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Natali5045456 [20]
3 years ago
6

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Mathematics
2 answers:
maw [93]3 years ago
4 0

Answer:

Convergent; \frac{1}{32}.

Step-by-step explanation:

We have been given an integral as \int _3^{\infty }\:\:\frac{1}{\left(x+1\right)^3}\:dx. We are asked to determine whether our given integral converges or diverges.

Let us integrate our given integral by u substitution as:

\int _3^{\infty }\:\:\frac{1}{\left(u)^3}\:dx

\int _3^{\infty }\:\:u^{-3}\:dx

\int _3^{\infty }\:\:u^{-3}\:dx=\frac{u^{-3+1}}{-3+1}

\int _3^{\infty }\:\:u^{-3}\:dx=\frac{u^{-2}}{-3+1}

\frac{(x+1)^{-2}}{-2}=-\frac{1}{2(x+1)^2}

Now, we will compute the boundaries.

-\frac{1}{2(\infty+1)^2}=-\frac{1}{\infty ^2}=0

-\frac{1}{2(3+1)^2}=-\frac{1}{2(4^2}=-\frac{1}{2*16}=-\frac{1}{32}

Our definite integral would be 0-(-\frac{1}{32})=\frac{1}{32}

Therefore, our given integral is convergent and its value is \frac{1}{32}.

Anika [276]3 years ago
3 0

Answer:

I=\dfrac{1}{32}

Step-by-step explanation:

Given,

Improper Integral I is given as  

I=\int^{\infty}_{3}\frac{1}{(x+1)^3}dx

integration of

\dfrac{1}{(x+1)^2}

now 

\int \dfrac{1}{(x+1)^3}dx = \dfrac{(x+1)^{-3+1}}{-3+1}

\int \dfrac{1}{(x+1)^3}dx =\dfrac{1}{-2(x+1)^2}

I=\left [\dfrac{1}{-2(x+1)^2}\right]^{\infty}_3

substituting value  

I=-\dfrac{1}{2}\left [ \dfrac{1}{\infty^2}-\dfrac{1}{4^2}\right ]

I=-\dfrac{1}{2}\left [-\dfrac{1}{16}\right ]

I=\dfrac{1}{32}

so the value of integral converges at \frac{1}{32}

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