Respuesta :
Answer:
[tex]I=\dfrac{1}{32}[/tex]
Step-by-step explanation:
Given,
Improper Integral I is given as
[tex]I=\int^{\infty}_{3}\frac{1}{(x+1)^3}dx[/tex]
integration of
[tex]\dfrac{1}{(x+1)^2}[/tex]
now
[tex]\int \dfrac{1}{(x+1)^3}dx = \dfrac{(x+1)^{-3+1}}{-3+1}[/tex]
[tex]\int \dfrac{1}{(x+1)^3}dx =\dfrac{1}{-2(x+1)^2}[/tex]
[tex]I=\left [\dfrac{1}{-2(x+1)^2}\right]^{\infty}_3[/tex]
substituting value
[tex]I=-\dfrac{1}{2}\left [ \dfrac{1}{\infty^2}-\dfrac{1}{4^2}\right ][/tex]
[tex]I=-\dfrac{1}{2}\left [-\dfrac{1}{16}\right ][/tex]
[tex]I=\dfrac{1}{32}[/tex]
so the value of integral converges at [tex]\frac{1}{32}[/tex]
Answer:
Convergent; [tex]\frac{1}{32}[/tex].
Step-by-step explanation:
We have been given an integral as [tex]\int _3^{\infty }\:\:\frac{1}{\left(x+1\right)^3}\:dx[/tex]. We are asked to determine whether our given integral converges or diverges.
Let us integrate our given integral by u substitution as:
[tex]\int _3^{\infty }\:\:\frac{1}{\left(u)^3}\:dx[/tex]
[tex]\int _3^{\infty }\:\:u^{-3}\:dx[/tex]
[tex]\int _3^{\infty }\:\:u^{-3}\:dx=\frac{u^{-3+1}}{-3+1}[/tex]
[tex]\int _3^{\infty }\:\:u^{-3}\:dx=\frac{u^{-2}}{-3+1}[/tex]
[tex]\frac{(x+1)^{-2}}{-2}=-\frac{1}{2(x+1)^2}[/tex]
Now, we will compute the boundaries.
[tex]-\frac{1}{2(\infty+1)^2}=-\frac{1}{\infty ^2}=0[/tex]
[tex]-\frac{1}{2(3+1)^2}=-\frac{1}{2(4^2}=-\frac{1}{2*16}=-\frac{1}{32}[/tex]
Our definite integral would be [tex]0-(-\frac{1}{32})=\frac{1}{32}[/tex]
Therefore, our given integral is convergent and its value is [tex]\frac{1}{32}[/tex].
