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doplovif

Answered 2021-06-06
Author has **28246** answers

asked 2021-09-07

Evaluate the integral by reversing the order of integration

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

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

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 2021-05-17

The integral represents the volume of a solid. Describe the solid.

\(\pi\int_{0}^{1}(y^{4}-y^{8})dy\)

a) The integral describes the volume of the solid obtained by rotating the region \(R=\{\{x,\ y\}|0\leq y\leq1,\ y^{4}\leq x\leq y^{2}\}\) of the xy-plane about the x-axis.

b) The integral describes the volume of the solid obtained by rotating the region \(R=\{\{x,\ y\}|0\leq y\leq1,\ y^{2}\leq x\leq y^{4}\}\) of the xy-plane about the x-axis.

c) The integral describes the volume of the solid obtained by rotating the region \(R=\{\{x,\ y\}|0\leq y\leq1,\ y^{4}\leq x\leq y^{2}\}\) of the xy-plane about the y-axis.

d) The integral describes the volume of the solid obtained by rotating the region \(R=\{\{x,\ y\}|0\leq y\leq1,\ y^{2}\leq x\leq y^{4}\}\) of the xy-plane about the y-axis.

e) The integral describes the volume of the solid obtained by rotating the region \(R=\{\{x,\ y\}|0\leq y\leq1,\ y^{4}\leq x\leq y^{8}\}\) of the xy-plane about the y-axis.

\(\pi\int_{0}^{1}(y^{4}-y^{8})dy\)

a) The integral describes the volume of the solid obtained by rotating the region \(R=\{\{x,\ y\}|0\leq y\leq1,\ y^{4}\leq x\leq y^{2}\}\) of the xy-plane about the x-axis.

b) The integral describes the volume of the solid obtained by rotating the region \(R=\{\{x,\ y\}|0\leq y\leq1,\ y^{2}\leq x\leq y^{4}\}\) of the xy-plane about the x-axis.

c) The integral describes the volume of the solid obtained by rotating the region \(R=\{\{x,\ y\}|0\leq y\leq1,\ y^{4}\leq x\leq y^{2}\}\) of the xy-plane about the y-axis.

d) The integral describes the volume of the solid obtained by rotating the region \(R=\{\{x,\ y\}|0\leq y\leq1,\ y^{2}\leq x\leq y^{4}\}\) of the xy-plane about the y-axis.

e) The integral describes the volume of the solid obtained by rotating the region \(R=\{\{x,\ y\}|0\leq y\leq1,\ y^{4}\leq x\leq y^{8}\}\) of the xy-plane about the y-axis.

asked 2022-01-03

How can I evaluate

\(\displaystyle{\int_{{-\infty}}^{\infty}}{\frac{{{\cos{{x}}}}}{{{\text{cosh}{{x}}}}}}{\left.{d}{x}\right.}\) and \(\displaystyle{\int_{{0}}^{\infty}}{\frac{{{\sin{{x}}}}}{{{e}^{{x}}-{1}}}}{\left.{d}{x}\right.}\)

\(\displaystyle{\int_{{-\infty}}^{\infty}}{\frac{{{\cos{{x}}}}}{{{\text{cosh}{{x}}}}}}{\left.{d}{x}\right.}\) and \(\displaystyle{\int_{{0}}^{\infty}}{\frac{{{\sin{{x}}}}}{{{e}^{{x}}-{1}}}}{\left.{d}{x}\right.}\)

asked 2021-08-17

Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.)

\(\displaystyle\int{37}{e}^{{{74}{x}}}{\arctan{{\left({e}^{{{37}{x}}}\right)}}}{\left.{d}{x}\right.}\)

Inverse Trigonometric Forms (92): \(\displaystyle\int{u}{{\tan}^{{-{1}}}{u}}\ {d}{u}={\frac{{{u}^{{{2}}}+{1}}}{{{2}}}}{{\tan}^{{-{1}}}{u}}-{\frac{{{u}}}{{{2}}}}+{C}\)

\(\displaystyle\int{37}{e}^{{{74}{x}}}{\arctan{{\left({e}^{{{37}{x}}}\right)}}}{\left.{d}{x}\right.}\)

Inverse Trigonometric Forms (92): \(\displaystyle\int{u}{{\tan}^{{-{1}}}{u}}\ {d}{u}={\frac{{{u}^{{{2}}}+{1}}}{{{2}}}}{{\tan}^{{-{1}}}{u}}-{\frac{{{u}}}{{{2}}}}+{C}\)

asked 2021-05-14

Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.)

\(\int37e^{74x}\arctan(e^{37x})dx\)

Inverse Trigonometric Forms (92): \(\int u\tan^{-1}u\ du=\frac{u^{2}+1}{2}\tan^{-1}u-\frac{u}{2}+C\)

\(\int37e^{74x}\arctan(e^{37x})dx\)

Inverse Trigonometric Forms (92): \(\int u\tan^{-1}u\ du=\frac{u^{2}+1}{2}\tan^{-1}u-\frac{u}{2}+C\)

asked 2021-06-01

The graph of f is shown. Evaluate each integral by interpreting it in terms of areas. integral.\(\int_0^2 f(x)dx\)