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Math Help - Antiderivative

  1. #1
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    Antiderivative

    Find f. (x>0) f ''(x) = x^-2
    x > 0
    f(1) = 0
    f(5) = 0

    i am having trouble finding f(x) because the denominator goes to infinity...
    Last edited by rushin25; May 14th 2010 at 12:08 PM.
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  2. #2
    Super Member Random Variable's Avatar
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     f'(x) = \frac{x^{2}}{2}  - 2x + C

     f(x) = \frac{x^{3}}{6} - x^{2} + Cx + D

     f(1) = 0 = \frac{1}{6} -1 + C + D

     f(5) = 0 = \frac{125}{6} - 25 + 5C + D

    now just solve the above two equations simultaneously to find the values of C and D
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  3. #3
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    i m sorry .... i guess i made a mistake while posting the question...it is

    f "(x)= x^-2
    x > 0
    f(1) = 0
    f(5) = 0

    find f(x)
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  4. #4
    Super Member Random Variable's Avatar
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    Quote Originally Posted by rushin25 View Post
    i m sorry .... i guess i made a mistake while posting the question...it is

    f "(x)= x^-2
    x > 0
    f(1) = 0
    f(5) = 0

    find f(x)
     f'(x) = -\frac{1}{x} + C

     f(x) = - \ln x + Cx + D


     f(1) = 0 = -\ln 1 + C + D = C + D

     f(5) = 0 = -\ln 5 + 5C + D


    so  C = \frac{\ln 5}{4}

    and  D = - \frac{\ln 5}{4}
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  5. #5
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    Quote Originally Posted by rushin25 View Post
    i m sorry .... i guess i made a mistake while posting the question...it is

    f "(x)= x^-2
    x > 0
    f(1) = 0
    f(5) = 0

    find f(x)
    It goes to infinity at x= 0 but that has nothing to do with this problem. It says explicitely (in your first post) that you are to find f(x) for x> 0.
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