$\displaystyle
\int_0^1 (1-t) e^{-st} dt
$
[tex]
\int_0^1 e^{2t-st} dt
[\math]
Would the be done by parts method? I am stuck please help.
$\displaystyle \int\limits_0^1 {\left( {1 - t} \right){e^{ - st}}dt} = - \frac{1}
{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}$$\displaystyle + \frac{1}{s}\int\limits_0^1 {{e^{ - st}}d\left( {1 - t} \right)} =$
$\displaystyle = \left. { - \frac{{{e^{ - st}}}}{s}\left( {1 - t} \right)} \right|_{t = 0}^{t = 1} - \frac{1}{s}\int\limits_0^1 {{e^{ - st}}dt} = \left. { - \frac{{{e^{ - st}}}}{s}\left( {1 - t} \right)} \right|_{t = 0}^{t = 1} + \frac{1}
{{{s^2}}}\int\limits_0^1 {d\left( {{e^{ - st}}} \right)} =$
$\displaystyle = \left. {\left( { - \frac{{{e^{ - st}}}}{s}\left( {1 - t} \right) + \frac{{{e^{ - st}}}}{{{s^2}}}} \right)} \right|_{t = 0}^{t = 1} = \frac{{{e^{ - s}}}}{{{s^2}}} + \frac{1}{s} - \frac{1}{{{s^2}}} = \frac{1}
{{{s^2}}}\left( {{e^{ - s}} + s - 1} \right).$