This ap calculus applications test answer key gives the full solution and the scoring guide for each of the 8 test questions on applications of integration.
Full solutions and a scoring guide for the applications of integration test. Points follow AP-style rubrics: one for the setup, one for the limits or integrand, and one for the final value, so partial credit is explained for every question.
AP Calculus Applications Test Answer Key: questions 1–5
Question 1: Average value (2 points)
- \(\int_{0}^{2}x^{3}\,dx=4\), so \(f_{\text{avg}}=\frac{4}{2}=2\).
- \(c^{3}=2\), 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: \(\frac{t^{3}}{3}-3t^{2}+8t\).
- \(\int_{0}^{5}v\,dt=\frac{125}{3}-75+40=\frac{20}{3}\), so \(s(5)=1+\frac{20}{3}=\frac{23}{3}\).
- \(\int_{0}^{2}v\,dt=\frac{20}{3}\), \(\int_{2}^{4}v\,dt=-\frac{4}{3}\), \(\int_{4}^{5}v\,dt=\frac{4}{3}\). 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)
\(A^{\prime}(t)=2t-6\).
- \(A(6)=50+\int_{0}^{6}(2t-6)\,dt=50+(36-36)=50\).
- \(A^{\prime}(t)=0\) 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 \(A(8)=66\)).
- \(A(3)=50+(9-18)=41\).
(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
20 points: add the points of every question of this ap calculus applications test answer key.
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Use this ap calculus applications test answer key to score the test fairly and to see which skills need more practice.
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AP Calculus AB Applications of Integration test answer key
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