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Thread: evaulate the integral...

  1. #1
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    evaulate the integral...

    the integral from 0 to infinity of x(e^x) dx
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by tiga killa View Post
    the integral from 0 to infinity of x(e^x) dx
    It diverges.

    $\displaystyle \int_0^{\infty} xe^x\,dx = xe^x-e^x\big|_0^{\infty} = \infty+1 = \infty$
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    It diverges.

    $\displaystyle \int_0^{\infty} xe^x\,dx = xe^x-e^x\big|_0^{\infty} = \infty+1 = \infty$
    I guess the "correct" question is $\displaystyle \int_0^{\infty} xe^{-x}\,dx $
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by tiga killa View Post
    the integral from 0 to infinity of x(e^{-x}) dx
    Use integration by parts:

    $\displaystyle \int_0^{\infty}xe^{-x}\,dx = -xe^{-x}-e^{-x}\big|_0^{\infty} = \lim_{a\to\infty}-ae^{-a}-e^{-a} + 1$

    $\displaystyle -ae^{-a} = -e^{\ln a}e^{-a} = -e^{\ln(a)-a} = -e^{\ln(a)-\ln(e^a)} = -e^{\ln(\frac{a}{e^a})}$

    $\displaystyle \lim_{a\to\infty}\frac{a}{e^a} = \lim_{a\to\infty}\frac{1}{e^a} = 0$

    So $\displaystyle \lim_{a\to\infty}-e^{\ln(\frac{a}{e^a})} = -e^{\ln(0)} = -e^{-\infty} = 0$
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