Unit 8 · Topic 8.4 · about 30 minutes · 7 parts · 11 scored checks, 3 of them multiple choice, plus 1 from earlier lessons

AP Calculus AB 8.4: Finding the Area Between Curves Expressed as Functions of x

Q: Does the x-axis count as one of the curves?

Yes. It is the line y = 0. For a region between a curve and the axis, the axis is the top when the curve is below it, and the bottom when the curve is above it.

Q: When do I need a calculator to find an area?

When f(x) = g(x) cannot be solved by hand, or the integrand has no antiderivative you can write. Then solve for the meeting points and evaluate the integral on the calculator, and show the integral you evaluated, with its limits and dx.

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 in the plane between two curves written as functions of x, with a definite integral.

Key vocabulary

Term	What it means	The exam trap
Region in the plane	A part of the coordinate plane closed off by graphs, such as the points between two curves from one x-value to another.	Read every boundary a question names, including the $x$ -axis or a vertical line, before you set up an integral.
Area between two curves	The size of the region enclosed by two graphs over an interval, found with a definite integral.	Find where the graphs meet and which one is higher before you write the integral, and check that it stays higher across the whole interval.
Limits of integration	The numbers $a$ and $b$ in $\int_{a}^{b}$ , which for an area in $x$ are where the region starts and stops along the $x$ -axis.	Make sure the limits and the differential describe the same variable.

The line f(x) = x + 2 and the parabola g(x) = x squared over 2 minus 2, meeting at (-2, 0) and (4, 6). The region between them, with the line on top, is shaded and partly below the x-axis. A thin strip at x = 1 runs from -1.5 up to 3, height 4.5, and the shaded area is 18.

Part 1 Think first A piece of glass between two edges

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

Predict first

A glass artist cuts a leaf-shaped piece of glass. On a grid marked in feet, its upper edge is the line $y = x$ and its lower edge is the curve $y = x^{2}$ , and the edges meet at $(0, 0)$ and $(1, 1)$ . The region under the line from $x = 0$ to $x = 1$ , down to the $x$ -axis, has area $\frac{1}{2}$ square foot. The region under the curve over the same interval has area $\frac{1}{3}$ square foot. What is the area of the piece of glass?

In short. The area of the region between $y = f (x)$ and $y = g (x)$ from $x = a$ to $x = b$ , where $f (x) \geq g (x)$ on $[a, b]$ , is the definite integral $\int_{a}^{b} (f (x) - g (x)) d x$ .

A thin vertical strip of the region has height top minus bottom, $f (x) - g (x)$ , and the integral adds up all the strips.
The limits are the $x$ -values where the region starts and stops: given vertical lines, or the solutions of $f (x) = g (x)$ .
Top minus bottom is a height even where the region dips below the $x$ -axis, so an area set up this way is never negative.

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

4a. Top minus bottom, one strip at a time

Cut the region between two curves into thin vertical strips, like the one in the graph at the top. A strip at $x$ runs from the bottom curve up to the top curve, so its height is top minus bottom, $f (x) - g (x)$ . Its width is a small $Δ x$ , so its area is about $(f (x) - g (x)) Δ x$ .

Adding the strips is a Riemann sum, and as the strips get thinner the sum becomes a definite integral. If $f (x) \geq g (x)$ for every $x$ in $[a, b]$ , the area between the curves from $x = a$ to $x = b$ is

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

In the graph, $f (x) = x + 2$ is on top of $g (x) = \frac{x^{2}}{2} - 2$ from $x = - 2$ to $x = 4$ , so the shaded area is

\int_{- 2}^{4} (x + 4 - \frac{x^{2}}{2}) d x = {[\frac{x^{2}}{2} + 4 x - \frac{x^{3}}{6}]}_{- 2}^{4} = \frac{40}{3} - (- \frac{14}{3}) = 18.

Think first's glass is the same idea in one integral: $\int_{0}^{1} (x - x^{2}) d x = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ square foot.

Quick check · Calculate

No calculator. From $x = 0$ to $x = 3$ , the graph of $y = x^{2} + 2$ lies above the graph of $y = x$ . What is the area of the region between the two graphs from $x = 0$ to $x = 3$ ?

Your answer

4b. Where the region starts and stops

Often no vertical lines are given, and the region is closed off where the two curves meet. Then the limits of integration are the $x$ -values of the meeting points, and you find them by setting the functions equal.

Take $y = x + 6$ and $y = x^{2}$ . Setting $x + 6 = x^{2}$ gives $x^{2} - x - 6 = 0$ , so $(x - 3) (x + 2) = 0$ , and the curves meet at $x = - 2$ and $x = 3$ . The meeting points are $(- 2, 4)$ and $(3, 9)$ , but the integral is in $x$ , so only $- 2$ and $3$ are limits. The 4 and the 9 play no part.

Next decide which curve is on top. Pick any $x$ between the limits and compare: at $x = 0$ , $x + 6 = 6$ and $x^{2} = 0$ , so the line is on top. The curves meet only at $x = - 2$ and $x = 3$ , so they cannot trade places in between, and one test value settles it for the whole interval. The area is

\int_{- 2}^{3} ((x + 6) - x^{2}) d x .

Be careful with a curve's own $x$ -intercepts. Where $y = x^{2}$ meets the $x$ -axis tells you nothing here, because the region's bottom is the parabola, not the axis.

Quick check · Multiple choice

No calculator. The region $R$ is bounded by the graphs of $y = 4 x - x^{2}$ and $y = x$ . Which integral gives the area of $R$ ?

4c. Regions below the x-axis

Top minus bottom does not depend on where the $x$ -axis is. In the graph at the top, the strip at $x = 1$ runs from $g (1) = - 1.5$ up to $f (1) = 3$ . Part of it is below the axis, and its height is still $3 - (- 1.5) = 4.5$ , a positive length.

The $x$ -axis can be one of the two curves: it is the line $y = 0$ . The region between $y = x^{2} - 1$ and the $x$ -axis lies below the axis from $x = - 1$ to $x = 1$ . Its top is $y = 0$ and its bottom is the parabola, so its area is

\int_{- 1}^{1} (0 - (x^{2} - 1)) d x = \int_{- 1}^{1} (1 - x^{2}) d x = \frac{4}{3} .

The integral $\int_{- 1}^{1} (x^{2} - 1) d x$ is $- \frac{4}{3}$ : the right size with the wrong sign, because it is bottom minus top. A definite integral can be negative, but an area cannot, so a negative answer to an area question means the subtraction is upside down.

Quick check · Decide and justify

No calculator. A classmate says the area of the region bounded by the graph of $y = x^{2} - 4$ and the $x$ -axis is $\int_{- 2}^{2} (4 - x^{2}) d x$ . Is the classmate right?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Subtracting top from bottom, and reporting a negative area.	The height of a strip is the upper value minus the lower one. The other order gives the same size with the wrong sign.	Test one $x$ between the limits to see which function is on top, and write that one first.
Using the $y$ -coordinates of the meeting points as the limits.	The integral is with respect to $x$ , so its limits are where the region starts and stops along the $x$ -axis.	Solve $f (x) = g (x)$ and use the solutions, which are $x$ -values. Leave the $y$ -coordinates out of the integral.
Integrating only the top function, as if the region reached down to the $x$ -axis.	That is the area under the top curve. The region stops at the bottom curve, which is the $x$ -axis only when the question says so.	Name both boundaries before you write anything, and subtract the bottom one inside the integral.

Part 3 Watch it done The area of a region closed off by two curves

Watch each step, then do the last one yourself.

Find the area of the region bounded by the graphs of $y = 2 x + 2$ and $y = x^{2} - 2 x - 3$ .

Step 1 · Find where the curves meet
$2 x + 2 = x^{2} - 2 x - 3$ gives $x^{2} - 4 x - 5 = 0$ , so $(x - 5) (x + 1) = 0$ , and $x = - 1$ or $x = 5$ .
Step 2 · Decide which curve is on top
At $x = 0$ , between the limits, $2 (0) + 2 = 2$ and $0^{2} - 2 (0) - 3 = - 3$ . The line is on top.
Why this step?
Why is one test value enough to know the line is on top for all of $[- 1, 5]$ ?
The curves meet only at $x = - 1$ and $x = 5$ . To trade places in between, the parabola would have to cross the line, and a crossing would be another meeting point. So the line stays on top for every $x$ between $- 1$ and $5$ .
Step 3 · Write the integral, top minus bottom
$A = \int_{- 1}^{5} ((2 x + 2) - (x^{2} - 2 x - 3)) d x = \int_{- 1}^{5} (- x^{2} + 4 x + 5) d x$
Step 4 · Find an antiderivative and its values at the limits
$F (x) = - \frac{x^{3}}{3} + 2 x^{2} + 5 x$ , so $F (5) = - \frac{125}{3} + 50 + 25 = \frac{100}{3}$ and $F (- 1) = \frac{1}{3} + 2 - 5 = - \frac{8}{3}$ .
Your step · Calculate

No calculator. Finish it. With $F (x) = - \frac{x^{3}}{3} + 2 x^{2} + 5 x$ , $F (5) = \frac{100}{3}$ and $F (- 1) = - \frac{8}{3}$ , what is the area of the region bounded by $y = 2 x + 2$ and $y = x^{2} - 2 x - 3$ ?

Your answer

Part 4 Try it A calculator region and a setup to sort

Try it · Calculate

Graphing calculator required. Let $R$ be the region bounded by the graphs of $y = \cos x$ and $y = x^{2}$ . Find the area of $R$ , with your calculator in radian mode. Give your answer to three decimal places.

Your answer

Try it · Sort it

No calculator. The region $R$ is bounded by the graphs of $y = 4 - x^{2}$ and $y = x + 2$ , which meet at $(- 2, 0)$ and $(1, 3)$ . Sort each expression: does it give the area of $R$ , or not? Tap a card, then a bin.

Gives the area of $R$

Does not give the area of $R$

Part 5 Explain it Answer it in free-response form

On the exam free response is handwritten, so work on paper first. Nothing you type here leaves this page. Then compare with the model answer and score each point.

Free response is 50% of the exam score. This one is shaped like Question 5, from free-response Part B (calculator not permitted).

Free-response practice · Question 5 style

No calculator. A garden bed is the region $R$ between two curved edges. On a grid marked in meters, one edge is the graph of $f (x) = 4 + 3 x - x^{2}$ and the other is the graph of $g (x) = x^{2} - 3 x + 4$ .

(a) Find the $x$ -coordinates of the two points where the edges meet. Show the work that leads to your answer.

(b) Find the area of the garden bed. Show the integral you use, and indicate units of measure.

(c) A gardener lays a row of bricks along the vertical line $x = 1.5$ . Verify that this line divides $R$ into two regions of equal area.

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

Model answer

(a) $f (x) = g (x)$ gives $4 + 3 x - x^{2} = x^{2} - 3 x + 4$ , so $6 x - 2 x^{2} = 0$ and $2 x (3 - x) = 0$ . The edges meet at $x = 0$ and $x = 3$ .

(b) At $x = 1$ , $f (1) = 6$ and $g (1) = 2$ , so $f$ is on top on $[0, 3]$ . The area is $\int_{0}^{3} (f (x) - g (x)) d x = \int_{0}^{3} (6 x - 2 x^{2}) d x = {[3 x^{2} - \frac{2 x^{3}}{3}]}_{0}^{3} = 27 - 18 = 9$ square meters.

(c) $\int_{0}^{1.5} (6 x - 2 x^{2}) d x = {[3 x^{2} - \frac{2 x^{3}}{3}]}_{0}^{1.5} = 6.75 - 2.25 = 4.5$ . The part of $R$ to the left of $x = 1.5$ has area 4.5 square meters, so the part to the right has area $9 - 4.5 = 4.5$ square meters. The two parts are equal.

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

Sets $f (x) = g (x)$ and gets $x = 0$ and $x = 3$ .
Shows $f$ is on top (for example $f (1) = 6 > g (1) = 2$ ) and writes $\int_{0}^{3} (f (x) - g (x)) d x$ .
Finds the antiderivative $3 x^{2} - \frac{2 x^{3}}{3}$ and gets 9 square meters.
Shows $\int_{0}^{1.5} (f (x) - g (x)) d x = 4.5$ , half of 9, so both parts have area 4.5.

Part 6 Check yourself A pond between two shores

The two case questions use the same pond. Answer them from the formulas and the table.

A pond in a park is drawn on a grid marked in meters. Its north shore is the graph of $f (x) = 8 + 2 x - x^{2}$ and its south shore is the graph of $g (x) = \frac{x^{2}}{2} - 4 x + 8$ . The table gives some values of both.

The two shores of the pond
$x$ (meters)	$f (x)$ (meters)	$g (x)$ (meters)
0	8	8
1	9	4.5
2	8	2
3	5	0.5
4	0	0

Case question 1 · Calculate

No calculator. The table shows where the two shores meet. What is the area of the pond, in square meters?

Your answer

Case question 2 · Multiple choice

No calculator. Which expression gives the area of the part of the pond between the vertical lines $x = 1$ and $x = 3$ ?

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 bounded by $y = 2 x$ and $y = x^{2}$ . Tap the first line with a mistake.

Check yourself · Fill the statement

No calculator. Tap a blank, then tap an answer.

The region bounded by the graph of $y = x^{2} - 9$ and the $x$ -axis runs from $x =$ to $x = 3$ . On that interval its top boundary is , so each vertical strip has height .

Part 7 Keep it Bring back earlier ideas

Every area in this lesson rested on evaluating a definite integral, and on reading one as an accumulation. Pulling both back up is what makes them last.

From 6.5 Interpreting the Behavior of Accumulation Functions Involving Area · Calculate

No calculator. The graph of $h$ is made of two line segments: one from $(0, 3)$ to $(3, 0)$ and one from $(3, 0)$ to $(6, - 2)$ . Let $k (x) = \int_{0}^{x} h (t) d t$ . What is $k (6)$ ?

Your answer

Bring it back · Multiple choice

No calculator. Cars enter a parking garage at a rate of $E (t)$ cars per hour, where $t$ is the number of hours since the garage opened. What does $\int_{1}^{4} E (t) d t$ give?

In three lines

The area between two curves is the integral of the top function minus the bottom one.
The limits are the $x$ -values where the region starts and stops, often the solutions of $f (x) = g (x)$ .
Top minus bottom is a height even below the $x$ -axis, so a negative result means the order is reversed.

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.

Before you integrate, show the equation you solved for the meeting points and one test value that shows which function is on top. Then write the integral with both limits, top minus bottom in parentheses, and $d x$ . On a calculator question, store the meeting points instead of retyping rounded values, give the area to three decimal places, and attach square units when there is a context.

Questions students ask

Do I have to graph the curves first?

It helps, but it is not required. What you need is where the curves meet and which one is on top, and both come from algebra: solve $f (x) = g (x)$ , then compare the two functions at one $x$ between the solutions.

What if my area comes out negative?

Then the integrand is bottom minus top. The size is right, but the expression you wrote is not the area. Rewrite it as top minus bottom, so what you show and what you report agree.

Does the x-axis count as one of the curves?

Yes. It is the line $y = 0$ . For a region between a curve and the axis, the axis is the top when the curve is below it, and the bottom when the curve is above it.

When do I need a calculator to find an area?

When $f (x) = g (x)$ cannot be solved by hand, or the integrand has no antiderivative you can write. Then solve for the meeting points and evaluate the integral on the calculator, and show the integral you evaluated, with its limits and $d x$ .

What if the curves cross more than twice?

Then the top and bottom curves trade places inside the region, and one integral of top minus bottom no longer works across the whole of it. Topic 8.6 takes up that case.

Course alignment, for teachers

AP Calculus AB topic 8.4, Unit 8: Applications of Integration.

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.

What you'll be able to do

Key vocabulary

Part 1 Think first A piece of glass between two edges

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

4a. Top minus bottom, one strip at a time

4b. Where the region starts and stops

4c. Regions below the x-axis

Part 3 Watch it done The area of a region closed off by two curves

Part 4 Try it A calculator region and a setup to sort

Part 5 Explain it Answer it in free-response form

Part 6 Check yourself A pond between two shores

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic