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2025 AP Calculus AB FRQ Question 2: Area and Volume Between f and g

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2025.

Question 2

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Area and Volume Between f and g

Area & Volume

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. The two graphs intersect at (0,0)(0,0) and (3,3)(3,3). (Calculator should be in radian mode.)
2025 AP Calculus AB FRQ Question 2 Graph — region R bounded by f and g
Region $R$, bounded by $y=f(x)$ and $y=g(x)$ on $[0,3]$.
Part AMedium2 points
Find the area of RR. Show the setup for your calculations.

Answer

Area=∫03(g(x)−f(x)) dx=5.136620≈5.137\text{Area} = \int_0^3 (g(x)-f(x))\,dx = 5.136620 \approx 5.137
Full Solution & Work

Set up the integral

On [0,3][0,3], g(x)≥f(x)g(x) \geq f(x), so:
Area=∫03(g(x)−f(x)) dx\text{Area} = \int_0^3 \big(g(x)-f(x)\big)\,dx

Evaluate with a calculator

∫03(g(x)−f(x)) dx=5.136620\int_0^3 (g(x)-f(x))\,dx = 5.136620

The area of RR is **5.137** (or 5.136).

AP Scoring — 2 Points

**P1**: Correct integrand g(x)−f(x)g(x)-f(x) (or ∣g(x)−f(x)∣|g(x)-f(x)|, f(x)−g(x)f(x)-g(x), ∣f(x)−g(x)∣|f(x)-g(x)|) in a definite integral over [0,3][0,3].
**P2**: Correct answer 5.137 (or 5.136).
Part BMedium2 points
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.

Answer

V=∫03x(g(x)−f(x)) dx=7.704930≈7.705V = \int_0^3 x\big(g(x)-f(x)\big)\,dx = 7.704930 \approx 7.705
Full Solution & Work

Identify the cross-sectional area

At each xx, the rectangle's base (in region RR) has length g(x)−f(x)g(x)-f(x), and its height is xx. So the cross-sectional area is:
A(x)=x(g(x)−f(x))A(x) = x\big(g(x)-f(x)\big)

Integrate over [0,3] and evaluate

V=∫03x(g(x)−f(x)) dx=7.704930V = \int_0^3 x\big(g(x)-f(x)\big)\,dx = 7.704930

The volume of the solid is **7.705** (or 7.704).

AP Scoring — 2 Points

**P3**: A definite integral with integrand presented as a product of two nonconstant factors, one of which is xx, g(x)−f(x)g(x)-f(x), or f(x)−g(x)f(x)-g(x).
**P4**: Correct answer 7.705 (or 7.704). The presence of dxdx is not required for scoring.
Part CHard3 points
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.

Answer

V=π∫03[(g(x)+2)2−(f(x)+2)2] dxV = \pi \int_0^3 \Big[\big(g(x)+2\big)^2 - \big(f(x)+2\big)^2\Big]\,dx
Full Solution & Work

Identify outer and inner radii

Rotating about y=−2y=-2, the outer radius (farther curve, gg) is g(x)−(−2)=g(x)+2g(x)-(-2) = g(x)+2, and the inner radius (closer curve, ff) is f(x)−(−2)=f(x)+2f(x)-(-2) = f(x)+2.

Apply the washer method

V=π∫03[(g(x)+2)2−(f(x)+2)2] dxV = \pi\int_0^3 \Big[\big(g(x)+2\big)^2 - \big(f(x)+2\big)^2\Big]\,dx

AP Scoring — 3 Points

**P5**: Definite integral with integrand of the form R2−r2R^2-r^2, where one of {R,r}\{R,r\} is correct or a difference between gg and a nonzero constant, and the other similarly for ff.
**P6**: The correct integrand (g(x)+2)2−(f(x)+2)2\big(g(x)+2\big)^2-\big(f(x)+2\big)^2 (or equivalent).
**P7**: Correct limits [0,3][0,3], the constant π\pi, and the differential — all present together. (A response missing π\pi can still earn P5–P6 but not P7.)
Part DMedium2 points
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.

Answer

x=0.675819≈0.676x = 0.675819 \approx 0.676
Full Solution & Work

Set the two derivatives equal

Tangent lines are parallel when the slopes match: f′(x)=g′(x)f'(x)=g'(x).
f′(x)=2x−2,2x−2=1+πcos⁡(πx)f'(x) = 2x-2, \qquad 2x - 2 = 1 + \pi\cos(\pi x)

Solve numerically

Using a calculator: x=0.675819x = 0.675819.

AP Scoring — 2 Points

**P8**: A general statement f′(x)=g′(x)f'(x)=g'(x), or the correct equation formed by substituting 2x−22x-2 for f′(x)f'(x), 1+πcos⁡(πx)1+\pi\cos(\pi x) for g′(x)g'(x), or both.
**P9**: Correct answer x=0.676x = 0.676 (or 0.675).

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