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Math Help - Integral by part

  1. #1
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    Integral by part

    Integral by part-ask5.jpg
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  2. #2
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    1. Isn't too bad have you done anything for this one?
    Last edited by dwsmith; January 5th 2011 at 01:53 PM. Reason: Changed that to this
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  3. #3
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    2. \displaystyle I_{n+1}=\int_0^1x^{n+1}e^{x-1}dx

    3. \displaystyle I_{3}=\int_0^1x^{3}e^{x-1}dx

    \displaystyle uv-\int vdu
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  4. #4
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    You are correct to say this is an integral by parts.

    First step is to find I_0 and I_1

    Recall that e^{1-x} = \frac{e^1}{e^x} = e\times e^{-x}
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  5. #5
    Super Member Random Variable's Avatar
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     \displaystyle I(n+1) = \int^{1}_{0} x^{n+1} e^{1-x} \ dx = - x^{n+1} e^{1-x} \Big|^{1}_{0} + \ (n+1) \int^{1}_{0} x^{n} e^{1-x} \ dx ( let  u = x^{n+1} and  dv = e^{1-x} \ dx )

     \displaystyle = (-1-0) +  (n+1) I(n) = (n+1) I(n) - 1
    Last edited by Random Variable; January 5th 2011 at 02:34 PM.
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  6. #6
    Super Member Random Variable's Avatar
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     I(0) = e-1

     I(1) = 1(e-1) - 1 = e-2

     I(2) = 2(e-2) - 1 = 2e-5

     I(3) = 3(2e-5) - 1 = 6e - 16
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