AP Calculus AB/BC Unit 8 Review: Applications of Integration

10–15% of AP multiple choice
13 topics
AP exam review verified for 2027

Unit 8 covers how to use definite integrals to find average values, analyze motion, measure net change, compute areas between curves, and calculate volumes of solids using cross sections, the disc method, and the washer method. Topic 8.13 adds arc length for BC students.

Fiveable's AP Calculus AB/BC Unit 8 review covers 13 topics, with guided practice questions for the unit.

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AP Calculus AB/BC unit 8 topics

8.1

Finding the Average Value of a Function on an Interval

Use (1/(b-a)) times the definite integral of f over [a, b] to find the average value. The Mean Value Theorem for Integrals guarantees a point c where f(c) equals that average.

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8.2

Connecting Position, Velocity, and Acceleration Using Integrals

Integrate velocity for displacement; integrate |v(t)| for total distance. Integrate acceleration with an initial condition to recover velocity, and integrate velocity with an initial condition to recover position.

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8.3

Using Accumulation Functions and Definite Integrals in Applied Contexts

The integral of a rate function over an interval gives net change. Use Q(b) = Q(a) + integral of r(t) dt to find a final quantity from an initial value and a rate.

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8.4

Finding the Area Between Curves Expressed as Functions of x

Integrate top minus bottom over the x-interval defined by intersection points. Solve f(x) = g(x) to find bounds, and verify which function is on top before writing the integrand.

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8.5

Finding the Area Between Curves Expressed as Functions of y

Rewrite curves as x = f(y), find y-bounds from intersection points, and integrate right minus left with respect to y. Use this approach when horizontal slices avoid splitting the region.

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8.6

Finding the Area Between Curves That Intersect at More Than Two Points

Find all intersection points, determine which function is on top in each subinterval, and sum the integrals. Alternatively, integrate |f(x) - g(x)| over the full interval.

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8.7

Volumes with Cross Sections: Squares and Rectangles

Integrate A(x) = s^2 for square cross sections or A(x) = l times w for rectangular cross sections, where the side length comes from the distance between boundary curves.

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8.8

Volumes with Cross Sections: Triangles and Semicircles

Use A = (sqrt(3)/4)s^2 for equilateral triangles, A = (1/2)s^2 for isosceles right triangles, and A = (pi/8)d^2 for semicircles, where s or d is the base length between curves.

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8.9

Volume with Disc Method: Revolving Around the x- or y-Axis

When a region with no gap from the axis is revolved, V = pi integral of [f(x)]^2 dx around the x-axis, or pi integral of [g(y)]^2 dy around the y-axis.

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8.10

Volume with Disc Method: Revolving Around Other Axes

For a shifted axis y = k or x = h, the radius is the perpendicular distance from the curve to the axis. Write the radius as |f(x) - k| or |g(y) - h| before squaring.

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8.11

Volume with Washer Method: Revolving Around the x- or y-Axis

When a gap exists between the region and the axis, use V = pi integral of (R^2 - r^2) dx. Identify the outer radius R (farther curve) and inner radius r (nearer curve) carefully.

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8.12

Volume with Washer Method: Revolving Around Other Axes

For axes like y = k or x = h, both radii become distance expressions. R = |f(x) - k| for the farther curve and r = |g(x) - k| for the nearer curve; then apply the washer formula.

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8.13

The Arc Length of a Smooth, Planar Curve and Distance Traveled (BC Only)

Arc length of y = f(x) from a to b is the integral of sqrt(1 + [f'(x)]^2) dx. Compute f'(x), substitute into the formula, and evaluate numerically when the integrand has no clean antiderivative.

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Practice AP Calculus AB/BC unit 8 questions

Try AP-style questions on unit 8, then use the review notes below for anything you miss.

Example AP-style MCQs

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MCQ

AP-style practice question

Question

A water tank is constructed by revolving the region bounded by $y = 3 - \frac{x^2}{4}$ and $y = 1$ around the horizontal line $y = 0.5$ from $x = -2$ to $x = 2$ . After setting up the washer method integral $\int_{-2}^{2} \pi[(3 - \frac{x^2}{4} - 0.5)^2 - (1 - 0.5)^2] dx$ and calculating its value to be approximately $18.8\pi$ cubic units, what does this result tell you about the tank's capacity?

The outer radius of the tank at its widest point is $18.8\pi$ units from the axis of rotation

The tank can hold approximately $18.8\pi$ cubic units of water when filled to capacity

The height of the tank from the bottom to the top is $18.8\pi$ units

The surface area of the tank walls that contact the water is $18.8\pi$ square units

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MCQ

AP-style practice question

Question

A student finds the volume of a solid with equilateral triangle cross sections, where the base is bounded by $y = x$ and $y = x^2$ from $x = 0$ to $x = 1$ . The student computes $V = \frac{\sqrt{3}}{4}\int_0^1 (x - x^2)^2\,dx = \frac{\sqrt{3}}{4}\int_0^1 (x^2 - 2x^3 + x^4)\,dx = \frac{\sqrt{3}}{4}\left[\frac{x^3}{3} - \frac{x^4}{2} + \frac{x^5}{5}\right]_0^1 = \frac{\sqrt{3}}{4}\left(\frac{1}{3} - \frac{1}{2} + \frac{1}{5}\right) = \frac{\sqrt{3}}{60}$ . Which statement correctly evaluates this solution?

The side length becomes negative for $x > 1$ , invalidating the calculation.

The curves intersect at endpoints, so the region has zero area.

The expansion $(x - x^2)^2 = x^2 - 2x^3 + x^4$ is incorrect.

The setup and expansion are correct, but the final arithmetic is wrong.

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Example FRQs

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FRQ

Average function value and enclosed area calculation

3. The function f is defined on the closed interval $-2 ≤ x ≤ 4$ . The graph of f consists of three pieces: a line segment from (-2,0) to (0,2), the upper semicircle of radius 1 centered at (1,2) from x = 0 to x = 2, and a line segment from (2,2) to (4,-2), as shown in Figure 1. Let g be the function defined by $g(x) = \int_{-2}^{x} f(t)\,dt$ for $-2 ≤ x ≤ 4$ .

Figure 1. Graph of f

Find the average value of f on the interval $-2 ≤ x ≤ 4$ . Show the work that leads to your answer.

Find the value of g(4). Show the work that leads to your answer.

Find the area of the region enclosed between the graph of f and the x-axis on the interval $2 ≤ x ≤ 4$ . Show the work that leads to your answer.

Region R is the region in the xy-plane enclosed between the graph of f and the x-axis on the interval $0 ≤ x ≤ 4$ . Region R is the base of a solid. For this solid, each cross-section perpendicular to the x-axis is a square. Find the volume of the solid. Show the work that leads to your answer.

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FRQ

Medication concentration over time and effective dosage levels

2. A patient receives a single dose of a medication at time $t = 0$ . For $0 ≤ t ≤ 8$ , the concentration of the medication in the patient's bloodstream is modeled by the function $C$ defined by $C(t) = 9te^{-0.5t}$ , where $C(t)$ is measured in milligrams per liter and $t$ is measured in hours. The graph of $C$ is shown in Figure 1.

Figure 1. Graph of the medication concentration C(t) = 9t e^{-0.5t} on 0 ≤ t ≤ 8 (t in hours, C(t) in mg/L).

Find the average concentration of the medication in the bloodstream over the time interval $0 ≤ t ≤ 8$ . Show the setup for your calculations.

Find all times $t$ in the interval $0 < t < 8$ at which the concentration $C(t)$ equals the average concentration found in part A. Show the setup for your calculations.

The medication is considered effective when $C(t) ≥ 3$ . Let $p$ and $q$ be the times, with $0 < p < q < 8$ , at which $C(t) = 3$ . Write an expression involving an integral that gives the average concentration of the medication during the time it is effective. Then find this average concentration.

A new function $M$ is defined by $M(x) = \frac{1}{x}\int_0^x C(t)\,dt$ for $0 < x ≤ 8$ , so that $M(x)$ is the average concentration from time $0$ to time $x$ . It can be shown that $M'(x) = \frac{C(x) - M(x)}{x}$ . Find the value of $x$ at which $M$ attains its maximum value on $0 < x ≤ 8$ . Justify your answer.

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Unit 8 review notes

8.1

Average Value of a Function

The average value of a continuous function f on [a, b] is (1/(b-a)) times the integral from a to b of f(x) dx. Geometrically, this is the height of a rectangle with width (b-a) whose area equals the area under the curve. The Mean Value Theorem for Integrals guarantees that a continuous function actually reaches its average value at some c in [a, b].

Average value formula: f_avg = (1/(b-a)) integral from a to b of f(x) dx
Geometric interpretation: The average value is the rectangle height that gives the same area as the integral over [a, b]
Mean Value Theorem for Integrals: There exists c in [a, b] such that f(c) equals the average value
Common confusion: Average value uses an integral divided by interval width; average rate of change uses (f(b)-f(a))/(b-a) with no integral

Given f(x) = x^2 on [0, 3], set up and evaluate the average value integral, then find the c guaranteed by the Mean Value Theorem for Integrals.

8.2

Position, Velocity, and Acceleration Using Integrals

For a particle in rectilinear motion, integrating velocity over [a, b] gives displacement (signed net change in position), while integrating speed (|v(t)|) gives total distance traveled. To recover position from velocity, integrate and apply the initial condition. Velocity sign changes are critical: where v(t) changes sign, the particle reverses direction, and you must split the distance integral at those points.

Displacement: Integral from a to b of v(t) dt; signed, can be negative or zero even if the particle moves
Total distance: Integral from a to b of |v(t)| dt; always nonnegative
Position from velocity: s(t) = s(t_0) + integral from t_0 to t of v(u) du
Velocity from acceleration: v(t) = v(t_0) + integral from t_0 to t of a(u) du
Speed: |v(t)|; the particle is at rest when v(t) = 0 and moving when |v(t)| > 0

A particle has v(t) = t^2 - 4 on [0, 3]. Find the displacement and the total distance traveled. Identify where the particle changes direction.

8.3

Accumulation Functions and Net Change in Applied Contexts

A definite integral of a rate of change function gives the net change of the underlying quantity over the interval. If you know an initial value Q(a) and a rate r(t), then Q(b) = Q(a) + integral from a to b of r(t) dt. This applies to any context: water flowing into a tank, population growing, cost accumulating from a marginal rate. Always track units: the integral of a rate (units per time) over time gives the quantity (units).

Net Change Theorem: Integral from a to b of f'(x) dx = f(b) - f(a); the integral of a rate gives net change
Final value formula: Q(b) = Q(a) + integral from a to b of r(t) dt
Accumulation function: F(x) = integral from a to x of f(t) dt accumulates the rate f from a up to x
Units check: If r(t) is in gallons per minute and t is in minutes, the integral gives gallons
Net vs. total: Net change can be negative if the rate is negative (outflow exceeds inflow); total accumulated amount requires |r(t)|

Water flows into a tank at r(t) = 3t - 1 gallons per minute. If the tank starts with 10 gallons, write an expression for the amount at time t = 5 and evaluate it.

8.4

Area Between Curves

To find the area of a region bounded by two curves, integrate the top function minus the bottom function (or right minus left when integrating with respect to y) over the interval defined by their intersection points. When curves cross more than twice, identify every intersection point, determine which function is on top in each subinterval, and add the integrals. Integrating with respect to y is preferable when the region has a single left and right boundary but would require splitting if integrated with respect to x.

Area with respect to x: Integral from a to b of [f(x) - g(x)] dx where f is the top curve and g is the bottom curve
Area with respect to y: Integral from c to d of [x_right(y) - x_left(y)] dy; rewrite curves as x = function of y
Intersection points: Solve f(x) = g(x) to find x-bounds; solve for y when integrating with respect to y
Multiple crossings: Split the integral at each crossing point and add absolute values of each piece, or integrate |f(x) - g(x)|
Choosing the variable: Use y-integration when horizontal slices avoid splitting a region that x-integration would require splitting

Find the total area enclosed between y = x^2 and y = x + 2. Then sketch a region where integrating with respect to y would be more efficient than integrating with respect to x.

Feature	Integrate with respect to x	Integrate with respect to y
Slice orientation	Vertical strips	Horizontal strips
Integrand	f(x) - g(x) (top minus bottom)	x_R(y) - x_L(y) (right minus left)
Bounds	x-values of intersection points	y-values of intersection points
Best when	Region has clear top and bottom functions	Region has clear left and right functions or x-integration would require splitting

8.7

Volumes with Known Cross Sections

When a solid has a known base region in the xy-plane and cross sections of a specified shape perpendicular to an axis, integrate the cross-sectional area function A(x) or A(y) over the appropriate bounds. The side length or diameter of each cross section is typically the distance between two boundary curves. Write the area formula for the named shape, substitute the side length expression, then integrate.

General volume formula: V = integral from a to b of A(x) dx, where A(x) is the area of the cross section at position x
Square cross section: A = s^2, where s = f(x) - g(x) is the side length
Equilateral triangle cross section: A = (sqrt(3)/4) s^2, where s is the base length between curves
Isosceles right triangle cross section: A = (1/2) s^2 when the legs equal the base length s
Semicircular cross section: A = (pi/8) d^2 = (pi/2) r^2, where d = f(x) - g(x) is the diameter

A solid has its base bounded by y = sqrt(x) and y = 0 on [0, 4]. Cross sections perpendicular to the x-axis are squares. Set up and evaluate the volume integral.

8.9

Disc Method: Revolving Around Any Axis

When a region is revolved around an axis and there is no gap between the region and the axis, each cross section is a full disc. The volume is pi times the integral of the squared radius. When revolving around the x-axis, the radius is f(x); when revolving around the y-axis, express x as a function of y. For a shifted axis like y = k or x = h, the radius becomes the perpendicular distance from the curve to that line, such as |f(x) - k|.

Disc method (x-axis): V = pi integral from a to b of [f(x)]^2 dx
Disc method (y-axis): V = pi integral from c to d of [g(y)]^2 dy, where x = g(y)
Shifted horizontal axis y = k: Radius = |f(x) - k|; V = pi integral of [f(x) - k]^2 dx
Shifted vertical axis x = h: Radius = |g(y) - h|; V = pi integral of [g(y) - h]^2 dy
Key step: Define the radius as a distance expression before squaring; do not square a negative radius

Find the volume of the solid formed by revolving y = sqrt(x) on [0, 4] around the line y = -1 using the disc method.

8.11

Washer Method: Revolving Around Any Axis

Use the washer method when there is a gap between the region and the axis of rotation, so each cross section is a ring (annulus) rather than a full disc. Subtract the squared inner radius from the squared outer radius, multiply by pi, and integrate. For axes other than the coordinate axes, both radii become distance expressions involving the shifted line.

Washer formula (x-axis): V = pi integral from a to b of [R(x)^2 - r(x)^2] dx, where R is the outer radius and r is the inner radius
Washer formula (y-axis): V = pi integral from c to d of [R(y)^2 - r(y)^2] dy
Outer radius: Distance from the axis to the farther curve
Inner radius: Distance from the axis to the nearer curve; if the region touches the axis, r = 0 and the washer reduces to a disc
Shifted axis: For y = k, R = |f(x) - k| and r = |g(x) - k| where f is farther from the axis than g

The region bounded by y = x^2 and y = x is revolved around the x-axis. Identify the outer and inner radii, set up the washer integral, and evaluate.

Method	When to use	Formula
Disc	No gap between region and axis	V = pi integral of R^2 dx (or dy)
Washer	Gap exists between region and axis	V = pi integral of (R^2 - r^2) dx (or dy)
Cross section	Solid is not a revolution; shape is specified	V = integral of A(x) dx (or A(y) dy)

8.13

Arc Length of a Smooth Planar Curve (BC Only)

The arc length of a smooth curve y = f(x) from x = a to x = b is the integral from a to b of sqrt(1 + [f'(x)]^2) dx. This formula comes from approximating the curve with short line segments and taking the limit. For a curve expressed as x = g(y), replace f'(x) with g'(y) and integrate with respect to y. On the BC exam, arc length integrals are almost always evaluated with a calculator because the integrand rarely simplifies to a standard antiderivative.

Arc length formula (y = f(x)): L = integral from a to b of sqrt(1 + [f'(x)]^2) dx
Arc length formula (x = g(y)): L = integral from c to d of sqrt(1 + [g'(y)]^2) dy
Smooth curve requirement: f' must be continuous on [a, b]; the curve cannot have corners or cusps
Calculator use: Most arc length integrands do not have elementary antiderivatives; set up the integral analytically, then evaluate numerically
Connection to distance traveled: For parametric motion, distance traveled = integral of speed = integral of sqrt((dx/dt)^2 + (dy/dt)^2) dt, which is the same structure

Set up (but do not evaluate by hand) the arc length integral for y = x^3 from x = 0 to x = 2. Identify f'(x) and write the complete integrand.

Key terms

Term	Definition
Displacement	The signed net change in position of a particle, computed as the integral of v(t) dt over a time interval; can be negative or zero even if the particle moves.
net displacement	The change in position from start to end, computed as the integral of v(t) dt without absolute value; differs from total distance when the particle changes direction.
Velocity Function	A function v(t) giving the instantaneous rate of change of position; its integral over an interval gives displacement, and the integral of its absolute value gives total distance.
Acceleration Function	The derivative of the velocity function; integrating it with an initial condition recovers the velocity function.
Position function	A function s(t) describing a particle's location at time t; recovered by integrating velocity and applying an initial position condition.
Area Between Curves	The region enclosed by two or more curves, computed by integrating the difference of the functions (top minus bottom or right minus left) over the interval defined by their intersection points.
Intersection Points	The x- or y-values where two curves meet; used as the limits of integration when finding area between curves or volumes of revolution.
Limits of Integration	The values a and b (or c and d) that define the interval over which a definite integral is evaluated; in area and volume problems, these come from intersection points or given boundaries.
Cross Sections	The two-dimensional shape formed when a solid is cut by a plane perpendicular to an axis; its area function A(x) is integrated to find the volume of the solid.
Solids of Revolution	A three-dimensional solid formed by rotating a two-dimensional region around an axis; its volume is found using the disc or washer method.
Outer radius	In the washer method, the distance from the axis of rotation to the farther boundary curve; squared and multiplied by pi in the volume integrand.
Area Under a Curve	The definite integral of a function over an interval, representing the accumulated value or net signed area between the curve and the horizontal axis.
Arc length formula	For y = f(x) on [a, b], arc length L = integral from a to b of sqrt(1 + [f'(x)]^2) dx; BC only topic in Unit 8.

How to review unit 8

Topics students miss most

Miss rates come from Fiveable multiple-choice practice.

8.13

The Arc Length of a Smooth, Planar Curve and Distance Traveled (BC Only)

57%180 tries

8.11

Volume with Washer Method: Revolving Around the x- or y-Axis

56%183 tries

8.5

Finding the Area Between Curves Expressed as Functions of y

54%293 tries

Common unit 8 mistakes

Confusing displacement with total distance

Integrating v(t) directly gives displacement, which can be zero or negative if the particle reverses. To get total distance, integrate |v(t)| and split the integral at every t where v(t) = 0 and changes sign.

Forgetting to find all intersection points

When curves cross more than twice, students often use only the outermost intersection points and miss interior crossings. Always solve f(x) = g(x) completely and check a sign chart for f(x) - g(x) before integrating.

Squaring the wrong expression in disc and washer problems

For a shifted axis like y = 2, the radius is f(x) - 2, not f(x). Students frequently square f(x) and then subtract 4 instead of squaring the entire distance expression (f(x) - 2)^2.

Using the wrong cross-section area formula

Each cross-section shape has its own area formula. For a semicircle, the diameter equals the base length between curves, so the radius is half that. Substituting the full base length as the radius doubles the area and produces an incorrect volume.

Applying the average value formula without dividing by (b-a)

The average value is (1/(b-a)) times the integral, not just the integral. Omitting the 1/(b-a) factor gives the net area, not the average height. This mistake is especially common when the integral evaluates to a clean number.

How this unit shows up on the AP exam

Setting up integrals from verbal or graphical descriptions

Free-response questions frequently present a region or solid through a graph, a table, or a written description and ask you to write a definite integral that represents area, volume, or net change. The skill being tested is translating the geometric or physical situation into a correct integrand and bounds, not just evaluating a given integral. Practice reading a problem, sketching the region, labeling radii or side lengths, and writing the integral before touching a calculator.

Distinguishing displacement from total distance in motion problems

Motion problems on both the multiple-choice and free-response sections often ask for both displacement and total distance from the same velocity function. The distinction between integrating v(t) and integrating |v(t)| is a recurring point of differentiation between correct and incorrect responses. Be prepared to locate sign changes in v(t) and split the distance integral accordingly.

Choosing and justifying the correct volume method

Volume problems may ask you to set up an integral using a specified method (disc, washer, or cross section) or may leave the method choice to you. On free-response questions, partial credit often depends on correctly identifying the outer and inner radii or the cross-section shape and writing a complete, correct integral expression even if the final numerical evaluation contains an arithmetic error. Label every component of the setup explicitly.

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Frequently Asked Questions

What's on the AP Calc Unit 8 progress check (MCQ and FRQ)?

The AP Calc Unit 8 progress check in AP Classroom includes both MCQ and FRQ parts drawn from the unit's 13 topics. The MCQ section tests skills like finding average value, computing area between curves, and setting up volume integrals. The FRQ part typically asks you to set up and evaluate integrals for volumes using the disc or washer method, or to analyze particle motion using accumulation functions. The progress check pulls heavily from these topics:
- Average value of a function (8.1)
- Position, velocity, and acceleration with integrals (8.2)
- Area between curves (8.4, 8.5, 8.6)
- Volumes with cross sections (8.7, 8.8)
- Disc and washer methods (8.9-8.12) Practice the same skills at AP Calc Unit 8 before you take the progress check.

How do I practice AP Calc Unit 8 FRQs?

AP Calc Unit 8 FRQs most often ask you to set up and evaluate integrals for area between curves, volumes using the disc or washer method, and particle motion problems involving net displacement or total distance. To practice, focus on writing the integral setup clearly before evaluating, since College Board awards points for the setup itself. Strong FRQ practice steps for this unit:
1. Work through area between curves problems where curves intersect at more than two points (8.6), since those setups trip a lot of students up.
2. Practice disc vs. washer method problems revolving around both the x-axis and other axes (8.9-8.12).
3. For particle motion (8.2, 8.3), practice distinguishing net displacement from total distance traveled.
4. After writing each setup, check your bounds and which function is on top or outside. Find FRQ-style practice problems at AP Calc Unit 8.

Where can I find AP Calc Unit 8 practice questions?

The best place to find AP Calc Unit 8 practice questions, including multiple-choice and FRQ-style problems, is AP Calc Unit 8. That page has practice aligned to all 13 topics, from average value and area between curves to disc, washer, and cross-section volume problems. For a practice-test feel, work through MCQ sets that mix topic types the way the real exam does, and time yourself on FRQ setups.

How should I study AP Calc Unit 8?

Start AP Calc Unit 8 by locking in the core idea: a definite integral measures accumulation, and every topic in this unit is just a different application of that. Once that clicks, the rest follows more naturally. A concrete study plan:
1. Start with average value (8.1) and particle motion (8.2, 8.3). These are the most straightforward and build your integral intuition.
2. Move to area between curves (8.4-8.6). Practice identifying which function is on top and setting up correct bounds, especially when curves intersect at more than two points.
3. Tackle volumes in order: cross sections (8.7, 8.8), then disc method (8.9, 8.10), then washer method (8.11, 8.12). Sketch every solid before writing the integral.
4. If you're in BC, finish with arc length (8.13).
5. After each topic, do a few timed FRQ setups without a calculator to simulate exam conditions. The biggest mistake students make is memorizing formulas without understanding when to use each one. Focus on recognizing the problem type first, then pulling the right setup. Practice at AP Calc Unit 8.

Ready to review Unit 8?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.

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