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Math Help - [SOLVED] Integrate by parts?

  1. #1
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    [SOLVED] Integrate by parts?

     \int xe^{-x} ~dx

    I came up with -xe^{-x} - e^{-x} for the answer.

    To get there I took  -xe^{-x} - \int -e^{-x} ~dx after evaluating that integral I came up with my answer.

    I don't know if its right.
    Last edited by redman223; July 2nd 2008 at 06:07 PM.
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  2. #2
    Newbie Msquared's Avatar
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    Quote Originally Posted by redman223 View Post
    Integrate xe^-x dx

    I came up with (-xe^-x) - (e^-x) for the answer.

    To get there I took -xe^-x - (integral sign) -e^-x dx after evaluating that integral I came up with my answer.

    Am I doing this right? Also, how can I input signs in my posts? Like the integral sign and what about if I was putting limits on there as well?

    Yes thats right.

    As for putting in stuff like integral signs, limits, etc.. you use or you can just highlight what you want "wrapped" and press the sigma button when done typing it all up.

    For this form: \int xe^{-x}dx

    You just type \int for the integral and for limits \int_{first limit}^{second limit} and then insert the rest of the integral. Theres also a "latex help" forum here with a thread concerning a tutorial that may help you out which is here: http://www.mathhelpforum.com/math-help/latex-help/
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  3. #3
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    Thanks for the help. I have another question though.

    What do I do when I have to integrate a product?

    For example:

    \int t^3e^t ~dt

    \int t^3e^t ~dt = t^3e^t - \int 3t^2e^t ~dt

    What do I do next? I don't know how to integrate that last part.
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  4. #4
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    You're going to have to apply integration by parts to that one as well (and possibly again )
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  5. #5
    Newbie Msquared's Avatar
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    Quote Originally Posted by redman223 View Post
    Thanks for the help. I have another question though.

    What do I do when I have to integrate a product?

    For example:

    \int t^3e^t ~dt

    \int t^3e^t ~dt = t^3e^t - \int 3t^2e^t ~dt

    What do I do next? I don't know how to integrate that last part.
    Apply integration by parts again to that last part.

    Depending on your values for u and dv for that integral, you should end up with:

     <br />
x^{3}e^{x} - (e^{x}3x^{2} - \int 6xe^{x}~dx)<br />

    Unfortunately, for that last integral, you have to apply another integration by parts and you should end up with.

     <br />
x^{3}e^{x} - e^{x}3x^{2} - (6xe^{x} - \int 6e^{x}~dx)<br />

    Hope that helps.
    Last edited by Msquared; July 2nd 2008 at 06:24 PM. Reason: Fixed a typo
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  6. #6
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    Ok, now I see. I thought I did something wrong with I integrated it by parts and came up with another integral that had to be integrated by parts.

    Thanks for the help.
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  7. #7
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    Another question.

    How do I use integration by parts to solve this? The question asks me to use integration by parts to prove the reduction formula.

    \int x^{n}e^{x} ~dx = x^{n}e^{x} - n\int x^{n-1}e^{x}~dx

    It looks like it is already done? Am I missing something?

    The formula I was taught was \int f(x)g'(x) = f(x)g(x) - \int f'(x)g(x)
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  8. #8
    Newbie Msquared's Avatar
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    Quote Originally Posted by redman223 View Post
    Another question.

    How do I use integration by parts to solve this? The question asks me to use integration by parts to prove the reduction formula.

    \int x^{n}e^{x} ~dx = x^{n}e^{x} - n\int x^{n-1}e^{x}~dx

    It looks like it is already done? Am I missing something?

    The formula I was taught was \int f(x)g'(x) = f(x)g(x) - \int f'(x)g(x)

    From the looks of it.. yeah. Just pretty much use integration by parts(just to show work involved) with u and dv and I'm pretty sure thats all you need to do.
    Last edited by Msquared; July 2nd 2008 at 07:26 PM. Reason: typo
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by redman223 View Post
    Another question.

    How do I use integration by parts to solve this? The question asks me to use integration by parts to prove the reduction formula.

    \int x^{n}e^{x} ~dx = x^{n}e^{x} - n\int x^{n-1}e^{x}~dx

    It looks like it is already done? Am I missing something?

    The formula I was taught was \int f(x)g'(x) = f(x)g(x) - \int f'(x)g(x)
    This is called a reduction formula. The premise is that for Engineers or others that don't wish to repeatedly calculate integrals they can just apply formulas like the one above.
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