AP Calculus AB Practice Quiz: Integrating Using Substitution
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 9 questions to check your progress.
Question 1 of 9
All Questions (9)
A) (1/5)(x² + 3)⁵ + C
B) (1/10)(x² + 3)⁵ + C
C) 2x(1/5)(x² + 3)⁵ + C
D) (x² + 3)⁵ + C
Correct Answer: A
This is a direct application of substitution. Let u = x² + 3. Then du = 2x dx. The integral becomes ∫u⁴ du, which evaluates to (1/5)u⁵ + C. Substituting back for u gives (1/5)(x² + 3)⁵ + C.
A) 7/9
B) 1/3
C) 7/3
D) 8/9
Correct Answer: A
Use substitution with u = x³ + 1, which means du = 3x² dx, or (1/3)du = x² dx. The limits of integration must be changed: when x=0, u=0³+1=1; when x=1, u=1³+1=2. The integral becomes (1/3)∫₁² u² du = (1/3)[(1/3)u³] from 1 to 2. This evaluates to (1/9)(2³ - 1³) = (1/9)(8 - 1) = 7/9.
A) The limits of integration remain unchanged.
B) The limits of integration must be converted to correspond to the new variable.
C) The limits of integration are ignored, and the final answer is evaluated with the original variable.
D) The upper limit is kept, but the lower limit is always changed to zero.
Correct Answer: B
According to the principles of integration by substitution for definite integrals, when the variable of integration is changed (e.g., from x to u), the limits of integration must also be changed to the corresponding values of the new variable.
A) -sin(x)e^(sin(x)) + C
B) e^(sin(x)) + C
C) e^(cos(x)) + C
D) sin(x)e^(sin(x)) + C
Correct Answer: B
To find the indefinite integral ∫cos(x)e^(sin(x)) dx, use the substitution u = sin(x). Then du = cos(x) dx. The integral transforms into ∫e^u du, which is e^u + C. Substituting back for u gives e^(sin(x)) + C.
A) 1/3
B) 1
C) e³/3
D) (e³-1)/3
Correct Answer: A
Let u = ln(x). Then du = (1/x) dx. We must change the limits of integration. When x=1, u = ln(1) = 0. When x=e, u = ln(e) = 1. The integral becomes ∫₀¹ u² du. Evaluating this gives [(1/3)u³] from 0 to 1, which is (1/3)(1³ - 0³) = 1/3.
A) ln|x²+4x| + C
B) 2 ln|x²+4x| + C
C) (1/2)ln|x²+4x| + C
D) -1/(x²+4x)² + C
Correct Answer: C
This integral requires substitution. Let u = x²+4x. Then du = (2x+4) dx, which can be written as du = 2(x+2) dx. Therefore, (1/2)du = (x+2) dx. The integral becomes (1/2)∫(1/u) du = (1/2)ln|u| + C. Substituting back for u gives (1/2)ln|x²+4x| + C.
A) ∫₁² 2eᵘ du
B) ∫₁⁴ eᵘ du
C) ∫₁² eᵘ/u du
D) ∫₁⁴ 2eᵘ du
Correct Answer: A
Given the substitution u = √x, we find du = (1/(2√x)) dx, which rearranges to 2du = (1/√x) dx. We must also change the limits of integration. When x=1, u=√1=1. When x=4, u=√4=2. Substituting u, du, and the new limits into the original integral gives ∫₁² eᵘ (2du), which is equivalent to ∫₁² 2eᵘ du.
A) 0
B) -1/2
C) 1
D) 1/2
Correct Answer: D
Use substitution with u = sin(x), so du = cos(x) dx. The limits must be changed: when x=0, u=sin(0)=0; when x=π/2, u=sin(π/2)=1. The integral becomes ∫₀¹ u du. Evaluating this gives [(1/2)u²] from 0 to 1, which equals (1/2)(1² - 0²) = 1/2. Alternatively, substituting u=cos(x) would also yield the same result.
A) (2/5)(x-1)⁵/² + (2/3)(x-1)³/² + C
B) (1/2)x²(2/3)(x-1)³/² + C
C) (2/3)(x-1)³/² + C
D) x(2/3)(x-1)³/² + C
Correct Answer: A
This requires a substitution where the original variable must also be substituted. Let u = x-1. Then du = dx and x = u+1. The integral becomes ∫(u+1)√u du = ∫(u³/² + u¹/²) du. Integrating term by term gives (2/5)u⁵/² + (2/3)u³/² + C. Substituting back u = x-1 gives the final answer.