# How do you find the antiderivative of e^(-2x)?

Jan 2, 2017

$\int {e}^{- 2 x} \mathrm{dx} = - \frac{1}{2} {e}^{- 2 x} + C$

#### Explanation:

it is important to remember that

$\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$

so let us see what happens if we differentiate the given function

$y = {e}^{- 2 x}$

$u = - 2 x \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 2$

$y = {e}^{u} \implies \frac{\mathrm{dy}}{\mathrm{du}} = {e}^{u}$

by the chain rule we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

giving us

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} \times - 2 {e}^{- 2 x} = - 2 {e}^{- 2 x}$

now integration is the reverse of differentiation, so comparing what we have after differentiating and the function we are given to integrate.

we have to adjust the function by a suitable constant to cancel the " -2

$\int {e}^{- 2 x} \mathrm{dx} = - \frac{1}{2} {e}^{- 2 x} + C$

if you now differentiate the resulting function you will see it gives the original function.