How can I use integration by parts to find int_{0}^{5}te^{-t}dt?

Jan 25, 2015

You can do this:

Using this approach you should get:
${\int}_{0}^{5} \left(t {e}^{- t}\right) \mathrm{dt} = - {e}^{-} t \cdot t - \int \left(- {e}^{-} t\right) \cdot 1 \mathrm{dt} =$
$= - {e}^{-} t \cdot t - {e}^{-} t =$
$= - {e}^{- t} \left(t + 1\right)$
Now you must use your extrema: $0 \mathmr{and} 5$
You get:
$- {e}^{-} 5 \cdot \left(5 + 1\right) - \left[- {e}^{0} \cdot \left(0 + 1\right)\right] =$
$= 1 - 6 {e}^{-} 5$