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Math Help - integral test.

  1. #1
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    integral test.

    I believe this problem involves the integral test, its very new to me. Please help me with this one.

    integral int 0 infinity

    xe^5x dx

    thankyou for any help.
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  2. #2
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    Quote Originally Posted by rcmango View Post
    I believe this problem involves the integral test, its very new to me. Please help me with this one.

    integral int 0 infinity

    xe^5x dx

    thankyou for any help.
    The integral diverges. Maybe you mean xe^(-5x)
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  3. #3
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    Yes I do mean xe^(-5x) dx sorry about that. where do I start?

    thanks alot.
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  4. #4
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    More generally:

    Let m\ge0,\,n\ge1.

    \int_0^\infty  {x^m e^{ - nx} \,dx}  = \frac{{m!}}<br />
{{n^{m + 1} }}.

    Proof:

    Set u=nx, then the integral becomes

    \frac{1}<br />
{{n^{m + 1} }}\int_0^\infty  {u^m e^{ - u} \,du}  = \frac{{\Gamma (m + 1)}}<br />
{{n^{m + 1} }} = \frac{{m!}}<br />
{{n^{m + 1} }}\quad\blacksquare

    Now plug (m,n)=(1,5) and we're done.
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  5. #5
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    Quote Originally Posted by rcmango View Post
    Yes I do mean xe^(-5x) dx sorry about that. where do I start?

    thanks alot.
    Have you learned integration by parts? Yes - then use that technique. No - then give up until you have.
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  6. #6
    GAMMA Mathematics
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    Quote Originally Posted by mr fantastic View Post
    Have you learned integration by parts? Yes - then use that technique. No - then give up until you have.
    A shortcut to integration by parts is tabular integration.
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  7. #7
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    Quote Originally Posted by colby2152 View Post
    A shortcut to integration by parts is tabular integration.
    I don't really see this as being a shortcut. However I will say that it does organize the process pretty well for a long chain of integration by parts.

    -Dan
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  8. #8
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    \int_{0}^{\infty}xe^{5x}dx

    \lim_{L\rightarrow{\infty}}\int_{0}^{L}xe^{5x}dx

    IBP: \int{u}dv=uv-\int{v}du

    Now, use parts by letting u=x, \;\ dv=e^{5x}dx, \;\ du=dx, \;\ v=\frac{1}{5}e^{5x}

    Putting it all together via parts gives us:

    \lim_{L\rightarrow{\infty}}\left[e^{5L}(\frac{L}{5}-\frac{1}{25})+\frac{1}{25}\right]

    Now, what's the limit as L\rightarrow{\infty}?.


    For what it's worth, here's a graph showing the Riemann sum. As you can see, it's pretty large just from 0 to 10.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  9. #9
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    \int_0^{\infty} xe^{-5x} dx = -\frac{1}{5}xe^{-5x}\bigg|_0^{\infty} + \frac{1}{5} \int_0^{\infty} e^{-5x} dx = - \frac{1}{25} e^{-5x} \bigg|_0^{\infty} = \frac{1}{25}
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