 2.3 is never a reciprocal. Second, do not claim an inverse before checking one-to-one: writing as "the inverse of " is only honest once the domain is cut to , since the full parabola has no inverse at all. When a problem hands you , read it as a reverse question, not a fraction, and confirm the function was one-to-one before you trust the answer.
(a) Find the inverse of , and verify it with one composition. (b) Which of and has an inverse over all real numbers, and why? (c) Using that is the inverse of tangent on , find and .
Hint
(a) Solve for , then rename (Appendix A.6); the composition should collapse to . (b) Apply the horizontal line test: a steadily rising graph passes, a graph that turns around fails. (c) Ask which angle in the range has the given tangent: and .
Show solution
(a) , and . (b) , because it is one-to-one; fails the horizontal line test. (c) and .
(a) Solve : add 3, divide by 6, giving , so . Check: .
(b) The cube climbs steadily and never repeats a value, so it passes the horizontal line test and has an inverse (the cube root). The square repeats every positive height, once on each side of the axis, so it fails and has no inverse until its domain is trimmed.
(c) is the angle in whose tangent is 0, namely 0. And is the angle whose tangent is 1, namely , since .
Exercises A.6
1. Find the inverse of each one-step function: (a) . (b) . Then verify . Warm up
Hint 1
An inverse undoes the rule, so undo the single step: adding 7 is undone by subtracting 7, and multiplying by 4 is undone by dividing by 4 (Appendix A.6).
Show solution
(a) . (b) , and .
(a) Adding 7 is reversed by subtracting 7, so .
(b) Multiplying by 4 is reversed by dividing by 4, so . Verify: , the machine run forward then backward.
2. A one-to-one function satisfies and . (a) Find and . (b) The reciprocal equals ; explain why is a different kind of quantity, and why it can be read from the data while cannot. Warm up
Hint 1
asks which input sent to 9, so scan the given outputs for a 9 (Appendix A.6). The reciprocal is just one divided by an output, unrelated to reversing the function.
Show solution
(a) and . (b) is the input that maps to 9 (an inverse), not a reciprocal; needs an input mapping to 2, which the data does not give.
(a) The statement says 2 is the input behind the output 9, so ; likewise gives .
(b) The notation reverses the function: it wants the input that produced 9, which the data supplies. The reciprocal instead divides 1 by an output and has nothing to do with reversal. And would need an input whose output is 2, but no listed value maps to 2, so it cannot be read from the data. Inverse and reciprocal are different questions, and the shared is a coincidence of notation.
3. Let . (a) Find . (b) Verify . (c) State the domain and range of . Core
Hint 1
Set and solve for , then rename the input, exactly as in Example A.6.1 (Appendix A.6).
Hint 2
Solving gives . For the domain and range, remember an inverse trades them with , and here is a line with all real inputs and outputs.
Show solution
(a) . (b) . (c) Domain all real numbers, range all real numbers.
(a) Solve : add 4, divide by 3, so and .
(b) Compose: .
(c) The line accepts every real input and produces every real output, so its inverse does the same: domain and range are both all real numbers.
4. Decide which functions are one-to-one on all real numbers, and say how you know: (a) . (b) . (c) . (d) . Then trim the domain of so it becomes one-to-one, and give the inverse there. Core
Hint 1
Run the horizontal line test on each (Appendix A.6): a graph that ever turns around repeats a height and fails, while a steadily rising or falling graph passes.
Hint 2
Both and send and to the same output. For the trim, keep ; the inverse of there is the square root.
Show solution
(a) Not one-to-one. (b) One-to-one. (c) Not one-to-one. (d) One-to-one. Trimmed to , has inverse .
(a) fails: 2 and both give 4, so a horizontal line cuts it twice.
(b) climbs steadily and never repeats a value, so it passes.
(c) fails: 3 and both give 3.
(d) The rising line is one-to-one. Trimming to keeps the right half, where the graph rises with no repeats, so it is one-to-one and .
5. Use that and undo each other (Appendix A.4) to solve, exact value first and then a decimal to three places: (a) . (b) . (c) . Core
Hint 1
To free from an exponent, apply the inverse ; to free it from a logarithm, apply the inverse (Appendix A.6).
Hint 2
(a) . (b) . (c) Rewrite first, then the exponents match.
Show solution
(a) . (b) . (c) .
(a) Take of both sides: .
(b) Raise to both sides: .
(c) Rewrite the right side as ; since and the exponential is one-to-one, the exponents must match, so . The answer is exact, so no decimal is needed.
6. A one-to-one function has , , , and . (a) Find and . (b) The point lies on ; give the matching point on and describe how the two points sit relative to . Core
Hint 1
The inverse reverses each pairing, so an output of becomes an input of (Appendix A.6).
Hint 2
For (b), the inverse swaps coordinates, turning into ; recall that a coordinate swap is a reflection across .
Show solution
(a) and . (b) , the mirror image of across .
(a) The pairing gives , and gives : each output of is read back to its input.
(b) The inverse swaps the coordinates of into , and swapping coordinates reflects a point across the line , so is the mirror image of .
7. Let . (a) Find . (b) Verify . (c) Explain why this function has an inverse over all real numbers while does not. Stretch
Hint 1
Solve for , peeling the operations off in reverse: subtract, divide, then undo the cube (Appendix A.6).
Hint 2
Undoing a cube is a cube root: . For (c), compare how a horizontal line meets a cubic against how it meets a parabola.
Show full solution
(a) . (b) . (c) The cube is one-to-one (steadily rising), so it passes the horizontal line test; the square repeats every positive height and fails.
(a) Solve : subtract 1, divide by 2, take the cube root:
So .
(b) Compose: .
(c) A cubic climbs steadily and meets every horizontal line exactly once, so it is one-to-one and invertible everywhere; a parabola turns around and meets high lines twice, so has no inverse until its domain is trimmed.
8. Let with domain . (a) Find and state its domain and range. (b) Explain why removing the restriction destroys the inverse. Stretch
Hint 1
On the inside is at least 0, so a square root recovers it cleanly without a sign worry (Appendix A.6).
Hint 2
Solve for with : take the positive root, . The inverse's domain is the range of .
Show full solution
(a) , domain , range . (b) Without the restriction, inputs and share an output, so is not one-to-one.
(a) On , the quantity is at least 0, so . Solve : , so and . The outputs of are , which become the inverse's domain, and the inputs become its range. Check: and .
(b) Drop the restriction and the full parabola reappears, where and both square to ; two inputs share an output, the function fails the horizontal line test, and no inverse exists.
9. The Appendix A.6 Dig In proved that each point on reflects to on across the line . Make the argument your own: pick a one-to-one function and two of its points, find the matching points on , and confirm each original point and its twin are mirror images across by checking the midpoint lands on the line and the connecting segment has slope . Dig In
Hint 1
Reread the Appendix A.6 Dig In: your job is one concrete instance of its two-part check, the coordinate swap and the perpendicular bisector. A simple line like keeps the arithmetic clean.
Hint 2
For each point , the midpoint of and is , which sits on ; the segment's slope is .
Show full solution
Worked with and its points and : the twins and lie on ; midpoints and sit on , and each connecting segment has slope .
Take , one-to-one because it rises steadily, with inverse . Its points and swap to and , and both twins satisfy the inverse: and .
Now the mirror check on the first pair. The midpoint of and is , whose coordinates are equal, so it lies on ; the segment's slope is , perpendicular to the line. The second pair behaves the same way: midpoint on the line, slope . Both original points and their twins are reflections across , exactly as the general argument promised.
Check your own version. Whatever one-to-one function and points you chose, two tests settle it: each twin must satisfy exactly (feed in, get out), and each original-to-twin segment must have a midpoint with equal coordinates and slope . A midpoint off the line usually means a coordinate got swapped in only one point of the pair.
10. The Appendix A.6 Dig Deeper restricts sine and tangent to make them one-to-one, producing and . Make it your own: choose or , state the restricted domain of sine or tangent that makes it one-to-one, evaluate one exact value from the unit circle (Appendix A.1), and name the Section 2.3 antiderivative form your inverse function appears in. Dig Deeper
Hint 1
Reread the Appendix A.6 Dig Deeper: sine is trimmed to and tangent to , the stretches where each is one-to-one.
Hint 2
For an exact value, ask which angle in the range gives the value you want: asks for the angle whose tangent is . Match your inverse to its integral form, or .
Show full solution
Worked with : tangent is one-to-one on , so has domain all real numbers; , and is the antiderivative in .
Restrict tangent to , the single stretch where it climbs from far below to far above without repeating a value, so it is one-to-one and has an inverse, , which accepts any real number and returns an angle in .
An exact value: , because (Appendix A.1). And this is the inverse function that Section 2.3 produces when a trigonometric substitution clears a sum : . The integral asks which angle builds the accumulated amount, and answers.
Check your own version. Whichever inverse you chose, two tests confirm it: the value you plug in must lie in that inverse's domain ( for , all real numbers for ), and your exact angle must return the value when you apply the original function ( or ). If the angle you found falls outside the restricted range, you picked a co-terminal or reflected angle; bring it back into or .
An inverse function runs backward: it reads each output to the input that produced it, so , and its graph is the mirror of across the line , every point reflected to . An inverse exists exactly when is one-to-one, which the horizontal line test checks at a glance: no horizontal line may cross the graph twice. The book's key pair is and , each undoing the other (Appendix A.4), the reversal Section 2.4 integrates. And restricting a repeating function's domain rescues an inverse: trimmed sine and tangent give and , the very antiderivatives the trigonometric substitution in Section 2.3 produces. The exponent always marks an inverse here, never a reciprocal.
Appendix A.7
Area, Surface Area, and Volume
- The areas of the three flat shapes the applications start from: a rectangle , a triangle , and a circle .
- The volume and surface area of a cylinder, reading its side as a rolled-up rectangle.
- The volume and surface area of a cone, and why its slant height is not its vertical height .
- The volume and surface area of a sphere, the shape whose and the calculus in Unit 3 rebuilds.
- Reading each formula as an answer key: Section 3.3 rebuilds a sphere's from thin bands, and Section 3.4 keeps going to shapes no formula here can reach.
Every solid in Unit 3 gets measured the same way: slice it into thin pieces, size up each piece, and add. Do that to a ball and the pieces stack up to ; do it to a bowl-shaped paraboloid and they stack up to . This page holds the shapes whose answers you already carry from geometry class: a can, a party hat, a ball. When the calculus in Unit 3 grinds one out from scratch, you will have something solid to check it against. If a problem sent you here for the volume of a cylinder or the skin of a sphere, you are in the right place; this page stands on its own.
Build the intuition
Two questions get asked of a shape, and they are not the same question. How much flat paint covers its skin? That is area for a flat shape, or surface area for a solid, and it is counted in square units. How much water fills its inside? That is volume, counted in cubic units. The square-versus-cubic split is the fastest error catcher there is: if you were after a volume and your units came out square, a formula went in wrong. Every formula on this page answers one of those two questions, and each one is a shortcut for the same slice-and-add that Unit 3 does the long way.
The area of a flat region is how much surface it covers, measured in square units (square centimeters, square meters, and so on). Area answers "how much paint," and it always carries a squared unit.
The volume of a solid is how much space it fills, measured in cubic units (cubic centimeters, cubic meters, and so on). Volume answers "how much water," and it always carries a cubed unit.
The surface area of a solid is the total area of its outside skin: everything you would have to paint. Like any area it is counted in square units, even though it wraps a three-dimensional object.
Figure A.7.1 A triangle is half its bounding rectangle. The rectangle has area ; the diagonal cuts it into two matching triangles, so each one has area .
A rectangle with base and height has area . A triangle with base and height (measured straight up from that base) has area . A circle with radius has area . Reason: a rectangle is rows of unit squares; a triangle is exactly half its bounding rectangle (the Dig In makes the cut); and a circle's is the round shape's slice-and-add answer. Every area here is in square units.
A cylinder with base radius and height has volume : the circular base area times the height. Its side unrolls into a flat rectangle wide and tall, so the lateral (side) surface area is ; adding the two circular caps gives the total surface area . Reason: stack worth of identical disks of area and you have filled the volume, and the side is one rectangle in disguise.
Figure A.7.2 The side of a cylinder unrolls into a flat rectangle wide (the base circle's circumference) and tall, so the side area is . Add the two circular caps, , for the total surface area.
A cone with base radius and height has volume , exactly one third of the cylinder that would box it in. Its slant height is , the distance straight up the slanted side; the lateral (slanted) surface area is , and adding the circular base gives the total . Reason: the is the one number here you cannot read off a picture, and the disk method of Section 3.4 is what proves it.
Figure A.7.3 A cone's slant height is the distance up the slanted side, not the vertical height . Radius, height, and slant form a right triangle, so . Volume uses the height ; the slanted surface uses the slant .
A sphere with radius has volume and surface area . A sphere has no flat base and no seam, so the is its entire skin. Reason: both are slice-and-add answers, and Unit 3 earns them honestly, the volume by stacking disks (Section 3.4) and the surface by wrapping thin bands (Section 3.3).
Find the area of (a) a rectangle 8 cm by 5 cm, (b) a triangle with base 8 cm and height 5 cm, and (c) a circle of radius 6 cm. Give the exact value, then round to three decimals where it helps.
Solution. (a) Base times height:
(b) Half of base times height:
exactly half the rectangle in part (a), as Figure A.7.1 promised.
(c) The circle carries a , so the exact answer keeps it:
Leave as the exact area and report only when a decimal is asked for.
Before you go on, predict each area, then check. Find (a) the area of a triangle with base 12 m and height 5 m, and (b) the area of a circle of radius 3 m.
Show solution
(a) . (b) .
(a) . (b) ; keep as the exact value.
A soup can is a cylinder of radius 4 cm and height 10 cm; a funnel is an open cone of radius 3 cm and height 4 cm. (a) Find the can's volume and its total surface area. (b) Find the funnel's volume and its slanted (lateral) surface area.
Solution. (a) Volume is base area times height:
The total surface area is the side plus the two caps:
(b) The funnel first needs its slant height, from the right triangle of radius and height:
Volume uses the vertical height , not the slant:
An open funnel has no base, so its surface is the slanted side alone, which uses the slant :
A ball has radius 6 cm. (a) Find its volume and surface area. (b) In Unit 3 the calculus rebuilds both of these from thin slices. What numbers must those calculations land on?
Solution. (a) The volume carries a cube of the radius:
The surface area carries a square of the radius:
(b) Section 3.3 wraps the sphere in thin bands and adds their areas; that sum has to come out to for this ball. Section 3.4 stacks thin disks and adds their volumes; that sum has to come out to . The same disk method also handles shapes with no line in any formula book: a paraboloid bowl comes out to exactly (that is Section 3.4's own worked example), a number you could never read off a picture. That is what this page is for. It is the answer key, and the calculus is the method that still works when there is no answer key.
A water tank's capacity, the sheet metal in a grain silo, the glaze on a bell, the foam in a ball: every one is a volume or a surface area, and every one gets reported in the physical units this page tracks (Appendix A.8 keeps those units straight). The formulas are quick because the shapes are common, a box, a can, a cone, a ball, so someone worked out the slice-and-add once and wrote the answer down. But the moving world is full of shapes nobody pre-solved: a curved dam face, the bell of a horn, a satellite dish, the paraboloid of a headlight. None of those has a line in a formula book, and that is exactly where the slicing of Unit 3 takes over. It reproduces the six answers on this page as a warm-up, then keeps going where the memorized formulas run out.
The one flat-shape formula worth proving by hand is the triangle's . Box any triangle inside a rectangle of the same base and the same height , then drop a straight line from the triangle's top corner down to the base. That line splits the triangle into two right triangles and splits the rectangle into two smaller rectangles, one around each. Inside each small rectangle, the triangle piece and the leftover piece are matching right triangles (same base, same height, same square corner), so the triangle fills exactly half of each small rectangle. Half plus half is half overall, so . Check it against Example A.7.1: the rectangle there was , and the triangle came out , exactly half.
Now the honest limit of scissors. A cone fills exactly of the cylinder that boxes it, and a sphere fills of its cylinder, but no straight cut turns three cones into a cylinder the way two triangles rebuild a rectangle. Those fractions are curved-shape facts, and the disk method in Section 3.4 is what earns them. The flat you can see; the round and you have to integrate for.
Dig Deeper The sphere in a can▸
Archimedes found a fact so clean he asked for it on his gravestone. Take a sphere of radius and drop it into the smallest can that holds it: a cylinder of radius and height , touching the ball at the top, the bottom, and all the way around the middle. Compare the two volumes:
The ball fills exactly two thirds of the can, whatever the radius. Now compare the skins. The can's curved side has area , which is the sphere's surface area on the nose: the ball and the can's wrapper cover the same amount. Try both facts at : the sphere is in volume and in surface, the can is in volume with a side, and . Neither fact is obvious, and neither can be cut out with scissors; the sphere's and are calculus results, waiting for you in Sections 3.3 and 3.4.
Two slips cost the most points on this page. First, radius versus diameter: every formula here takes the radius , so if a problem hands you a diameter, halve it before anything else. A circle of diameter 12 has area , not , which is four times too big. Second, height versus slant in a cone: the volume uses the vertical height , while the slanted surface uses the slant . They are different numbers (the slant is always the longer one), and swapping them is the classic cone mistake. A third check catches many errors for free: volume comes out in cubic units and area in square units, so if you wanted a volume and your answer reads square centimeters, a formula went in wrong.
(a) Find the area of a circle of radius 10 m. (b) A cylinder has radius 5 cm and height 8 cm; find its volume and its total surface area. (c) A ball has radius 3 m; find its volume and its surface area, and notice what the two numbers do.
Hint
(a) Circle area is , a squared unit. (b) Volume is ; the total surface is the side plus the two caps . (c) A sphere has and ; when you write down the two answers, keep an eye on the units, not just the numbers.
Show solution
(a) . (b) ; . (c) and ; the numbers match but the units do not.
(a) .
(b) The volume:
The total surface, side plus two caps:
(c) The sphere:
Both land on , but one is cubic meters and one is square meters. For the sphere's volume and surface area happen to share the number , a coincidence of that one radius, not a rule; the units keep the two quantities apart.
Exercises A.7
1. Find the area of (a) a rectangle 9 in by 4 in, (b) a triangle with base 9 in and height 4 in, and (c) a circle of radius 7 in. Then say in one sentence how (b) compares with (a). Warm up
Hint 1
Use the three flat-shape formulas from Appendix A.7: rectangle , triangle , circle . Keep the in the circle's exact answer and round only for the decimal.
Show solution
(a) . (b) , exactly half of (a). (c) .
(a) .
(b) , half the rectangle, since the triangle shares its base and height.
(c) .
2. A cylindrical mug has radius 3 cm and height 7 cm. Find its volume, exact then rounded to three decimals. Warm up
Hint 1
A cylinder's volume is the base area times the height (Appendix A.7): . Square the radius before multiplying by the height.
Show solution
.
The unit is cubic centimeters, the tell that this is a volume and not an area.
3. A cylindrical storage tank has radius 5 m and height 12 m. Find (a) its volume and (b) its total surface area. Core
Hint 1
Two formulas from Appendix A.7: the volume is , and the total surface is the side plus the two circular caps .
Hint 2
For part (b), compute the side area and the cap area separately, then add. Keep everything as a multiple of until the last step.
Show solution
(a) . (b) .
(a) .
(b) Side plus two caps:
4. A solid cone has radius 5 cm and height 12 cm. Find (a) its slant height, (b) its volume, and (c) its total surface area (slanted side plus base). Core
Hint 1
Start with the slant height from the right triangle of radius and height (Appendix A.7): . Volume uses the vertical height ; the slanted surface uses the slant .
Hint 2
Here , a whole number. Then and the total surface is .
Show solution
(a) . (b) . (c) .
(a) .
(b) Volume uses the vertical height:
(c) Slanted side plus base:
5. A ball has radius 9 cm. Find (a) its volume and (b) its surface area. Core
Hint 1
A sphere has and (Appendix A.7). The volume cubes the radius; the surface squares it.
Hint 2
Compute for the volume and for the surface, then carry the to the end.
Show solution
(a) . (b) .
(a) .
(b) .
6. A cone and a cylinder share the same base radius 4 cm and the same height 9 cm. (a) Find each volume. (b) Confirm the cone is exactly one third of the cylinder, the fact the Dig In said calculus has to earn. Core
Hint 1
Both volumes use the same base radius and height (Appendix A.7): the cylinder is and the cone is . The only difference is the .
Hint 2
Compute once. The cone is a third of that, so divide by 3; the ratio of the two answers must reduce to .
Show solution
(a) Cylinder ; cone . (b) is one third of .
(a) The cylinder: . The cone:
(b) The ratio is : the cone fills exactly a third of the cylinder that boxes it, no scissors involved.
7. A sphere has surface area . (a) Find its radius. (b) Find its volume. Stretch
Hint 1
Run the surface-area formula backward (Appendix A.7): set and solve for . Once you have the radius, the volume is .
Hint 2
Dividing by gives , so is the positive square root. Then cube that radius for the volume.
Show full solution
(a) . (b) .
(a) Solve the surface equation:
(b) Cube the radius in the volume formula:
8. A grain silo is a cylinder of radius 3 m and height 10 m, capped by a hemisphere of radius 3 m on top. Find the total volume it holds. (A hemisphere is half a sphere.) Stretch
Hint 1
Split the silo into the two shapes you know (Appendix A.7): a cylinder plus a hemisphere. Find each volume, then add them.
Hint 2
The cylinder is . A hemisphere is half of , so its volume is . Both pieces use .
Show full solution
.
The cylinder body:
The hemisphere cap, half of a full sphere:
Add the two pieces:
9. The Appendix A.7 Dig In boxed a triangle inside a rectangle to show its area is exactly half. Make the argument yours: pick your own whole-number base and height (not the 8 by 5 the box used), draw the triangle in its bounding rectangle, compute both areas, and confirm the triangle is half. Then explain in one or two sentences why a cone's cannot be shown the same way. Dig In
Hint 1
Reread the Dig In in Appendix A.7: your job is one concrete instance of the same cut. Whole-number sides keep the areas clean.
Hint 2
Compute the rectangle and the triangle for your numbers, then confirm the second is half the first. For the explanation, think about whether any straight cut turns three cones into a cylinder.
Show full solution
Worked with base 10 and height 6: the rectangle is and the triangle is , exactly half; the cone's is a curved-shape fact no straight cut produces, so it needs the disk method of Section 3.4.
Boxing a triangle of base 10 and height 6 inside its rectangle:
The triangle is exactly half, matching the Dig In's cut. The cone is different: three cones do not reassemble into a cylinder by any straight cut, so the is not something scissors can show. It is a statement about a curved solid, and only the slicing of Section 3.4 proves it.
Check your own version. Two tests: your triangle area must be exactly your rectangle area divided by 2 (a stray factor means a base or height got miscounted), and your one or two sentences should name why scissors settle the flat triangle but not the round cone, that no straight cut turns three cones into their cylinder.
10. The Appendix A.7 Dig Deeper showed a sphere fills of its smallest cylinder and shares that cylinder's curved side area. Make it your own: pick a radius (your own number), then compute the sphere's volume and surface area and the enclosing cylinder's volume and lateral surface area, and confirm both of Archimedes' facts, exact values then rounded. Dig Deeper
Hint 1
Reread the Dig Deeper in Appendix A.7: the enclosing cylinder has the same radius as the sphere and height . Compute the sphere's and , and the cylinder's and lateral side .
Hint 2
The volume ratio should reduce to with the and the canceling, and the sphere's should match the cylinder's lateral term for term.
Show full solution
Worked with : sphere and ; cylinder with lateral side ; the volume ratio is and the two side areas match.
The sphere, radius 4:
The enclosing cylinder, radius 4 and height 8:
Both facts land: , and the sphere's skin equals the cylinder's side.
Check your own version. Two audits: the volume ratio must reduce to exactly (if it does not, make sure your cylinder's height is , not ), and the sphere's must equal the cylinder's lateral for your radius.
Two questions, two families of formula. Flat area (square units): a rectangle is , a triangle is half of that at , a circle is . Solids carry both a volume (cubic units) and a surface area (square units): a cylinder holds with side ; a cone holds with slanted side , where ; a sphere holds with skin . Guard the two common slips: use the radius, never the diameter, and keep a cone's height and slant apart. These six answers are the ones Unit 3 rebuilds by slicing, and the reason to trust that machinery is that it lands right here first: Section 3.3 rebuilds a sphere's , and Section 3.4 rebuilds the volumes and then keeps going, to shapes like a paraboloid's that no formula on this page can reach.
Appendix A.8
Basic Physics
- Reading density as amount packed into each unit of size: mass density (mass per volume) and linear density (mass per length).
- The newton, the unit of force, and how an object's weight is its mass times the pull of gravity.
- Work as force times distance for a constant force, measured in joules or foot-pounds.
- The weight density of water, about 9800 newtons per cubic meter, or about 62.4 pounds per cubic foot.
- Hooke's law , the physical fact that a spring's pull grows in proportion to how far it is stretched.
You have felt all of this physics in your hands, even without naming it. A full grocery bag pulls down on your arm, and that pull is a force. Drag a heavy box across a room and the effort you spend moving it is work, and the farther you push, the more work it takes. A wooden baseball bat is heavier in the fat barrel than in the thin handle because the wood is packed tighter there, and packed-tighter is density. A spring fights back harder the farther you stretch it. This page gives those everyday feelings their names and their units, so that when Section 3.5 and Section 3.6 add them up with calculus, the words are already yours. Nothing earlier in the book is needed here. If a problem sent you to this page in the middle of a chapter, you are in the right place.
Build the intuition
Three plain quantities carry this whole section: density, force, and work. Density answers "how much stuff is packed in here?" A cubic meter of water holds 1000 kilograms of it, so we say water has a mass density of 1000 kilograms per cubic meter. Force is a push or a pull, and the one you meet every second is your own weight, the downward pull of gravity on your mass. Work is what a force accomplishes as it moves something: lift a crate, push a cart, stretch a spring, and you have done work on it. The rest of the page is just pinning down the units so the numbers mean something. One anchor to carry: a one liter bottle of water has a mass of about 1 kilogram and weighs about 9.8 newtons, roughly the pull you feel holding it in your hand.
The density of an object is the amount of it packed into each unit of size. When the size is measured by volume, this is mass density, the mass per unit volume, in units like kilograms per cubic meter. When an object is long and thin, like a rod or a cable, it is often easier to track its mass per unit length instead, its linear density, in units like kilograms per meter (the calculus version, where the density changes along the rod, lives in Section 3.5). Whatever the flavor, density times size gives mass: mass density times volume, or linear density times length.
A force is a push or a pull. In the metric system its unit is the newton (N): one newton is the force that speeds up a 1 kilogram mass by 1 meter per second every second. The force you meet constantly is weight, the downward pull of gravity, and weight is mass times the gravitational strength :
So a 1 kilogram mass weighs about 9.8 N, and a small apple, roughly 0.1 kilogram, weighs about 1 N. In US customary units the unit of force is the pound (lb), which already measures a weight, so you rarely multiply by there.
The weight density of a substance is its weight per unit volume, the force with which each unit of volume is pulled down by gravity. Because weight is mass times , weight density is mass density times . For water, the mass density 1000 kilograms per cubic meter gives a weight density of
about 9800 newtons per cubic meter. The same water, measured the US customary way, weighs about , that is 62.4 pounds per cubic foot. These are the two numbers the pumping problems in Section 3.6 lean on.
When a constant force moves an object a distance along the direction of the force, the work done is
In metric units, a force in newtons times a distance in meters gives work in newton-meters, called joules (J): one joule is one newton acting through one meter. In US customary units, a force in pounds times a distance in feet gives foot-pounds (ft-lb). Reason: work measures force paid out over distance, so both the size of the push and the length of the trip count. Fine print: this clean product holds only while the force stays constant. The moment the force changes as the object moves (a stiffening spring, a draining tank), the work becomes an integral instead, which is exactly the story of Section 3.6.
A spring stretched (or compressed) a distance from its natural length pulls back with a force proportional to :
This is Hooke's law, and the constant is the spring's stiffness, called the spring constant, measured in newtons per meter (or pounds per foot). Reason: near its natural length a spring resists twice as hard when pulled twice as far, which is what "proportional" means. One measurement fixes : if a known force holds the spring stretched a known distance , then . Section 3.6 spends this fact to compute the work a spring stores.
| Quantity | Metric (SI) | US customary |
|---|---|---|
| Force | newton (N) | pound (lb) |
| Work | joule (J), one newton-meter | foot-pound (ft-lb) |
| Weight density of water | about 9800 newtons per cubic meter | about 62.4 pounds per cubic foot |
Take . (a) A full grocery bag has a mass of 6 kilograms. Find its weight in newtons. (b) A small apple has a mass of about 0.1 kilogram. Find its weight, and compare it with one newton.
Solution. (a) Weight is mass times :
(b) The apple's weight is N, which rounds to about 1 N. That is the handy mental picture for a newton: it is roughly the weight of a small apple. Notice that kilograms measured the mass and newtons measured the weight; the factor is what carried one to the other.
Before you compute, predict which is heavier and by how much. Take . Find the weight, in newtons, of (a) a 2 kilogram textbook and (b) a 10 kilogram suitcase.
Show solution
(a) N. (b) N.
Each weight is mass times : the textbook is N and the suitcase is N. The suitcase is 5 times the mass, so it weighs 5 times as much, which the numbers confirm.
(a) A constant force of 12 N pushes a cart 3 meters along a level floor. Find the work done, in joules. (b) A worker lifts a 25 pound box straight up 4 feet. Find the work done, in foot-pounds. Figure A.8.1 shows part (a) as an area.
Solution. (a) The force is constant, so work is force times distance:
(b) Now in US customary units, pounds times feet:
Both answers carry their units, because a bare number of newtons or pounds is a force that forgot to travel. The joule and the foot-pound each bundle a force with the distance it acted through.
Figure A.8.1 For a constant force, work is the area of a rectangle. A steady N push over m does J of work. When the force changes as the object moves, this rectangle becomes the area under a curve, the integral of Section 3.6.
(a) A tank holds water filling a volume of 2 cubic meters. Using the weight density 9800 newtons per cubic meter, find the weight of the water. (b) A uniform steel rod has a linear density of 4 kilograms per meter and is 3 meters long. Find its mass.
Solution. (a) Weight is weight density times volume:
(b) The rod's density does not change, so mass is linear density times length:
Each answer is "density times size," the pattern from Definition A.8.1: weight density times volume for the water, linear density times length for the rod. This works only because each object is uniform. When the density varies, the product turns into the integral of Section 3.5. Rusty on a tank's volume? Appendix A.7 has the shapes.
A force of 20 N holds a spring stretched 0.1 meter beyond its natural length. (a) Find the spring constant . (b) What force is needed to hold the spring stretched 0.25 meter?
Solution. (a) Hooke's law says , and one measurement pins down :
(b) With , the force at any stretch is , so at meter,
Stretching the spring two and a half times as far took two and a half times the force, exactly the proportionality Hooke's law promises.
These four words are the working vocabulary of anything that gets built or moved. A crane is rated by the force it can lift, so a hoist raising a 500 kilogram load is fighting a weight of N, and its motor is billed for the work that force does over the height of the lift. A dam is designed around the weight density of water, because the deeper water pushes harder on the wall, and that growing push is what the concrete has to hold back. The springs in a car's suspension follow Hooke's law, stiff enough to carry the body yet soft enough to swallow a bump. Every one of these starts as a constant-force product like the ones on this page, and then Section 3.6 lets the force change as the object moves, turning force times distance into an integral. The units never change; only the arithmetic gets richer.
Where does the number 9800 come from? It is not a fact to memorize; it is one multiplication you can rebuild any time. Weight is mass times , so the weight of each cubic meter of water is the mass of that cubic meter times . A cubic meter of water has a mass of 1000 kilograms (a metric tonne), so its weight is
and since that is the weight of one cubic meter, the weight density is 9800 newtons per cubic meter. The rule behind it is general: for any substance,
You can even feel it a slice at a time. One kilogram of water weighs N, and a full cubic meter is 1000 of those kilograms, so it weighs 1000 times as much, which is the same 9800 N. The US customary figure, 62.4 pounds per cubic foot, needs no such multiplication, because the pound is already a weight; it is simply how much a cubic foot of water weighs. Exercise 9 rebuilds the metric weight density for a liquid of your own choosing.
Dig Deeper Mass and weight are not the same thing▸
Everyday speech blurs two different quantities, and physics keeps them apart. Mass is the amount of matter in an object, measured in kilograms; it does not change no matter where the object goes. Weight is the force gravity exerts on that mass, measured in newtons, and it does change, because it depends on the local gravitational strength .
Carry a 6 kilogram rock to the Moon and it is still 6 kilograms of rock. But the Moon pulls with only about , against Earth's , so the rock's weight changes:
The Moon weight is of the Earth weight, close to one sixth, which is why astronauts bounce. The mass never budged; only the force did. This is also why the pound can be confusing: it names a weight, so an object's "weight in pounds" is really a force, while its mass in kilograms is a separate, unchanging number. Exercise 10 sends an object of your choosing to another world and compares.
The most common slip on this page is treating mass and weight as one thing. A 2 kilogram melon does not "weigh 2 kilograms"; it has a mass of 2 kilograms and a weight of about 19.6 newtons, and the factor is what separates them. Kilograms measure mass, while newtons and pounds measure force, so a weight always carries a force unit. The second slip is dropping the unit off a work answer: force times distance is joules or foot-pounds, never bare newtons, and an answer of "36" with no unit hides whether you meant a force or the work it did. The third is reaching for when the force is not constant. That product is only the flat, unchanging-force case; a spring or a draining tank makes the force move, and then the honest total is the integral of Section 3.6, not a single multiplication.
(a) Take and find the weight, in newtons, of a 15 kilogram object. (b) A constant force of 8 N pushes a box 5 meters; find the work done. (c) A force of 40 N holds a spring stretched 0.2 meter; find its spring constant.
Hint
(a) Weight is mass times (Definition A.8.2). (b) Constant force means . (c) Hooke's law gives , the measured force over the measured stretch.
Show solution
(a) N. (b) J. (c) N/m.
(a) N.
(b) The force is constant, so J.
(c) From one measurement, N/m: the spring pulls back with 200 newtons for every meter it is stretched.
Exercises A.8
1. Take . Find the weight, in newtons, of an 8 kilogram object. Warm up
Hint 1
Weight is mass times the gravitational strength (Definition A.8.2 in Appendix A.8). The answer is a force, so its unit is newtons.
Show solution
N.
The mass was 8 kilograms; the weight it produces on Earth is 78.4 newtons.
2. A constant force of 15 N pushes a cart 4 meters along a level floor. Find the work done. Warm up
Hint 1
The force is constant, so work is force times distance, (Appendix A.8). Newtons times meters gives joules.
Show solution
J.
The push and the distance both count, so the work is their product, in joules.
3. A uniform copper rod has a linear density of 5 kilograms per meter and is 3 meters long. Find its mass, and name the unit of your answer. Core
Hint 1
The rod is uniform, so its density does not change. Then mass is density times size (Definition A.8.1 in Appendix A.8), here linear density times length.
Hint 2
Multiply the kilograms per meter by the number of meters; the "per meter" and the meters cancel, leaving kilograms.
Show solution
kilograms.
The unit check does the bookkeeping: kilograms per meter, times meters, leaves kilograms. Density times size gives mass because the rod is uniform; a rod whose density changed would need the integral of Section 3.5.
4. A container holds water filling a volume of 0.5 cubic meter. Using the weight density 9800 newtons per cubic meter, find the weight of the water. Core
Hint 1
Weight density is weight per unit volume (Definition A.8.3 in Appendix A.8), so weight is weight density times volume.
Hint 2
Multiply 9800 newtons per cubic meter by 0.5 cubic meter; the cubic meters cancel, leaving newtons.
Show solution
N.
Half a cubic meter of water weighs 4900 newtons, half of the 9800 newtons a full cubic meter would weigh.
5. A worker lifts a 25 pound toolbox straight up 6 feet at a steady speed. Find the work done, in foot-pounds. Core
Hint 1
The toolbox weighs a constant 25 pounds and rises the same 6 feet the whole way, so this is the constant-force case (Appendix A.8).
Hint 2
In US customary units, pounds times feet gives foot-pounds. No factor of is needed, because pounds already measure a weight.
Show solution
ft-lb.
The weight was constant and the lift was a fixed height, so the work is a plain product, in foot-pounds.
6. A force of 30 N holds a spring stretched 0.15 meter beyond its natural length. (a) Find the spring constant . (b) What force holds the spring stretched 0.25 meter? Core
Hint 1
Hooke's law is , and one measured force-and-stretch pair gives (Appendix A.8).
Hint 2
Once you have , the force at any stretch is ; substitute .
Show solution
(a) N/m. (b) N.
(a)
(b) With , at meter the force is N. The stretch grew and the force grew right along with it, in proportion.
7. A sealed container holds 0.4 cubic meter of water (weight density 9800 newtons per cubic meter). (a) Find the weight of the water. (b) A crane lifts the full container straight up 5 meters at a steady speed. Treating the container itself as weightless, find the work done against the water's weight, and explain why no integral is needed. Stretch
Hint 1
Part (a) is weight density times volume (Definition A.8.3 in Appendix A.8). Part (b) is a lift of a constant weight over a fixed height.
Hint 2
For (b), the whole sealed container rises the same 5 meters, so every part of the water travels the same distance and the force is constant. That is the case, unlike a pumping problem where each layer rises a different distance.
Show full solution
(a) N. (b) J; no integral is needed because the whole load rises the same distance, so the force is constant.
(a) The water's weight is weight density times volume:
(b) The sealed container rises as one piece, so all 3920 newtons travel the same 5 meters, a constant force over a fixed distance:
Because every drop moves the same distance, the work is a single product. In a pumping problem the water leaves the top a layer at a time, each layer rising a different distance, and only then does the total become the integral of Section 3.6.
8. A spring stretches 0.2 meter beyond its natural length under a force of 50 N. (a) Find the spring constant. (b) How far does the spring stretch under a force of 125 N? (c) How far does it stretch under 100 N, and what does comparing (a) and (c) show about doubling the force? Stretch
Hint 1
Hooke's law runs both directions (Appendix A.8): a measured force-and-stretch pair gives , and then any force gives its stretch by .
Hint 2
For (b) and (c), solve with the from part (a). Compare the stretch under 50 N with the stretch under 100 N.
Show full solution
(a) N/m. (b) meter. (c) meter; doubling the force from 50 N to 100 N doubles the stretch from 0.2 m to 0.4 m.
(a)
(b) Solving for :
(c) Under 100 N,
The original 50 N gave a stretch of 0.2 m, and doubling the force to 100 N gave 0.4 m, exactly double. Force and stretch move in lockstep, which is Hooke's law in action.
9. The Appendix A.8 Dig In rebuilt water's weight density as (mass density), getting 9800 newtons per cubic meter. Make the calculation your own: pick a liquid other than pure water, look up or choose a reasonable mass density for it in kilograms per cubic meter, and compute its weight density using . State the weight of one cubic meter of your liquid, with units, and say whether it is heavier or lighter than water. Dig In
Hint 1
Reread the Dig In in Appendix A.8: weight density is mass density times , and the weight of one cubic meter is just that weight density (per cubic meter) times one cubic meter.
Hint 2
Multiply your chosen mass density by 9.8. Compare the result with water's 9800 to decide heavier or lighter.
Show full solution
Worked with seawater at a mass density of 1025 kilograms per cubic meter: weight density newtons per cubic meter, so one cubic meter weighs 10045 N, heavier than water.
Take seawater, whose mass density is about 1025 kilograms per cubic meter (the dissolved salt adds mass). Its weight density is
so one cubic meter of seawater weighs 10045 N. That is more than water's 9800 N, so seawater is heavier, which is part of why objects float a little higher in the ocean than in a freshwater lake.
Check your own version. Two audits settle it. Your weight density must equal your mass density times 9.8 exactly, with units of newtons per cubic meter, and it must exceed 9800 if and only if your liquid's mass density exceeds 1000. If your number is smaller than 9800 but you called the liquid heavier than water, recheck which mass density you started from.
10. The Appendix A.8 Dig Deeper kept mass and weight apart: mass does not change from place to place, but weight does, because weight is mass times the local . Build your own comparison. Pick an object and a mass for it in kilograms, then find its weight on Earth () and on the Moon (). State both weights with units, and state the object's mass in each place. Dig Deeper
Hint 1
Reread the Dig Deeper in Appendix A.8: weight is mass times , so the same mass gives two different weights when differs. The mass itself is unchanged.
Hint 2
Compute (mass) for the Earth weight and (mass) for the Moon weight. The mass stays exactly what you chose, in both places.
Show full solution
Worked with a 5 kilogram object: Earth weight N, Moon weight N, and the mass is 5 kilograms in both places.
Take a 5 kilogram object. On Earth its weight is
and on the Moon, where gravity is weaker, its weight is
The mass is 5 kilograms in both places, because mass is the amount of matter and does not depend on where the object sits. Only the weight fell, from 49 N to 8 N, when the pull of gravity weakened.
Check your own version. Whatever mass you chose, three things must hold: your two weights must be (mass) and (mass) in newtons, the Moon weight must be about one sixth of the Earth weight (since ), and the mass you report must be identical in both places. If the mass changed between the two, you multiplied where you should not have; only the weight uses .
Density is amount per unit of size, and "density times size" gives mass: mass density times volume, or linear density times length, whenever the object is uniform. Force is a push or pull, measured in newtons (or pounds), and an object's weight is its mass times gravity, with , so mass and weight are different quantities in different units. Work is force times distance for a constant force, , measured in joules (newton-meters) or foot-pounds; when the force changes as the object moves, that product becomes an integral. The weight density of water is about 9800 newtons per cubic meter, or 62.4 pounds per cubic foot, and it is just water's mass density times . And Hooke's law says a spring's pull grows in proportion to its stretch, with the spring constant read off a single measurement. Section 3.5 and Section 3.6 take each of these and let the density, the force, or the distance vary, which is where the integral earns its keep.
Appendix A.9
Basic Probability
- Listing a sample space and reading the probability of an event as the share of equally likely outcomes it fills, .
- Using the rules every probability obeys: it lands from 0 to 1, the whole sample space has probability 1, and the complement rule .
- Reading a random variable's distribution and computing its expected value as the weighted sum .
- Seeing why this discrete counting is the bridge Section 3.7 crosses, where the sum turns into an integral of a density.
Roll a fair die. Before it lands you cannot name the face, but you can name every face it could show and say each is equally likely. That short list, and the even split across it, is all of basic probability in one breath: probability is the fraction of the possibilities that count as a win. Add a payout to each face and you can also say what you expect to make per roll in the long run, a weighted average called the expected value. This page is the two-minute refresher on both ideas from your Probability and Matrices course, and it is the launch pad for Section 3.7, where the outcomes stop being a short list and spread into a smooth range, and the sum you are about to meet becomes an integral.
Build the intuition
Start with a list of every outcome an experiment can produce, all equally likely: the six faces of a die, the two sides of a coin, the eight numbers on a spinner. That list is the sample space. An event is just some of those outcomes gathered together, the ones you care about, like "the roll is even." Its probability is the plainest fraction there is: how many outcomes are in the event, over how many are in the whole list. Figure A.9.1 reads it straight off a die: three of the six faces are even, so the probability of an even roll is .
Figure A.9.1 The six faces of a die are equally likely. The three even faces are shaded, so : the share of the sample space the event fills.
The sample space of an experiment is the set of all outcomes it can produce; each member is one outcome. An event is any collection of outcomes, that is, a subset of . The count is how many outcomes are in the whole space, and is how many are in the event.
When the outcomes in are equally likely, the probability of an event is
the number of winning outcomes over the total. Every probability lands from 0 to 1: the impossible event (no outcomes) has probability 0, the certain event has probability 1, and the complement rule reads . Reason: each of the equally likely outcomes carries the same share of the whole, and collects of those shares; since and "not " together fill , their probabilities add to 1.
A fair die is rolled once, so and . Find (a) the probability the roll is even, and (b) the probability the roll is 5 or more.
Solution. (a) The event "even" is , so :
(b) The event "5 or more" is , so :
Both answers are fractions of the same six equally likely faces. Count the winners, count the total, divide.
Before you read on, predict: a spinner has eight equal wedges numbered to . (a) What is the probability it lands on a multiple of 3? (b) What is the probability it lands on a number greater than 5?
Show solution
(a) . (b) .
(a) The multiples of 3 in the list are , two of the eight wedges, so .
(b) Greater than 5 means , three of the eight, so .
Now attach a number to each outcome, not just a label. Call one roll of the die ; it can come out 1, 2, 3, 4, 5, or 6, each with probability . A quantity like , whose value depends on the outcome of a random experiment, is a random variable, and the list of its values with their probabilities is its distribution. The one number that sums up a distribution is its expected value: what each value is worth, weighted by how often you expect it.
A random variable is a quantity whose value is set by the outcome of a random experiment. Its probability distribution lists the possible values alongside their probabilities , which are each from 0 to 1 and add to 1 across the whole list.
The expected value of a random variable , written and also called its mean, is the weighted sum
each value multiplied by its own probability, then added. Reason: over many trials, the value turns up about a fraction of the time, so the long-run average per trial is exactly that weighted sum. Read it as the balance point of the distribution, the place it would tip level, not a value any single trial has to hit.
Let be the number showing on one roll of a fair die. Find .
Solution. Each of the six values carries probability , so the weighted sum is
When every probability is the same, the weighted sum collapses to the plain average of the values, here . Notice 3.5 is not a face the die can show: the expected value is a long-run average, not a forecast of any one roll.
You roll one fair die. If it shows a 6 you win 4 dollars; on any other face you lose 1 dollar. Let be your net result. Find and say what it means.
Solution. The value 4 dollars happens on one face and the value dollar on the other five, so the distribution is the table below.
| Roll | Net result | Probability |
|---|---|---|
| shows a 6 | ||
| 1 through 5 |
The probabilities add to 1, a healthy sign. Now weight each value by its probability and add:
The expected value is exactly of a dollar, about dollars. Over many plays you would average a loss near 17 cents a roll, so the game quietly favors the house.
The moving world rarely hands you a short list of outcomes. A bus arrives somewhere from 0 to 10 minutes from now; a battery lasts some number of hours you cannot pin to a face. There are infinitely many possibilities and each single one has probability 0, so has nothing to count. Section 3.7 keeps both ideas from this page and swaps only the machine. A density curve replaces the list, "total probability is 1" becomes "the total area under the curve is 1," and the probability of an interval is the area over it, . The weighted sum you just built, , becomes the integral , the same balance-point mean that returns as a function's average value in Section 2.1. Summing dots and integrating a curve are one habit at two resolutions.
Figure A.9.2 The number of heads in three coin flips takes the values 0, 1, 2, 3 with probabilities , which sum to 1. The distribution balances at its expected value, . In Section 3.7 these bars smooth into a curve and the sum becomes an area.
Two facts about every distribution are worth pinning down, because both cross the bridge to Section 3.7 whole. First, the probabilities of all the outcomes add to 1. For the heads count in Figure A.9.2, . That is just again: the values are the entire sample space, so their shares must fill it. In the continuous world this same total becomes , total area 1.
Second, the expected value is a weighted average, and equal weights make it the ordinary average. For the fair die,
which is exactly , the plain mean of the faces. Pull the weights apart, as in the game of Example A.9.3, and the weighted sum starts to lean toward the heavier values. That leaning weighted sum is the discrete ancestor of .
Dig Deeper Watch the sum become an integral▸
Here is the bridge built one plank at a time. Chop the interval from 0 to 1 into equally likely outcomes, the points , each with probability . The expected value is the weighted sum
using the sum of the first whole numbers, . Feed it larger and watch: , , . The values close in on .
That limit is no accident. The weighted sum is exactly a right Riemann sum for , the kind you first met in Section 1.1: value times width, added across the interval. As it converges to the definite integral of Section 1.3,
which is the mean of the continuous uniform distribution on . The discrete weighted sum has become the integral , right in front of you. That single move, letting the outcomes crowd together until the sum turns into an area, is the whole of what Section 3.7 does.
The most common slip is averaging the values and forgetting the weights: multiplies each value by its own probability first, so it is neither the plain average of the values nor the average of the probabilities. In the die game of Example A.9.3, the plain average of and is , but the expected value is , because the losing outcome carries five times the weight. A second slip is expecting the answer to be an outcome that can really occur: a fair die's expected value is , a face no die owns, and that is fine, because expected value is a long-run average, not a single prediction. And if a probability ever comes out above 1 or below 0, stop: recount the sample space or check that you divided by , since every probability must sit from 0 to 1.
(a) A drawer holds 12 socks: 5 black, 4 white, and 3 gray. You grab one without looking. Find and . (b) A spinner pays 5 dollars with probability , 2 dollars with probability , and 0 dollars with probability . Find the expected value of one spin.
Hint
(a) Count winners over total for ; for the complement rule is faster than adding black and white. (b) Multiply each dollar value by its own probability, then add, following the expected-value rule above.
Show solution
(a) ; . (b) dollars.
(a) Five of the twelve socks are black, so . Three are gray, so by the complement rule .
(b) Weight each payout by its probability and add:
The probabilities add to 1, a quick check that the distribution is complete before you trust the mean.
Exercises A.9
1. A fair die is rolled once. (a) Find the probability the roll is odd. (b) Find the probability the roll is greater than 4. Warm up
Hint 1
Write out , list the winning outcomes for each event, and use with .
Show solution
(a) . (b) .
(a) The odd faces are , three of six: .
(b) Greater than 4 means , two of six: .
2. A spinner has 5 equal wedges: 2 red, 2 blue, and 1 green. Find (a) and (b) . Warm up
Hint 1
The sample space has equally likely wedges. Count the red wedges for (a); for (b), the complement rule is the quick road.
Show solution
(a) . (b) .
(a) Two of the five wedges are red: .
(b) One wedge is green, so . Adding the two red and two blue wedges gives the same .
3. A fair eight-sided die shows the numbers 1 through 8, each equally likely. Let be the number rolled. (a) Confirm the eight probabilities add to 1. (b) Find . Core
Hint 1
Each face has probability . For (b), use the expected-value rule ; with equal weights it becomes the plain average of the eight values.
Hint 2
Since every probability is the same , add the values 1 through 8 and divide by 8; that shortcut is the plain average.
Show solution
(a) Eight copies of add to 1. (b) .
(a) Adding eight times gives , so the distribution is complete.
(b) With equal weights the expected value is the plain average:
4. A game costs 2 dollars to play. You draw one ball from an urn holding 10 balls: one labeled 10 dollars, three labeled 2 dollars, and six labeled 0 dollars, and you win the labeled amount. (a) Find the expected prize. (b) Find your expected net gain per play, and say whether the game is worth playing. Core
Hint 1
The prize is a random variable with three values. Build its distribution: probability for 10 dollars, for 2 dollars, for 0 dollars, then apply .
Hint 2
For (b), the net gain is the expected prize minus the 2 dollar cost. A negative expected net means the game loses money on average.
Show full solution
(a) Expected prize dollars. (b) Expected net dollars per play, so it is not worth playing.
(a) Weight each prize by its probability:
(b) Subtract the cost: dollars. On average each play loses 40 cents, so a player who repeats it long enough goes steadily into the red.
5. A fair coin is flipped three times, and every sequence of heads and tails is equally likely. (a) How many outcomes are in the sample space? (b) Find the probability of getting at least one head. Core
Hint 1
Each flip has 2 results, so three flips make equally likely sequences. For (b), "at least one head" is the complement of a single outcome.
Hint 2
The only way to avoid a head is all tails, one sequence. Use .
Show solution
(a) 8 outcomes. (b) .
(a) Three flips give equally likely sequences.
(b) Exactly one sequence has no heads (all tails), so by the complement rule .
6. Flip a fair coin three times and let be the number of heads. Its distribution is , , , . Find , the mean number of heads. Core
Hint 1
Apply directly: multiply each head count by its probability, then add. This is the distribution drawn in Figure A.9.2.
Hint 2
The weights are not equal, so you cannot just average 0, 1, 2, 3. Compute over the common denominator 8.
Show solution
heads.
Weight each value by its probability:
The symmetry of the distribution puts the balance point dead center, matching the fulcrum in Figure A.9.2.
7. Two fair dice are rolled and the numbers are added, so the 36 ordered pairs are equally likely. Find (a) the probability the sum is 7, and (b) the probability the sum is 10 or more. Stretch
Hint 1
The sample space is all ordered pairs with each of and from 1 to 6, so . List the pairs whose sum matches each event and count carefully.
Hint 2
For a sum of 7, the pairs run up to . For 10 or more, collect the sums 10, 11, and 12 separately, then add their counts before dividing by 36.
Show full solution
(a) . (b) .
(a) The sum is 7 for , six pairs: .
(b) Sum 10 comes from ; sum 11 from ; sum 12 from . That is pairs, so .
8. A raffle sells 100 tickets at 1 dollar each. One ticket wins 50 dollars, one wins 20 dollars, and the rest win nothing. (a) Find the expected prize for a single ticket. (b) Find the expected net gain for a ticket you buy, and say whether it is favorable. Stretch
Hint 1
Your ticket is equally likely to be any of the 100, so the prize has probability for 50 dollars, for 20 dollars, and for 0 dollars. Apply the expected-value rule.
Hint 2
For (b), subtract the 1 dollar ticket price from the expected prize. If the expected net is negative, the raffle is not favorable, though people still play for the small chance and the cause.
Show full solution
(a) Expected prize dollars. (b) Expected net dollars, so it is not favorable.
(a) Weight each prize by where it applies:
(b) Subtract the price: dollars per ticket. The expected net is negative, so on pure expected value it is not favorable; a buyer is paying 30 cents on average for the thrill and the cause.
9. The Dig In showed that total probability is always 1 and that equal weights turn the expected value into the plain average. Make it yours: invent a probability distribution on at least four values that are NOT equally likely (a small game, a weighted spinner, anything). Confirm your probabilities add to 1, compute the expected value, then recompute it with the same values made equally likely and confirm you get their plain average instead. Dig In
Hint 1
Reread the Dig In box: you are building one concrete instance of "total probability 1" and "equal weights give the plain mean." Pick probabilities with a common denominator like 8 so they add to 1 cleanly.
Hint 2
Run once with your uneven weights, then again with every probability set to (for four values). The second answer should equal the values added up and divided by 4.
Show full solution
Worked with values and probabilities : the probabilities sum to 1, the weighted mean is , and the equal-weight mean is , the plain average.
The probabilities add to 1: .
The weighted expected value:
With every probability reset to , the mean becomes the plain average . The uneven weights pulled the mean down toward 1, the value carrying the most probability.
Check your own version. Three tests must pass: your probabilities must add to exactly 1, your weighted mean must land between your smallest and largest value, and your equal-weight recomputation must equal the plain average of the values. If the weighted and equal-weight means come out identical, your weights were secretly all equal; make one value likelier than another and they must differ.
10. The Dig Deeper watched a discrete weighted sum turn into an integral. Retrace it: put equally likely outcomes at the points , each with probability . (a) Show the expected value is , using . (b) Evaluate it for , , and . (c) In one sentence, say what value it approaches and how that previews the continuous mean in Section 3.7. Dig Deeper
Hint 1
Reread the Dig Deeper box: the weighted sum is . Factor the constant out front before you apply the sum formula.
Hint 2
After factoring, ; cancel one . For (c), take the limit of your as grows and compare it with from the Dig Deeper.
Show full solution
(a) . (b) , , . (c) It approaches , the continuous uniform mean, so the discrete weighted sum becomes the integral .
(a) Factor the constant and apply the sum formula:
(b) Substituting: , , .
(c) The values close in on . Since is a right Riemann sum for , the weighted sum becomes the integral exactly as it does in Section 3.7.
Check your own version. If you regrouped the points differently or used a different clean density, confirm two things: your weighted sum must match a Riemann sum for the matching integral (value times width, added up), and the limit you claim must equal that integral's exact value. A quick numerical test: your formula at should sit within a hundredth of the limit.
A sample space lists every equally likely outcome, an event is a subset of it, and its probability is the share it fills, , always a number from 0 to 1 with . A random variable pins a number to each outcome, and its expected value is the weighted average: each value times its own probability, added, the balance point of the distribution and its long-run mean. Equal weights collapse that sum to the plain average, uneven weights lean it toward the likelier values, and the answer need not be an outcome the variable can hit. Every idea here has a continuous twin one bridge away: in Section 3.7 the list becomes a density curve, total probability 1 becomes total area 1, and the weighted sum becomes the integral . You summed dots here; there you integrate a curve, and it is the same habit at a finer resolution.
Appendix A.10
Tables of Derivatives and Integrals
- Reading a derivative fact backward into the antiderivative it names, since is the same statement as .
- Looking up the standard antiderivatives this course uses: powers, , , , , the basic trig functions, and the and forms.
- Stretching the short table to any sum of its rows with the constant-multiple and sum rules.
- Checking any antiderivative you write, from the table or not, by differentiating it back to the integrand, and never dropping the .
A derivative fact and an integral fact are one sentence read from opposite ends. The Derivatives volume built a table saying "the rate of is ." Turn that same sentence around and it says "a function whose rate is is ," which is exactly what means. Nothing new has to be discovered; every antiderivative in this book is a derivative fact you already own, spoken backward. This page is the reference card that lays the two directions side by side, one row per fact, so you can read a row left to right when you differentiate and right to left when you integrate. If a problem sent you here mid-course, you are in the right place; this section stands on its own.
Build the intuition
Differentiating takes a function down to its rate. Integrating climbs back from the rate to the function. Because the two moves undo each other, one table can carry both: write the function and its rate next to each other, and you have written a derivative fact and an integral fact at the same time. The power rule is the clean example. In the Derivatives volume it read . Solve that for the antiderivative once and it becomes the row this book leans on, : raise the exponent by one, divide by the new exponent, done.
The return trip adds exactly one thing the forward trip did not have: the . Differentiating erases a constant, since the rate of a flat piece is zero, so integrating cannot know which constant was there and reports the whole family at once. Two functions with the same rate differ only by a constant, the family fact from Section 1.2, and the is how the table remembers it. Figure A.10.1 draws the two directions as one road.
Figure A.10.1 One road, two directions. Differentiating drops the total to its rate ; integrating climbs back from to , and the only thing the climb adds is the .
A table of integrals is a reference list of standard antiderivatives. Each entry is a derivative fact read from the other end: because , the reverse statement holds, and the table simply collects the ones worth keeping at hand. A table of derivatives lists the same facts in the forward direction, . The one table in this section carries both, since each row is a single fact you can read either way.
If on an interval, then . Reason: an antiderivative is a derivative fact named from the far end, so knowing the rate of is is the same as knowing that is an antiderivative of . The is not decoration: two functions with the same rate differ only by a constant (Section 1.2), so the indefinite integral has to report every one of them at once. This single rule is why the whole Derivatives volume doubles as an integral table.
| Antiderivative | Its rate | Integral fact |
|---|---|---|
Three notes keep the rows honest. In the power row, is any constant except ; the excluded case is the very next row, , which fills the one gap the power rule leaves. The absolute value in is required, not optional: is defined on both sides of zero, and is its antiderivative on each side (Section 2.4 and Appendix A.4 say why). In the last two rows, is a constant, and in the row above, is any nonzero constant; the and forms are the ones a trigonometric substitution produces (Section 2.3), and Appendix A.6 is the refresher on and themselves.
A constant slides across the integral sign, and a sum splits into separate integrals: and . Reason: differentiation obeys these same two rules, so reversing it does too. Together they turn the short table into a long one: any sum of constant multiples of table rows integrates term by term. What you may never do is pull a variable across the sign, only a constant.
Use Table A.10.1 only. (a) The tan row says ; read it backward to write the matching integral fact. (b) Find . (c) Name the derivative fact that reverses.
Solution. (a) Reading a row right to left turns a rate into its antiderivative and adds the :
(b) Hunt the row whose rate column is . That is the row, and reading it across gives
(c) Read the integral fact forward instead of backward: the antiderivative has rate , so the derivative fact behind it is
Each answer was one row of the table, read in one direction or the other.
Before you go on, predict, then check. (a) The row says ; write the integral fact it becomes. (b) The integral fact reverses which derivative fact?
Show solution
(a) . (b) .
(a) The row reads the same both ways, so backward it says , the added on the return. (b) The antiderivative has rate , which is the derivative fact the integral fact turns around.
Find .
Solution. The sum rule splits it into three integrals, and the constant-multiple rule pulls each number out front:
Now look up each piece. The power row with gives ; the row gives ; the row gives . Reassemble:
One constant of integration covers the whole expression; there is no need for a separate on each term. Differentiate the result and every term lands back on the integrand, the check the Dig In makes a habit.
The last two rows carry a constant . Match each integral to its row, read off , and write the antiderivative. (a) . (b) .
Solution. (a) This matches with , so :
(b) This matches with , so , and this row carries a out front:
Confirm (b) by differentiating: , back on the integrand. The only care point is reading from , not grabbing the 9 or the 4 raw.
Every total this book builds ends at this table. The odometer is the first case: it turns the speedometer's rate into a distance by finding the speed's antiderivative, and that antiderivative lives in one of these rows. Unit 3 does the same move over and over, on a spring's work, a tank's volume, a wait time's probability: slice the quantity, read off an integrand, then look its antiderivative up right here. The table stays short on purpose. A handful of rows, plus the two rules that pass constants and sums through, plus the substitution of Section 2.2, reaches every function this course meets. A pharmacologist tracking a dose that decays like and an engineer sizing a curved tank reach for the same short list you are reading now.
The table is only worth trusting if every row survives one test: differentiate the antiderivative and you must land exactly on the rate beside it. Most rows pass at a glance. The two parameter rows earn a full check, since they hide a chain-rule step, and running them here is the same move you will run on your own answers.
The row. Differentiate , where and the inside function contributes :
using for . It lands on the rate beside it, so the row is honest.
The row. Differentiate , where and again contributes :
Again it matches. Differentiating any antiderivative in the table returns its integrand, and that is precisely the check to run on a homework answer: if the derivative of your answer is not the thing you started with, the answer is wrong, table or no table.
Dig Deeper The One Table That Refuses to Be Finished▸
The two columns are not equal partners, and the difference is deep. Differentiating is an algorithm that never fails. Hand it any function built from powers, roots, exponentials, logarithms, and trig functions, and the product, quotient, and chain rules grind out a derivative built from those same pieces. The table of derivatives is complete: forward, you can always get an answer, and the answer never leaves the family of functions you started in.
Integrating has no such guarantee. Some functions built from the very same pieces have no elementary antiderivative, meaning no antiderivative you can write with a finite combination of powers, roots, exponentials, logarithms, and trig functions. The famous one is , the shape behind the bell curve. It has an antiderivative in the honest sense, since it is continuous, so the accumulation function from Section 1.4 is a perfectly good function. That function just cannot be written with any row of this table or any finite mix of them. A computer will hand you a name for it, but the name is a brand-new function defined by the integral, not built from the old ones.
This is why Section 2.1 has to bracket between bounds instead of evaluating it exactly, and why statisticians read normal-curve areas from a table of values rather than a closed formula. The table of integrals is a genuine reference precisely because the reverse trip is not guaranteed to succeed. The table of derivatives never needed to be a reference at all, because going forward always works.
Three traps live on this page. First, the is part of the answer, not a flourish: an indefinite integral names a whole family, so is missing its family and is wrong the same way a missing unit is wrong. Second, match a row exactly. The table has and , but there is no plain row, so do not invent one; that integral needs a trick this table does not carry. Third, the return trip is not always a bare copy. Reading backward is a clean copy, but and each need a constant fix on the way back (a and a ), the move Section 2.2 formalizes. Differentiating your answer catches a wrong constant or a wrong sign at once, but it will never catch a dropped , since a constant differentiates to zero, so the needs its own discipline: write it every time.
(a) Read backward to write its integral fact. (b) Name the derivative fact that reverses. (c) Use the power row and the two pass-through rules to find .
Hint
(a) Reading a row right to left turns the rate into its antiderivative and adds the . (b) Read the same integral fact forward: which antiderivative has rate ? (c) Split with the sum rule, integrate by the power row (), and remember .
Show solution
(a) . (b) . (c) .
(a) The row read backward gives .
(b) The antiderivative has rate , so the derivative fact behind the integral is .
(c) Split and look up each piece:
Differentiating returns , so the answer checks.
Exercises A.10
1. Read each derivative fact backward and write the integral fact it becomes, with the . (a) . (b) . (c) . Warm up
Hint 1
Reading a row of Table A.10.1 (Appendix A.10) right to left turns the rate into its antiderivative; the only thing you add on the way back is the .
Show solution
(a) . (b) . (c) .
Each is one row read backward: the rate on the right of the derivative fact becomes the integrand, its function becomes the antiderivative, and the joins on the return. All three rows read the same both directions, so no sign or constant fix is needed.
2. Each integral fact reverses one derivative fact. Name the derivative fact behind each. (a) . (b) . (c) . Warm up
Hint 1
Read each integral fact forward instead: the antiderivative on the right has a rate, and that rate equals the integrand. Differentiate the antiderivative to name it (Appendix A.10).
Show solution
(a) . (b) . (c) .
(a) The antiderivative has rate . (b) The antiderivative has rate , which is why the row carries the minus sign. (c) Differentiating drops the exponent by one and cancels the denominator, leaving : the power rule reversed.
3. Use the power-rule row to integrate, and write the each time. (a) . (b) . (c) . Core
Hint 1
The power row (Appendix A.10) says raise the exponent by one and divide by the new exponent. Rewrite roots and reciprocals as powers first: .
Hint 2
In (b), the new exponent is , so divide by . In (c), the new exponent is , and dividing by multiplies by .
Show solution
(a) . (b) . (c) .
(a) Raise 5 to 6 and divide by 6: .
(b) Raise to and divide by :
(c) Write , raise to , and divide by :
4. Use the two pass-through rules and three table rows to find . Core
Hint 1
The sum rule splits the integral into three, and the constant-multiple rule pulls each number out front (Appendix A.10). Then it is three lookups.
Hint 2
The pieces are the power row with , the row, and the row. Keep one for the whole expression.
Show solution
.
Split and pull the constants out:
Look up each row: , , . Reassemble:
Differentiating returns , so the answer checks.
5. Each integral matches one of the two parameter rows. Read off and write the antiderivative, with the . (a) . (b) . Core
Hint 1
Match (a) to and (b) to (Appendix A.10). Read from , so take the square root of the constant.
Hint 2
In (a), so . In (b), so , and the arctan row carries a out front.
Show solution
(a) . (b) .
(a) With , , and the row gives .
(b) With , , and the row carries the : .
6. Look up the antiderivative in the table, then evaluate the definite integral . Remember a definite integral is a number, with no . Core
Hint 1
The row gives the antiderivative (Appendix A.10). A definite integral evaluates that antiderivative with the bar and subtracts.
Hint 2
Use the evaluation bar , with and .
Show solution
.
The antiderivative of is . Evaluate with the bar:
The number 2 is the area under one hump of the sine curve. The drops out of a definite integral, since it would cancel in the subtraction.
7. Match each integrand to its row and integrate, with the . Watch for the constant fix. (a) . (b) . (c) . Stretch
Hint 1
Part (a) is the row with , part (b) is the row with , and part (c) is the row read straight across (Appendix A.10).
Hint 2
The row divides by ; the row carries a out front. Part (c) needs no fix at all.
Show solution
(a) . (b) . (c) .
(a) The row with : .
(b) The row with carries the : .
(c) The row reads straight across: . No constant fix, since that row is a bare reversal.
8. Verify a table entry by differentiating its antiderivative. Show that differentiates to , and state which row, and which value of , this confirms. Stretch
Hint 1
Differentiate with the chain rule: with , and the inside function contributes a (Appendix A.10).
Hint 2
You will collect two factors of and a in the denominator. Multiply that denominator by 4 to clear the inner fraction and watch appear.
Show full solution
, confirming the row with , so .
Differentiate, taking the outer , the arctan rate , and the inside rate :
The result is the integrand of , which is the row at .
9. The Appendix A.10 Dig In checked the two parameter rows by differentiating their antiderivatives back to the integrand. Make the check yours: pick any row the Dig In did not fully work, differentiate its antiderivative, and confirm you land exactly on the rate beside it. Dig In
Hint 1
Reread the Dig In in Appendix A.10: the whole test is "differentiate the antiderivative and compare with the rate column." A row like or makes a clean target.
Hint 2
For the row, differentiate using ; the in the denominator should cancel.
Show full solution
Worked with the row: , exactly the rate beside it, so is honest.
The antiderivative in the row is , with a constant. Differentiate, using :
The cancels, and the derivative is , the rate column exactly. The row survives its own check.
Check your own version. Whichever row you chose, one test settles it: the derivative of the antiderivative in the left column must equal the rate in the middle column, character for character. If a stray constant or sign survives, the row you wrote down does not match the table, and the usual cause is a dropped chain-rule factor or a missed constant multiple.
10. The Appendix A.10 Dig Deeper showed that differentiating never leaves the family of elementary functions, while integrating sometimes has no elementary answer. Make the contrast yours: (a) pick a function by differentiating any table antiderivative, so its own antiderivative is guaranteed to be elementary, and explain why that is guaranteed. (b) Name one integrand, such as , whose antiderivative is not elementary, and say what "not elementary" means in one sentence. Dig Deeper
Hint 1
Reread the Dig Deeper in Appendix A.10: any derivative of a table entry is elementary by construction, so integrating it just walks back up the same row. The trouble is only with integrands nobody assembled from a derivative.
Hint 2
For (a), if you differentiate to get , then itself is an elementary antiderivative of , so success is guaranteed. For (b), "not elementary" means it cannot be written with a finite combination of powers, roots, exponentials, logarithms, and trig functions.
Show full solution
(a) Worked example: differentiate to get ; then is guaranteed elementary, because you already hold an elementary antiderivative, namely . (b) has no elementary antiderivative, meaning none can be written with a finite combination of powers, roots, exponentials, logarithms, and trig functions.
(a) Start from any table antiderivative and differentiate it. If , then , and . Success is guaranteed the moment you build this way, because the function you differentiated, , is itself an elementary antiderivative of ; the reverse trip cannot fail when you already carried the answer up. This is why the whole derivative table doubles as an integral table.
(b) The integrand is continuous, so it does have an antiderivative, the accumulation function . What it lacks is an elementary one: no finite combination of the standard functions equals that antiderivative. That is exactly why statisticians read bell-curve areas from a table of values, not a formula, and why Section 2.1 brackets its between bounds instead of evaluating it.
Check your own version. Two audits confirm your split. For your part (a) function, differentiate your chosen , then integrate the result and you must return to plus a constant, no dead ends. For your part (b) integrand, it should be continuous (so an antiderivative exists) yet resist every row of the table and every substitution you try; if you found an elementary antiderivative, the integrand was not one of the hard ones after all.
One table carries both directions: read a row left to right and you have a derivative fact, read the same row right to left and you have an integral fact, and the return trip adds only the . The rows cover the powers, , , , , the basic trig functions, and the and forms, and the two pass-through rules (constant multiple and sum) stretch that short list to any combination of its rows. Two care points never lapse: write the on every indefinite integral, and keep the absolute value in . The deepest fact is the honest asymmetry: differentiating always works and never leaves the family, but integrating sometimes has no answer inside it, which is why this table of integrals is worth keeping open through Units 2 and 3. When in doubt, differentiate your antiderivative and check that it lands back on the integrand.
Back matter · study tool
Glossary
Every technical term, in plain language, with its symbol and a link back to where it is taught. Search it, filter it by unit, or flip it into flashcards to quiz yourself.
The total area of the rectangles used to estimate the area under a curve: for a regular partition of into strips of width with a sample point in each, it is . It sharpens as the strips get thinner.
Unit 1Section 1.1
A cut of into pieces of equal width , with cut points . Every strip in a Riemann sum has the same width.
Unit 1Section 1.1
The one input chosen in each strip to set that rectangle's height . Choosing the left edge, right edge, or midpoint of every strip gives the left, right, and midpoint sums.
Unit 1Section 1.1
The Riemann sum that reads each strip's height at its left edge. For a curve that only rises it under-estimates the area; for a falling curve it over-estimates.
Unit 1Section 1.1
The Riemann sum that reads each strip's height at its right edge. For a curve that only rises it over-estimates the area; for a falling curve it under-estimates.
Unit 1Section 1.1
The Riemann sum that reads each strip's height at its middle. It usually beats both the left and right sums, since each strip's overshoot roughly cancels its undershoot.
Unit 1Section 1.1
The average of the left and right sums, . It replaces each flat-topped rectangle with a slanted-topped trapezoid, so it is exact for a straight line and close for a gentle curve.
Unit 1Section 1.1
The compact way to write a long sum: says run the counter from 1 to and add every term along the way. A Riemann sum lives in this notation, , so an estimate built from hundreds of rectangles fits on one line.
Unit 1Section 1.1
The common width of the strips when is cut into equal pieces (a regular partition). It multiplies each sampled height in a Riemann sum, and as grows the strips thin toward the inside the integral.
Unit 1Section 1.1
A function whose derivative is on an interval: differentiation run backward. Because a constant has derivative 0, antiderivatives come in families, so the general antiderivative is , the whole family whose derivative is .
Unit 1Section 1.2
The general antiderivative of , written . It names a whole family of functions, so the is part of the answer every time; it evaluates to no single number.
Unit 1Section 1.2
The free constant in a general antiderivative . It stands for the one piece of information a rate of change cannot carry, like a trip's starting position, and it is the entire difference between any two antiderivatives of the same function on an interval.
Unit 1Section 1.2
A problem that gives a function's rate of change together with one known value and asks for the function. The rate fixes the general antiderivative up to ; the known value pins to a single number, selecting one function from the family.
Unit 1Section 1.2
The whole family of antiderivatives of a function, written : every function whose derivative is , all differing only by the constant . A single antiderivative is one member of the family; the general antiderivative is all of them at once.
Unit 1Section 1.2
The single number the Riemann sums of a function close in on as the number of strips runs to infinity: . It equals the net signed area between the curve and the x-axis on , and it is a number, never carrying a .
Unit 1Section 1.3
A function is integrable on when its Riemann sums close on one number no matter how the sample points are chosen. Every function continuous on a closed interval is integrable there, so the definite integral exists.
Unit 1Section 1.3
What a definite integral measures: area above the x-axis minus area below it. A strip where is negative subtracts, so the integral can be zero or negative. Total geometric area instead uses .
Unit 1Section 1.3
The two numbers (bottom) and (top) on the integral sign that mark where the region starts and stops: integrate from to . Swapping them flips the sign of the integral.
Unit 1Section 1.3
The function inside a definite integral : the height of the strip at position . The variable it uses is a dummy, so and give the same integral.
Unit 1Section 1.3
The variable of integration, a placeholder with no meaning outside the integral: and are the same number. Rename it freely; it never survives into the answer.
Unit 1Section 1.3
An even function has , a graph mirrored across the y-axis (like or ); an odd function has , a graph turned through the origin (like or ). On a symmetric interval the even one's integral is twice the half and the odd one's cancels to 0.
Unit 1Section 1.3
A definite integral with a fixed left endpoint and a moving right endpoint : the net signed area under from out to , read as a function of where you stop. The variable inside is a dummy , and . By FTC Part 1 its derivative is the integrand, .
Unit 1Section 1.4
The theorem that ties the derivative and the integral together as inverse operations. Part 1: the derivative of an accumulation function is its integrand. Part 2 (evaluation): for continuous and any antiderivative , the definite integral is the change in , namely . It turns a limit of Riemann sums into one subtraction.
Unit 1Section 1.4
The bracket shorthand for the subtraction in FTC Part 2, written . Find an antiderivative , write it inside the bar with the two limits, then evaluate at the top and the bottom and subtract. The constant of integration drops because it cancels.
Unit 1Section 1.4
The statement that the integral of a rate of change is the net change in the quantity: for a rate that is continuous on (as every rate in this book is), the net change from to is . It is the Fundamental Theorem read as a sentence about rates.
Unit 1Section 1.5
The signed change in an object's position over a time interval, the integral of its velocity. Forward motion counts positive and backward motion negative, so net displacement can be zero or negative even when the object moved a lot.
Unit 1Section 1.5
The whole length of the path an object travels, the integral of its speed . Every stretch counts as positive, forward or backward. Compute it by finding where , splitting the interval there, and adding the magnitudes of the pieces.
Unit 1Section 1.5
The height of the equal-area rectangle: , the area under the curve on divided by the width. It is the one steady value that fences off the same total, not the average of the two endpoints unless the graph is a line.
Unit 2Section 2.1
A definite integral over an infinite interval, defined as a limit: . It converges when that limit is a finite number and diverges when the running total grows without bound. For example converges while diverges.
Unit 2Section 2.1
The technique that reverses the chain rule (also called u-substitution): name an inside function , trade for , integrate in the single variable , then rename back. It applies exactly when the integrand is a composition times the inside's derivative , give or take a constant.
Unit 2Section 2.2
The technique that reverses the product rule: . Split the integrand into a part to differentiate, , and a part to integrate, , and you trade the original integral for an easier one. It is the move for a product like , where substitution stalls.
Unit 2Section 2.2
The inverse sine: is the angle in whose sine is . It is the antiderivative form of , appearing as .
Unit 2Section 2.3
The inverse tangent: is the angle in whose tangent is . It is the antiderivative form of , appearing as .
Unit 2Section 2.3
A substitution that replaces with a trig function of a new angle so a Pythagorean identity clears a root or a sum of squares: for an expression carrying , or for one carrying . It brings the and antiderivative forms within reach.
Unit 2Section 2.3
A way to integrate a ratio of polynomials by splitting it into a sum of simpler fractions, each with an easy antiderivative. For a factorable bottom, ; solve for and , then each piece integrates to a logarithm. It turns one hard fraction into two log integrals.
Unit 2Section 2.4
The true length of a curve measured along the curve itself, not the straight-line distance between its endpoints. Chop the curve into tiny near-straight pieces; each piece is the hypotenuse , and summing them and passing to the limit gives .
Unit 3Section 3.1
The length of one infinitesimal slanted piece of the curve, the hypotenuse of a right triangle with horizontal run and vertical rise . Summed by the integral, it rebuilds the whole arc length, and carried once around an axis it becomes the slant width of a surface band in Section 3.3.
Unit 3Section 3.1
A curve whose derivative is continuous on the interval, so it has no corners or breaks. Smoothness is what the arc-length formula needs: the integrand has to be a well-behaved function before you can integrate it.
Unit 3Section 3.1
The area of the region caught between two graphs over an interval. Slice it into thin strips: a vertical strip at position has height , so the area is . Because the height is a difference, the rule holds even when both curves sit below the x-axis.
Unit 3Section 3.2
The curved skin swept out when a curve is spun once around an axis. Its area counts only the curved skin, not the flat end caps. Slicing that skin into thin bands gives about the x-axis.
Unit 3Section 3.3
A cone with its tip sliced off: a short collar with a small circle of radius at one end and a bigger circle of radius at the other, joined by a slant of length . Its lateral area is , the average circumference times the slant. One thin band of a surface of revolution is a frustum.
Unit 3Section 3.3
The flat face you see when you cut a solid with a plane. When the cuts are perpendicular to the x-axis, the cross-section at position has area , and stacking those slices gives the volume .
Unit 3Section 3.4
The solid you get by spinning a flat region all the way around a fixed line, the axis of revolution. Its cross-sections perpendicular to that axis are circles or rings, which is what makes the disk, washer, and shell methods work.
Unit 3Section 3.4
The volume of a solid of revolution when the region touches the axis, so each slice is a full circle of radius : , and . The radius is the distance from the axis to the curve.
Unit 3Section 3.4
The volume of a solid of revolution when a gap opens between the region and the axis, so each slice is a ring: with outer radius and inner radius , both measured from the axis. Square each radius separately, then subtract; a disk is the special case .
Unit 3Section 3.4
A thin nested tube used to build a solid of revolution about the y-axis. Unrolled, one shell is a flat sheet of width (its circumference), height , and thickness , so . Shells run parallel to the axis and avoid solving the curve for .
Unit 3Section 3.4
The mass per unit length at each position along a rod, written as a function with units of mass over length (kilograms per meter, grams per centimeter). A rod is uniform when is constant and non-uniform when it varies, which is why its mass is an integral rather than a single product.
Unit 3Section 3.5
The total mass of a non-uniform object found by the slice-sum-integrate recipe: cut it into pieces so thin each is nearly uniform, give each slice the mass (density)(size), and integrate. For a rod this is ; for a plate with radial density it is .
Unit 3Section 3.5
The balance point of an object. For a rod with linear density on , it is : each position is weighted by the mass sitting there, so a heavy end pulls the balance point toward itself, past the geometric center.
Unit 3Section 3.5
The energy a force delivers as it acts through a distance. When the force varies with position, work is not force times distance but the area under the force curve, . It is measured in joules (newton-meters) or foot-pounds.
Unit 3Section 3.6
A spring pulls back with a force proportional to how far it is stretched or compressed, , where is the displacement from the spring's natural length. The farther you pull, the harder it resists, which is exactly why stretching a spring takes an integral rather than a single multiplication.
Unit 3Section 3.6
The stiffness in Hooke's law : the force needed per unit of stretch. A larger means a stiffer spring. You find it from one measured force-and-stretch pair, then use it to set up the work integral.
Unit 3Section 3.6
A curve that describes a continuous random variable: it is never negative, , and encloses total area 1, . The probability of an interval is the area over it, . A density is a height, not a probability, so it may exceed 1; only its area must stay at 1.
Unit 3Section 3.7
A quantity whose outcome can be any real number in a range, such as a wait time, a height, or a battery's life. Its outcomes cannot be listed one by one, so probability is read from a density curve as an area rather than summed from a list. Any single value has probability 0, since a point has no width.
Unit 3Section 3.7
The balance point of a continuous distribution: , each value weighted by the density there. It is the continuous version of the weighted-sum expected value from discrete probability. For a uniform density the mean is the midpoint; for the exponential it is .
Unit 3Section 3.7
How spread out a continuous distribution sits around its mean: the variance is the density-weighted average squared distance from the mean, , with the shortcut form . The standard deviation is its square root, a typical distance from the mean in the variable's own units.
Unit 3Section 3.7
The point where the angle 's terminal ray meets the unit circle, with coordinates . The whole unit circle is a table of this point at the landmark angles.
Appendix AAppendix A.1
The circle of radius 1 centered at the origin. For an angle theta measured counterclockwise from the positive x-axis, the point where the terminal ray meets the circle has coordinates (cosine theta, sine theta), so sine and cosine are read straight off the circle and tangent is their ratio.
Appendix AAppendix A.1
The angle unit calculus runs on: on the unit circle, an angle of theta radians cuts off an arc of length theta from the positive x-axis. Pi radians equal 180 degrees, the whole conversion factor between the two units, and every integral in this book measures angles this way.
Appendix AAppendix A.1
The acute angle between an angle's terminal ray and the x-axis, always between 0 and pi over 2. Every angle in this course reduces to one of the five landmarks (0, pi over 6, pi over 4, pi over 3, pi over 2), and the reference angle sets the magnitude of sine, cosine, and tangent; ASTC then supplies the sign.
Appendix AAppendix A.1
The y-coordinate of the point swept out by angle on the unit circle, : bounded between negative 1 and 1, with period .
Appendix AAppendix A.2
The x-coordinate of that same point, : bounded between negative 1 and 1, period , a quarter turn ahead of sine.
Appendix AAppendix A.2
The ratio , the slope of the radius to the point: unbounded, period , undefined wherever .
Appendix AAppendix A.2
The smallest positive length after which a repeating graph returns to itself exactly. Sine and cosine have period ; tangent repeats twice as often, with period , because a half lap already cancels in its ratio.
Appendix AAppendix A.2
Half the vertical distance between a wave's crest and trough, the reach above and below its middle line. Sine and cosine both have amplitude 1; tangent has none, since its graph has no highest or lowest point.
Appendix AAppendix A.2
A vertical line a graph races toward without crossing, as the output climbs or falls without bound nearby. Tangent has one at every angle where the unit-circle radius stands vertical, since a vertical radius has no slope to report.
Appendix AAppendix A.2
The reciprocal of cosine, ; its square is the antiderivative fact .
Appendix AAppendix A.3
The reciprocal of sine, , not the reciprocal of cosine despite the shared co: the letters cross over.
Appendix AAppendix A.3
An equation that holds for every value of the variable where both sides are defined, not only at a few special values, unlike a puzzle such as that pins down one answer. Every boxed rule in this section is an identity.
Appendix AAppendix A.3
A power-reduction identity that rewrites a squared sine or cosine as a first power of a doubled angle: and . Each comes from solving the double-angle identity . It is the rewrite Section 2.3 makes before integrating a squared trig function.
Appendix AAppendix A.3
The base- logarithm , the exponent is raised to in order to give ; every base-b log is a rescaled natural log, , the change-of-base formula.
Appendix AAppendix A.4
The natural logarithm of a positive number is the exponent must be raised to in order to give : means . It is the inverse of , so it is defined only for , with and .
Appendix AAppendix A.4
A function of the form with base , , and positive multiplier . It grows by multiplying: equal steps in multiply the output by equal factors, since . This is the family whose integrals Section 2.4 handles.
Appendix AAppendix A.5
The behavior of when the base satisfies : each step right multiplies the output by more than 1, so the outputs climb without any ceiling and eventually outrun every power of .
Appendix AAppendix A.5
The behavior of when the base satisfies : each step right multiplies the output by less than 1, so the outputs shrink toward the asymptote while staying positive forever.
Appendix AAppendix A.5
The multiplier in . Since , it is the output at zero, , so the graph crosses the y-axis at the initial value, not necessarily at 1.
Appendix AAppendix A.5
A horizontal line a graph closes in on as runs far out to one side: the gap shrinks toward zero, though the graph never needs to touch the line. Every exponential has the x-axis, , as its horizontal asymptote.
Appendix AAppendix A.5
The fixed base , the limit of as grows. It is the one base whose exponential is its own rate of change, so is also its own running total; that is why Section 2.4 leans on it for the cleanest exponential integral.
Appendix AAppendix A.5
For a decaying quantity, the time it takes to fall to half of its current amount. Because an exponential multiplies by a fixed factor per step, the half-life is constant: each half-life halves what remains again, so the quantity fades toward without ever reaching it.
Appendix AAppendix A.5
For a growing quantity, the time it takes to double. Because the growth factor is fixed, the doubling time is constant: a quantity with a doubling time of 3 years is four times its start after 6 years and eight times after 9.
Appendix AAppendix A.5
The inverse of a function runs backward: exactly when , so . It exists only when is one-to-one, its graph is the mirror of across , and the symbol is an inverse, never the reciprocal .
Appendix AAppendix A.6
A function is one-to-one when different inputs always give different outputs, so forces . This is the exact condition for an inverse to exist, and the horizontal line test checks it: no horizontal line may cross the graph more than once.
Appendix AAppendix A.6
The distance from the center of a circle, cylinder, cone, or sphere out to its edge, the input every area and volume formula on this page takes; if a problem gives a diameter, halve it to get .
Appendix AAppendix A.7
How much surface a flat region covers, measured in square units. A rectangle is , a triangle is , and a circle is . Area always carries a squared unit, the fast way to tell it apart from a volume.
Appendix AAppendix A.7
How much space a solid fills, measured in cubic units. A cylinder holds , a cone holds , and a sphere holds . These are the totals Section 3.4 rebuilds by stacking thin disks.
Appendix AAppendix A.7
The total area of a solid's outside skin, everything you would paint, counted in square units even though it wraps a solid. A sphere's is , the value Section 3.3 rebuilds by adding up thin bands.
Appendix AAppendix A.7
A cone's slant height : the straight distance up the slanted side from the base rim to the apex, the hypotenuse of the right triangle made by the radius and the vertical height. The lateral surface uses it, not the vertical height.
Appendix AAppendix A.7
The amount of an object packed into each unit of size. Mass density is mass per unit volume (kilograms per cubic meter); linear density is mass per unit length (kilograms per meter). For a uniform object, density times size gives mass: mass density times volume, or linear density times length.
Appendix AAppendix A.8
A push or a pull, measured in newtons (N) in the metric system or pounds (lb) in US customary units. One newton speeds up a 1 kilogram mass by 1 meter per second every second. Weight is the force of gravity on a mass, equal to with , so a 1 kilogram mass weighs about 9.8 N.
Appendix AAppendix A.8
The weight per unit volume of a substance, equal to its mass density times . Water's mass density of 1000 kilograms per cubic meter gives a weight density of about 9800 newtons per cubic meter, or about 62.4 pounds per cubic foot. This is the number the pumping problems of Section 3.6 use to weigh a slice of liquid.
Appendix AAppendix A.8
The set of all outcomes an experiment can produce, each one an equally likely possibility when you roll a die, flip a coin, or spin a spinner. Its size is written the count of S, the total number of outcomes.
Appendix AAppendix A.9
Any collection of outcomes gathered from the sample space, that is, a subset of S, such as the roll is even. Its count is the number of outcomes it contains.
Appendix AAppendix A.9
For equally likely outcomes, the share of the sample space an event fills: the count of the event over the count of the whole space. Every probability lands from 0 to 1, and the complement rule gives P of not A as 1 minus P of A.
Appendix AAppendix A.9
A quantity whose value is set by the outcome of a random experiment, like the number showing on a die or the number of heads in three flips. Its distribution lists the possible values with their probabilities, which add to 1.
Appendix AAppendix A.9
The weighted sum of a random variable's values, each multiplied by its own probability: the mean of the distribution and its long-run average per trial. It is the balance point of the distribution, not a value any single trial must hit.
Appendix AAppendix A.9
A single possible result of a random experiment, one element of the sample space, such as a 4 on a die or heads on a coin. The sample space is the set of all the outcomes, and an event is a collection of them.
Appendix AAppendix A.9
The list of a random variable's possible values paired with their probabilities, which always add to 1. It shows how the whole probability of 1 is spread across the values.
Appendix AAppendix A.9
The probability that an event does not happen is 1 minus the probability that it does: . Reason: an event and its complement together fill the whole sample space, whose total probability is 1.
Appendix AAppendix A.9
A reference list of standard antiderivatives. Each entry is a derivative fact read from the other end: because , the reverse statement holds. One such table carries both a table of derivatives (read forward) and a table of integrals (read backward), since each row is a single fact you can read either way.
Appendix AAppendix A.10
An antiderivative that can be written with a finite combination of powers, roots, exponentials, logarithms, and trig functions. Some continuous functions, such as , have an antiderivative (an accumulation function) but no elementary one, which is why the integral table is a genuine reference while the derivative table always works.
Appendix AAppendix A.10
Back matter · study tool
Which Tool? Read the Shape
Every section taught its tool with the label attached; a test does not. This page collects the reading cues the course already taught, each drawn from its own section, so choosing a method is a lookup, not a memory feat. Print it with the Print this section button above, and keep the Formula Sheet and the Rules & Theorems bank beside it.
Run these questions in order and stop at the first yes.
- Is it a table form? A power of , , an exponential, a basic trig function, or an inverse-trig form reads straight off Table A.10.1, and every row read right to left adds the . Constant multiples and sums pass through, so any sum of rows integrates term by term (Appendix A.10).
- A composition, with the inside's derivative nearby? Substitution pays off "when the inside's derivative is already in the integrand, give or take a constant, and not otherwise" (Section 2.2). Name the inside , write , and for a definite integral change the limits.
- The top of a fraction is the derivative of the bottom? The logarithm pattern: (Section 2.4). Its simplest case is the power the reverse power rule skips: , absolute value required.
- A squared trig function? It "has no direct antiderivative and no substitution; rewrite it with a half-angle identity first" (Section 2.3).
- A root of , or a sum ? Trigonometric substitution: "match a root of to arcsine and a sum to arctangent, read off the expression, and the answer is one line" (Section 2.3).
- Whatever you chose, check it. Differentiate the answer back; the check "settles every substitution in one line" (Section 2.2), and it settles every table row the same way.
Every Unit 3 setup is "the same four-word recipe, slice the thing into thin pieces, approximate each piece, add them, and take the limit, pointed at a different quantity" (Unit 3 Recap). Name the quantity first, and the slice follows.
- A length along a curve: (Section 3.1).
- An area between two curves: integrate top minus bottom, and split at each crossing (Section 3.2).
- The skin of a revolved curve: , an arc length carried once around a circle (Section 3.3).
- A solid's volume: every disk, washer, and shell is a specific in (Section 3.4; the box below picks between them).
- A mass from a density: (Section 3.5).
- A work from a varying force: (Section 3.6).
- A probability: the area under the density, (Section 3.7).
Section 3.4's own rule: "pick the method that fits the solid: disks and washers cut perpendicular to the axis, shells wrap parallel to it." Ask every time whether the region reaches the axis; if it does not, the slice is a washer, you owe a , and each radius squares separately. And its Watch Out says the tiebreaker plainly: "pick the method whose slice you can describe without solving the curve backward, and the variable comes along for free. Always name the axis first, and the rest follows."
Back matter · study tool
Formula Sheet
The course on one printed page: every antiderivative row with its side conditions, the evaluation statement, the average value, and the Unit 3 setups, each exactly as the Rules & Theorems bank states it. To print just this page, open ↓ Print / PDF in the top bar and choose Formula sheet.
Antiderivatives · Table A.10.1
- ()
- (absolute value required)
- (constant )
- (, )
- ()
- ()
Angles in radians. Every indefinite integral carries the .
Evaluating · Units 1 and 2
- FTC evaluation (Part 2) For continuous on and any antiderivative :
- Evaluation bar
- Average value
Setups · Unit 3
- Arc length
- Surface of revolution (about the x-axis; about the y-axis the radius becomes )
- Volume by cross-sections
- Disk (slice touches the axis)
- Washer (slice with a gap; radii are distances to the axis)
- Cylindrical shell (about the y-axis, ; always state the axis of revolution)
- Work
- Mass from linear density
- Probability density totals 1 and
- Mean of a density
Back matter · study tool
Rules & Theorems
Every named rule, theorem, and formula from the course, with its statement and a link back to where it is shown. The Glossary tests whether you remember what something is called; this tests whether you remember what it says. Search it, filter it by unit, or flip it into flashcards to quiz yourself.
On a regular partition of width , the four estimates are , , , and . Same strips, same width; only the sample height changes.
Unit 1Section 1.1
If only rises on then ; if only falls, the roles swap. Reason: a rising curve is sampled low on the left and high on the right, so the left rectangles fall short and the right ones overshoot. A curve that turns inside the interval gets no such guarantee.
Unit 1Section 1.1
for : raise the exponent by one and divide by the new exponent. Constant multiples and sums pass straight through. The one exponent it cannot touch is , the case , which waits for the logarithm.
Unit 1Section 1.2
If and are both antiderivatives of on an interval, then is a constant. Two functions with the same derivative differ only by an added number, because their difference has derivative 0 and a zero slope across an interval means a fixed value. This is why one letter holds the whole freedom.
Unit 1Section 1.2
The definite integral is the limit of the Riemann sums as the strips get thinner. For a continuous on , the sum built from each strip's shortest height and the sum built from its tallest height squeeze together, so the limit exists and does not depend on the sample points.
Unit 1Section 1.3
Four working rules: zero width ; reversing flips the sign ; linearity ; and additivity .
Unit 1Section 1.3
On a symmetric interval : if is even (), then ; if is odd (), then . It follows from splitting at 0 and reflecting.
Unit 1Section 1.3
For continuous , the accumulation function is differentiable with : differentiating an integral hands back the integrand. With a variable upper limit , the chain rule adds a toll: . One consequence: every continuous function has an antiderivative.
Unit 1Section 1.4
For continuous on and any antiderivative (any function with ), the definite integral is . The choice of antiderivative does not matter: two differ by a constant, which cancels in the subtraction. Requires continuous across the whole interval.
Unit 1Section 1.4
The bracket shorthand for the subtraction in FTC Part 2: . Antidifferentiate, then evaluate at the top and bottom limits and subtract. The constant of integration is dropped because it cancels.
Unit 1Section 1.4
The net change in a quantity is the integral of its rate of change (for a rate that is continuous on , as every rate in this book is): , equivalently . Reason: it is the Fundamental Theorem of Calculus with the integrand viewed as a rate, so the integral adds the rate over every instant into one net change.
Unit 1Section 1.5
Net displacement is (forward and backward cancel); total distance is (nothing cancels). Find total distance by locating each time , splitting there, integrating each piece, and adding the magnitudes. You cannot recover distance by taking the absolute value of the final displacement.
Unit 1Section 1.5
The average value of on is its integral divided by the width. Read it as area over width: the definite integral is the area, and dividing by flattens that area into a rectangle of the same width whose height is .
Unit 2Section 2.1
If on , then : the integral is trapped between its shortest and tallest rectangle. Also, on forces . This brackets integrals you cannot evaluate exactly, such as .
Unit 2Section 2.1
If is continuous on , there is at least one point in with , equivalently . Reason: the comparison bounds put the average value between the curve's minimum and maximum, and a continuous curve passes through every height in between. The equal-area rectangle's height is a height the curve genuinely reaches.
Unit 2Section 2.1
With and , where is continuous and is an antiderivative of : . For a definite integral, change the limits: . Reason: it is the chain rule read backward, since differentiating rebuilds .
Unit 2Section 2.2
, , , and . Each is a derivative fact from the Derivatives volume read backward, checkable by differentiating the answer. The minus on sine's antiderivative is the one sign to guard, and all four hold in radians.
Unit 2Section 2.3
, found by writing and substituting . The absolute value is required. The same move gives .
Unit 2Section 2.3
A squared trig function has no direct antiderivative and no substitution; rewrite it with a half-angle identity first. and , from and . The average of over a full cycle is .
Unit 2Section 2.3
and , reached by the trigonometric substitutions and . Match a root of to arcsine and a sum to arctangent, read off the expression, and the answer is one line.
Unit 2Section 2.3
; for a nonzero constant ; and for a base with . Each is the matching derivative fact read backward: the and the are what dividing undoes.
Unit 2Section 2.4
, the case the reverse power rule skips at . The absolute value is required: is defined for negative too, and is the one formula whose derivative is on both sides of zero. The bars may drop only when the argument is provably positive.
Unit 2Section 2.4
: when the top of a fraction is the derivative of the bottom, the integral is a logarithm. Reason: the substitution , turns it into . A constant mismatch on top can be supplied; this is how falls out.
Unit 2Section 2.4
For a smooth curve on , the length is . The reason: a tiny piece of the curve is the hypotenuse ; summing the hypotenuses and taking the limit turns the sum into this integral. Keep the inside the root, and never split into .
Unit 3Section 3.1
With the top curve at or above the bottom on , the area is , the integral of the strip height. If the curves cross inside the interval, split at each crossing and add the pieces. When a boundary in splits into branches, slice sideways instead and integrate right minus left in : .
Unit 3Section 3.2
The area of the skin swept out by revolving about the x-axis, with continuous. Each band is a circle of circumference times the slant width , so a surface area is an arc length carried once around a circle. About the y-axis the radius becomes .
Unit 3Section 3.3
One thin band of a surface of revolution is a frustum of area . As the band thins, its average radius becomes and its slant becomes , so the band area is . Using the flat instead of the slant undercounts a sloping band.
Unit 3Section 3.3
If a solid runs from to with cross-sectional area at each position, its volume is the integral of that area. Reason: a slab at holds about , and summing the slabs and taking the limit gives the integral. Every disk, washer, and shell formula is just a specific .
Unit 3Section 3.4
Revolving about an axis, a slice that touches the axis is a disk of area ; a slice with a gap is a washer of area . Reason: each slice is a circle (or a circle with a hole), and its area feeds . Radii are distances to the axis, so name the axis first.
Unit 3Section 3.4
Revolving the region under on , with , about the y-axis, peel the solid into tubes: one shell of radius unrolls into a sheet wide and tall. Reason: circumference times height times thickness is , and summing gives the integral. Use it when washers would force solving for .
Unit 3Section 3.4
The mass of a rod along is the integral of its linear density, the area under the density curve. Each thin slice at position has mass ; summing and taking the limit gives the integral. When is constant it collapses to density times length.
Unit 3Section 3.5
The mass of a flat circular plate whose density depends only on distance from the center. Slice into thin rings: the ring at radius has area (circumference times thickness, not ), so its mass is . This is the same shell-style slicing used for volumes.
Unit 3Section 3.5
Over a tiny displacement the force is nearly constant, so the little bit of work is ; summing the pieces and taking the limit gives . Constant-force work is just the special case where never changes.
Unit 3Section 3.6
A spring's restoring force is , so the work to stretch it from to is . Find first from a measured force-and-stretch pair.
Unit 3Section 3.6
To pump liquid out of a tank, slice it into thin horizontal layers. A layer at depth has weight (weight density times its area times its thickness ) and must rise its own lifting distance, so its work is (weight)(distance). The total is the integral over all the layers; each layer travels a different distance, so that distance lives inside the integral, never as a constant out front.
Unit 3Section 3.6
A function is a probability density exactly when everywhere and its total area is 1. The area is the probability of the outcome landing anywhere at all, which is certainty. Setting this integral equal to 1 is what fixes the normalizing constant of a candidate density.
Unit 3Section 3.7
The probability that a continuous variable lands between and is the area under its density over that interval. A single point has width 0, so ; only an interval carries probability. Every such probability is a slice of the total area 1, so it stays between 0 and 1.
Unit 3Section 3.7
The mean weights each outcome by its density and adds them as an integral, the balance point of the area under the curve. It is the weighted-sum expected value with the sum turned into an integral. A lopsided density balances off its midpoint, pulled toward the heavier side.
Unit 3Section 3.7
The spread of a continuous distribution: each squared distance from the mean, weighted by the density there. The shortcut is the same quantity rearranged, and the standard deviation is the square root. Reason: squaring keeps near and far misses from canceling and weighs far-out values harder, so a sprawling density earns a bigger than a bunched one with the same mean.
Unit 3Section 3.7
Sine, cosine, and tangent at 0, pi over 6, pi over 4, pi over 3, and pi over 2, always rationalized (root 3 over 3, never 1 over root 3). Reason: they fall out of the 30-60-90 and 45-45-90 right triangles, and every other angle in the course reduces to one of these five.
Appendix AAppendix A.1
Theta in radians equals theta in degrees times pi over 180, and the reverse multiplies by 180 over pi. Reason: pi radians and 180 degrees measure the identical half turn, so the ratio only changes the units, never the angle itself.
Appendix AAppendix A.1
Reflecting an angle into a new quadrant changes the sign of sine, cosine, and tangent, never their magnitude. All Students Take Calculus names which stays positive: All three in Quadrant I, Sine in Quadrant II, Tangent in Quadrant III, Cosine in Quadrant IV. Reason: the axes act as mirrors on the point's coordinates.
Appendix AAppendix A.1
For every real , and , since both are coordinates on a circle of radius 1. Tangent is a slope, not a coordinate, so it has no such leash and races toward infinity near its breaks.
Appendix AAppendix A.2
In radians, , , and . Section 2.3 reads each backward to an antiderivative (, and so on); the minus on cosine's derivative is the one sign to guard, since it lands on sine's antiderivative.
Appendix AAppendix A.2
for every angle . Reason: on the unit circle, the point at angle forms a right triangle with legs and and hypotenuse 1, and this is just the Pythagorean theorem on that triangle.
Appendix AAppendix A.3
(divide by ) and (divide by ). Reason: dividing every term of the Pythagorean identity by the same nonzero quantity keeps the equation true; it just relabels the ratios. The first clears an in Section 2.3's trigonometric substitution.
Appendix AAppendix A.3
For , , and upside down . Reason: on the unit circle, rise over run from the origin to the point at angle is exactly this ratio, the definition of tangent.
Appendix AAppendix A.3
, , and , each name built to flip one basic ratio. Cosecant's partner is sine, not cosine, and secant's partner is cosine, not sine: the "co" prefix crosses over rather than matching.
Appendix AAppendix A.3
and , each trading a square for a doubled angle. Reason: solve the double-angle identity for the square. This is the rewrite Section 2.3 needs to integrate and .
Appendix AAppendix A.3
The graph of passes through and , is defined only for , is negative for and positive for , climbs ever more slowly, and drops toward as nears 0, so the y-axis is a vertical asymptote. Reason: it is the mirror of across the line , so every exponential fact reflects into a logarithm fact.
Appendix AAppendix A.4
For positive , and any power : , , and . Reason: a logarithm is an exponent, so these are the exponent rules read back through the inverse. Each trades a harder operation for an easier one: a product for a sum, a quotient for a difference, a power for a multiple.
Appendix AAppendix A.4
The graph of passes through and , stays positive and never reaches zero, climbs for and falls for , and hugs the asymptote on its low side. Reason: equal steps multiply, since , and multiplying a positive number by a positive factor never makes it zero or negative. A multiplier scales the intercept to and leaves the rest untouched.
Appendix AAppendix A.5
For any base and any fixed power , the exponential eventually passes and stays ahead forever. Reason: each step multiplies by the fixed factor , while it multiplies by , a factor sliding toward 1; once that factor drops below , the exponential gains ground on every remaining step.
Appendix AAppendix A.5
A function has an inverse exactly when it is one-to-one, which its graph shows at a glance: is one-to-one when no horizontal line crosses the graph more than once. Reason: a horizontal line meets the graph wherever , so a second crossing is a second input sharing an output. A steadily rising or falling graph passes; one that turns around fails.
Appendix AAppendix A.6
The graph of is the mirror of the graph of across the line : every point on reflects to on , and the domain and range trade places. Reason: the inverse swaps each point's coordinates, and a coordinate swap is a reflection across ; the compositions and confirm it.
Appendix AAppendix A.6
A rectangle with base and height has area ; a triangle with the same base and height has area ; a circle of radius has area . Reason: the triangle is exactly half its bounding rectangle, and the circle's is the round shape's slice-and-add answer. Every area is in square units.
Appendix AAppendix A.7
A cylinder of radius and height has volume , the base area times the height. Its side unrolls into a rectangle wide and tall, so the lateral surface is ; adding the two caps gives the total .
Appendix AAppendix A.7
A cone of radius and height has volume , one third of its boxing cylinder. Its slant height is ; the lateral surface is , and adding the base gives the total . Volume uses , the slanted surface uses .
Appendix AAppendix A.7
A sphere of radius has volume and surface area , with no flat base and no seam. Section 3.4 rebuilds the volume by stacking disks and Section 3.3 rebuilds the surface by wrapping thin bands.
Appendix AAppendix A.7
A constant force moving an object a distance does work , measured in joules (newton-meters) or foot-pounds. Reason: work is force paid out over distance, so both the size of the push and the length of the trip count. This product holds only while the force stays constant; a changing force makes work an integral instead.
Appendix AAppendix A.8
A spring stretched a distance from its natural length pulls back with a force proportional to : , where the spring constant is its stiffness in newtons per meter. Find from one measured force-and-stretch pair, . Pull twice as far and it resists twice as hard.
Appendix AAppendix A.8
For equally likely outcomes, , the winning outcomes over the total. Reason: each of the outcomes carries an equal share , and A collects of them; every probability lands from 0 to 1, with .
Appendix AAppendix A.9
The mean of a random variable, : each value weighted by its probability, then summed. Reason: over many trials each value turns up about a share of the time, so the long-run average per trial is that weighted sum. It is the balance point of the distribution, and Section 3.7 turns the sum into the integral for a continuous variable.
Appendix AAppendix A.9
If on an interval, then . An antiderivative is a derivative fact named from the far end, so the whole derivative table doubles as an integral table. The collects every antiderivative, since two functions with the same rate differ only by a constant.
Appendix AAppendix A.10
and . Reason: differentiation obeys these two rules, so reversing it does too. Together they integrate any sum of constant multiples of table rows term by term. Only a constant may cross the integral sign, never a variable.
Appendix AAppendix A.10
Back matter
Sources, and how this book was built
This textbook is an original synthesis. The mathematics is standard and is drawn from three excellent, freely available sources, but the prose, examples, ordering, and the dashboard throughline were written fresh for this course. Nothing here is copied from those books; they are the shoulders this one stands on, and each is worth reading directly.
The three source books
1. Strang, G., & Herman, E. (2016). Calculus Volume 1. OpenStax, Rice University. The primary spine for the calculus itself. Free at openstax.org/details/books/calculus-volume-1. Used across all three units: the definite integral and the Fundamental Theorem of Calculus (Ch 5), the applications of integration (Ch 6), antiderivatives (Ch 4), and the logarithm as an integral (Ch 6).
2. Boelkins, M., Austin, D., & Schlicker, S. (2024). Active Calculus: Single Variable (2nd ed.). Grand Valley State University. A free, inquiry-first calculus text at activecalculus.org/single2e. Used as a second voice for the definite integral and the Fundamental Theorem (Ch 4), techniques of integration (Ch 5), and the applications of integration (Ch 6).
3. Strang, G. (2017). Calculus (3rd ed.). Wellesley-Cambridge Press. Free through MIT OpenCourseWare as RES.18-001. The source of the most vivid framings in calculus teaching, including the speedometer and odometer pairing this book opens with (his "function 1 and function 2").
How the pieces were compiled
The course syllabus already pointed to specific sections of these books. This book keeps that exact order and stitches the sections into one continuous narrative, so a student never feels the seams between three different authors. Where the books cover the same idea, the clearest treatment was chosen and rewritten in a single consistent voice, with one running example style and one notation system (see the Notation table). The dashboard framing, the low-floor Picture This hooks, the Dig Deeper extensions, and every exercise are new to this book.
| Part of this book | OpenStax Calculus Vol 1 | Active Calculus 2e | Strang, Calculus |
|---|---|---|---|
| Unit 1: approximating area, antiderivatives, the definite integral, the Fundamental Theorem, net change | 4.10, 5.1 to 5.4 | 4.1 to 4.4 | 5.1 to 5.7 |
| Unit 2: integral properties and average value, substitution, trigonometric integrals, exponential and logarithmic integrals | 5.5, 5.6, 6.7 | 5.3 | 5.6, 7.1 to 7.3 |
| Unit 3: arc length, areas between curves, surface area, volume, mass and density, work, probability | 6.1 to 6.5 | 6.1 to 6.4 | 8.1 to 8.6 |
| Appendix A: prerequisite toolkit (trig, logs, exponentials, inverse functions, geometry, physics, probability) | 1.4, 1.5, App A to C | 1.3 | 6.1 to 6.4 |
Citations follow a standard author-date style. All three books were freely available from the linked pages at the time this edition was compiled. Please support the authors by citing them in any work that builds on this book.
Back matter
About this book
Written for the Calculus Integrals course, 2026-2027. The book is one connected argument: a total is built from infinitely many small pieces, the rectangles under a curve close in on an exact amount, the Fundamental Theorem turns any rate into its total, and a small set of techniques puts that total to work measuring the lengths, areas, volumes, masses, work, and probabilities of a moving world.
The course itself is Megan Warren's design: she chose the three source books, decided what to teach and in what order, and shaped the flow that carries a student from a first stack of rectangles to the exact lengths, volumes, and probabilities of a moving world, in one connected argument. This book was then written by Claude to that design. The mathematics is original and self-contained. Topic order follows her course outline exactly. Every worked example and exercise solution is independently checked before it ships, including cross-checks against a computer algebra system for anything numeric.
Design and voice: hers as well, navy with a berry accent for this course. Math is typeset with KaTeX and fully embedded, so the book renders with no internet connection.
Like what you've seen here? Megan takes on a small number of commissions at a time. Visit megan-warren.com for more.