AP Calculus BC Practice Quiz: Integral Test for Convergence
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 7 questions to check your progress.
Question 1 of 7
All Questions (7)
A) $\int_{1}^{\infty} \frac{1}{x^2+4} dx$
B) $\int_{1}^{\infty} \frac{x}{x^2+4} dx$
C) $\int_{0}^{1} \frac{x}{x^2+4} dx$
D) $\int_{1}^{\infty} \frac{n}{n^2+4} dn$
Correct Answer: B
To apply the Integral Test to a series $\sum a_n$, we must define a corresponding function $f(x)$ such that $f(n) = a_n$. For the series $\sum_{n=1}^{\infty} \frac{n}{n^2+4}$, the corresponding function is $f(x) = \frac{x}{x^2+4}$. The test then compares the convergence of the series to the convergence of the improper integral $\int_{1}^{\infty} f(x) dx$. Therefore, the correct integral is $\int_{1}^{\infty} \frac{x}{x^2+4} dx$.
A) The series converges to a sum of 12.5.
B) The series converges, but its sum is not necessarily 12.5.
C) The series diverges.
D) The convergence of the series cannot be determined from the given information.
Correct Answer: B
The Integral Test states that if the conditions on $f(x)$ are met, the series $\sum a_n$ and the improper integral $\int_{1}^{\infty} f(x) dx$ either both converge or both diverge. Since the integral converges (to a finite value of 12.5), the series must also converge. However, the value of the integral is not equal to the sum of the series. Therefore, the series converges, but not necessarily to 12.5.
A) $\sum_{n=1}^{\infty} \frac{1}{n^2+1}$
B) $\sum_{n=1}^{\infty} ne^{-n}$
C) $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$
D) $\sum_{n=2}^{\infty} \frac{1}{n \ln(n)}$
Correct Answer: C
The Integral Test requires the corresponding function $f(x)$ to be positive, continuous, and decreasing for $x \ge N$. The series $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ is an alternating series, so its terms are not all positive. Therefore, the condition of positivity is not met, and the Integral Test cannot be applied.
A) The series converges because $\lim_{n \to \infty} \frac{1}{n(\ln n)^3} = 0$.
B) The series converges because the integral $\int_{2}^{\infty} \frac{1}{x(\ln x)^3} dx$ converges.
C) The series diverges because the integral $\int_{2}^{\infty} \frac{1}{x(\ln x)^3} dx$ diverges.
D) The test is inconclusive.
Correct Answer: B
To test the series, we evaluate the improper integral $\int_{2}^{\infty} \frac{1}{x(\ln x)^3} dx$. Let $u = \ln x$, so $du = \frac{1}{x} dx$. The integral becomes $\int_{\ln 2}^{\infty} \frac{1}{u^3} du = \lim_{b \to \infty} [-\frac{1}{2u^2}]_{\ln 2}^{b} = \lim_{b \to \infty} (-\frac{1}{2b^2} + \frac{1}{2(\ln 2)^2}) = 0 + \frac{1}{2(\ln 2)^2}$. Since the integral converges to a finite value, the series also converges by the Integral Test.
A) $f(x)$ is positive and continuous for $x \ge 1$, and it is decreasing for all $x \ge 1$.
B) $f(x)$ is positive and continuous for $x \ge 1$, and it is ultimately decreasing.
C) $f(x)$ is not continuous for all $x \ge 1$.
D) $f(x)$ is not a positive function for all $x \ge 1$.
Correct Answer: B
The function $f(x) = \frac{10x}{x^2+1}$ is clearly positive and continuous for $x \ge 1$. To check if it is decreasing, we find the derivative: $f'(x) = \frac{(x^2+1)(10) - (10x)(2x)}{(x^2+1)^2} = \frac{10x^2+10-20x^2}{(x^2+1)^2} = \frac{10-10x^2}{(x^2+1)^2}$. The derivative $f'(x)$ is negative when $10-10x^2 < 0$, which means $10 < 10x^2$, or $1 < x^2$. For $x \ge 1$, this is true for $x>1$. Since the function is decreasing for all $x>1$, it is 'ultimately decreasing', which is a sufficient condition for the Integral Test to apply.
A) Converges, because $\int_{1}^{\infty} (3x+1)^{-1/2} dx$ converges.
B) Diverges, because $\int_{1}^{\infty} (3x+1)^{-1/2} dx$ diverges.
C) Converges, because it is a p-series with p=1/2.
D) Diverges, because $\lim_{n \to \infty} \frac{1}{\sqrt{3n+1}} \neq 0$.
Correct Answer: B
We evaluate the improper integral $\int_{1}^{\infty} \frac{1}{\sqrt{3x+1}} dx = \int_{1}^{\infty} (3x+1)^{-1/2} dx$. This evaluates to $\lim_{b \to \infty} [\frac{2}{3}(3x+1)^{1/2}]_{1}^{b} = \lim_{b \to \infty} (\frac{2}{3}\sqrt{3b+1} - \frac{2}{3}\sqrt{4})$. As $b \to \infty$, the expression $\frac{2}{3}\sqrt{3b+1}$ approaches infinity. Since the integral diverges, the series also diverges by the Integral Test. Note that the limit of the terms is 0, so option D is incorrect, and it is not a p-series, so option C is incorrect.
A) $\sum_{n=1}^{\infty} \frac{1}{n^{0.99}}$
B) $\sum_{n=1}^{\infty} \frac{1}{n}$
C) $\sum_{n=1}^{\infty} e^{n}$
D) $\sum_{n=1}^{\infty} \frac{1}{n^{1.01}}$
Correct Answer: D
All four options are examples of series where the Integral Test can be applied. The test relates the convergence of the series $\sum \frac{1}{n^p}$ to the convergence of the integral $\int_{1}^{\infty} \frac{1}{x^p} dx$. This integral is known to converge if and only if $p > 1$. A) $p=0.99 \le 1$, so it diverges. B) $p=1 \le 1$, so it diverges (this is the harmonic series). C) The integral $\int_{1}^{\infty} e^x dx$ clearly diverges. D) $p=1.01 > 1$, so the corresponding integral converges, and therefore the series converges.