Drill 1 · Math · u-Substitution
AP Calculus AB: u-Substitution (Drill 1) is a Math practice drill covering u-Substitution. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice u-substitution for both indefinite and definite integrals, including recognizing the correct substitution, changing bounds, and working with trigonometric and exponential functions. These AP Calculus AB integration skills are tested on both the multiple-choice and free-response sections.
Question 1. Let \( u = x^2 + 1 \). What is \( \int 2x(x^2 + 1)^4 \, dx \)?
Explanation: Choice A is correct. Let \( u = x^2+1 \), so \( du = 2x\,dx \). The integral becomes \( \int u^4\,du = \dfrac{u^5}{5}+C = \dfrac{(x^2+1)^5}{5}+C \). Choice B is incorrect because the student used \( du = x\,dx \) (missing the factor of 2), producing a spurious \( \frac{1}{2} \). Choice C is incorrect because the student left the \( 2x \) factor in the result instead of absorbing it into \( du \). Choice D is incorrect because the student forgot to divide by the new exponent after integrating.
Question 2. What is \( \int_0^1 x \, e^{x^2} \, dx \)?
Explanation: Choice A is correct. Let \( u = x^2 \), so \( du = 2x\,dx \), meaning \( x\,dx = \frac{du}{2} \). Bounds: \( x=0 \Rightarrow u=0 \); \( x=1 \Rightarrow u=1 \). The integral becomes \( \frac{1}{2}\int_0^1 e^u\,du = \frac{1}{2}[e^u]_0^1 = \frac{e-1}{2} \). Choice B is incorrect because the student forgot the \( \frac{1}{2} \) factor when resolving \( du = 2x\,dx \). Choice C is incorrect because the student changed the upper bound to \( u=2 \) instead of \( u=1 \). Choice D is incorrect because the student multiplied by 2 rather than dividing when resolving \( du = 2x\,dx \).
Question 3. What is \( \int \cos(3x) \, dx \)?
Explanation: Choice C is correct. Let \( u = 3x \), so \( du = 3\,dx \), meaning \( dx = \frac{du}{3} \). The integral becomes \( \frac{1}{3}\int \cos u\,du = \frac{\sin u}{3}+C = \frac{\sin(3x)}{3}+C \). Choice A is incorrect because the student integrated as if the argument were simply \( x \), ignoring the chain rule factor of 3. Choice B is incorrect because the student applied the wrong sign, confusing the antiderivative of \( \cos \) with that of \( \sin \), while also ignoring the \( \frac{1}{3} \) factor. Choice D is incorrect because the student multiplied by 3 rather than dividing when accounting for the substitution.
Question 4. A student evaluates \( \int_0^2 x^2 e^{x^3} \, dx \) using the substitution \( u = x^3 \). Which of the following correctly shows the integral after substituting and changing the bounds?
Explanation: Choice A is correct. Let \( u = x^3 \), so \( du = 3x^2\,dx \), meaning \( x^2\,dx = \frac{du}{3} \). Bounds: \( x=0 \Rightarrow u=0 \); \( x=2 \Rightarrow u=8 \). The integral becomes \( \frac{1}{3}\int_0^8 e^u\,du \). Choice B is incorrect because the student correctly adjusted the \( \frac{1}{3} \) factor but failed to change the upper bound, the most common u-substitution error on definite integrals. Choice C is incorrect because the student multiplied by 3 instead of dividing when resolving \( du = 3x^2\,dx \). Choice D is incorrect because the student left \( x^2 \) in the integrand rather than converting it through the substitution, the entire integrand must be expressed in \( u \).
Question 5. Which of the following integrals can both be evaluated using u-substitution with \( u = \ln x \)?
(I) \( \int \dfrac{\ln x}{x} \, dx \) (II) \( \int \dfrac{1}{x \ln x} \, dx \) (III) \( \int x \ln x \, dx \)
Explanation: Choice D is correct. For (I): with \( u = \ln x \), \( du = \frac{dx}{x} \), so the integral becomes \( \int u\,du = \frac{u^2}{2}+C \), substitution works. For (II): the integral becomes \( \int \frac{1}{u}\,du = \ln|u|+C \), substitution also works. For (III): with \( u = \ln x \), the remaining factor \( x\,dx \) is not a clean multiple of \( du = \frac{dx}{x} \); this integral requires integration by parts. Choice A is incorrect because (II) also works with \( u = \ln x \). Choice B is incorrect because (I) also works with \( u = \ln x \). Choice C is incorrect because (III) cannot be evaluated by u-substitution alone; it requires integration by parts.