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wornoutwomanC

Answered 2021-08-13
Author has **17581** answers

asked 2021-08-14

Convert the indefinite integral to into a definite integral using the interval [0,1], and solve it.

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

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

asked 2021-08-12

Convert the indefinite integral to into a definite integral using the interval [0,1], and solve it.

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

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

asked 2021-11-01

Convert the indefinite integral into definate integral using the interval [0,1]

\(\displaystyle\int{\left({x}^{{{5}}}+{\frac{{{1}}}{{{5}}}}{x}^{{{5}}}+{\frac{{{1}}}{{{4}}}}{x}^{{{4}}}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle\int{\left({x}^{{{5}}}+{\frac{{{1}}}{{{5}}}}{x}^{{{5}}}+{\frac{{{1}}}{{{4}}}}{x}^{{{4}}}\right)}{\left.{d}{x}\right.}\)

asked 2021-10-30

Convert the indefinite integral into definate integral using the interval [0,1]

\(\displaystyle\int{\left({3}{x}^{{{2}}}-{6}{x}+{5}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle\int{\left({3}{x}^{{{2}}}-{6}{x}+{5}\right)}{\left.{d}{x}\right.}\)

asked 2021-11-18

Evaluate the definite integral.

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

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

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 2021-11-20

Evaluate the definite integral.

\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{5}}}{{{3}{x}+{1}}}}{\left.{d}{x}\right.}\)

\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{5}}}{{{3}{x}+{1}}}}{\left.{d}{x}\right.}\)