Given:

Question

asked 2021-08-07

Does the following improper integral converge?

\(\displaystyle{\int_{{{0}}}^{{{3}}}}{\frac{{{1}}}{{{\left({x}-{2}\right)}^{{{3}}}}}}{\left.{d}{x}\right.}\)

\(\displaystyle{\int_{{{0}}}^{{{3}}}}{\frac{{{1}}}{{{\left({x}-{2}\right)}^{{{3}}}}}}{\left.{d}{x}\right.}\)

asked 2021-06-12

Explain why each of the following integrals is improper.

(a) \(\int_6^7 \frac{x}{x-6}dx\)

-Since the integral has an infinite interval of integration, it is a Type 1 improper integral.

-Since the integral has an infinite discontinuity, it is a Type 2 improper integral.

-The integral is a proper integral.

(b)\(\int_0^{\infty} \frac{1}{1+x^3}dx\)

Since the integral has an infinite interval of integration, it is a Type 1 improper integral.

Since the integral has an infinite discontinuity, it is a Type 2 improper integral.

The integral is a proper integral.

(c) \(\int_{-\infty}^{\infty}x^2 e^{-x^2}dx\)

-Since the integral has an infinite interval of integration, it is a Type 1 improper integral.

-Since the integral has an infinite discontinuity, it is a Type 2 improper integral.

-The integral is a proper integral.

d)\(\int_0^{\frac{\pi}{4}} \cot x dx\)

-Since the integral has an infinite interval of integration, it is a Type 1 improper integral.

-Since the integral has an infinite discontinuity, it is a Type 2 improper integral.

-The integral is a proper integral.

(a) \(\int_6^7 \frac{x}{x-6}dx\)

-Since the integral has an infinite interval of integration, it is a Type 1 improper integral.

-Since the integral has an infinite discontinuity, it is a Type 2 improper integral.

-The integral is a proper integral.

(b)\(\int_0^{\infty} \frac{1}{1+x^3}dx\)

Since the integral has an infinite interval of integration, it is a Type 1 improper integral.

Since the integral has an infinite discontinuity, it is a Type 2 improper integral.

The integral is a proper integral.

(c) \(\int_{-\infty}^{\infty}x^2 e^{-x^2}dx\)

-Since the integral has an infinite interval of integration, it is a Type 1 improper integral.

-Since the integral has an infinite discontinuity, it is a Type 2 improper integral.

-The integral is a proper integral.

d)\(\int_0^{\frac{\pi}{4}} \cot x dx\)

-Since the integral has an infinite interval of integration, it is a Type 1 improper integral.

-Since the integral has an infinite discontinuity, it is a Type 2 improper integral.

-The integral is a proper integral.

asked 2021-08-08

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges.

\(\displaystyle{\int_{{{0}}}^{{\infty}}}{\frac{{{e}^{{{\frac{{-{1}}}{{{x}}}}}}}}{{{x}^{{{2}}}}}}{\left.{d}{x}\right.}\)

\(\displaystyle{\int_{{{0}}}^{{\infty}}}{\frac{{{e}^{{{\frac{{-{1}}}{{{x}}}}}}}}{{{x}^{{{2}}}}}}{\left.{d}{x}\right.}\)

asked 2021-09-29

Evaluate the improper integral \(\displaystyle{\int_{{{0}}}^{{\infty}}}{\frac{{{\left.{d}{x}\right.}}}{{{e}^{{{x}}}+{e}^{{-{x}}}}}}\)