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Math Help - integration of parts

  1. #1
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    integration of parts

    Hi,

    Can someone please tell me if this is correct? I'm trying to solve this integral with the integration of parts method.

    The integral xe^(-x)dx

    udv=uv-the integral vdu

    u=x
    dv=e^-xdx

    du=1dx (derivative of u)
    v=-e^-x (antiderivative of du)

    xe^-x-the integral -e^-x=
    e^-x+e^-x=
    e^-x(x+1)+c

    Thank you very much
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by chocolatelover View Post
    Hi,

    Can someone please tell me if this is correct? I'm trying to solve this integral with the integration of parts method.

    The integral xe^(-x)dx

    udv=uv-the integral vdu

    u=x
    dv=e^-xdx

    du=1dx (derivative of u)
    v=-e^-x (antiderivative of du)

    xe^-x-the integral -e^-x=
    e^-x+e^-x=
    e^-x(x+1)+c

    Thank you very much
    You need to sort out how you are expressing things in ASCII.


    You should have:

    int xe^(-x)dx = (-x e^{-x}) + int e^{-x} dx = -x e^{-x} - e^{-x} + C

    which you can now simplify, and check by differentiating

    RonL
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  3. #3
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    Hello, chocolatelover!

    You dropped a minus-sign . . .


    Is this correct?

    Integrate: . \int xe^{\text{-}x}dx

    By parts: . \int u\,dv\;=\;uv - \int v\,du

    . \begin{array}{ccccccc}u & = & x & \quad & dv & = & e^{\text{-}x}dx \\<br /> <br />
du & = & dx & \quad & v & = & \text{-}e^{\text{-}x}\end{array}

    Then we have: . -xe^{\text{-}x} + \int e^{\text{-}x}dx \;=\;-xe^{\text{-}x} - e^{\text{-}x} + C \;=\;-e^{\text{-}x}(x +1) + C

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  4. #4
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    Hi,

    I looked at it again. What would the antiderivative of -e^-x be? Is that just "-e^-x"?

    Thank you
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Hi,

    I looked at it again. What would the antiderivative of -e^-x be? Is that just "-e^-x"?

    Thank you
    Technically it's \frac{-e^{-x}}{-1} = e^{-x}. So not quite.

    In general \frac{d}{dx}e^{kx} = ke^x. So \int e^{kx}~dx = \frac{e^{kx}}{k} + C.

    -Dan
    Last edited by topsquark; August 31st 2007 at 03:35 PM.
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