Integrate $e^{-ax}$ and $xe^{-ax}$?

We assume $a>0$:

$$\int_0^\infty e^{-ax}dx=-\frac{1}{a}e^{-ax}\bigg|_0^\infty=\frac{1}{a}$$

$$\int_0^\infty xe^{-ax}dx=\int_0^\infty\frac{t}{a}e^{-t}\frac{dt}{a}=\frac{1}{a^2}\Gamma(2)=\frac{1}{a^2}$$

Hints For the first one, use a substitution.

For the second one, integrate by parts.

$A$ can be taken outside of the integral and hence just multiplies the final answer. The second integral can be expressed as the first using integration by parts. For the first, note that the derivative of $-ae^{-ax}$ is $e^{-ax}$. Edit: The derivative of $e^{-ax}$ is $-ae^{-ax}$.