AP Calculus BC 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) u = x²
B) u = x³ + 5
C) u = (x³ + 5)⁴
D) u = x
Correct Answer: B
According to the principles of substitution, we look for a function and its derivative within the integrand. If we let u = x³ + 5, then its derivative du = 3x² dx. The term x² dx is present in the integrand, making this an effective substitution to simplify the integral into the form ∫(1/3 * u⁴) du.
A) -cos⁶(x)/6 + C
B) sin⁶(x)/6 + C
C) cos⁶(x)/6 + C
D) 5sin⁴(x) + C
Correct Answer: B
This integral requires substitution. Let u = sin(x). Then, du = cos(x) dx. The integral becomes ∫u⁵ du. The antiderivative is u⁶/6 + C. Substituting back for u gives sin⁶(x)/6 + C.
A) e - 1
B) e
C) 1 - e
D) 2e - 2
Correct Answer: A
Use the substitution u = x². Then du = 2x dx. When using substitution for a definite integral, the limits of integration must be changed. The lower limit x=0 becomes u=0²=0. The upper limit x=1 becomes u=1²=1. The integral transforms to ∫ from 0 to 1 of e^u du. The antiderivative is e^u, evaluated from 0 to 1, which is e¹ - e⁰ = e - 1.
A) Finding the derivative of a composite function.
B) Calculating the limit of a function.
C) Finding antiderivatives.
D) Determining the concavity of a curve.
Correct Answer: C
As stated in the provided content, substitution of variables is a technique for finding antiderivatives (or indefinite integrals). It is essentially the reverse of the chain rule for differentiation.
A) ∫ from 1 to 9 of ((u-1)/2 * u^(-1/2)) du
B) ∫ from 0 to 4 of (1/2 * u^(-1/2)) du
C) ∫ from 1 to 9 of (1/4 * (u-1) * u^(-1/2)) du
D) ∫ from 0 to 9 of (1/2 * (u-1) * u^(-1/2)) du
Correct Answer: C
Given the substitution u = 1 + 2x, we must change the limits, the variable x, and dx. The new limits are u(0) = 1+2(0)=1 and u(4) = 1+2(4)=9. From the substitution, du = 2dx, so dx = du/2. Also, we must rearrange for x: u-1 = 2x, so x = (u-1)/2. Substituting all parts gives ∫ from 1 to 9 of (((u-1)/2) / √u) * (du/2), which simplifies to ∫ from 1 to 9 of (1/4 * (u-1) * u^(-1/2)) du.
A) The limits of integration remain unchanged.
B) The limits must be converted to correspond to the new variable of integration.
C) The original limits are used after substituting the original variable back into the antiderivative.
D) The limits are always changed to 0 and 1.
Correct Answer: B
The provided content explicitly states that for a definite integral, substitution of variables requires corresponding changes to the limits of integration. This means the original x-values of the limits must be plugged into the u-substitution equation to find the new u-values for the limits.
A) ln|1 + tan(x)| + C
B) -1 / (1 + tan(x))² + C
C) sec(x)tan(x) * ln|1 + tan(x)| + C
D) arctan(tan(x)) + C
Correct Answer: A
This integral can be solved using substitution. Let u = 1 + tan(x). The derivative is du = sec²(x) dx. The integral transforms into ∫(1/u) du. The antiderivative of 1/u is ln|u| + C. Substituting back for u gives ln|1 + tan(x)| + C.
A) 1/2
B) 1
C) e²/2
D) e - 1
Correct Answer: A
Let u = ln(x). Then du = (1/x) dx. We must change the limits of integration. The lower limit x=1 becomes u=ln(1)=0. The upper limit x=e becomes u=ln(e)=1. The integral becomes ∫ from 0 to 1 of u du. The antiderivative is u²/2. Evaluating from 0 to 1 gives (1²/2) - (0²/2) = 1/2.
A) From 0 to 1
B) From 1 to 0
C) From 0 to π/2
D) From -1 to 1
Correct Answer: B
For a definite integral, substitution requires changing the limits. The new lower limit is found by substituting the original lower limit into the substitution equation: u = cos(0) = 1. The new upper limit is found by substituting the original upper limit: u = cos(π/2) = 0. Therefore, the new limits of integration are from 1 to 0.