Practice Problems – Definite & Indefinite Integrals – AP Calculus

Below are 20 practice problems covering all topics from Module 03A: Definite & Indefinite Integrals. Topics include: antiderivatives, basic integration rules, Riemann sums, definite integrals, and the Fundamental Theorem of Calculus. Each problem includes a hidden solution — click the button to reveal the step-by-step answer.

Problem 01 Antiderivative

Find $\displaystyle \int (3x^2 - 4x + 5) \, dx$.

Solution:
$\int 3x^2 dx = x^3$, $\int -4x dx = -2x^2$, $\int 5 dx = 5x$
$\int (3x^2 - 4x + 5) dx = x^3 - 2x^2 + 5x + C$

Problem 02 Power Rule

Find $\displaystyle \int \sqrt{x} \, dx$.

Solution:
$\sqrt{x} = x^{1/2}$
$\int x^{1/2} dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C$

Problem 03 Special Integral

Find $\displaystyle \int \frac{1}{x} \, dx$.

Solution:
$\int \frac{1}{x} dx = \ln|x| + C$

Problem 04 Exponential

Find $\displaystyle \int 4e^x \, dx$.

Solution:
$\int 4e^x dx = 4e^x + C$

Problem 05 Trigonometric

Find $\displaystyle \int (\cos x - \sin x) \, dx$.

Solution:
$\int \cos x dx = \sin x$, $\int -\sin x dx = \cos x$
$\int (\cos x - \sin x) dx = \sin x + \cos x + C$

Problem 06 Left Riemann Sum

Approximate $\displaystyle \int_0^4 x^2 dx$ using a left Riemann sum with $n=4$.

Solution:
$\Delta x = 1$, left endpoints: $0,1,2,3$
$f(0)=0$, $f(1)=1$, $f(2)=4$, $f(3)=9$
$\text{Area} \approx 1(0+1+4+9) = 14$

Problem 07 Right Riemann Sum

Approximate $\displaystyle \int_0^4 x^2 dx$ using a right Riemann sum with $n=4$.

Solution:
$\Delta x = 1$, right endpoints: $1,2,3,4$
$f(1)=1$, $f(2)=4$, $f(3)=9$, $f(4)=16$
$\text{Area} \approx 1(1+4+9+16) = 30$

Problem 08 Midpoint Riemann Sum

Approximate $\displaystyle \int_0^4 x^2 dx$ using a midpoint Riemann sum with $n=4$.

Solution:
$\Delta x = 1$, midpoints: $0.5,1.5,2.5,3.5$
$f(0.5)=0.25$, $f(1.5)=2.25$, $f(2.5)=6.25$, $f(3.5)=12.25$
$\text{Area} \approx 1(0.25+2.25+6.25+12.25) = 21$

Problem 09 Definite Integral (Geometry)

Find $\displaystyle \int_2^5 4 \, dx$ using geometry.

Solution:
Rectangle: width $= 5-2 = 3$, height $= 4$
Area $= 3 \times 4 = 12$

Problem 10 Definite Integral (Geometry)

Find $\displaystyle \int_0^3 x \, dx$ using geometry.

Solution:
Triangle: base $= 3$, height $= 3$
Area $= \frac{1}{2} \times 3 \times 3 = 4.5$

Problem 11 FTC Part 1

Find $\displaystyle \frac{d}{dx} \int_1^x \sin t \, dt$.

Solution:
By FTC Part 1, $\frac{d}{dx} \int_1^x \sin t \, dt = \sin x$

Problem 12 FTC Part 1

Find $\displaystyle \frac{d}{dx} \int_0^{x^2} e^{t} \, dt$.

Solution:
By FTC Part 1 with Chain Rule: $= e^{x^2} \cdot 2x = 2x e^{x^2}$

Problem 13 FTC Part 2

Evaluate $\displaystyle \int_1^2 (3x^2) \, dx$.

Solution:
Antiderivative: $F(x) = x^3$
$\int_1^2 3x^2 dx = [x^3]_1^2 = 8 - 1 = 7$

Problem 14 FTC Part 2

Evaluate $\displaystyle \int_0^{\pi/2} \cos x \, dx$.

Solution:
Antiderivative: $\sin x$
$\int_0^{\pi/2} \cos x dx = [\sin x]_0^{\pi/2} = 1 - 0 = 1$

Problem 15 FTC Part 2

Evaluate $\displaystyle \int_1^4 \frac{1}{x} \, dx$.

Solution:
Antiderivative: $\ln|x|$
$\int_1^4 \frac{1}{x} dx = [\ln|x|]_1^4 = \ln 4 - \ln 1 = \ln 4$

Problem 16 Properties

If $\int_0^2 f(x) dx = 5$ and $\int_0^2 g(x) dx = 3$, find $\int_0^2 [2f(x) - 3g(x)] dx$.

Solution:
$= 2\int_0^2 f(x) dx - 3\int_0^2 g(x) dx = 2(5) - 3(3) = 10 - 9 = 1$

Problem 17 Additivity

If $\int_0^3 f(x) dx = 8$ and $\int_3^5 f(x) dx = 4$, find $\int_0^5 f(x) dx$.

Solution:
$\int_0^5 f(x) dx = \int_0^3 f(x) dx + \int_3^5 f(x) dx = 8 + 4 = 12$

Problem 18 Reversing Limits

If $\int_1^3 f(x) dx = 6$, find $\int_3^1 f(x) dx$.

Solution:
$\int_3^1 f(x) dx = -\int_1^3 f(x) dx = -6$

Problem 19 Trapezoidal Rule

Use the left and right sums from Problems 6 and 7 to find the trapezoidal rule approximation for $\int_0^4 x^2 dx$.

Solution:
Left sum $= 14$, Right sum $= 30$
Trapezoidal $= \frac{14 + 30}{2} = 22$

Problem 20 Definite Integral

Find $\displaystyle \int_{-2}^2 \sqrt{4 - x^2} \, dx$ using geometry.

Solution:
$y = \sqrt{4 - x^2}$ is the top half of a circle of radius $2$.
Area of semicircle $= \frac{1}{2} \pi (2)^2 = 2\pi$
$\int_{-2}^2 \sqrt{4 - x^2} dx = 2\pi$

← Back to Module Page

📝 Practice Problems