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Math Help - evaluating infinite integral

  1. #1
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    evaluating infinite integral

    f'(x)=x(1-f(x))

    Evaluate \int_0^{\infty} x(1-f(x))

    f(0)= 10
    Last edited by yoman360; March 27th 2010 at 01:23 PM.
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  2. #2
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    Quote Originally Posted by yoman360 View Post
    f'(x)=x(1-f(x))

    Evaluate \int_0^{\infty} x(1-f(x))
    f(0)=10
    heres my attempt

    \lim_{b \to \infty} \int_0^b x(1-f(x))

    then after that i don't know what to do.
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  3. #3
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    the integral equals f(b)-f(0)=f(b)-10, and in order to have convergence, we require that \lim_{b\to\infty}f(b) exists.
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    MHF Contributor Drexel28's Avatar
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    f'(x)=x(1-f(x))\implies \ln\left|\frac{1}{f(x)-1}\right|=\frac{x^2}{2}+C
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    f'(x)=x(1-f(x))\implies \ln\left|\frac{1}{f(x)-1}\right|=\frac{x^2}{2}+C
    Good job seeing that! But why isn't it -\ln\left|{1-f(x)}\right|, instead of \ln\left|\frac{1}{f(x)-1}\right|?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathemagister View Post
    Good job seeing that! But isn't it \ln\left|\frac{1}{1-f(x)}\right|, not \ln\left|\frac{1}{f(x)-1}\right|
    Isn't |-1|=1
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  7. #7
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    Sorry, that's not what I meant. Please refer to my modified post. Thanks!
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathemagister View Post
    Sorry, that's not what I meant. Please refer to my modified post. Thanks!
    Isn't -\ln|1-f(x)|=-\ln|f(x)-1|=\ln\left(\frac{1}{|f(x)-1|}\right)=\ln\left(\frac{|1|}{|f(x)-1|}\right)=\ln\left|\frac{1}{f(x)-1}\right|?
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  9. #9
    Member mathemagister's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Isn't -\ln|1-f(x)|=-\ln|f(x)-1|=\ln\left(\frac{1}{|f(x)-1|}\right)=\ln\left(\frac{|1|}{|f(x)-1|}\right)=\ln\left|\frac{1}{f(x)-1}\right|?
    Haha! Wow, I'm really not into logarithms today... Thanks for clearing that up. Conventionally, are both forms equally simplified?
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathemagister View Post
    Haha! Wow, I'm really not into logarithms today... Thanks for clearing that up. Conventionally, are both forms equally simplified?
    Not being a conventionalist, I wouldn't know. I sometimes put it in a convoluted form to make the OP work. It's medicine for their soul.
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  11. #11
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    Quote Originally Posted by Drexel28 View Post
    f'(x)=x(1-f(x))\implies \ln\left|\frac{1}{f(x)-1}\right|=\frac{x^2}{2}+C
    I don't understand how you got that
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by yoman360 View Post
    I don't understand how you got that
    Are you acquainted with the idea of separable ODEs?
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  13. #13
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    Quote Originally Posted by Drexel28 View Post
    Are you acquainted with the idea of separable ODEs?
    I learned separable equations. Is that the same as separable ODEs?
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  14. #14
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by yoman360 View Post
    I learned separable equations. Is that the same as separable ODEs?
    Don't ask me.

    f'(x)=x(1-f(x))\implies \frac{f'(x)}{1-f(x)}=x. Integrate both sides with respect to x.
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  15. #15
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    Quote Originally Posted by Drexel28 View Post
    Don't ask me.

    f'(x)=x(1-f(x))\implies \frac{f'(x)}{1-f(x)}=x. Integrate both sides with respect to x.
    I actually see how you got that now

    f'(x)=x(1-f(x))

    dy/dx=x(1-y) where y=f(x)

    \int \frac{1}{1-y} dy = \int x dx

    -ln|1-y| = \frac{1}{2} x^2 + C

    -ln|1-f(x)| = \frac{1}{2} x^2 + C

    then using the initial condition f(0) = 10 I can find C

    -ln|1-f(0)| = \frac{1}{2} (0)^2 + C
    -ln|9|= C

    thanks for the help i think i can finish the problem from here but if i get stuck i will post it.
    Last edited by yoman360; March 27th 2010 at 03:14 PM.
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