Question

How do you find the integral of `e^(-x)dx`?

Answer 1

it's `-e^(-x) + c`

Explanation:

let `u = -x` then `(du)/dx = -1` which means `dx = -du`
then
`int (e^(-x))dx = int (-e^u)du = -int (e^u)du = -e^u + c= -e^(-x) + c`