# Improper Integral Practice Problems with Answers

> 12 improper integral practice problems with answers: infinite limits, vertical asymptotes, convergence or divergence and the comparison test.

Source: https://practicecalculus.com/practice-problems/improper-integrals
Updated: 2026-11-06

$$ \int_a^{\infty} f(x)\,dx = \lim_{b\to\infty}\int_a^{b} f(x)\,dx $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 2 and AP Calculus BC, Unit 6
- **You need:** [Definite integrals](/practice-problems/definite-integrals), limits at infinity and [integration by parts](/practice-problems/integration-by-parts)
- **Main idea:** Replace the bad endpoint with a variable, integrate, then take a limit
- **Practice with new numbers:** [Infinite limits of integration lesson](/learn/calculus-2/improper-integrals-and-applications/infinite-limits-of-integration)

## Key takeaways

- An integral is improper if a limit is infinite or the integrand blows up somewhere in $[a, b]$, including inside the interval.
- $\int_1^{\infty}\frac{1}{x^p}\,dx$ converges when $p > 1$; $\int_0^{1}\frac{1}{x^p}\,dx$ converges when $p < 1$.
- Use comparison when you cannot integrate: smaller than a convergent integral converges, larger than a divergent one diverges.

These **improper integral practice problems** cover infinite limits, vertical asymptotes at an endpoint and inside the interval, and the comparison test. For each one, decide whether it converges, and if it does, find its value. The answers come first, then full solutions.

The definition for an infinite limit:

$$\int_a^{\infty} f(x)\,dx = \lim_{b\to\infty}\int_a^{b} f(x)\,dx$$

For an asymptote at $x = a$, use $\lim_{t\to a^+}\int_t^{b} f(x)\,dx$ instead. For more examples, read the [improper integrals guide](/blog/improper-integrals).

## Problems

**Infinite limits**

1. $\displaystyle\int_1^{\infty}\frac{1}{x^2}\,dx$
2. $\displaystyle\int_1^{\infty}\frac{1}{x}\,dx$
3. $\displaystyle\int_1^{\infty}x^{-3/2}\,dx$
4. $\displaystyle\int_0^{\infty}e^{-2x}\,dx$
5. $\displaystyle\int_0^{\infty}x\,e^{-x}\,dx$
6. $\displaystyle\int_{-\infty}^{\infty}\frac{1}{1 + x^2}\,dx$

**Vertical asymptotes**

7. $\displaystyle\int_0^{1}\frac{1}{\sqrt{x}}\,dx$
8. $\displaystyle\int_0^{1}\frac{1}{x}\,dx$
9. $\displaystyle\int_0^{1}\ln x\,dx$
10. $\displaystyle\int_{-1}^{1}\frac{1}{x^2}\,dx$
11. $\displaystyle\int_0^{3}\frac{1}{(x - 1)^{2/3}}\,dx$

**Comparison**

12. Does $\displaystyle\int_1^{\infty}\frac{2 + \cos x}{x^2}\,dx$ converge?

## Answers

| Problem | Answer |
| --- | --- |
| 1 | Converges to $1$ |
| 2 | Diverges |
| 3 | Converges to $2$ |
| 4 | Converges to $\frac{1}{2}$ |
| 5 | Converges to $1$ |
| 6 | Converges to $\pi$ |
| 7 | Converges to $2$ |
| 8 | Diverges |
| 9 | Converges to $-1$ |
| 10 | Diverges |
| 11 | Converges to $3 + 3\sqrt[3]{2} \approx 6.78$ |
| 12 | Converges, by comparison with $\frac{3}{x^2}$ |

## Step-by-step solutions

**1.** $\displaystyle\lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^{b} = \lim_{b\to\infty}\left(1 - \frac{1}{b}\right) = 1$

**2.** $\displaystyle\lim_{b\to\infty}\big[\ln x\big]_1^{b} = \lim_{b\to\infty}\ln b = \infty$, so it diverges.

**3.** An antiderivative of $x^{-3/2}$ is $-2x^{-1/2}$:

$$\lim_{b\to\infty}\left(2 - \frac{2}{\sqrt{b}}\right) = 2$$

This is a $p$-integral with $p = \frac{3}{2} > 1$, so convergence is expected.

**4.** $\displaystyle\lim_{b\to\infty}\left[-\frac{1}{2}e^{-2x}\right]_0^{b} = \lim_{b\to\infty}\left(\frac{1}{2} - \frac{1}{2}e^{-2b}\right) = \frac{1}{2}$

**5.** By parts, $\int x\,e^{-x}\,dx = -(x + 1)e^{-x}$. Then:

$$\lim_{b\to\infty}\left(1 - (b + 1)e^{-b}\right) = 1$$

The term $(b + 1)e^{-b} \to 0$ because the exponential grows much faster than $b + 1$ (use L'Hôpital's rule if you want to check).

**6.** Split at $0$. Each half is $\lim\arctan b = \frac{\pi}{2}$, so the total is $\pi$. Both halves must converge for the whole integral to converge.

**7.** The integrand blows up at $x = 0$:

$$\lim_{t\to 0^+}\big[2\sqrt{x}\big]_t^{1} = 2 - 0 = 2$$

**8.** $\displaystyle\lim_{t\to 0^+}\big(\ln 1 - \ln t\big) = \lim_{t\to 0^+}(-\ln t) = \infty$, so it diverges.

**9.** An antiderivative is $x\ln x - x$. At $x = 1$ it is $-1$. As $t \to 0^+$, $t\ln t \to 0$, so:

$$-1 - \lim_{t\to 0^+}(t\ln t - t) = -1 - 0 = -1$$

**10.** The asymptote at $x = 0$ is inside the interval, so split there. The right half is $\lim_{t\to 0^+}\left(\frac{1}{t} - 1\right) = \infty$, so the whole integral diverges. The tempting answer $\left[-\frac{1}{x}\right]_{-1}^{1} = -2$ is impossible, since the integrand is positive.

**11.** The asymptote is at $x = 1$, inside $[0, 3]$. An antiderivative is $3(x - 1)^{1/3}$, which is continuous at $x = 1$:

$$\int_0^1 = 3(0) - 3(-1) = 3, \qquad \int_1^3 = 3\sqrt[3]{2} - 0 = 3\sqrt[3]{2}$$

Both halves converge, so the total is $3 + 3\sqrt[3]{2} \approx 6.78$.

**12.** Since $-1 \le \cos x \le 1$, the integrand is between $\frac{1}{x^2}$ and $\frac{3}{x^2}$. And $\int_1^{\infty}\frac{3}{x^2}\,dx = 3$ converges. A positive function below a convergent one also converges.

## Common mistakes on these problems

- **Missing an asymptote inside the interval.** Problems 10 and 11 look like ordinary definite integrals. Always check where the integrand is undefined.
- **Writing $\infty$ as a number.** Write the limit. Expressions like $\frac{1}{\infty}$ hide mistakes.
- **Comparing in the wrong direction.** Being smaller than a divergent integral proves nothing. Being larger than a convergent one proves nothing.
- **Splitting only one side.** In problem 6, both halves must converge separately.

## Keep practicing

For unlimited problems with instant feedback, try the [infinite limits lesson](/learn/calculus-2/improper-integrals-and-applications/infinite-limits-of-integration), the [infinite discontinuities lesson](/learn/calculus-2/improper-integrals-and-applications/infinite-discontinuities) and the [comparison test lesson](/learn/calculus-2/improper-integrals-and-applications/comparison-test). The same $p$ rule comes back for series in the [convergence test problems](/practice-problems/convergence-tests).

## Frequently asked questions

### How do you know if an improper integral converges?

Rewrite it as a limit and evaluate. If the limit is a finite number, it converges to that number. If the limit is infinite or does not exist, it diverges. When you cannot integrate directly, compare with a $p$-integral.

### Why does the integral of 1/x from 1 to infinity diverge?

Its antiderivative is $\ln x$, and $\ln b \to \infty$ as $b \to \infty$. The area keeps growing without bound, even though the curve gets close to the $x$-axis. Problem 2 shows this.

### Why is the integral of 1/x² from -1 to 1 not -2?

Because $\frac{1}{x^2}$ blows up at $x = 0$, inside the interval. You must split at $0$, and each half diverges. Plugging in the endpoints directly ignores the asymptote. Problem 10 shows this.