Applications of Integration: AP Calculus AB Math Test Answer Key – PDF Download

AP Calculus Applications Test Answer Key: questions 1–5

Question 1: Average value (2 points)

1. \(\int_{0}^{2}x^{3}\,dx=4\), so \(f_{\text{avg}}=\frac{4}{2}=2\).
2. \(c^{3}=2\), so \(c=2^{1/3}\approx 1.260\).

(a) 2; (b) \(c\approx 1.260\)

Scoring: 1 point for the average value 2; 1 point for \(c\approx 1.260\).

Question 2: Particle motion (3 points)

\(v=(t-2)(t-4)\). Antiderivative: \(\frac{t^{3}}{3}-3t^{2}+8t\).

1. \(\int_{0}^{5}v\,dt=\frac{125}{3}-75+40=\frac{20}{3}\), so \(s(5)=1+\frac{20}{3}=\frac{23}{3}\).
2. \(\int_{0}^{2}v\,dt=\frac{20}{3}\), \(\int_{2}^{4}v\,dt=-\frac{4}{3}\), \(\int_{4}^{5}v\,dt=\frac{4}{3}\). Distance \(=\frac{20}{3}+\frac{4}{3}+\frac{4}{3}=\frac{28}{3}\).

(a) \(\frac{23}{3}\); (b) \(\frac{28}{3}\)

Scoring: (a) 1 point. (b) 1 point for splitting at \(t=2\) and \(t=4\); 1 point for \(\frac{28}{3}\).

Question 3: Area between curves (2 points)

Intersections: \(6x-x^{2}=x^{2}\) gives \(2x(3-x)=0\), so \(x=0\) and \(x=3\). The parabola \(6x-x^{2}\) is on top.

\(A=\int_{0}^{3}(6x-2x^{2})\,dx=[3x^{2}-\frac{2x^{3}}{3}]_{0}^{3}=27-18=9\).

9

Scoring: 1 point for the correct limits and integrand; 1 point for the value 9.

Question 4: Area using y (2 points)

\(y^{2}=6-y\) gives \(y^{2}+y-6=0\), so \(y=-3\) or \(y=2\). The line \(x=6-y\) is the right curve.

\(A=\int_{-3}^{2}(6-y-y^{2})\,dy=[6y-\frac{y^{2}}{2}-\frac{y^{3}}{3}]_{-3}^{2}=\frac{22}{3}-(-\frac{27}{2})=\frac{125}{6}\).

\(\frac{125}{6}\)

Scoring: 1 point for the limits and integrand; 1 point for \(\frac{125}{6}\).

Question 5: Semicircle cross sections (3 points)

The diameter at \(x\) is \(4-x^{2}\), for \(-2\le x\le 2\). The area is \(\frac{\pi}{2}(\frac{4-x^{2}}{2})^{2}=\frac{\pi}{8}(4-x^{2})^{2}\).

\(V=\frac{\pi}{8}\int_{-2}^{2}(4-x^{2})^{2}\,dx=\frac{\pi}{8}\cdot\frac{512}{15}=\frac{64\pi}{15}\approx 13.404\).

\(\frac{64\pi}{15}\)

Scoring: 1 point for the semicircle area \(\frac{\pi}{8}(4-x^{2})^{2}\); 1 point for the integral setup with limits; 1 point for the value.

AP Calculus Applications Test Answer Key: questions 6–8

Question 6: Washer method (3 points)

Intersections: \(\sqrt{x}=\frac{x}{2}\) at \(x=0\) and \(x=4\). Outer radius \(R=\sqrt{x}\), inner radius \(r=\frac{x}{2}\).

\(V=\pi\int_{0}^{4}(x-\frac{x^{2}}{4})dx=\pi(8-\frac{16}{3})=\frac{8\pi}{3}\).

\(\frac{8\pi}{3}\)

Scoring: 1 point for the limits; 1 point for \(\pi(R^{2}-r^{2})\) with correct radii; 1 point for the value.

Question 7: Rates of change (3 points)

\(A^{\prime}(t)=2t-6\).

1. \(A(6)=50+\int_{0}^{6}(2t-6)\,dt=50+(36-36)=50\).
2. \(A^{\prime}(t)=0\) at \(t=3\); \(A^{\prime}<0\) before and \(A^{\prime}>0\) after, so \(A\) is least at \(t=3\) (the endpoints are \(A(0)=50\) and \(A(8)=66\)).
3. \(A(3)=50+(9-18)=41\).

(a) 50; (b) \(t=3\); (c) 41 hundred people

Scoring: 1 point for each part.

Question 8: Area with three intersections (2 points)

\(x^{3}-9x=x(x-3)(x+3)\), so the graph meets the axis at \(x=-3,0,3\). On \([-3,0]\) it is above the axis; on \([0,3]\) below.

\(\int_{-3}^{0}(x^{3}-9x)\,dx=0-(\frac{81}{4}-\frac{81}{2})=\frac{81}{4}\). \(\int_{0}^{3}(x^{3}-9x)\,dx=\frac{81}{4}-\frac{81}{2}=-\frac{81}{4}\). Area \(=\frac{81}{4}+\frac{81}{4}=\frac{81}{2}\).

\(\frac{81}{2}\)

Scoring: 1 point for the three intersections and the split; 1 point for \(\frac{81}{2}\).

Scoring summary

How to use this ap calculus applications test answer key

Use this ap calculus applications test answer key to score the test fairly and to see which skills need more practice.

Printable PDF · Test Answer Key

AP Calculus AB Applications of Integration test answer key

• 📄 US Letter
• 💾 211 KB
• 🖨️ Black and white friendly

Download the PDF

See every chapter on the AP Calculus AB math page.