The area between two curves as the integral of top minus bottom, why the axis is irrelevant, splitting where the curves cross, integrating with respect to y using horizontal strips, and choosing the variable that avoids splitting.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 6 — Applications of Integration
Areas between Curves
Objectives
Five outcomes. The integration is routine; every one of these is about setting the integral up.
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 540-550 — the section these objectives are drawn from
Warm-up
Chapter 5 computed the area between a curve and the axis as a definite integral, splitting where the curve crossed the axis.
Discussion prompt
What is the area between the line y equals 2x and the parabola y equals x squared, from where they meet to where they meet again?
Hint: Think of a thin vertical strip.
Answer:
A thin vertical strip at position x runs from the parabola up to the line, so its height is the difference of the two functions and its width is dx. Adding the strips is integrating that difference.
\[ A = \int_{0}^{2}\left(2x - x^{2}\right)dx = \left[x^{2}-\frac{x^{3}}{3}\right]_{0}^{2} = \frac43 \]
The curves meet where 2x equals x squared, at 0 and 2 — which is where the limits came from. That is the whole method, and the rest of the section is about the situations where identifying top, bottom and limits takes more care.
Concept
The area between two curves is the integral of the upper function minus the lower one, over the interval where the region lies. A thin strip's height is that difference and its width is the differential.
the area between curves — The integral of the difference between the upper and lower boundary functions, taken over the interval on which the region lies.
\[ A = \int_{a}^{b}\left[f(x)-g(x)\right]dx, \quad f \ge g \text{ on } [a,b] \]
Because both boundaries appear as a difference, the position of the axis is irrelevant — a fact that removes all of Chapter 5's care about regions below the axis.
Figure (svg): The region between two curves, with a representative vertical strip
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 597-606
Section
Section 1
Concept
A representative vertical strip has height equal to the upper function minus the lower and width equal to the differential. Integrating adds the strips, exactly as in Section 5.1.
representative strip — A thin rectangle standing in for all the strips making up the region. Its height and width, written in terms of the variable, give the integrand and the differential.
\[ dA = \left[f(x)-g(x)\right]dx \]
Drawing one strip on a sketch and labelling its height is the most reliable way to get the integrand right. Every setup in Chapter 6 uses the same device.
Figure (svg): The region between two curves, with a representative vertical strip
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 597-605 — area between two curves
Picture it
The region between a line and a parabola.
Figure (svg): The region between two curves, with a representative vertical strip
The strip's height is the top function minus the bottom, and its width is the differential. Everything else in the setup follows from getting that one rectangle right.
Worked example
Example 6.1. Find the limits from the intersections.
\[ \text{Find the area between } y=2x \text{ and } y=x^{2}. \]
Find where the curves meet
Why: Set them equal.
\[ 2 x = x ^{2}\text{ at } x = 0\text{ and } 2 \]
Decide which is on top
Why: Test a point between.
\[ \text{at } x = 1: 2 > 1,\text{ the line} \]
Write the integrand
Why: Top minus bottom.
\[ 2 x - x ^{2} \]
Integrate
Why: Term by term.
\[ x ^{2} - x ^{3} / 3 \]
Evaluate from 0 to 2
Why: Substitute.
\[ 4 - \frac{8}{3} = \frac{4}{3} \]
Figure (svg): The region between two curves, with a representative vertical strip
\[ A = \frac43 \]
Verify: sanity-check against a bounding rectangle
Why: The region sits inside a box 2 wide and 4 tall, area 8, and it is a thin lens shape occupying well under a quarter of that — so 1.33 is plausible. Note that testing a point between the intersections was essential: assuming the line is on top without checking is how the sign gets reversed, and a negative area is the symptom.
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 600-601
Fill the middle
The limits of integration come from where the curves meet.
Fill in the blanks
2x = x^2 \;\Longrightarrow\; x = 0 \text___ x = ___
Why: Setting the functions equal and solving gives the interval's ends. The limits of integration in this section almost always come from intersections rather than being given.
Worked example
Checkpoint 6.1. The same region, shifted down.
\[ \text{Find the area between } y=2x-3 \text{ and } y=x^{2}-3. \]
Find the intersections
Why: The shift cancels.
\[ \text{still } x = 0\text{ and } 2 \]
Write the integrand
Why: Top minus bottom.
\[ (2 x - 3) - (x ^{2} - 3) \]
Simplify
Why: The constants cancel.
\[ 2 x - x ^{2} \]
Note it is identical
Why: Same integrand, same limits.
State
Why: The area.
\[ \frac{4}{3} \]
Figure (svg): Why the formula works even below the axis
\[ A = \frac43 \text{ again} \]
Verify: see why translating changes nothing
Why: Both boundaries moved down by the same amount, so their difference is unaffected — and the difference is all the integrand contains. Part of this region lies below the axis, and none of Chapter 5's splitting was needed, because the axis plays no role in a region bounded by two curves. That is worth internalising early, since the reflex from Chapter 5 is to check for sign changes.
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 601-603
Trap
\[ A = \int_{0}^{2}\left(x^{2}-2x\right)dx = -\frac43 \]
Subtract the top from the bottom
Why: The student takes the functions in the order given.
A negative answer to an area question is the symptom. The strip's height must be the upper value minus the lower, whichever function that happens to be.
\[ A = \int_{0}^{2}\left(2x-x^{2}\right)dx = \frac43 \]
Test a point between the intersections to see which is on top
Why: The order in which the problem lists the curves means nothing.
A sketch settles it instantly, and a single test point settles it without one. The negative sign is a reliable warning, so an area that comes out negative should always send you back to the setup.
Ranking
An area between two curves.
Put in order
Why: Step a makes steps b and c almost automatic, which is why it is worth the thirty seconds. Skipping step c is what produces negative areas, and skipping step a is what produces the wrong number of integrals.
Two truths and a lie
All three are about the basic setup.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Which function is on top is a fact about the region, discoverable by a sketch or a test point, and the order in which a problem happens to name the curves carries no information about it.
Prediction
Commit before reasoning.
Predict first
Chapter 5 required splitting where a curve crossed the axis. Why is no such splitting needed here?
Correct: Because both boundaries appear as a difference.
\[ (f+c)-(g+c) = f-g \quad \text{for any } c \]
Why: The integrand is the top function minus the bottom, and translating the whole region vertically adds the same constant to both — which cancels in the difference. So a region entirely below the axis has exactly the same integral as its translate above it. Chapter 5's care was needed because one boundary there was the axis itself, which does not move with the region.
Section
Section 2
Concept
When curves cross inside the interval, which is on top changes there. Integrating the difference across a crossing lets the two parts cancel, so the region must be split.
splitting at a crossing — Dividing the interval at every point where the curves intersect, integrating the appropriate difference on each piece, and adding the results.
\[ A = \int_{a}^{c}(g-f) + \int_{c}^{b}(f-g) \]
This is Chapter 5's total-against-net distinction in a new setting. A single integral gives a signed quantity in which the pieces cancel; the area needs each piece taken positively.
Figure (svg): Curves that cross: which is on top changes, so the region splits
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 603-611 — regions where the curves cross
Picture it
A line and a cubic meeting three times.
Figure (svg): Curves that cross: which is on top changes, so the region splits
Left of the origin the cubic is above; right of it the line is. Integrating one difference across the whole interval would let the two symmetric halves cancel to zero.
Worked example
Example 6.3. Split at the crossing.
\[ \text{Find the area between } y=x \text{ and } y=x^{3} \text{ on } [-1,1]. \]
Find the crossings
Why: Set them equal.
\[ x = -1, 0, 1 \]
Test left of zero
Why: At x = -0.5.
Test right of zero
Why: At x = 0.5.
Integrate each piece
Why: The correct difference on each.
\[ \frac{1}{4}\text{ and } \frac{1}{4} \]
Add
Why: Both taken positively.
\[ \frac{1}{2} \]
Figure (svg): Curves that cross: which is on top changes, so the region splits
\[ A = \frac14+\frac14 = \frac12 \]
Verify: see what a single integral would have given
Why: Integrating the line minus the cubic across the whole interval from negative one to one gives zero, because the integrand is odd and the interval symmetric — Section 5.4's shortcut applying in a way that destroys the answer. The region plainly has area, so zero is visibly wrong, but a less symmetric example would give a plausible wrong number instead. Finding the crossings first is what prevents it.
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 605-607
Sorting
Do the curves cross inside the interval?
Sort into buckets
Sort each pair on the interval given.
Crossings AT the endpoints are fine and are usually where the limits come from; only crossings strictly inside force a split. That distinction is what makes the second and third rows differ despite involving the same two curves.
Worked example
Checkpoint 6.3. Three intersections.
\[ \text{Where do } y=x^{3}-4x \text{ and } y=x \text{ meet?} \]
Set the functions equal
Why: Standard.
\[ x ^{3} - 4 x = x \]
Collect on one side
Why: Subtract.
\[ x ^{3} - 5 x = 0 \]
Factor
Why: Common factor first.
\[ x(x ^{2} - 5) = 0 \]
Solve
Why: Three roots.
\[ x = 0\text{ and plus or minus } \sqrt{5} \]
Note the consequence
Why: Two crossings inside.
Figure (svg): The solution to Worked example crossings found by factoring shown as a ladder of expressions, one row per legal move
\[ x = 0, \; \pm\sqrt5 \]
Verify: check that every root was found
Why: A cubic equation has at most three roots and three were found, so none is missing — which matters, because an unnoticed crossing means an unsplit piece and a wrong answer. Factoring out the common x first was what made the rest immediate; expanding and hunting for roots numerically would have been slower and less certain. Whenever a polynomial equation arises here, factoring completely is worth the moment it takes.
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 607-609
Error analysis
A student computes an area over an interval containing a crossing.
Annotate
On: \( A = \int_{-1}^{1}\left(x-x^{3}\right)dx = 0 \)
An area of zero for a visibly non-empty region is the warning. Solving for the intersections before integrating reveals every crossing, and the interior ones are exactly the split points.
Fill the middle
A line and a cubic on a symmetric interval.
Fill in the blanks
x = x^1 \;\Longrightarrow\; x = 0, \; \pm___
Why: The three roots are 0 and plus or minus one. On the interval from negative one to one the endpoints supply the limits and the origin supplies the split point.
Ranking
Curves crossing inside the interval.
Put in order
Why: Step d must be done separately on each piece, since the answer changes from piece to piece — that is the whole reason for splitting. Doing it once and assuming it holds throughout is equivalent to not splitting at all.
Prediction
Commit before reasoning.
Predict first
Integrating one difference across a crossing gives what?
Correct: A signed quantity in which the pieces cancel.
\[ \int_{-1}^{1}(x-x^{3})dx = 0 \quad \text{but the area is } \tfrac12 \]
Why: Beyond the crossing the chosen difference becomes negative, so that part subtracts instead of adding — exactly the net-against-total distinction of Section 5.4 appearing again. For a symmetric example the cancellation is complete and the answer is zero, which is visibly wrong; for an asymmetric one it is partial and the answer is plausible, which is more dangerous.
Section
Section 3
Concept
When a region is bounded on the left and right by curves, horizontal strips are natural: each has width the right function minus the left, expressed in terms of y, and height dy.
horizontal strips — Thin rectangles running left to right, whose width is the difference of the right and left boundary functions written as functions of y, and whose height is the differential in y.
\[ A = \int_{c}^{d}\left[u(y)-v(y)\right]dy \]
Nothing new is being introduced. The same strip argument runs sideways, with the roles of the axes exchanged, and the boundaries must be rewritten as functions of y.
Figure (svg): Integrating with respect to y: horizontal strips, right minus left
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 608-616 — integrating with respect to y
Picture it
A region bounded left and right.
Figure (svg): Integrating with respect to y: horizontal strips, right minus left
The strip's width is the right boundary minus the left, both written in terms of y. Everything about the method is unchanged except which letter plays which role.
Worked example
Example 6.5. Horizontal strips.
\[ \text{Find the area between } x=y^{2} \text{ and } x=y+2. \]
Find where they meet
Why: Set them equal.
\[ y ^{2} = y + 2\text{ at } y = -1\text{ and } 2 \]
Decide which is on the right
Why: Test y = 0.
\[ x = 2\text{ beats } x = 0:\text{ the line} \]
Write the strip's width
Why: Right minus left.
\[ (y + 2) - y ^{2} \]
Integrate with respect to y
Why: Term by term.
\[ y ^{2} / 2 + 2 y - y ^{3} / 3 \]
Evaluate from -1 to 2
Why: Substitute.
\[ \frac{9}{2} \]
Figure (svg): Integrating with respect to y: horizontal strips, right minus left
\[ A = \frac92 \]
Verify: consider what vertical strips would have required
Why: With vertical strips the top boundary is the upper half of the parabola until the line takes over, so the region would split into two integrals, each needing the parabola solved for y as a square root. Horizontal strips need no split and no radicals. Choosing the variable well turned two awkward integrals into one straightforward one, which is the point of the next idea.
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 611-613
Fill the middle
A horizontal strip between two boundaries.
Fill in the blanks
\texty^2 = (y+2) - ___
Why: Right minus left, both written as functions of y. The differential must then be dy and the limits must be y values, or the integral is inconsistent.
Worked example
Checkpoint 6.5. The conversion step.
\[ \text{Express } y=\sqrt{x} \text{ and } y=x/2 \text{ as functions of } y. \]
Solve the first for x
Why: Square both sides.
\[ x = y ^{2} \]
Note the branch
Why: The square root is non-negative.
\[ y \ge 0 \]
Solve the second for x
Why: Multiply by two.
\[ x = 2 y \]
Identify which is on the right
Why: Test y = 1.
\[ 2\text{ beats } 1:\text{ the line} \]
Write the strip's width
Why: Right minus left.
\[ 2 y - y ^{2} \]
Figure (svg): The solution to Worked example rewriting boundaries as functions of y shown as a ladder of expressions, one row per legal move
\[ A = \int_{0}^{2}\left(2y-y^{2}\right)dy = \frac43 \]
Verify: check the answer against the vertical-strip computation
Why: With vertical strips the integrand is the square root of x minus x over 2, integrated from 0 to 4, giving 16 over 3 minus 4, which is 4 over 3 — matching. Both routes work here and neither needs splitting, so the choice is a matter of which integrand is more pleasant. Note that solving a square root for x requires attention to the branch, since squaring can introduce solutions the original did not have.
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 613-615
Trap
\[ A = \int_{-1}^{2}\left[(y+2)-y^{2}\right]dx \]
Integrate a function of y with respect to x
Why: The student changes the integrand but not the differential.
The integrand is written in y and the differential says x, so the expression can be integrated with respect to neither.
\[ A = \int_{-1}^{2}\left[(y+2)-y^{2}\right]dy \]
Match the differential to the variable, and the limits too
Why: Horizontal strips need y limits and a dy.
This is Section 5.5's leftover-variable error in a new setting. The limits must also be y values — here negative one and two, not the corresponding x values of one and four.
Matching
The two orientations.
Match the pairs
Why: The two columns are mirror images with the roles of the axes exchanged. Nothing new has to be learned for horizontal strips beyond rewriting the boundaries as functions of y.
Two truths and a lie
All three are about horizontal strips.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The integrand, the differential and the limits must all refer to the same variable, exactly as in Section 5.5's substitutions. An integrand in y with a dx can be integrated with respect to neither variable and has no value.
Prediction
Commit before reasoning.
Predict first
Switching from vertical to horizontal strips, what must be converted?
Correct: All three together.
\[ y=\sqrt x, \; x\in[0,4] \;\longrightarrow\; x=y^{2}, \; y\in[0,2] \]
Why: The boundaries must be rewritten as functions of y, the differential becomes dy, and the limits become the y values at which the region begins and ends — which are generally different numbers from the x limits. Converting some but not all produces an expression that cannot be integrated at all, which is the same failure Section 5.5's incomplete substitutions produced.
Section
Section 4
Concept
A region whose top boundary changes needs several integrals with vertical strips but may need only one with horizontal strips. The choice can turn three integrals into one.
choosing the direction — Slicing in the direction along which each boundary is a single function throughout, so the region needs no splitting.
\[ \text{one integral} \;\text{ vs }\; \text{three} \]
The rule of thumb is to slice perpendicular to the direction in which the boundaries stay simple. A sketch shows this immediately and nothing else does.
Figure (svg): Choosing the variable: one integral, or three
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 612-620 — choosing the variable of integration
Picture it
The same region, two orientations.
Figure (svg): Choosing the variable: one integral, or three
The right-hand column is not merely tidier: it is a different amount of work entirely. Deciding which applies takes one sketch and saves the difference.
Worked example
Example 6.6. Compare the two setups.
\[ \text{Find the area bounded by } y=x-1, \; y^{2}=2x+6. \]
Find the intersections
Why: Substitute the line into the parabola.
\[ y = -2\text{ and } y = 4 \]
Consider vertical strips
Why: The top boundary changes.
Consider horizontal strips
Why: Left is the parabola, right is the line.
Write the width
Why: Right minus left, in y.
\[ (y + 1) - \frac{y ^{2} - 6}{2} \]
Integrate from -2 to 4
Why: One integral.
\[ 18 \]
Figure (svg): Choosing the variable: one integral, or three
\[ A = 18 \]
Verify: count what the other route would have cost
Why: With vertical strips the region splits at x equal to negative 3, where the parabola's two branches meet: the left piece runs between the two branches and the right piece between a branch and the line — two integrals, both containing square roots, and the split point itself has to be found. The horizontal route needed one integral of a quadratic. The sketch was what revealed the difference, and it took half a minute.
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 615-617
Sorting
Which boundaries stay simple?
Sort into buckets
Sort each region.
A sideways-opening parabola is the reliable signal for horizontal strips, because its two branches are what force a vertical-strip split. Spotting that shape in a sketch decides the question immediately.
Worked example
Checkpoint 6.6. The decision rule.
\[ \text{How do you decide which variable to integrate in?} \]
Sketch the region
Why: Always first.
Scan left to right
Why: Does the top or bottom boundary change?
Scan bottom to top
Why: Does the left or right boundary change?
Choose the direction with no change
Why: Or the fewest.
If both are equal
Why: Choose the pleasanter integrand.
Figure (svg): The solution to Worked example how to tell which direction shown as a ladder of expressions, one row per legal move
\[ \text{fewest pieces} \to \text{simplest integrand} \]
Verify: apply the rule to a case where both work
Why: For the region between the square root of x and x over 2, neither direction splits — so the tie is broken on the integrand, and the y version is a polynomial while the x version contains a square root. The polynomial is easier, so horizontal strips win on the second criterion rather than the first. Having two criteria in order, splitting first and simplicity second, resolves nearly every case.
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 617-619
Error analysis
A student sets up an area without drawing the region.
Annotate
On: \( A = \int_{-3}^{5}\left[(x-1)-\sqrt{2x+6}\right]dx \)
Almost every error in this section is prevented by a sketch, and almost none is prevented without one. It is the cheapest step in the whole procedure and the one most often skipped.
Fill the middle
A region needing two integrals one way and one the other.
Fill in the blanks
\text1 2 \text___; \quad \text___ ___
Why: Choosing the direction along which the boundaries do not change removes the split entirely. The saving is not cosmetic — it also removes the radicals that the other route's boundaries would require.
Two truths and a lie
All three are about choosing.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Vertical strips are always POSSIBLE but can require several integrals with awkward radicals, while horizontal strips may need one integral of a polynomial. The habit of always slicing vertically costs real work on exactly the problems where it matters.
Prediction
Commit before reasoning.
Predict first
Why is sketching the region the first step, every time?
Correct: Because it shows which boundaries change.
\[ \text{sketch} \to \text{crossings} + \text{which is on top} + \text{direction} \]
Why: The sketch makes visible where the curves cross, which function is on top, and whether the top or the left boundary changes along the region — the three facts the setup depends on. Every error in this section comes from getting one of those wrong, and none of them is reliably visible in the algebra alone. Thirty seconds of drawing prevents them all.
Section
Section 5
Concept
Every difficulty in this section is in the setup: finding the crossings, choosing the direction, identifying which boundary is which. The integral itself is Chapter 5 routine.
the setup — The sequence of decisions preceding the integration: sketch, intersections, direction, boundary identification. Errors here are invisible in the arithmetic that follows.
\[ \text{sketch} \to \text{crossings} \to \text{direction} \to \text{integrand} \]
This pattern holds for the whole of Chapter 6. Volumes, arc length and work all require a representative slice to be identified and described before any integration begins.
Figure (svg): The setup, with the step that decides everything
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 597-620 — setting up area problems
Picture it
The procedure, with the sketch first.
Figure (svg): The setup, with the step that decides everything
The last step is the only one Chapter 5 taught, and it is the only one that is routine. The four before it are where areas are won or lost.
Worked example
Example 6.4. All five steps.
\[ \text{Find the area bounded by } y=x^{2} \text{ and } y=x+2. \]
Sketch
Why: A parabola and a line cutting it.
Find the intersections
Why: Set equal and factor.
\[ x ^{2} - x - 2 = 0\text{ at } x = -1, 2 \]
Choose the direction
Why: The top and bottom are single functions.
Identify top and bottom
Why: Test x = 0.
\[ \text{the line at } 2,\text{ the parabola at } 0 \]
Integrate
Why: Top minus bottom.
\[ \frac{9}{2} \]
Figure (svg): The setup, with the step that decides everything
\[ A = \int_{-1}^{2}\left(x+2-x^{2}\right)dx = \frac92 \]
Verify: sanity-check with a rough geometric estimate
Why: The region is roughly a lens 3 wide and at most 2.25 tall at its widest, so an area between 3 and 6 is expected — and 4.5 sits in the middle. Each of the five steps did something: the sketch showed a single region, the intersections gave the limits, the direction avoided a split, and the test point fixed the sign. Only the last step used Chapter 5.
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 609-611
Ranking
An area between curves, from start to finish.
Put in order
Why: Only step e uses Chapter 5, and it is the only routine one. The first four are where the answer is determined, and skipping step a makes each of the others substantially harder.
Worked example
Checkpoint 6.4. More than two curves.
\[ \text{Find the area bounded by } y=x, \; y=x/2 \text{ and } x=4. \]
Sketch
Why: A triangle with a vertex at the origin.
Find the relevant corners
Why: Where each pair meets.
\[ (0, 0), (4, 4), (4, 2) \]
Choose vertical strips
Why: Top and bottom are single functions.
Write the height
Why: Top minus bottom.
\[ x - \frac{x}{2} = \frac{x}{2} \]
Integrate from 0 to 4
Why: The power rule.
\[ 4 \]
Figure (svg): The solution to Worked example an area with three boundaries shown as a ladder of expressions, one row per legal move
\[ A = \int_{0}^{4}\frac{x}{2}\,dx = 4 \]
Verify: check against the triangle's area formula
Why: The triangle has vertices at the origin and at the points four comma four and four comma two, giving a vertical side of length 2 and a horizontal extent of 4 — so its area is half of 4 times 2, which is 4. The integral reproduces elementary geometry exactly, which is the check worth making whenever a region happens to be a polygon. A third boundary changed nothing about the method: it simply supplied one of the limits.
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 611-612
Trap
\[ A = \int_{0}^{4}\left(x+2-x^{2}\right)dx \]
Use limits that appear in the problem statement
Why: The student takes numbers from the wording rather than the intersections.
The curves meet at negative one and two, so the region does not extend to four. Beyond two the parabola is above and the integrand goes negative.
\[ A = \int_{-1}^{2}\left(x+2-x^{2}\right)dx \]
Take the limits from where the region actually begins and ends
Why: Usually the intersections, unless a vertical boundary is given.
A third boundary such as a vertical line does supply a limit, but the intersections supply the rest. The sketch makes clear which numbers bound the region and which merely appear in the problem.
Fill the middle
The intersections supply the interval.
Fill in the blanks
x^2=x+2 \;\Longrightarrow\; x=-1 \text___ x=___
Why: The limits come from where the region begins and ends, which for two curves means their intersections. Numbers appearing elsewhere in a problem are limits only if they are boundaries of the region.
Sorting
The region's ends, not the problem's wording.
Sort into buckets
Sort each source of a limit.
The test is always the same: does the region actually stop there? The sketch answers it in a second, and no amount of reading the problem statement does so reliably.
Prediction
Commit before reasoning.
Predict first
Which step in an area problem is most often the source of error?
Correct: The setup.
\[ \text{setup errors are silent; arithmetic errors are not} \]
Why: The integration is Chapter 5 material and is usually a polynomial. What goes wrong is a missed crossing, a reversed difference, limits taken from the wording rather than the region, or a direction forcing an unnecessary split — and none of those is visible in the arithmetic afterwards. A sketch prevents all four, which is why it is the first step every time.
Comparison
Fill the blanks. The methods are mirror images.
Comparison matrix
| Vertical strips | Horizontal strips | |
|---|---|---|
| Integrate in | x, with dx | y, with dy |
| The strip's size | top minus bottom | right minus left |
| Boundaries written as | functions of x | functions of y |
| Choose it when | the top and bottom never change | the left and right never change |
The last row is the decision rule, and a sketch is what answers it. Choosing badly does not give a wrong answer, only several times as much work.
Pattern
Given a region bounded by curves.
Only the last step is Chapter 5 material. The four before it determine the answer and none of their errors shows up in the arithmetic, which is why the sketch is worth the time it takes.
Stewart, Calculus: Early Transcendentals 8e, §6.1 Areas Between Curves §6.1, pp. 428-437
Check
The basic setup.
Check your understanding
What is the area between y = 2x and y = x^2?
Answer: A
Why: The line is above between 0 and 2, giving the integral of 2x minus x squared.
Check
Crossings.
Check your understanding
Integrating x minus x cubed from -1 to 1 gives zero. Is that the area?
Answer: A
Why: Which curve is on top changes at the origin, so the interval must be split there.
Check
Choosing the direction.
Check your understanding
A region is bounded by a sideways-opening parabola and a line. Which strips are usually better?
Answer: A
Why: The parabola's two branches would force a vertical-strip split.
Real world
A civil engineer is designing a drainage channel whose cross-section is bounded below by a parabolic bed and above by the water surface at a given level. The flow capacity depends on the cross-sectional area of the water.
Discussion prompt
Explain how the area is set up, which variable is natural, and how the area changes as the water level rises.
Hint: The bed is a parabola opening upward and the surface is a horizontal line.
Answer:
The region is bounded below by the bed and above by a horizontal line at the water level, so vertical strips are natural: each has height the level minus the bed, and the limits are where the surface meets the bed.
\[ A(h) = \int_{-\sqrt{h/k}}^{\sqrt{h/k}}\left(h - kx^{2}\right)dx = \frac{4}{3}h\sqrt{\frac{h}{k}} \]
The area grows faster than the depth, as the three-halves power — because a deeper channel is also wider. Doubling the depth multiplies the area by about 2.83, not by 2, which is why a channel that is adequate in normal conditions can carry far more than expected in a flood.
Note the limits depend on the water level, so this is not a fixed integral but a function of it — and differentiating that function gives the rate at which capacity grows with depth, which is what a designer actually needs when sizing for a rare event.
The choice of vertical strips matters here: horizontal strips would need the parabola solved for x, producing square roots and a width of twice a root rather than a difference of polynomials. The sketch settles it immediately, and the setup is the whole difficulty as usual.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why does a region between two curves need no splitting where it crosses the x-axis?
Correct: Because the integrand is a difference.
\[ (f+c)-(g+c) = f-g \]
Why: Adding a constant to both boundaries leaves their difference unchanged, so a region entirely below the axis has the same integral as its translate above. Chapter 5 needed splitting because one boundary there was the axis itself, which does not move with the region. Splitting is still needed here, but at crossings of the two CURVES rather than of the axis.
Explain it
They computed an area and got a negative number.
Discussion prompt
In four sentences or fewer, tell them what happened.
Hint: Ask which curve is on top.
Answer:
A negative answer means the difference was taken the wrong way round: they subtracted the top from the bottom rather than the other way. Which curve is on top is a fact about the region, not about the order the problem happens to list them in.
Have them test a point between the intersections and see which function gives the larger value — that one goes first. A quick sketch shows it even faster, and it also reveals whether the curves cross inside the interval, which would need a split.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For intersections, set the functions equal and factor completely, counting the roots. For which is on top, test a point between. For horizontal strips, rewrite the boundaries as functions of y and convert the limits too. For choosing, sketch and ask which boundaries change. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, sketch two curves with a labelled vertical strip between them, and write the integral formula beside it with the strip's height and width marked. Below, draw the same region translated so part of it lies below the axis, and write one line on why the area is unchanged. In the middle of the page, sketch two curves crossing inside an interval, shade the two pieces differently, and write both the correct split computation and what a single integral would have given. Beside it, sketch a region bounded left and right, draw a labelled horizontal strip, and write the corresponding integral with dy. In the lower half, sketch a region bounded by a sideways parabola and a line, and set it up both ways — counting the integrals each needs. At the bottom, list the five steps of the procedure with the sketch first.
If your two setups at the bottom took the same amount of work, look again — the vertical one should need two integrals containing square roots and the horizontal one a single polynomial, and seeing that gap once is what makes the choice automatic afterwards.
Recap
Five things, and only the last uses Chapter 5.
| If you see | Then |
|---|---|
| Two curves and no interval | The intersections are the limits |
| A region below the axis | Nothing changes: it is still top minus bottom |
| Curves crossing inside | Split there and add the pieces |
| A negative area | The difference was reversed |
| A sideways-opening curve | Consider horizontal strips |
| A boundary that changes along the region | Slice the other way |
| Any area problem at all | Sketch it first |
Section 6.2 replaces the strip with a slice of a solid. The same argument — describe one representative piece, then integrate — gives volumes, and the whole of the rest of this chapter is that one idea applied to new quantities.
OpenStax Calculus Volume 1, §6.1 Areas between Curves §6.1, pp. 540-550 — everything on these slides traces back here
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