# What is the antiderivative of e^-x?

I found: $- {e}^{-} x + c$
$\int {e}^{-} x \mathrm{dx} = - {e}^{-} x + c$ which is your antiderivative.
$\frac{d}{\mathrm{dx}} \left(- {e}^{-} x + c\right) = - \left(- 1 \cdot {e}^{-} x\right) + 0 = {e}^{-} x$