AP Calculus AB Fundamental Theorem of Calculus and Antiderivatives Test Solutions and Scoring (PDF)

AP Calculus AB math test solutions: Fundamental Theorem of Calculus and Antiderivatives – Maths PDF

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This ftc test answers for AP Calculus AB covers fundamental theorem of calculus and antiderivatives (Common Core AP CALC AB 6.4, AP CALC AB 6.5, AP CALC AB 6.6, AP CALC AB 6.7, AP CALC AB 6.8), with a free PDF to print.

Use this answer key to score the fundamental theorem test out of 20 points. Each solution shows the work and a scoring guide that says how the points are awarded. Give partial credit for a correct setup, a correct derivative or integral, and a correct final statement with units where they apply.

AP Calculus AB · Common Core State Standards: AP CALC AB 6.4, AP CALC AB 6.5, AP CALC AB 6.6, AP CALC AB 6.7, AP CALC AB 6.8

Ftc test answers: worked solutions

Question 1 – Indefinite integral [2 pts]

\(2x^{4}-3x^{2}+\ln|x|+C\).

Scoring: 1 point for the polynomial terms; 1 point for ln|x| and + C.

Question 2 – Definite integral [2 pts]

\([\sin x+2x]_{0}^{\pi/2}=\left(1+\pi\right)-0=1+\pi\).

Final answer: \(1+\pi\).

Scoring: 1 point for the antiderivative; 1 point for the value.

Question 3 – FTC with the chain rule [3 pts]

\(G^{\prime}(x)=\sqrt{1+(x^{3})^{2}}\cdot 3x^{2}=3x^{2}\sqrt{1+x^{6}}\).
\(G^{\prime}(1)=3\sqrt{2}\).

Scoring: (a) 1 point for the integrand at the upper limit; 1 point for the chain factor 3x²; (b) 1 point.

Question 4 – Accumulation from a graph [3 pts]

\(g(2)=2\cdot 2=4\). \(g(-2)=-\int_{-2}^{0}f=-\frac{1}{2}(2)(2)=-2\).
\(g^{\prime}=f\) changes from positive to negative at \(x=3\) (where \(f=0\) on the falling segment: \(2-2(x-2)=0\) gives \(x=3\)). So x = 3.

The graph of f.

Scoring: (a) 1 point per value; (b) 1 point for the sign change of f.

Question 5 – Initial value problem [2 pts]

\(f=x^{4}-3x+C\). \(f(2)=16-6+C=10\), so \(C=0\).

Final answer: \(f(x)=x^{4}-3x\).

Scoring: 1 point for the antiderivative; 1 point for C.

Question 6 – Motion with integrals [3 pts]

\(s(t)=12t-\frac{3}{2}t^{2}+5\).
\(\int_{0}^{6}(12-3t)dt=72-54=18\) ft.
\(v=0\) at \(t=4\). \(\int_{0}^{4}=48-24=24\) and \(\left|\int_{4}^{6}\right|=\left|(72-54)-(48-24)\right|=6\), so the distance is \(30\) ft.

Velocity in feet per second.

Scoring: 1 point per part.

Question 7 – Net change in context [3 pts]

\(\int_{0}^{10}20e^{-0.2t}\,dt\).
\(\left[-100e^{-0.2t}\right]_{0}^{10}=-100e^{-2}+100=100(1-e^{-2})\approx 86.47\) gallons.

Leak rate in gallons per hour.

Scoring: 1 point for the integral; 1 point for the antiderivative; 1 point for the value.

Question 8 – Properties [2 pts]

\(F(6)-F(2)=9\) gives \(F(6)=13\). \(\int_{6}^{2}3f=-3(9)=-27\).

Scoring: 1 point for F(6); 1 point for the integral.

Total score

Total

20 points

Key facts behind this ftc test answers

Key facts

FTC Part 1: d/dx of the integral from a to x of f(t) dt equals f(x).
FTC Part 2: the integral from a to b of f equals F(b) – F(a) for any antiderivative F.
For an upper limit g(x), multiply by g′(x) (chain rule).
Every indefinite integral needs + C; an initial condition determines C.

How to use this ftc test answers

Score your test with the point values shown for each question. Then read the worked solution of every question where you lost points and note the method to review.

All AP Calculus AB chapters: AP Calculus AB math.

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