$\displaystyle e^{3-x} = \frac{e^{3}}{e^{x}} = e^{3}e^{-x}$
It is all the same. Please ponder it until you see that it is not scary.
$\displaystyle \int a\cdot e^{-x}dx = a\cdot\int e^{-x}dx$
Feel free simply to remove constants from the integral argument. $\displaystyle e^{3}$ is a constant.
By the way, "the derivative of e" is zero. Be more careful with words.
perhaps it will make more sense if you realize that $\displaystyle \frac{d}{dx}(e^{-x}) = -e^{-x}$.
this is a consequence of the chain rule: $\displaystyle e^{-x}$ is really the composition:
$\displaystyle exp\circ g(x)$, where g(x)= -x = (-1)x.
so when we integrate, we need a minus sign to go inside the integral, and another minus sign to go outside (to balance, (-1)(-1) = 1). the minus sign inside the integral disappears, since when we integrate:
$\displaystyle \int \ -e^{-x} dx = e^{-x} + C$.
the minus sign outside the integral stays, however.