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.
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 $a\neq 0$: $$\int e^{ax} \, \mathrm{d}x = \frac{e^{ax}}{a} + c$$
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:
$$\frac{1}{\pi}\left[ -\frac{1}{u} \right]^{u=e^{2 \pi}}_{u=1} = \frac{-1}{\pi e^{2 \pi}} +\frac{1}{\pi}  $$