# Question #942b1

May 23, 2015

Mike, all three can be done by substitution. Let $u$ (or $g \left(x\right)$ depending on the notation you've been taught) be the exponent on $e$.

I'll go through the first one, {EDIT: I'll use notation and an approach that may be more familiar to you

${\int}_{0}^{3} {e}^{- 3 x} \mathrm{dx}$

First let's just find the indefinite integral:

$\int {e}^{- 3 x} \mathrm{dx}$

Let $u = - 3 x$,

That makes $\mathrm{du} = - 3 \mathrm{dx}$.

I don't see a $- 3$ in front of the $\mathrm{dx}$ in the integrand, so We'll solve for $\mathrm{dx}$.

$\mathrm{dx} = - \frac{1}{3} \mathrm{du}$, now here is what we have:

$\int {e}^{{\overbrace{\textcolor{red}{- 3 x}}}^{\textcolor{red}{\text{u}}}} {\overbrace{\textcolor{g r e e n}{\mathrm{dx}}}}^{\textcolor{g r e e n}{- \frac{1}{3} \mathrm{du}}} = \int {e}^{\textcolor{red}{u}} \cdot \textcolor{g r e e n}{\left(- \frac{1}{3}\right) \mathrm{du}} = - \frac{1}{3} \int {e}^{u} \mathrm{du} = - \frac{1}{3} {e}^{u} + C$

Now use $u = - 3 x$ to finish:

$\int {e}^{- 3 x} \mathrm{dx} = - \frac{1}{3} {e}^{- 3 x} + C$

For the definite integral, we don't need the $+ C$ (it subtracts out)

${\int}_{0}^{3} {e}^{- 3 x} \mathrm{dx} = - \frac{1}{3} \left({e}^{- 3 x} {|}_{0}^{3}\right)$

$= - \frac{1}{3} \left({e}^{- 3 \left(3\right)} - {e}^{- 3 \left(0\right)}\right)$

$= - \frac{1}{3} \left({e}^{- 9} - {e}^{0}\right)$.

$= - \frac{1}{3} \left({e}^{- 9} - 1\right)$

Now do whatever algebra you like.

I prefer:

$= \frac{1}{3} \left(1 - \frac{1}{e} ^ 9\right)$ or $= \frac{1}{3} - \frac{1}{3 {e}^{9}}$ or $= \frac{{e}^{9} - 1}{3 {e}^{9}}$