Calculus-Master
Start Free Trial

Free Response · Integrating between curves and rotating around axes

Area and Volume AP Calculus AB FRQ: Explained with Practice & Scoring Tips

You are given two functions, f(x) and g(x), which bound a region R (or S). You will be asked to find the area of the region, the volume of a solid with given cross-sections, and the volume of a solid of revolution using the Washer or Disc method.

On this page

Key Formulashover to see usage
  • Area Between Two CurvesArea=∫ab[Top(x)−Bottom(x)] dxArea = \int_a^b [Top(x) - Bottom(x)]\,dx
    When to useIdentify which function is on top. If the functions intersect, find the intersection points (limits of integration a and b).
  • Volume with Known Cross-SectionsV=∫abA(x) dxV = \int_a^b A(x)\,dx
    When to useThe area A(x) depends on the shape. For a square: s2s^2; For a semi-circle: 1/2∗pi∗r21/2*pi*r^2; For a rectangle: base∗heightbase * height.
  • Washer Method (Rotation around y = k)V=π∫ab([Rout(x)]2−[Rin(x)]2) dxV = \pi \int_a^b ([R_{out}(x)]^2 - [R_{in}(x)]^2)\,dx
    When to useUsed when there is a 'hole'. Outer Radius Rout=∣farfunction−axis∣R_out = |far function - axis|;
    Inner Radius     Rin=∣nearfunction−axis∣R_in = |near function - axis|.
  • Disc Method (Rotation around y = k)V=π∫ab[R(x)]2 dxV = \pi \int_a^b [R(x)]^2\,dx
    When to useA special case of Washer Method where the region is flushed against the axis of rotation.
Step-by-Step SOP
  1. 1

    1. Sketch and Identify the Region

    Graph f(x) and g(x). Label intersection points as a and b. Determine which is 'Top' and which is 'Bottom' (or 'Right' vs 'Left').
  2. 2

    2. Set up the Integrand for Area

    Always use integral of (Top - Bottom) dx. If you must integrate with respect to y, use integral of (Right - Left) dy.
  3. 3

    3. Build the Cross-Sectional Area A(x)

    Find the 'length' of the base in the region, usually s = f(x) - g(x). Plug this s into the geometry formula (e.g., if cross-section is a rectangle of height H, A(x) = s * H).
  4. 4

    4. Apply the 'Big R, Little r' Strategy

    For rotations, draw a line from the axis of rotation into the region. The further curve is R(x), the closer curve is r(x). Radius = |Function - Axis|.

Practice Questions

12

Question 1

2025 AP Calculus AB Exam — FRQ 2

Hard
The shaded region RR is bounded by the graphs of the functions ff and gg, where f(x)=x2−2xf(x) = x^2 - 2x and g(x)=x+sin⁡(πx)g(x) = x + \sin(\pi x), as shown in the figure. (Note: Your calculator should be in radian mode.)
Shaded region R bounded by f(x) and g(x)
The shaded region R is bounded by f(x) = x^2 - 2x and g(x) = x + sin(πx).
APart A
Easy
Find the area of RR. Show the setup for your calculations.
Need a Hint?
Area = ∫ab(Top−Bottom) dx\int_a^b (\text{Top} - \text{Bottom}) \, dx. Looking at the graph, g(x)g(x) is above f(x)f(x) from x=0x=0 to x=3x=3.
Show Solution

Set up the integral

Area=∫03(g(x)−f(x)) dx=∫03(x+sin⁡(πx)−(x2−2x)) dx\text{Area} = \int_{0}^{3} (g(x) - f(x)) \, dx = \int_{0}^{3} (x + \sin(\pi x) - (x^2 - 2x)) \, dx

Evaluate the definite integral

Using a calculator: ∫03(g(x)−f(x)) dx≈5.137\int_{0}^{3} (g(x) - f(x)) \, dx \approx 5.137 (or 5.136).

AP Scoring Note — P1 + P2

P1 (Form): Earned for a correct integrand of g(x)−f(x)g(x)-f(x) or ∣f(x)−g(x)∣|f(x)-g(x)| with limits 0 to 3.

P2 (Answer): Correct numerical value to 3 decimal places. Note: If you swapped top/bottom but took the absolute value to get 5.137, you still earn both points.
BPart B
Medium
Region RR is the base of a solid. For this solid, at each xx the cross section perpendicular to the xx-axis is a rectangle with height xx and base in region RR. Find the volume of the solid. Show the setup for your calculations.
Need a Hint?
Volume = ∫abA(x) dx\int_a^b A(x) \, dx. Here, A(x)=base×heightA(x) = \text{base} \times \text{height}. The base is the vertical distance (g(x)−f(x))(g(x) - f(x)) and the height is given as xx.
Show Solution

Set up the area function A(x)

A(x)=(base)⋅(height)=(g(x)−f(x))⋅xA(x) = (\text{base}) \cdot (\text{height}) = (g(x) - f(x)) \cdot x

Integrate to find Volume

V=∫03x(g(x)−f(x)) dxV = \int_{0}^{3} x(g(x) - f(x)) \, dx

Evaluate

V≈7.705V \approx 7.705 (or 7.704).

AP Scoring Note — P3 + P4

P3 (Form): Earned for an integrand that is a product of xx and (g(x)−f(x))(g(x)-f(x)).

P4 (Answer): Correct numerical value. Missing the xx factor or using (g−f)2(g-f)^2 (square cross-section) will lose these points.
CPart C
Hard
Write, but do not evaluate, an integral expression for the volume of the solid generated when the region RR is rotated about the horizontal line y=−2y = -2.
Need a Hint?
Use the Washer Method: V=π∫[R(x)2−r(x)2] dxV = \pi \int [R(x)^2 - r(x)^2] \, dx. The axis y=−2y = -2 is below the region. Outer radius R(x)=g(x)−(−2)R(x) = g(x) - (-2), Inner radius r(x)=f(x)−(−2)r(x) = f(x) - (-2).
Show Solution

Identify Radii

Router=g(x)−(−2)=g(x)+2R_{\text{outer}} = g(x) - (-2) = g(x) + 2 \\ rinner=f(x)−(−2)=f(x)+2r_{\text{inner}} = f(x) - (-2) = f(x) + 2

Set up the Washer integral

V=π∫03((g(x)+2)2−(f(x)+2)2) dxV = \pi \int_{0}^{3} ((g(x) + 2)^2 - (f(x) + 2)^2) \, dx

AP Scoring Note — P5 + P6 + P7

P5 (Inner/Outer): Earned for correct R2−r2R^2 - r^2 form.

P6 (Integrand): Earned for the specific terms (g(x)+2)2(g(x)+2)^2 and (f(x)+2)2(f(x)+2)^2.

P7 (Limits/Constant): Must include π\pi, limits 00 to 33, and dxdx. If you forget π\pi, you cannot earn P7.
DPart D
Medium
It can be shown that g′(x)=1+πcos⁡(πx)g'(x) = 1 + \pi \cos(\pi x). Find the value of xx, for 0<x<10 < x < 1, at which the line tangent to the graph of ff is parallel to the line tangent to the graph of gg.
Need a Hint?
Parallel tangent lines mean the slopes are equal: f′(x)=g′(x)f'(x) = g'(x). Find f′(x)f'(x) and set it equal to the given g′(x)g'(x).
Show Solution

Find f'(x)

f(x)=x2−2x  ⟹  f′(x)=2x−2f(x) = x^2 - 2x \implies f'(x) = 2x - 2

Set slopes equal

2x−2=1+πcos⁡(πx)2x - 2 = 1 + \pi \cos(\pi x)

Solve for x

Using a calculator to find the intersection in the interval (0,1)(0, 1): x≈0.676x \approx 0.676 (or 0.675).

AP Scoring Note — P8 + P9

P8 (Equation): Earned for setting f′(x)=g′(x)f'(x) = g'(x).

P9 (Answer): Correct xx value. You must show the equation first to earn the answer point.

Question 2

2024 AP Calculus AB Exam — FRQ 6

Hard
The functions ff and gg are defined by f(x)=x2+2f(x) = x^2 + 2 and g(x)=x2−2xg(x) = x^2 - 2x, as shown in the graph. Region RR is bounded by ff and gg from x=0x=0 to x=2x=2. Region SS is bounded by gg and the xx-axis from x=2x=2 to x=5x=5.
Regions R and S bounded by f(x) and g(x)
The functions f and g are defined by f(x) = x^2 + 2 and g(x) = x^2 - 2x.
APart A
Easy
Let RR be the region bounded by the graphs of ff and gg, from x=0x = 0 to x=2x = 2, as shown in the graph. Write, but do not evaluate, an integral expression that gives the area of region RR.
Need a Hint?
Area = ∫ab(Top−Bottom) dx\int_a^b (\text{Top} - \text{Bottom}) \, dx. Looking at the graph, f(x)f(x) is strictly above g(x)g(x) between x=0x=0 and x=2x=2.
Show Solution

Identify Top and Bottom functions

For 0≤x≤20 \le x \le 2, f(x)=x2+2f(x) = x^2 + 2 is the upper curve and g(x)=x2−2xg(x) = x^2 - 2x is the lower curve.

Set up the integral

Area=∫02(f(x)−g(x)) dx=∫02((x2+2)−(x2−2x)) dx\text{Area} = \int_{0}^{2} (f(x) - g(x)) \, dx = \int_{0}^{2} ((x^2 + 2) - (x^2 - 2x)) \, dx

AP Scoring Note — P1 + P2

P1 (Integrand): Earned for f(x)−g(x)f(x)-g(x) or ∣f(x)−g(x)∣|f(x)-g(x)|.

P2 (Answer): Earned for the complete definite integral with correct limits 0 to 2.
BPart B
Hard
Let SS be the region bounded by the graph of gg and the xx-axis, from x=2x = 2 to x=5x = 5. Region SS is the base of a solid. For this solid, at each xx the cross section perpendicular to the xx-axis is a rectangle with height equal to half its base in region SS. Find the volume of the solid. Show the work that leads to your answer.
Need a Hint?
Volume = ∫A(x) dx\int A(x) \, dx. The base bb of the rectangle is the distance from g(x)g(x) to the xx-axis, which is g(x)−0g(x) - 0. The height is 12b\frac{1}{2}b. Thus A(x)=b⋅(12b)=12(g(x))2A(x) = b \cdot (\frac{1}{2}b) = \frac{1}{2}(g(x))^2.
Show Solution

Determine the Area of Cross-Section A(x)

Base b=g(x)=x2−2xb = g(x) = x^2 - 2x. Height h=12g(x)h = \frac{1}{2}g(x). A(x)=base⋅height=g(x)⋅12g(x)=12(x2−2x)2A(x) = \text{base} \cdot \text{height} = g(x) \cdot \frac{1}{2}g(x) = \frac{1}{2}(x^2 - 2x)^2.

Set up and Expand the Integral

V=∫2512(x2−2x)2 dx=12∫25(x4−4x3+4x2) dxV = \int_{2}^{5} \frac{1}{2}(x^2 - 2x)^2 \, dx = \frac{1}{2} \int_{2}^{5} (x^4 - 4x^3 + 4x^2) \, dx

Find the Antiderivative and Evaluate

12[x55−x4+4x33]∣25\frac{1}{2} [\frac{x^5}{5} - x^4 + \frac{4x^3}{3}] \Big|_2^5 After plugging in 5 and 2, the result is 4145\frac{414}{5} (or 82.8).

AP Scoring Note — P1 to P4

P1 (Integrand): Needs k(g(x))2k(g(x))^2.

P2 (Limits): Needs 2 to 5.

P3 (Antiderivative): Correct integration of the polynomial.

P4 (Answer): Correct numerical value.
CPart C
Hard
Write, but do not evaluate, an integral expression that gives the volume of the solid generated when region SS is rotated about the horizontal line y=20y = 20.
Need a Hint?
Use Washer Method: π∫(Rout2−Rin2) dx\pi \int (R_{out}^2 - R_{in}^2) \, dx. The axis y=20y=20 is above region SS. RoutR_{out} is the distance from y=20y=20 to the xx-axis (y=0y=0). RinR_{in} is the distance from y=20y=20 to g(x)g(x).
Show Solution

Identify Outer and Inner Radii

Outer Radius R(x)=20−0=20R(x) = 20 - 0 = 20. \\ Inner Radius r(x)=20−g(x)=20−(x2−2x)r(x) = 20 - g(x) = 20 - (x^2 - 2x).

Set up the Washer integral

V=π∫25[202−(20−g(x))2] dxV = \pi \int_{2}^{5} [20^2 - (20 - g(x))^2] \, dx

AP Scoring Note — P1 to P3

P1 (Form): Must be a difference of squares R2−r2R^2 - r^2.

P2 (Integrand): Correct expression 202−(20−g(x))220^2 - (20-g(x))^2.

P3 (Constant/Limits): Must include π\pi and limits 2 to 5.

Timed Practice

Ready to challenge yourself?

Take a 20-minute timed mock exam to test your understanding.

Start Timed Test (20m)
Common Pitfalls
  • ⚠
    Forgetting the pi or the SquaresIn Volume of Revolution, the most common error is forgetting pi outside the integral or writing (R-r)^2 instead of the correct R^2 - r^2.
  • ⚠
    Incorrect Axis of RotationIf rotating around y = 20 (above the region), the 'far' function is actually the one at the bottom. Check: R_out = 20 - g(x) and R_in = 20 - f(x) if f is above g.
  • ⚠
    Mixing up dx and dyIf cross-sections are perpendicular to the x-axis, use dx. If perpendicular to the y-axis, use dy. Ensure your limits and functions match the variable of integration.
  • ⚠
    Interchanging Area and VolumeRead carefully! If it asks for 'Area', don't square the functions or add pi. If it asks for 'Volume', identify if it's 'Cross-section' (No pi) or 'Revolution' (Needs pi).
  • ⚠
    Calculator PrecisionIf this is in the calculator section, store your intersection points as variables. Do not round until the very last step. Final answers must be accurate to 3 decimal places.

You Might Also Like