Answer
$$ - {e^{ - x}} + C$$
Work Step by Step
$$\eqalign{
& {\text{The derivative of }}{e^{ - x}}{\text{ is }}\frac{d}{{dx}}\left[ {{e^{ - x}}} \right] = - {e^{ - x}},{\text{ then}} \cr
& {\text{the antiderivatices of }}{e^{ - x}}{\text{ are:}} \cr
& - \int {\left( { - {e^{ - x}}} \right)dx = } - \left( {{e^{ - x}}} \right) + C \cr
& = - {e^{ - x}} + C,{\text{ where }}C{\text{ is an arbitrary constant}}{\text{.}} \cr
& - {e^{ - x}} + C \cr} $$