This ap calculus integration test answer key gives the full solution and the scoring guide for each of the 8 test questions on integration and accumulation of change.
Worked solutions and scoring guides for the Unit 6 test. Each answer shows the setup an AP reader expects: the Riemann sum terms, the antiderivative, the limits of integration and a clear final statement with units. Award points exactly as the scoring note describes and total them out of 20.
AP Calculus Integration Test Answer Key: questions 1–5
Question 1: Estimating from a table (3 points)
(a) The widths are \(1,2,3,2\). Left sum: \(12(1)+18(2)+20(3)+15(2)=12+36+60+30=138\).
(b) Trapezoidal: \(\frac{12+18}{2}(1)+\frac{18+20}{2}(2)+\frac{20+15}{2}(3)+\frac{15+9}{2}(2)=15+38+52.5+24=129.5\).
(c) \(\int_{0}^{8} R(t)\,dt\) is the total number of cars that enter the garage between 6 A.M. and 2 P.M., measured in cars.
Scoring: (a) 1 point for the correct sum 138 (setup with the widths \(1,2,3,2\)). (b) 1 point for 129.5. (c) 1 point for the meaning (total or net number of cars entering) with the time span and the unit “cars”.
Question 2: Properties of integrals (2 points)
(a) \(\int_{0}^{7} f=\int_{0}^{4} f+\int_{4}^{7} f=10+(-3)=7\).
(b) \(3(7)-6+2(7-0)=21-6+14=29\).
Scoring: (a) 1 point for 7. (b) 1 point for 29; the constant term must be integrated as \(2\cdot 7=14\).
Question 3: Accumulation from a graph (3 points)
(a) \(\int_{0}^{2}f=2(2)=4\); \(\int_{2}^{4}f=\frac{2+(-2)}{2}(2)=0\); \(\int_{4}^{6}f=-2(2)=-4\). So \(g(6)=4+0-4=0\).
(b) \(g^{\prime}(5)=f(5)=-2\).
(c) \(g^{\prime}=f\), which is positive on \([0,3)\), negative on \((3,7)\) and positive on \((7,8]\). So \(g\) has candidates \(g(0)=0\), \(g(3)=4+\frac{1}{2}(1)(2)=5\), \(g(7)=0+(-\frac{1}{2}(1)(2))=-1\) and \(g(8)=0\). The absolute maximum is at \(x=3\).
Scoring: (a) 1 point for 0 with signed areas shown. (b) 1 point for \(-2\). (c) 1 point for \(x=3\) with a justification (sign change of \(g^{\prime}\) or comparison of candidate values).
Question 4: Fundamental Theorem evaluation (2 points)
An antiderivative is \(2x^{3/2}-\frac{2}{x}\). Then \([2x^{3/2}-\frac{2}{x}]_{1}^{4}=(16-\frac{1}{2})-(2-2)=\frac{31}{2}\).
Scoring: 1 point for the correct antiderivative; 1 point for evaluating the limits to get \(\frac{31}{2}\).
Question 5: Substitution (2 points)
(a) Let \(u=x^{2}-5\), so \(du=2x\,dx\) and \(4x\,dx=2\,du\). Then \(\int 2u^{3}\,du=\frac{u^{4}}{2}+C=\frac{(x^{2}-5)^{4}}{2}+C\).
(b) Let \(u=\ln x\), so \(du=\frac{1}{x}dx\), with limits \(u=0\) to \(u=1\). Then \(\int_{0}^{1}u^{2}\,du=\frac{1}{3}\).
Scoring: (a) 1 point for \(\frac{(x^{2}-5)^{4}}{2}+C\); omit the point if \(+C\) is missing. (b) 1 point for \(\frac{1}{3}\).
AP Calculus Integration Test Answer Key: questions 6–8
Question 6: Water in a tank (3 points)
(a) \(V(t)=30+\int_{0}^{t}(12-2s)\,ds=30+12t-t^{2}\).
(b) \(V^{\prime}(t)=r(t)=12-2t=0\) at \(t=6\). \(r\) is positive before \(t=6\) and negative after, so \(V\) increases then decreases and the maximum is at \(t=6\): \(V(6)=30+72-36=66\) gallons.
(c) \(V(8)=30+96-64=62\) gallons. Since \(r(8)=-4 < 0\), the amount of water is decreasing at \(t=8\).
Scoring: (a) 1 point for \(30+12t-t^{2}\). (b) 1 point for \(t=6\) with the sign-change justification and 66 gallons. (c) 1 point for 62 gallons and the explanation using \(r(8)<0\).
Question 7: Division and completing the square (3 points)
(a) Divide: \(x^{2}-1=(x+3)(x-3)+8\), so the integrand is \(x-3+\frac{8}{x+3}\). Then \(\int(x-3+\frac{8}{x+3})dx=\frac{x^{2}}{2}-3x+8\ln|x+3|+C\).
(b) \(x^{2}+6x+10=(x+3)^{2}+1\), so \(\int\frac{1}{(x+3)^{2}+1}\,dx=\tan^{-1}(x+3)+C\).
Scoring: (a) 1 point for the correct division; 1 point for the antiderivative with \(+C\). (b) 1 point for \(\tan^{-1}(x+3)+C\).
Question 8: Differentiating an integral (2 points)
(a) By the Fundamental Theorem and the chain rule, \(G^{\prime}(x)=(x^{2}+3)\cdot 2x\).
(b) \(G^{\prime}(2)=(4+3)(4)=28\).
Scoring: (a) 1 point for \(2x(x^{2}+3)\); the factor \(2x\) must appear. (b) 1 point for 28.
Scoring summary
20 points: add the points of every question of this ap calculus integration test answer key.
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AP Calculus AB Integration and Accumulation of Change test answer key
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