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How do you find the integral of #e^((-1/2)*x)#?

1 Answer

Sep 26, 2015

#-2e^(-1/2x)+C#

Explanation:

#t=-1/2x => dt=-1/2dx => dx=-2dt#

#I=int e^(-1/2x)dx = int e^t*(-2dt) = -2 inte^tdt = -2e^t + C#

#I=-2e^(-1/2x)+C#

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