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How do you find the integral of #e^(-x)dx#?

1 Answer

May 15, 2018

it's #-e^(-x) + c#

Explanation:

let #u = -x# then #(du)/dx = -1# which means #dx = -du#
then
#int (e^(-x))dx = int (-e^u)du = -int (e^u)du = -e^u + c= -e^(-x) + c#

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