# How do you use substitution to integrate e^ (-5x) dx?

May 25, 2015

Here are two solutions:

The "usual way" (as far as my experience), is to turn this into an integral of the form $\int {e}^{u} \mathrm{du}$.

Let $u = - 5 x$, this makes $\mathrm{du} = - 5 \mathrm{dx}$, so the integral becomes:

$\int {e}^{- 5 x} \mathrm{dx} = - \frac{1}{5} \int {e}^{u} \mathrm{du} = - \frac{1}{5} {e}^{u} + C = - \frac{1}{5} {e}^{- 5 x} + C$

If you want to be different, turn it into $\int {u}^{r} \mathrm{du}$

We know that ${e}^{- 5 x} = {\left({e}^{x}\right)}^{-} 5$.

Alternative Method
If we want to make $u = {e}^{x}$ so that $\mathrm{du} = {e}^{x} \mathrm{dx}$, we need to change the exponent on ${\left({e}^{x}\right)}^{-} 5$. We write:

$\int {e}^{- 5 x} \mathrm{dx} = \int {\left({e}^{x}\right)}^{-} 6 {x}^{x} \mathrm{dx}$.

Now with the substitution: $u = {e}^{x}$ so that $\mathrm{du} = {e}^{x} \mathrm{dx}$,

we get

$\int {e}^{- 5 x} \mathrm{dx} = \int {u}^{-} 6 \mathrm{du} = \frac{{u}^{- 5}}{-} 5 + C = - \frac{1}{5} {\left({e}^{x}\right)}^{-} 5 + C$

Which is equal to the answer by the more usual method.