Improper Integrals and Convergence Cheat Sheet

A printable reference covering improper integrals, infinite intervals, vertical asymptotes, convergence tests, and p-integrals for grades 11-12.

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Improper integrals extend definite integration to intervals that are infinite or functions that become unbounded. Students need this cheat sheet to know when an integral must be rewritten as a limit before evaluating it. It also helps separate the idea of finding an antiderivative from deciding whether the integral converges.

Key Facts

An integral over an infinite interval is improper and must be written as $\int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx$ .
An integral with a vertical asymptote at $x = c$ inside $[a,b]$ must be split as $\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$ , and both limits must converge.
If $f(x)$ is unbounded at the right endpoint $b$ , then $\int_a^b f(x)\,dx = \lim_{t \to b^-} \int_a^t f(x)\,dx$ .
If the limit that defines an improper integral exists as a finite number, the integral converges.
If the limit that defines an improper integral is infinite or does not exist, the integral diverges.
The p-integral $\int_1^{\infty} \frac{1}{x^p}\,dx$ converges when $p > 1$ and diverges when $p \le 1$ .
The p-integral $\int_0^1 \frac{1}{x^p}\,dx$ converges when $p < 1$ and diverges when $p \ge 1$ .
For nonnegative functions, if $0 \le f(x) \le g(x)$ and $\int_a^{\infty} g(x)\,dx$ converges, then $\int_a^{\infty} f(x)\,dx$ also converges.

Vocabulary

Improper Integral: An integral that involves an infinite interval of integration or an integrand that becomes unbounded.
Convergence: An improper integral converges when its defining limit exists and equals a finite real number.
Divergence: An improper integral diverges when its defining limit is infinite or fails to exist.
Vertical Asymptote: A vertical line $x = c$ where a function grows without bound or is not defined in a way that affects integration.
p-Integral: An improper integral involving a power function such as $\int_1^{\infty} \frac{1}{x^p}\,dx$ or $\int_0^1 \frac{1}{x^p}\,dx$ .
Comparison Test: A convergence test that compares a nonnegative improper integral to another integral with known convergence behavior.

Common Mistakes to Avoid

Evaluating an improper integral like an ordinary definite integral is wrong because the endpoint or interval must first be replaced with a limit.
Ignoring a discontinuity inside the interval is wrong because an integral such as $\int_0^2 \frac{1}{x-1}\,dx$ must be split at $x = 1$ before testing convergence.
Assuming a positive and negative infinity cancel is wrong because improper integrals require each separate one-sided limit to converge as a finite number.
Using the p-integral rule on the wrong interval is wrong because $\int_1^{\infty} \frac{1}{x^p}\,dx$ and $\int_0^1 \frac{1}{x^p}\,dx$ have opposite convergence conditions.
Forgetting the direction of one-sided limits is wrong because an endpoint singularity at $b$ uses $\lim_{t \to b^-}$ while one at $a$ uses $\lim_{t \to a^+}$ .

Practice Questions

1 Determine whether $\int_1^{\infty} \frac{1}{x^3}\,dx$ converges, and if it does, find its value.
2 Determine whether $\int_0^1 \frac{1}{\sqrt{x}}\,dx$ converges, and if it does, find its value.
3 Rewrite $\int_0^4 \frac{1}{x-2}\,dx$ as limits that correctly test for convergence or divergence.
4 Explain why finding an antiderivative is not enough to prove that an improper integral converges.

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