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Math Help - liate rule i think

  1. #1
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    liate rule i think

    integral from 0 to 8 of (5xe^-x)(dx)

    so far i think

    5(e^-x)

    so then do i plug in 8
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  2. #2
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    Just in case a picture helps...

    Trial and error will quickly establish (and liate confirm) that you need the legs-crossed version of...



    ... the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x.

    The general drift is...



    In this case...



    And the rest...
    Spoiler:

    _________________________________________
    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; May 6th 2010 at 05:34 AM. Reason: I meant crossed not un-crossed
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  3. #3
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    so is it

    (5x)(e^-x) + (5) (e^-x) + C
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  4. #4
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    Quote Originally Posted by cummings15 View Post
    so is it

    (5x)(e^-x) + (5) (e^-x) + C
    The negative of that, as you can maybe see above now that I've fixed the second pic (in the spoiler).

    I think you missed that we need the chain rule for e^(-x). I didn't bother zooming in on that, but (gimme a minute)...



    ... where



    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    I uncrossed the legs for clarity - i.e. switched the 5x and e^-x.

    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; May 6th 2010 at 05:16 AM.
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  5. #5
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    Hello, cummings15!

    5\int^8_0 xe^{-x}\,dx
    Integrate by parts . . .

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


    We have: . 5\bigg[-xe^{-x} + \int e^{-x}\,dx\bigg] \;=\;5\bigg[-xe^{-x} - e^{-x}\bigg] + C \;=\;-5e^{-x}(x+1) + C


    Evaluate: . -\frac{5}{e^x}(x+1)\,\bigg]^8_0 \;=\;\bigg[-\frac{5}{e^8}(8+1) \bigg] - \bigg[-\frac{5}{e^0}(0+1)\bigg] \;=\;5 - \frac{45}{e^8}

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  6. #6
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    Quote Originally Posted by cummings15 View Post
    integral from 0 to 8 of (5xe^-x)(dx)

    so far i think

    5(e^-x)

    so then do i plug in 8
    tabular integration ...

    sign ........... u ........... dv

    +............... 5x .......... e^{-x}

    - ................ 5 .......... -e^{-x}

    ................... 0 .......... e^{-x}

     <br />
\left[-5xe^{-x} - 5e^{-x}\right]_0^8<br />

     <br />
\left[-5e^{-x}(x+1) \right]_0^8<br />

     <br />
5 -\frac{45}{e^8}<br />
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