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Math Help - divergent/convergent integral,simple Q icant solve

  1. #1
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    divergent/convergent integral,simple Q icant solve

    Let f(x) be a continuous function defined on the interval [2, infinity) such that





    determine the value of




    it seems like an easy Q but I can't find the trick to it.
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  2. #2
    Super Member wingless's Avatar
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    Quote Originally Posted by monokill View Post
    Let f(x) be a continuous function defined on the interval [2, infinity) such that





    determine the value of




    it seems like an easy Q but I can't find the trick to it.


    Here's the trick:

    \int f(x)e^{-x/2}~dx = -f(x)e^{-x/2}+2\int f'(x)e^{-x/2}~dx
    (Integration by parts )
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  3. #3
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    hmm i still dont haf the right answer.

    from what u suggested:


    <br />
\int f(x)e^{-x/2}~dx = -f(x)e^{-x/2}+2\int f'(x)e^{-x/2}~dx<br />
= -4


    so

    { -f(x)e^(-x/2) } evaluated from 3 to infinity + 2x = -4

    then i got

    -infinity e^(-infinity/2) + 10e^(-3/2) = -4

    but infinity x 0 = ??
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  4. #4
    Super Member wingless's Avatar
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    \lim_{x\to \infty} f(x)e^{-x/2}= 0 because it says \int_3^{\infty} f(x)e^{-x/2}~dx = -4. The function can approach nowhere but 0.

    So,

    -f(x)e^{-x/2} |^{\infty}_{3} + 2\int^{\infty}{3}f'(x)e^{-x/2}~dx = -4

    Let's call \int^{\infty}{3}f'(x)e^{-x/2}~dx = t

    -f(x)e^{-x/2} |^{\infty}_{3} + 2t = -4


    (\lim_{x\to \infty} -f(x)e^{-x/2})-(-f(3)e^{-3/2}) + 2t = -4

    f(3)e^{-3/2} + 2t = -4...
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  5. #5
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    That's what I did. So the answer should be



    but its incorrect
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  6. #6
    Super Member wingless's Avatar
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    Sorry, I mistyped the integral and went wrong..

    \int f(x)e^{-x/2}~dx = -2f(x)e^{-x/2}+2\int f'(x)e^{-x/2}~dx

    -2f(x)e^{-x/2}+2\int f'(x)e^{-x/2}~dx~|^{\infty}_{3} = -4

    t = \int^{\infty}_{3}f'(x)e^{-x/2}~dx

    -2f(x)e^{-x/2}+2t~|^{\infty}_{3} = -4

    \lim_{x\to \infty} f(x)e^{-x/2}= 0

    (0) - (-2f(3)e^{-3/2}) + 2t = -4

    20e^{-3/2} + 2t = -4

    t = -2 - 10e^{-3/2}

    I hope this one is true..
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  7. #7
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    it is! thanks so much
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