AP Calculus BC 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) 8
B) -8
C) 1/8
D) 0
Correct Answer: B
This question uses the property of reversal of limits of integration, which states that ∫ₐᵇ f(x) dx = -∫ᵇₐ f(x) dx. Therefore, ∫₅² f(x) dx = -∫₂⁵ f(x) dx = -8. [cite: 2655]
A) 1
B) -12
C) -3/4
D) 7
Correct Answer: B
This question uses the property of the integral of a constant times a function, which states that ∫ₐᵇ c*f(x) dx = c * ∫ₐᵇ f(x) dx. In this case, ∫₀⁷ 4g(x) dx = 4 * ∫₀⁷ g(x) dx = 4 * (-3) = -12. [cite: 2655]
A) 4
B) 8
C) 10
D) 14
Correct Answer: D
This question uses the properties of the integral of a sum/difference and a constant multiple. ∫₁⁴ [2f(x) - g(x)] dx = ∫₁⁴ 2f(x) dx - ∫₁⁴ g(x) dx = 2∫₁⁴ f(x) dx - ∫₁⁴ g(x) dx. Substituting the given values: 2(6) - (-2) = 12 + 2 = 14. [cite: 2655]
A) 3
B) 7
C) -7
D) Cannot be determined
Correct Answer: B
This question combines the properties of adjacent intervals and reversal of limits. First, reverse the limits of the second integral: ∫₃⁸ h(x) dx = -∫₈³ h(x) dx = -(-2) = 2. Now, use the adjacent intervals property: ∫₀⁸ h(x) dx = ∫₀³ h(x) dx + ∫₃⁸ h(x) dx. Substituting the values: 5 + 2 = 7. [cite: 2655]
A) π
B) 2π
C) 4π
D) 8π
Correct Answer: B
The definite integral ∫₀⁴ f(x) dx represents the area under the curve of f(x) from x=0 to x=4. The described graph is a semicircle with radius 2. The area of a full circle is πr², so the area of this semicircle is (1/2)π(2)² = 2π. This demonstrates evaluating a definite integral using geometry. [cite: 2654]
A) 1
B) 4
C) 8
D) 10
Correct Answer: B
The function has a jump discontinuity at x=3. The integral can be split into two parts: ∫₀⁵ f(x) dx = ∫₀³ f(x) dx + ∫₃⁵ f(x) dx. For the first interval, f(x)=2, so ∫₀³ 2 dx = 2 * (3-0) = 6. For the second interval, f(x)=-1, so ∫₃⁵ -1 dx = -1 * (5-3) = -2. The total value is 6 + (-2) = 4. This shows the integral definition extended to functions with jump discontinuities. [cite: 2656]
A) Reversal of limits of integration
B) Integral of the sum of two functions
C) Integral of a function over adjacent intervals
D) Integral of a constant times a function
Correct Answer: C
The relationship ∫₂⁹ f(x) dx = ∫₂⁵ f(x) dx + ∫₅⁹ f(x) dx is an application of the property of integrating a function over adjacent intervals. By rearranging this equation, one can solve for the unknown integral ∫₂⁵ f(x) dx. [cite: 2655]
A) A single rectangle with height 3 and width 5
B) A semicircle with radius 2.5
C) Two triangles
D) A trapezoid with height 5
Correct Answer: C
The graph of y=|x| from x=-2 to x=3 forms a 'V' shape with the vertex at the origin. The area under this graph consists of two separate triangles. The first triangle is in the second quadrant with vertices at (-2,0), (0,0), and (-2,2). The second triangle is in the first quadrant with vertices at (0,0), (3,0), and (3,3). The integral is the sum of the areas of these two triangles. [cite: 2654]
A) 10 - A
B) 4 - A
C) 20 - A
D) 14 - 2A
Correct Answer: A
First, use the properties of integrals on the given equation: ∫₁⁵ [2f(x) - 3] dx = 2∫₁⁵ f(x) dx - ∫₁⁵ 3 dx = 8. The integral of the constant is ∫₁⁵ 3 dx = 3(5-1) = 12. So, 2∫₁⁵ f(x) dx - 12 = 8, which means 2∫₁⁵ f(x) dx = 20, and ∫₁⁵ f(x) dx = 10. Now, use the adjacent intervals property: ∫₁⁵ f(x) dx = ∫₁³ f(x) dx + ∫₃⁵ f(x) dx. Substitute the known values: 10 = A + ∫₃⁵ f(x) dx. Solving for the desired integral gives ∫₃⁵ f(x) dx = 10 - A. [cite: 2655]
A) 25
B) 5
C) -5
D) -25
Correct Answer: D
First, break apart the given integral: ∫ₐᵇ [3f(x) + 5] dx = 3∫ₐᵇ f(x) dx + ∫ₐᵇ 5 dx = 10. Substitute the given value for the integral of f(x): 3(-5) + ∫ₐᵇ 5 dx = 10, which simplifies to -15 + ∫ₐᵇ 5 dx = 10. Therefore, ∫ₐᵇ 5 dx = 25. The question asks for ∫ᵇₐ 5 dx. Using the reversal of limits property, ∫ᵇₐ 5 dx = -∫ₐᵇ 5 dx = -25. [cite: 2655]