Section 7.4 - Improper Integrals
Instructions
- This page includes exercises that you should attempt to solve yourself. You can check your answers and watch the videos explaining how to solve the exercises.
Concepts
- Using a limit to determine if improper integrals diverge or converge (and the value to which convergent integrals converge)
- Recognizing type 2 improper integrals and using limits to evaluate the integrals
- Using the Comparison Theorem to determine if an improper integral converges or diverges
Exercises
Directions: You should try to solve each problem first, and then click "Reveal Answer" to check your answer. You can click "Watch Video" if you need help with a problem.
1. Explain why the definite integral is improper. If the integral converges, evaluate the integral. If it diverges, explain why. \(\displaystyle \int_2^{\infty} \frac{1}{\sqrt{x+7}}\,dx\)
It is an improper integral because the upper limit of integration is \(\infty.\)
The integral diverges to \(\infty.\)
The integral diverges to \(\infty.\)
2. Explain why the definite integral is improper. If the integral converges, evaluate the integral. If it diverges, explain why. \(\displaystyle \int_1^{\infty} \frac{1}{5x-1}\,dx\)
It is an improper integral because the upper limit of integration is \(\infty.\)
The integral diverges to \(\infty.\)
The integral diverges to \(\infty.\)
3. Explain why the definite integral is improper. If the integral converges, evaluate the integral. If it diverges, explain why. \(\displaystyle \int_1^{\infty} \frac{dx}{(4x+1)^2}\)
It is an improper integral because the upper limit of integration is \(\infty.\)
The integral converges to \(\displaystyle \frac{1}{20}.\)
The integral converges to \(\displaystyle \frac{1}{20}.\)
4. Explain why the definite integral is improper. If the integral converges, evaluate the integral. If it diverges, explain why. \(\displaystyle \int_{e^3}^{\infty} \frac{dx}{x \ln x}\)
It is an improper integral because the upper limit of integration is \(\infty.\)
The integral diverges to \(\infty.\)
The integral diverges to \(\infty.\)
5. Explain why the definite integral is improper. If the integral converges, evaluate the integral. If it diverges, explain why. \(\displaystyle \int_{0}^{\infty} x^2e^{-x^3}\,dx\)
It is an improper integral because the upper limit of integration is \(\infty.\)
The integral converges to \(\displaystyle \frac{1}{3}.\)
The integral converges to \(\displaystyle \frac{1}{3}.\)
6. Explain why the definite integral is improper. If the integral converges, evaluate the integral. If it diverges, explain why. \(\displaystyle \int_{6}^{\infty} \frac{1}{x^2-7x+12}\,dx\)
It is an improper integral because the upper limit of integration is \(\infty.\)
The integral converges to \(-\displaystyle \ln \frac{2}{3}.\)
The integral converges to \(-\displaystyle \ln \frac{2}{3}.\)
7. Explain why the definite integral is improper. If the integral converges, evaluate the integral. If it diverges, explain why. \(\displaystyle \int_{-\infty}^{0} x e^{4x}\,dx\)
It is an improper integral because the lower limit of integration is \(-\infty.\)
The integral converges to \(\displaystyle -\frac{1}{16}.\)
The integral converges to \(\displaystyle -\frac{1}{16}.\)
8. Explain why the definite integral is improper. If the integral converges, evaluate the integral. If it diverges, explain why. \(\displaystyle \int_{5}^{\infty} \frac{\ln x}{x^{10}}\,dx\)
It is an improper integral because the upper limit of integration is \(\infty.\)
The integral converges to \( \displaystyle \frac{\ln 5}{9(5^9)}.\)
The integral converges to \( \displaystyle \frac{\ln 5}{9(5^9)}.\)
9. Explain why the definite integral is improper. If the integral converges, evaluate the integral. If it diverges, explain why. \(\displaystyle{\int_9^{36} \displaystyle{{1}\over{\root 3 \of {x-9}}}\,dx}\)
It is an improper integral because \(x=9\) is a vertical asymptote of \(\displaystyle \frac{1}{\sqrt[3]{x-9}}.\)
The integral converges to \(\displaystyle \frac{27}{2.}\)
The integral converges to \(\displaystyle \frac{27}{2.}\)
10. Explain why the definite integral is improper. If the integral converges, evaluate the integral. If it diverges, explain why. \(\displaystyle{\int_{0}^{5/4} \displaystyle{{1}\over{4x-5}}\,dx}\)
It is an improper integral because \(x=\displaystyle \frac{5}{4}\) is a vertical asymptote of \(\displaystyle{{1}\over{4x-5}}.\)
The integral diverges to \(-\infty .\)
The integral diverges to \(-\infty .\)
11. Explain why the definite integral is improper. If the integral converges, evaluate the integral. If it diverges, explain why. \(\displaystyle{\int_{0}^9 \displaystyle{{1}\over{\sqrt[3]{x-1}}}\,dx}\)
It is an improper integral because \(x=1\) is a vertical asymptote of \(\displaystyle\frac{1}{\sqrt[3]{x-1}}.\)
The integral converges to \( \displaystyle \frac{9}{2}.\)
The integral converges to \( \displaystyle \frac{9}{2}.\)
12. Use the Comparison Theorem to determine whether the improper integral converges or diverges. \(\displaystyle{\int_1^{\infty} \displaystyle{{x}\over{x^5+1}}\,dx}\)
The integral converges by comparison to \(\displaystyle \int_1^\infty \frac{dx}{x^4}.\)
13. Use the Comparison Theorem to determine whether the improper integral converges or diverges. \(\displaystyle{\int_2^{\infty} \displaystyle{{1}\over{x+e^{2x}}}\,dx}\)
It converges by comparison to \(\displaystyle \int_2^\infty \frac{1}{e^{2x}}\,dx.\)
14. Use the Comparison Theorem to determine whether the improper integral converges or diverges.\(\displaystyle{\int_1^{\infty} \displaystyle{{\sin^2 x}\over{x^4}}\,dx}\)
The integral converges by comparison to \(\displaystyle \int_1^\infty \frac{1}{x^4}\, dx.\)