How do you evaluate the integral #inte^(-x) dx#?
1 Answer
Aug 20, 2014
The answer is
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:
For simple looking integrands, you should try a quick check to see if substitution works before trying harder integration methods.