# How do you integrate int e^(3-x)dx from [3,4]?

May 12, 2017

${\int}_{3}^{4} \left({e}^{3 - x}\right) \mathrm{dx} = 1 - \frac{1}{e}$

#### Explanation:

${\int}_{3}^{4} \left({e}^{3 - x}\right) \mathrm{dx}$

Do a $u$-substitution:
$u = 3 - x$
$\frac{\mathrm{du}}{\mathrm{dx}} = - 1$
$\mathrm{dx} = - \mathrm{du}$

Remember to change the bounds of the integral according to the $u$ values
${\int}_{0}^{- 1} \left({e}^{u}\right) \left(- 1\right) \mathrm{du}$

${\int}_{0}^{- 1} \left(- {e}^{u}\right) \mathrm{du}$

$= {\left[- {e}^{u}\right]}_{0}^{- 1}$

$= - {e}^{- 1} + 1$

$= 1 - \frac{1}{e}$