Answer
$\frac{1}{\pi}(1-e^{-2\pi})$
Work Step by Step
$\int_0^2\frac{dx}{e^{\pi x}}$
$=\int_0^2 e^{-\pi x}dx$
Let $u=-\pi x$. Then $du=-\pi dx$, and $\frac{1}{-\pi}du=dx$.
$=\int_{-\pi*0}^{-\pi*2}e^u* \frac{1}{-\pi} du$
$=\frac{1}{-\pi}\int_0^{-2\pi} e^u du$
$=-\frac{1}{\pi}(e^{-2\pi}-e^0)$
$=-\frac{1}{\pi}(e^{-2\pi}-1)$
$=\frac{1}{\pi}(1-e^{-2\pi})$