Question

How do you evaluate the integral `inte^(-x) dx`?

Answer 1

The answer is `I=-e^(-x)+C`.

This integral can be solved by a substitution:

`u=-x`
`du=-dx`
`-du=dx`

So, now we can substitute:

`int e^(-x)dx = int e^u (-du)`
`=-int e^u du`
`=-e^u + C`

and substitute back:
`=-e^(-x) + C`

For simple looking integrands, you should try a quick check to see if substitution works before trying harder integration methods.