Evaluate the Integral: $\int_0^2 \frac{dx}{e^{\pi x}}$ 
Evaluate the definite integral: $$\int_0^2 \frac{\mathrm{d}x}{e^{\pi x}}$$

My attempt:
$u=e^{\pi x}$
$du=\pi e^{\pi x}\ dx$
$\int_0^2 \frac{1}{u}\ dx$ 
So at this point do I divide $\pi e^{\pi x}$ by dx?
Thus, $\frac{du}{\pi\ e^{\pi x}}=dx$
and $\int_0^2 \frac{1}{u}\cdot \frac{du}{\pi\ e^{\pi x}}$
and $\frac{1}{u}\int_0^2\frac{du}{\pi\ e^{\pi x}} $ 
Find the new values for integral? I have no idea how to do this. 
I basically would like to know if my process of solving this problem is correct and how to complete solving the problem.  
 A: You can simply do $$ \int_0^2 \frac{\mathrm{d}x}{e^{\pi x}} = \int_0^2 e^{-\pi x} \, \mathrm{d}x = \left[-\frac{e^{-\pi x}}{\pi}\right]_0^2 = \frac{1}{\pi} - \frac{e^{-2\pi}}{\pi} = \frac{1}{\pi} \left(1 - \frac{1}{e^{2\pi} }\right)$$
This is because we have, for $a\neq 0$: $$\int e^{ax} \, \mathrm{d}x = \frac{e^{ax}}{a} + c$$
A: The steps presented are on the correct path, but are slighlty off. Consider
\begin{align}
I &= \int_{0}^{2} \frac{dx}{e^{a x}} = \int_{0}^{2} e^{-a x} \, dx
\end{align}
let $u = e^{-ax}$ for which $du = -a \, e^{-a x} \, dx$, $x = 0 \to u = 1$, $x = 2 \to u = e^{-2 a}$ and leads to
\begin{align}
I &= \int_{1}^{e^{-2 a}} \frac{du}{(-a)} = - \frac{1}{a} \, \int_{1}^{e^{-2 a}} \, du = - \frac{1}{a} \, \left[ u \right]_{1}^{e^{-2 a}} \\
&= \frac{1 - e^{-2 a}}{a} = \frac{2}{a} \, e^{- a} \, \sinh(a).
\end{align}
When $a = \pi$ the desired integral value is obtained. 
A: note that in $\int_0^2 \frac{1}{u}\cdot \frac{du}{\pi\ e^{\pi x}}$ we can substitute $\pi\ e^{\pi x}$ with $\pi u$. Thus $$\int_0^2 \frac{1}{u}\cdot \frac{du}{\pi\ e^{\pi x}} = \int_0^2 \frac{1}{u}\cdot \frac{du}{\pi\ u} = \int_0^2 \frac{du}{\pi u^2} = \frac{1}{\pi}\left[ -\frac{1}{u} \right]^{u=2}_{u=0}$$.
However you forgot to change the range of the integral with the substitution with $u=e^{\pi x}$ thus this does not work. $x=0 \rightarrow u= 1$ and $x=2 \rightarrow u= e^{2 \pi}$. You have to put these in so we get:
$$\frac{1}{\pi}\left[ -\frac{1}{u} \right]^{u=e^{2 \pi}}_{u=1} = \frac{-1}{\pi e^{2 \pi}} +\frac{1}{\pi}  $$
