AP Calculus AB Practice Quiz: Applying Properties of Definite Integrals
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 10 questions to check your progress.
Question 1 of 10
All Questions (10)
A) 4
B) 12
C) 15
D) 36
Correct Answer: D
This question uses the property of the integral of a constant times a function: ∫ c*f(x) dx = c * ∫ f(x) dx. Therefore, ∫ from 2 to 7 of 3f(x) dx = 3 * ∫ from 2 to 7 of f(x) dx = 3 * 12 = 36. [cite: 2655]
A) -24
B) 5
C) 11
D) -5/3
Correct Answer: B
This question uses the property of the integral of the sum of two functions: ∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx. Therefore, the value is 8 + (-3) = 5. [cite: 2655]
A) 6
B) 1/6
C) -6
D) 0
Correct Answer: C
This question uses the property of reversal of limits of integration: ∫ from b to a of f(x) dx = - ∫ from a to b of f(x) dx. Therefore, ∫ from 4 to -1 of h(t) dt = - ∫ from -1 to 4 of h(t) dt = -6. [cite: 2655]
A) -5
B) 5
C) 13
D) 36
Correct Answer: C
This question uses the property of the integral of a function over adjacent intervals: ∫ from a to c of f(x) dx = ∫ from a to b of f(x) dx + ∫ from b to c of f(x) dx. So, ∫ from 0 to 7 of f(x) dx = ∫ from 0 to 3 of f(x) dx + ∫ from 3 to 7 of f(x) dx = 4 + 9 = 13. [cite: 2655]
A) 8
B) 12
C) 16
D) 24
Correct Answer: C
The integral represents the area under the line y = x + 2 from x = 0 to x = 4. This area is a trapezoid with bases of length f(0) = 2 and f(4) = 6, and a height of 4. The area of a trapezoid is (1/2) * (base1 + base2) * height. So, the area is (1/2) * (2 + 6) * 4 = (1/2) * 8 * 4 = 16. [cite: 2654]
A) 25π
B) 50π
C) 25π/2
D) 5π
Correct Answer: C
The integral represents the area of the region bounded by the curve y = √(25 - x^2) and the x-axis. This equation describes the top half of a circle centered at the origin with a radius of 5. The area of a full circle is πr^2, so the area of this semicircle is (1/2)πr^2 = (1/2)π(5)^2 = 25π/2. [cite: 2654]
A) 8
B) 15
C) 16
D) 22
Correct Answer: C
The function has a jump discontinuity at x = 2. The integral can be split at the point of discontinuity: ∫ from 0 to 4 of f(x) dx = ∫ from 0 to 2 of f(x) dx + ∫ from 2 to 4 of f(x) dx. For the first integral, f(x) = 3, so ∫ from 0 to 2 of 3 dx = 3 * (2 - 0) = 6. For the second integral, f(x) = 5, so ∫ from 2 to 4 of 5 dx = 5 * (4 - 2) = 10. The total value is 6 + 10 = 16. [cite: 2656, 2655]
A) 10
B) 14
C) 18
D) 22
Correct Answer: C
First, find ∫ from 5 to 8 of f(x) dx using the adjacent intervals property: ∫ from 2 to 8 = ∫ from 2 to 5 + ∫ from 5 to 8. So, 10 = 3 + ∫ from 5 to 8 of f(x) dx, which means ∫ from 5 to 8 of f(x) dx = 7. Now, apply the sum and constant multiple properties: ∫ from 5 to 8 of [2f(x) - g(x)] dx = 2 * ∫ from 5 to 8 of f(x) dx - ∫ from 5 to 8 of g(x) dx = 2 * (7) - (-4) = 14 + 4 = 18. [cite: 2655]
A) 2π - 2
B) 2π + 2
C) 4π - 2
D) π - 2
Correct Answer: A
The integral is the net area from x = -4 to x = 2. This can be split into two parts: ∫ from -4 to 0 of f(x) dx and ∫ from 0 to 2 of f(x) dx. The first part is the area of a semicircle with radius 2, which is (1/2)π(2)^2 = 2π. The second part is the area of a triangle below the x-axis with base 2 and height 2, which is -(1/2) * 2 * 2 = -2. The total value of the integral is the sum of these areas: 2π - 2. [cite: 2653, 2654]
A) 8
B) 10
C) 13
D) 21
Correct Answer: C
According to the Fundamental Theorem of Calculus, f(5) = f(1) + ∫ from 1 to 5 of f'(x) dx. The integral ∫ from 1 to 5 of f'(x) dx can be calculated as the area under the graph of f'. The graph of f' from x=1 to x=5 is a triangle with base 5 - 1 = 4 and height 4. The area of this triangle is (1/2) * base * height = (1/2) * 4 * 4 = 8. Therefore, f(5) = f(1) + 8 = 5 + 8 = 13. [cite: 2653, 2654]