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Math Help - Step Functions

  1. #1
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    Step Functions

    Another stupid question!



    Now, I know that we use "h(c-a)(F(c)-F(a)) - h(b-a)((F(b)-F(a))"

    But, from what I can tell, that only applys for when there is only one step function. Is someone able to direct me on how to do this/post up the first couple of lines to help me p[lease? As I said, it's the inclusion of that second step function that has me in a twist
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  2. #2
    MHF Contributor chisigma's Avatar
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    First it is useful remember that \sin (t-\frac{\pi}{2}) = - \cos t... and after that you can compute the integral as sum of two integrals ...

    \int _{\frac{\pi}{2} } ^{\infty} \cos t \cdot e^{-pt}\cdot dt - \int _{\frac{\pi}{4} } ^{\infty} \cos t \cdot e^{-pt}\cdot dt

    Kind regards

    \chi \sigma
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  3. #3
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    Being particularly poor at the multiple integration by parts, I've got no idea where to go.

    I've done it a couple of times, but end up back where I started (that is, with an integral of e^{-pt}cos(t)dt on both sides.

    Im so stuck
    Last edited by exphate; August 29th 2009 at 04:10 PM.
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  4. #4
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    Quote Originally Posted by exphate View Post
    Being particularly poor at the multiple integration by parts, I've got no idea where to go.

    I've done it a couple of times, but end up back where I started (that is, with an integral of e^{-pt}cos(t)dt on both sides.

    Im so stuck
    <br />
\int \cos t\; e^{-pt} \ dt = \frac{e^{-pt}\;(\sin t - p \cos t)}{1+p^2}<br />
    Last edited by mr fantastic; September 18th 2009 at 09:30 AM. Reason: Restored original reply
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