Exercises: Improper Integrals

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Various exercises relating to improper integrals.

Evaluate the improper integral: $\displaystyle \int _0^\infty e^{5-2x}\ dx = \answer {(e^5)/2}.$

Evaluate the given improper integral: $\displaystyle \int _{-\infty }^\infty \frac {1}{x^2+9}\ dx = \answer {\pi /3}.$

Evaluate the integral: $\int _0^1 \frac {e^{\sqrt {x}}}{\sqrt {x}} \ dx = \answer {2(e-1)}.$ This integral is not improperimproper because of the behavior of the integrand near $x = 0$ .

Evaluate the given improper integral. $\displaystyle \int _{3}^\infty \frac {1}{x^2-4}\ dx = \answer {\frac {\ln 5}{4}}.$

Use the Direct Comparison Test or the Limit Comparison Test to determine whether the integral converges or diverges: $\displaystyle \int _{10}^\infty \frac {3}{\sqrt {3x^2+2x-5}} \ dx.$ Answer: the integral convergesdiverges by directlimit comparison with the function $\displaystyle \frac {1}{x^{\answer {1}}}.$

Use the Direct Comparison Test or the Limit Comparison Test to determine whether the integral converges or diverges: $\displaystyle \int _{2}^\infty \frac {2}{\sqrt {7x^3-x}} \ dx.$ Answer: the integral convergesdiverges by directlimit comparison with the function $\displaystyle \frac {1}{x^{\answer {3/2}}}$ (select the largest exponent for the denominator which makes the statement true).

Use the Direct Comparison Test or the Limit Comparison Test to determine whether the integral converges or diverges: $\displaystyle \int _{1}^\infty e^{-x} \ln x \ dx.$ Answer: the integral convergesdiverges by direct comparison with the function

$e^{-x}$ $x e^{-x}$ $e^{-x}/x$

Use the Direct Comparison Test or the Limit Comparison Test to determine whether the integral converges or diverges: $\displaystyle \int _{1}^\infty e^{-x^2 + 3x + 1} \ dx.$ Answer: the integral convergesdiverges by direct comparison with the function

$e^{-x^2}$ $e^{-x}$ $e^{3x+1}$

Use the Direct Comparison Test or the Limit Comparison Test to determine whether the integral converges or diverges: $\displaystyle \int _{1}^\infty \frac {x}{x^2+\cos x} \ dx.$ Answer: the integral convergesdiverges by directlimit comparison with the function

$1/x$ $x/\cos x$ $1/(x^2+\cos x)$

Use the Direct or Limit Comparison Test to determine whether the integral converges or diverges: $\int _0^{1/e} \frac {(\ln x)^2-1}{x^3 + x^2 + x} dx$ Answer: The integral convergesdiverges by directlimit comparison with the function

$\displaystyle \frac {(\ln x)^2}{x^3}$ $\displaystyle \frac {(\ln x)^2}{x^2}$ $\displaystyle \frac {(\ln x)^2}{x}$ $\displaystyle \frac {1}{x^3}$ $\displaystyle \frac {1}{x^2}$

Use the Direct or Limit Comparison Test to determine whether the integral converges or diverges: $\int _0^{1/e} \frac {(\ln x)^2-1}{x + \sqrt {x} + e^{-1/x}} dx$ Answer: The integral convergesdiverges by direct comparison with the function

$\displaystyle \frac {(\ln x)^2}{x}$ $\displaystyle \frac {(\ln x)^2}{\sqrt {x}}$ $\displaystyle \frac {(\ln x)^2}{e^{-1/x}}$ $\displaystyle \frac {1}{x}$ $\displaystyle \frac {1}{\sqrt {x}}$ $\displaystyle \frac {1}{e^{-1/x}}$

Sample Quiz Questions

Which of the following improper integrals is convergent? Show how you used comparison tests to justify your answer. $\mathrm {I}: \ \int _0^1\frac {\sqrt {{1}+{x^2}{e^{-x}}}}{{(\cos x)}{x^2}}~dx \qquad \mathrm {II}: \ \int _{2}^\infty \frac {{e^{x}}}{{x}{e^{x}}+{x^2}}~dx \qquad \mathrm {III}: \ \int _{2}^\infty \frac {{x^2}}{{x^4}+{1}}~dx$

only $\mathrm {I}$ converges only $\mathrm {II}$ converges only $\mathrm {III}$ converges $\mathrm {I}$ and $\mathrm {II}$ converge $\mathrm {II}$ and $\mathrm {III}$ converge $\mathrm {I}$ and $\mathrm {III}$ converge

Integral $\mathrm {I}$ is divergent by direct comparison to the function $\displaystyle \frac {{1}}{{x^2}}$ . Integral $\mathrm {II}$ is divergent by limit comparison to the function $\displaystyle \frac {{1}}{{x}}$ . Integral $\mathrm {III}$ is convergent by direct comparison to the function $\displaystyle \frac {{1}}{{x^2}}$ .

Which of the following improper integrals is convergent? Show how you used comparison tests to justify your answer. $\mathrm {I}: \ \int _0^1\frac {\sqrt {{e^{2x}}+{x^3}}}{{x}}~dx \qquad \mathrm {II}: \ \int _0^1\frac {{x^2}}{{x^2}{\sqrt {x}}+{x^3}}~dx \qquad \mathrm {III}: \ \int _{2}^\infty \frac {{x^2}{\ln x}}{-{x}+{x^4}}~dx$

Integral $\mathrm {I}$ is divergent by direct comparison to the function $\displaystyle \frac {{1}}{{x}}$ . Integral $\mathrm {II}$ is convergent by direct comparison to the function $\displaystyle \frac {{1}}{{\sqrt {x}}}$ . Integral $\mathrm {III}$ is convergent by limit comparison to the function $\displaystyle \frac {{\ln x}}{{x^2}}$ .

Sample Exam Questions

Only one of the following four improper integrals diverges. Choose that improper integral and justify why it diverges. (You need NOT justify why the other integrals converge.)

$\displaystyle \int ^{\infty }_{2} \frac {\arctan x}{1+x^3} dx$ $\displaystyle \int ^{\infty }_{2} \frac {1}{\sqrt {x^4+x^2}} dx$ $\displaystyle \int ^{\infty }_{2} \frac {1+\sin x}{x^2} dx$ $\displaystyle \int _{2}^{\infty } \frac {1}{\sqrt [3]{x^2-1}} dx$