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

1 Answer
Aug 20, 2014

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.