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

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

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This ftc practice solutions 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.

This answer key gives a full worked solution for every problem in the fundamental theorem worksheet. Compare each step, not only the final answer, because AP graders award credit for a correct setup and clear justification. Final answers are in bold. Redo any problem you missed before moving on to the test.

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 practice solutions: exercises 1 to 10

Exercise 1 – Polynomial antiderivative

Use the power rule term by term: $\int x^{n}dx=\frac{x^{n+1}}{n+1}$.

Check by differentiating $x^{3}-2x^{2}+5x$: you get $3x^{2}-4x+5$.

Final answer: $x^{3}-2x^{2}+5x+C$.

Exercise 2 – Negative powers

Rewrite $\frac{2}{x^{2}}=2x^{-2}$. Then $\int 2x^{-2}dx=\frac{2x^{-1}}{-1}=-\frac{2}{x}$.

Final answer: $\frac{x^{4}}{4}-\frac{2}{x}+C$.

Exercise 3 – Roots

Rewrite as $x^{1/2}+x^{-1/2}$. Then $\frac{x^{3/2}}{3/2}+\frac{x^{1/2}}{1/2}$.

Final answer: $\frac{2}{3}x^{3/2}+2\sqrt{x}+C$.

Exercise 4 – Sine and cosine

$\int\sin x\,dx=-\cos x$ and $\int\cos x\,dx=\sin x$.

Final answer: $-\cos x-2\sin x+C$.

Exercise 5 – Exponential and reciprocal

$\int e^{x}dx=e^{x}$ and $\int\frac{1}{x}dx=\ln|x|$.

Final answer: $e^{x}+3\ln|x|+C$.

Exercise 6 – Trig and inverse trig forms

$\tan x+C$ since $(\tan x)^{\prime}=\sec^{2}x$.
$\tan^{-1}x+C$ since $(\tan^{-1}x)^{\prime}=\frac{1}{1+x^{2}}$.

Exercise 7 – Definite integral, linear

$F(x)=x^{2}+x$. $F(3)-F(1)=12-2=10$.

Geometry: a trapezoid with parallel sides $3$ and $7$ and width $2$ has area $\frac{3+7}{2}\cdot 2=10$.

Final answer: $10$.

Exercise 8 – Definite integral, cubic

$F=\frac{x^{4}}{4}-\frac{x^{2}}{2}$. $F(2)=4-2=2$ and $F(0)=0$.

Final answer: $2$.

Signed areas: −1/4 and 9/4 give 2.

Exercise 9 – Definite integral, root

$F=\frac{2}{3}x^{3/2}$. $F(4)=\frac{2}{3}(8)=\frac{16}{3}$ and $F(1)=\frac{2}{3}$.

Final answer: $\frac{14}{3}$.

Exercise 10 – Definite integral, sine

$[-\cos x]_{0}^{\pi}=-\cos\pi-(-\cos 0)=1+1=2$.

Final answer: $2$.

Exercises 11 and up

Exercise 11 – Exponential and logarithm integrals

$[e^{x}]_{0}^{1}=e-1\approx 1.718$.
$[\ln x]_{1}^{e}=1-0=1$.

Two regions: area e − 1 and area 1.

Exercise 12 – Absolute value

Split at $0$: $\int_{-1}^{0}(-x)dx=\frac{1}{2}$ and $\int_{0}^{2}x\,dx=2$.

Final answer: $\frac{1}{2}+2=\frac{5}{2}$.

Exercise 13 – FTC part 1, direct

By FTC Part 1, $F^{\prime}(x)=\sqrt{x^{2}+3}$.

$F^{\prime}(1)=\sqrt{4}=2$.

Final answer: $F^{\prime}(x)=\sqrt{x^{2}+3},\ F^{\prime}(1)=2$.

Exercise 14 – FTC part 1, chain rule

Substitute the upper limit and multiply by its derivative: $\cos(x^{2})\cdot 2x$.

Final answer: $2x\cos(x^{2})$.

Exercise 15 – FTC part 1, variable lower limit

Swap the limits to put the variable on top: $\int_{x}^{5}=-\int_{5}^{x}$.

So the derivative is $-(x^{3}+1)$.

Final answer: $-(x^{3}+1)$.

Exercise 16 – Accumulation from a graph

$g(2)=2\cdot 2=4$. From $2$ to $3$ the average height is $\frac{2+0}{2}=1$, so $g(3)=5$. From $3$ to $4$ the average height is $-1$: $g(4)=4$. From $4$ to $6$: $-2\cdot 2=-4$, so $g(6)=0$.

$g^{\prime}=f$ changes from positive to negative at $t=3$, so $g$ is largest at $x=3$.

Final answer: $g(2)=4,\ g(3)=5,\ g(4)=4,\ g(6)=0;\ \text{maximum at }x=3$.

Exercise 17 – Initial value problem

$f(x)=2x^{3}-2x+C$. $f(1)=0+C=5$, so $C=5$.

Final answer: $f(x)=2x^{3}-2x+5$.

The particular solution passes through (1, 5).

Exercise 18 – Second-order initial value problem

$f^{\prime}=6x^{2}-6x+C_{1}$ and $f^{\prime}(0)=1$ gives $C_{1}=1$.

$f=2x^{3}-3x^{2}+x+C_{2}$ and $f(0)=2$ gives $C_{2}=2$.

Final answer: $f(x)=2x^{3}-3x^{2}+x+2$.

Exercise 19 – Position from velocity

$s(t)=t^{3}-6t^{2}+9t+C$ with $s(0)=C=2$.

$s(4)=64-96+36+2=6$.

Final answer: $s(t)=t^{3}-6t^{2}+9t+2;\ s(4)=6\text{ ft}$.

Exercise 20 – Net change of water

Net change: $\int_{0}^{6}(12-2t)\,dt=[12t-t^{2}]_{0}^{6}=72-36=36$.

Check with geometry: a triangle with base $6$ and height $12$ has area $36$.

Final answer: $50+36=86\text{ gallons}$.

Exercise 21 – Accumulation function analysis

$F^{\prime}(x)=x^{2}-4=(x-2)(x+2)$. Signs: $+$ for $x < -2$, $-$ for $-2 < x < 2$, $+$ for $x > 2$.

$F(x)=\frac{x^{3}}{3}-4x$, so $F(-2)=\frac{16}{3}$ and $F(2)=-\frac{16}{3}$.

Final answer: $\text{increasing on }(-\infty,-2),(2,\infty);\ \text{relative max at }x=-2;\ \text{relative min at }x=2$.

Exercise 22 – Area with a sign change

$t^{2}-4$ changes sign at $t=2$. On $[0,2]$: $\int_{0}^{2}(4-t^{2})dt=8-\frac{8}{3}=\frac{16}{3}$.

On $[2,3]$: $\int_{2}^{3}(t^{2}-4)dt=\left(9-12\right)-\left(\frac{8}{3}-8\right)=\frac{7}{3}$.

Final answer: $\frac{16}{3}+\frac{7}{3}=\frac{23}{3}$.

Exercise 23 – Total cost from marginal cost

$\int_{50}^{100}(40-0.2x)\,dx=[40x-0.1x^{2}]_{50}^{100}=(4000-1000)-(2000-250)=3000-1750=1250$.

Final answer: $\$1250$.

Key facts behind this ftc practice solutions

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 practice solutions

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

Download the ftc practice solutions as a PDF with the button below and keep it with your notes.

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