# definite integral

• Dec 5th 2011, 01:15 PM
icelated
definite integral
I would like to think i have a simple problem. Can you see if i am doing it right?

$\displaystyle \int_2^{\infty}\frac {2} { v^2 - v }\,dv$

assign b = infinity?

Then, bring the 2 to the outside and bring the denominator up!

$\displaystyle 2 \int_2^{\infty}\frac {1} { v^2 - v }\,dv$

$\displaystyle 2 \int_2^{\infty} { (v^2 - v )}^{-1}\,dv$

from here

$\displaystyle 2 \frac{ (v^2 - v )^{1/2}} {1/2} |_2^{\infty}$

$\displaystyle 4 (v^2 - v )^{1/2} |_2^{\infty}$

finally, b = infinity

$\displaystyle 4 ({\infty}^2 - {\infty} )^{1/2} - (2^2 - 2 )^{1/2}$

im not sure how to handle the 1/2

is this somewhat close? and how would i finish it?
• Dec 5th 2011, 01:24 PM
wnvl
Re: definite integral
Quote:

Originally Posted by icelated
...

Then, bring the 2 to the outside and bring the denominator up!

$\displaystyle 2 \int_2^{\infty}\frac {1} { v^2 - v }\,dv$

$\displaystyle 2 \int_2^{\infty} { (v^2 - v )}^\frac {-1}{2}\,dv$

It should be

$\displaystyle 2 \int_2^{\infty} { (v^2 - v )}^{-1}\,dv$
• Dec 5th 2011, 01:24 PM
Plato
Re: definite integral
Quote:

Originally Posted by icelated
I would like to think i have a simple problem. Can you see if i am doing it right?

$\displaystyle \int_2^{\infty}\frac {2} { v^2 - v }\,dv$

assign b = infinity?

Then, bring the 2 to the outside and bring the denominator up!

$\displaystyle 2 \int_2^{\infty}\frac {1} { v^2 - v }\,dv$

$\displaystyle \color{red}2 \int_2^{\infty} { (v^2 - v )}^\frac {-1}{2}\,dv$

The part in red is wrong.

Here is the idea
$\displaystyle \frac{2}{v^2-v}=\frac{-2}{v}+\frac{2}{v-1}$
• Dec 5th 2011, 04:48 PM
icelated
Re: definite integral
@plato - excellent!!!