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Math Help - definite integral

  1. #1
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    definite integral

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

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

    assign b = infinity?

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

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

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

    from here

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

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

    finally, b = infinity

    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?
    Last edited by icelated; December 6th 2011 at 01:42 PM.
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  2. #2
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    Re: definite integral

    Quote Originally Posted by icelated View Post
    ...

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

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

    2 \int_2^{\infty} { (v^2 - v )}^\frac {-1}{2}\,dv
    It should be

    2 \int_2^{\infty} { (v^2 - v )}^{-1}\,dv
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  3. #3
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    Re: definite integral

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

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

    assign b = infinity?

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

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

    \color{red}2 \int_2^{\infty} { (v^2 - v )}^\frac {-1}{2}\,dv
    The part in red is wrong.

    Here is the idea
    \frac{2}{v^2-v}=\frac{-2}{v}+\frac{2}{v-1}
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  4. #4
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    Re: definite integral

    @plato - excellent!!!
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