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Math Help - Integration by Parts?

  1. #1
    Member diddledabble's Avatar
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    Post Integration by Parts?

    <br />
\int_0^1 (1-t) e^{-st} dt<br /> <br />

    [tex]
    \int_0^1 e^{2t-st} dt
    [\math]

    Would the be done by parts method? I am stuck please help.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by diddledabble View Post
    <br />
\int_0^1 (1-t) e^{-st} dt<br /> <br />
    [tex]
    \int_0^{\infty} e^{(2-s)t} dt
    [\math]
    Would the be done by parts method? I am stuck please help.
    u  = 1-t
    \frac{du}{dt} = -1

    \frac{dv}{dt} = e^{st}
    v = \frac{1}{s}e^{st}

    That -1 should mean only one iteration is needed
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by diddledabble View Post
    <br />
\int_0^1 (1-t) e^{-st} dt

    \int_0^1 e^{2t-st} dt

    Would the be done by parts method? I am stuck please help.
    \int\limits_0^1 {\left( {1 - t} \right){e^{ - st}}dt}  =  - \frac{1}<br />
{s}\int\limits_0^1 {\left( {1 - t} \right)d\left( {{e^{ - st}}} \right)}  = \left. { - \frac{{{e^{ - st}}}}{s}\left( {1 - t} \right)} \right|_{t = 0}^{t = 1} + \frac{1}{s}\int\limits_0^1 {{e^{ - st}}d\left( {1 - t} \right)}  =

    = \left. { - \frac{{{e^{ - st}}}}{s}\left( {1 - t} \right)} \right|_{t = 0}^{t = 1} - \frac{1}{s}\int\limits_0^1 {{e^{ - st}}dt}  = \left. { - \frac{{{e^{ - st}}}}{s}\left( {1 - t} \right)} \right|_{t = 0}^{t = 1} + \frac{1}<br />
{{{s^2}}}\int\limits_0^1 {d\left( {{e^{ - st}}} \right)} =

    = \left. {\left( { - \frac{{{e^{ - st}}}}{s}\left( {1 - t} \right) + \frac{{{e^{ - st}}}}{{{s^2}}}} \right)} \right|_{t = 0}^{t = 1} = \frac{{{e^{ - s}}}}{{{s^2}}} + \frac{1}{s} - \frac{1}{{{s^2}}} = \frac{1}<br />
{{{s^2}}}\left( {{e^{ - s}} + s - 1} \right).
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  4. #4
    Member diddledabble's Avatar
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    I tried typing this one in Latex but it wouldn't come out.

    [tex]
    \int_0^1 e^{(2-s)t} dt
    [\math]
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  5. #5
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    Quote Originally Posted by diddledabble View Post
    I tried typing this one in Latex but it wouldn't come out.

    [tex]
    \int_0^1 e^{(2-s)t} dt
    [\math]
    /math not \math
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  6. #6
    Member diddledabble's Avatar
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    Is this also by parts?

    <br />
\int_0^1 e^{(2-s)t} dt<br />
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  7. #7
    Senior Member DeMath's Avatar
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    Quote Originally Posted by diddledabble View Post
    <br />
\int_0^1 e^{(2-s)t} dt<br />
    It's not by parts

    \int\limits_0^1 {{e^{(2 - s)t}}dt}  = \frac{1}{{2 - s}}\int\limits_0^1 {d\left( {{e^{(2 - s)t}}} \right)}  = \left. {\frac{{{e^{(2 - s)t}}}}{{2 - s}}} \right|_{t = 0}^{t = 1} = \frac{{{e^{2 - s}} - 1}}{{2 - s}}.
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