Unit 8 · Topic 8.6 · about 30 minutes · 7 parts · 11 scored checks, 3 of them multiple choice, plus 1 from earlier lessons

AP Calculus AB 8.6: Finding the Area Between Curves That Intersect at More Than Two Points

Q: Should I split the integral or use the absolute value?

Both give the same area. By hand, split at the meeting points, because the absolute value changes formula there: on each piece it is just top minus bottom, and that is what you can find an antiderivative for. On a graphing calculator, the absolute value form is one entry, and the calculator handles the switch for you.

Q: How do I know which graph is on top on a piece?

Pick any input strictly between two meeting points and compare f and g there. Between meeting points the difference never reaches 0, so it cannot change sign, and one test settles the whole piece.

Q: What if a piece comes out negative?

Then you subtracted top from bottom on that piece. Swap the two functions there; the size of the piece stays the same and the sign becomes positive.

Q: Why does a single integral of f(x) - g(x) sometimes give 0?

A definite integral adds signed values. If the pieces where f is on top have the same total area as the pieces where g is on top, they cancel exactly, even though the region has positive area.

Q: Do I need to round the meeting points before I integrate?

No. Store them on the calculator and use the stored values as limits. Rounding them first can change the third decimal place of the area.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 8: 10 to 15% of the multiple-choice section

What you'll be able to do

Calculate the area of a region between two curves that cross inside the interval, either as a sum of definite integrals, one for each piece, or as one definite integral of the absolute value of the difference of the two functions.

Key vocabulary

Term	What it means	The exam trap
Area between two curves	The area of the region bounded by two graphs, calculated with one or more definite integrals.	Before you write a single integral, check whether the top graph stays on top across the whole interval.
Point of intersection	A point where the graphs of $f$ and $g$ meet, at an input where $f (x) = g (x)$ .	On a calculator, find every one in the interval, not only the ones in the first window you see.
Absolute value of the difference, $\| f (x) - g (x) \|$	The vertical distance between the graphs of $f$ and $g$ at $x$ , whichever one is on top.	Read where the bars sit before you evaluate: bars around the integrand and bars around the whole integral are different expressions.
Sum of definite integrals	An area written as several definite integrals over pieces of the interval that are added together.	Each piece needs its own top and bottom; one that comes out negative has them reversed.

Graphs of f(x) = x cubed minus 6x squared plus 9x and g(x) = x for x from 0 to 4. They meet at (0, 0), (2, 2) and (4, 4). The region between them is shaded in two colors: from x = 0 to x = 2 the graph of f is on top, and from x = 2 to x = 4 the graph of g is on top.

Part 1 Think first How much metal is in the two pieces?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A sign maker cuts two leaf-shaped pieces from a sheet of metal. On her plan, measured in feet, each piece lies between the line $y = x$ and the curve $y = x^{3}$ for $- 1 \leq x \leq 1$ , and the two graphs meet at $x = - 1$ , 0 and 1. Her apprentice works out $\int_{- 1}^{1} (x^{3} - x) d x = 0$ . How many square feet of metal do the two pieces use together?

In short. When the graphs of $f$ and $g$ cross, the area between them is a sum of definite integrals of top minus bottom, one for each piece between meeting points, which equals $\int_{a}^{b} | f (x) - g (x) | d x$ .

Solve $f (x) = g (x)$ to find where the graphs meet; those inputs split the interval into pieces.
On each piece one function stays on top, so integrate top minus bottom there and add the results.
One integral of $| f (x) - g (x) |$ gives the same area, and a graphing calculator evaluates it directly.

Part 2 Learn it Three short ideas, each checked as you go

2a. Find where the graphs meet, then which is on top

In Topic 8.4 the area between two curves was one integral of top minus bottom, $\int_{a}^{b} (f (x) - g (x)) d x$ , with $f$ on top across the whole interval. When the graphs cross, the top and bottom curves trade places, so first you need to know where they cross.

Solve $f (x) = g (x)$ . Each solution in the interval is a point where the graphs meet, and those inputs cut the interval into pieces.

Test one input in each piece. Between two meeting points $f - g$ is continuous and never 0, so it keeps one sign there. Its sign at any one input of the piece tells you which graph is on top across the whole piece.

The graph at the top of this page shows $f (x) = x^{3} - 6 x^{2} + 9 x$ and $g (x) = x$ on $[0, 4]$ . Setting them equal gives $x^{3} - 6 x^{2} + 8 x = 0$ , which factors as $x (x - 2) (x - 4) = 0$ , so the graphs meet at $x = 0$ , 2 and 4. At $x = 1$ , $f (1) = 4$ and $g (1) = 1$ , so $f$ is on top on $(0, 2)$ . At $x = 3$ , $f (3) = 0$ and $g (3) = 3$ , so $g$ is on top on $(2, 4)$ .

Quick check · Sort it

No calculator. Let $f (x) = x^{3}$ and $g (x) = x^{2} + 2 x$ . Their graphs meet at $x = - 1$ , $x = 0$ and $x = 2$ . On each interval, which graph is on top? Tap a card, then a bin.

The graph of $f$ is on top

The graph of $g$ is on top

2b. One integral for each piece, then add

For the pair in the graph, a single integral over $[0, 4]$ gives $\int_{0}^{4} (f (x) - g (x)) d x = 0$ , though the shaded region plainly has area. On $(2, 4)$ the difference $f - g$ is negative, so that piece counts as a negative amount and cancels the first piece. A definite integral of $f - g$ adds signed values; an area adds positive ones.

So write one integral per piece, each with top minus bottom:

Area = \int_{0}^{2} (f (x) - g (x)) d x + \int_{2}^{4} (g (x) - f (x)) d x .

Here $f (x) - g (x) = x^{3} - 6 x^{2} + 8 x$ , with antiderivative $D (x) = \frac{x^{4}}{4} - 2 x^{3} + 4 x^{2}$ . Then $D (0) = 0$ , $D (2) = 4$ and $D (4) = 0$ . The first piece is $D (2) - D (0) = 4$ . The second is $- (D (4) - D (2)) = - (0 - 4) = 4$ . The area is $4 + 4 = 8$ .

Every piece of an area comes out positive when it is set up as top minus bottom. If a piece comes out negative, the top and bottom are reversed on that piece.

Quick check · Calculate

No calculator. Let $f (x) = x^{3} - 4 x^{2}$ and $g (x) = 2 x^{2} - 5 x$ . Their graphs meet at $x = 0$ , $x = 1$ and $x = 5$ . Find the area of the region between the graphs from $x = 0$ to $x = 5$ . Give an exact decimal.

Your answer

2c. Or one integral of the absolute value

The height of the region at $x$ is the distance between the graphs, whichever is on top, and that distance is $| f (x) - g (x) |$ . So one integral gives the whole area:

Area = \int_{a}^{b} | f (x) - g (x) | d x .

On each piece the absolute value is just top minus bottom, so this integral equals the sum from 2b. For the pair in the graph, $\int_{0}^{4} | f (x) - g (x) | d x = 8$ .

Which form you use depends on the tool. By hand you still split at the meeting points, because the absolute value changes formula there: on each piece it is just top minus bottom, and that is what you can find an antiderivative for. On a graphing calculator you can enter $\int_{a}^{b} | f (x) - g (x) | d x$ as it stands. When the region runs between meeting points the calculator finds by solving $f (x) = g (x)$ , store those values and use them as the limits, so no rounding creeps into the answer.

Watch where the bars go. In $| \int_{a}^{b} (f (x) - g (x)) d x |$ the bars come after the integral has added the signed pieces, so the canceling has already happened. For the pair in the graph that expression is $| 0 | = 0$ .

Quick check · Multiple choice

No calculator. The functions $f$ and $g$ are continuous, and their graphs cross at $x = 1$ and at $x = 3$ , with no other meeting points between $x = 0$ and $x = 5$ . Which expression gives the area of the region between the graphs from $x = 0$ to $x = 5$ ?

Common mistakes

The mistake	Why it is wrong	Do this instead
Writing one integral of $f (x) - g (x)$ over the whole interval when the graphs cross inside it.	On the pieces where $g$ is on top, $f - g$ is negative, so those pieces subtract and the result can be far too small, or even 0 or negative.	Solve $f (x) = g (x)$ , split at every meeting point, and integrate top minus bottom on each piece; or integrate $\| f (x) - g (x) \|$ .
Putting the absolute value around the whole integral, $\| \int_{a}^{b} (f (x) - g (x)) d x \|$ .	The integral has already let the pieces cancel before the bars act, so the bars only fix the sign of what is left.	Put the bars around the integrand: $\int_{a}^{b} \| f (x) - g (x) \| d x$ .
Finding only the meeting points that show in the calculator's first window.	A missed meeting point leaves a piece where the top and bottom switch partway, and its integral cancels part of itself.	Graph both functions in a window that covers the whole interval, and solve $f (x) = g (x)$ for every meeting point in it.

Part 3 Watch it done An area that needs the calculator

Watch each step, then do the last one yourself.

Graphing calculator required. Let $f (x) = x^{3} - 4 x$ and $g (x) = \cos x$ . The graphs of $f$ and $g$ meet at three points. In radian mode, find the area of the region enclosed between the graphs from the first meeting point to the last.

Step 1 · Find the meeting points
Solve $x^{3} - 4 x = \cos x$ on the calculator. The solutions are $x = A$ , $B$ and $C$ , with $A \approx - 2.056$ , $B \approx - 0.246$ and $C \approx 1.952$ . Store all three; do not retype the rounded values.
Step 2 · See which is on top
At $x = - 1$ , between $A$ and $B$ : $f (- 1) = 3$ and $g (- 1) = \cos (- 1) \approx 0.540$ , so $f$ is on top. At $x = 1$ , between $B$ and $C$ : $f (1) = - 3$ and $g (1) \approx 0.540$ , so $g$ is on top.
Step 3 · Write the area
$Area = \int_{A}^{B} (x^{3} - 4 x - \cos x) d x + \int_{B}^{C} (\cos x - x^{3} + 4 x) d x$ , which equals $\int_{A}^{C} | x^{3} - 4 x - \cos x | d x$ .
Why this step?
Why not enter $\int_{A}^{C} (x^{3} - 4 x - \cos x) d x$ and stop there?
Between $B$ and $C$ the integrand $x^{3} - 4 x - \cos x$ is negative, because $g$ is on top there, so that piece would subtract. The calculator gives $\int_{A}^{C} (x^{3} - 4 x - \cos x) d x \approx - 1.817$ : the two pieces partly cancel, and an area cannot be negative.
Your step · Calculate

Graphing calculator required. Finish it. With $A$ and $C$ the first and last solutions of $x^{3} - 4 x = \cos x$ , stored on your calculator, evaluate $\int_{A}^{C} | x^{3} - 4 x - \cos x | d x$ in radian mode. Give your answer to three decimal places.

Your answer

Part 4 Try it A pendant and a classmate's shortcut

Try it · Calculate

Graphing calculator required. A jeweler cuts a pendant from silver sheet. On her plan, in centimeters, the pendant is the region between $y = x \cos x + 2$ and $y = 2 - 0.25 x$ from $x = 0$ , where the graphs meet, to the last point where they meet for $0 \leq x \leq 5$ . In radian mode, find the area of the pendant in square centimeters. Give your answer to three decimal places.

Your answer

Try it · Decide and justify

No calculator. The functions $f$ and $g$ are continuous, and their graphs meet only at $x = 2$ for $0 \leq x \leq 5$ . A classmate says $\int_{0}^{5} | f (x) - g (x) | d x = | \int_{0}^{2} (f (x) - g (x)) d x | + | \int_{2}^{5} (f (x) - g (x)) d x |$ . Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Answer it on paper, then score it

Free response is handwritten on the exam, so write your answer on paper first. Nothing you type here leaves this page. Then open the model answer and score yourself on each point.

Free response is 50% of the exam score. This one is shaped like Question 1, from free-response Part A (graphing calculator required).

Free-response practice · Question 1 style

Graphing calculator required. A garden plan, drawn with both axes in meters, has a curved path edge $y = f (x)$ and a straight path edge $y = g (x)$ , where $f (x) = 1 + 2 \cos x$ and $g (x) = 0.5 x - 1$ for $0 \leq x \leq 8$ . Between the points where the two edges meet lie two flower beds. Use radian mode.

(a) Find the $x$ -coordinates of all points where the graphs of $f$ and $g$ meet for $0 \leq x \leq 8$ . Give each to three decimal places.

(b) Find the total area of the two flower beds, the region between the graphs of $f$ and $g$ from the first meeting point to the last. Show the setup you use on your calculator, and give your answer to three decimal places. Indicate units of measure.

(c) A helper evaluates $\int_{P}^{R} (f (x) - g (x)) d x$ , where $P$ and $R$ are the first and last meeting points, gets $- 1.761$ , and reports the beds' total area as 1.761 square meters. Explain why the helper's value is not the total area.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) Solve $1 + 2 \cos x = 0.5 x - 1$ on the calculator, storing the solutions as $P$ , $Q$ and $R$ : $x = 2.073$ , $x = 4.953$ and $x = 7.004$ .

(b) $Area = \int_{P}^{R} | f (x) - g (x) | d x = \int_{P}^{R} | 1 + 2 \cos x - (0.5 x - 1) | d x = 4.228$ square meters.

(c) The graph of $g$ is on top of the graph of $f$ from $P$ to $Q$ , so $f (x) - g (x) < 0$ there, and that bed enters the helper's integral as about $- 2.994$ . The second bed, where $f$ is on top, enters as about $1.233$ . The helper's integral adds these signed values, so the first bed cancels part of the second, giving $- 1.761$ ; taking the absolute value afterward does not undo that. The area needs each bed counted as positive, which gives 4.228 square meters.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Sets up $1 + 2 \cos x = 0.5 x - 1$ (or $f (x) = g (x)$ ) and gives all three meeting points, 2.073, 4.953 and 7.004.
Writes an integral with the integrand $| f (x) - g (x) |$ , limits at the first and last meeting points and $d x$ ; or a sum of two integrals of top minus bottom split at 4.953.
Gets 4.228 (or 4.227) square meters.
Explains that $f - g$ is negative on the first bed, so that bed subtracts in the helper's integral and partly cancels the second.

Part 6 Check yourself Two graphs from a table of integrals

The two case questions use the same table. Answer them from it.

The functions $f$ and $g$ are continuous on $[0, 6]$ , and their graphs meet only at $x = 1$ and $x = 4$ on that interval. The table gives the definite integral of $f (x) - g (x)$ over each piece.

The integral of f minus g over each piece
Interval	$\int (f (x) - g (x)) d x$ over the interval
$0 \leq x \leq 1$	2.5
$1 \leq x \leq 4$	$- 6$
$4 \leq x \leq 6$	1.5

Case question 1 · Calculate

No calculator. What is the area of the region between the graphs of $f$ and $g$ from $x = 0$ to $x = 6$ ?

Your answer

Case question 2 · Multiple choice

No calculator. Which sum of integrals is equal to the area of the region between the graphs of $f$ and $g$ from $x = 0$ to $x = 6$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds the area of the region between the graphs of $f (x) = x^{3} - x^{2}$ and $g (x) = 2 x$ from $x = - 1$ to $x = 2$ . Tap the first line with a mistake.

Check yourself · Put it in order

No calculator. Put the steps for finding the area between two graphs that cross, by hand, in order. Tap them in sequence.

Part 7 Keep it Bring back earlier ideas

Today's lesson stands on area between curves that do not cross, from Topic 8.4, and on reading a definite integral as an accumulation, from Topic 6.4.

Bring it back · Multiple choice

No calculator. The graphs of $y = 4 - x^{2}$ and $y = x + 2$ meet at $x = - 2$ and $x = 1$ . What is the area of the region between them?

From 6.4 The Fundamental Theorem of Calculus and Accumulation Functions · Calculate

No calculator. Let $f (t) = 6 - 2 t$ , and let $g (x) = \int_{1}^{x} f (t) d t$ . What is $g (3)$ ?

Your answer

In three lines

When two graphs cross, solve $f (x) = g (x)$ to find where they meet; those points split the region into pieces.
On each piece integrate top minus bottom, and add the pieces; every piece of an area is positive.
One integral of $| f (x) - g (x) |$ gives the same area, with the bars around the integrand, not the integral.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

On a calculator part, write the equation you solve for the meeting points and the integral you evaluate, with the integrand, both limits and $d x$ , before the number. Store the meeting points and use the stored values as limits, and give the result to three decimal places with units. Without a calculator, show the split at each meeting point and an antiderivative for each piece.

Questions students ask

Should I split the integral or use the absolute value?

Both give the same area. By hand, split at the meeting points, because the absolute value changes formula there: on each piece it is just top minus bottom, and that is what you can find an antiderivative for. On a graphing calculator, the absolute value form is one entry, and the calculator handles the switch for you.

How do I know which graph is on top on a piece?

Pick any input strictly between two meeting points and compare $f$ and $g$ there. Between meeting points the difference never reaches 0, so it cannot change sign, and one test settles the whole piece.

What if a piece comes out negative?

Then you subtracted top from bottom on that piece. Swap the two functions there; the size of the piece stays the same and the sign becomes positive.

Why does a single integral of

f (x) - g (x)

sometimes give 0?

A definite integral adds signed values. If the pieces where $f$ is on top have the same total area as the pieces where $g$ is on top, they cancel exactly, even though the region has positive area.

Do I need to round the meeting points before I integrate?

No. Store them on the calculator and use the stored values as limits. Rounding them first can change the third decimal place of the area.

Course alignment, for teachers

AP Calculus AB topic 8.6, Unit 8: Applications of Integration.

Skill 2.B: Identify mathematical information from graphical, numerical, analytical, and/or verbal representations.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.