What is the antiderivative of #e^-x#?

1 Answer
mason m
Jan 27, 2016

#-e^(-x)+C#

Explanation:

The problem, written out, is:

#inte^(-x)dx#

The following rule will be used:

#inte^udu=e^u+C#

Using substitution:
Set #u=-x#, so we know #(du)/dx=-1# and #-du=dx#.

This gives us a simplified integral of

#=inte^u(-du)#

We can bring the negative sign out:

#=-inte^udu=-e^u+C=-e^(-x)+C#

Using intuition:

We know that the derivative of #e^(-x)# is #-e^(-x)#, but we want a derivative of the positive version #e^(-x)#. So, we make the antiderivative negative, causing the derivative of be positive:

#d/dx-e^(-x)=e^(-x)#

Then, we simply add #C#, the constant of integration.

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