How do you evaluate #int_1^2 e^(1-x)dx#?

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Answer 1 · 2017-03-08T01:09:24

#1-e^(-1) ~= 0.63212#

#int_1^2 e^(1-x)dx#

Let #u = 1-x -> (du)/dx=-1#

#I = int-e^u du#

#= -e^u#

Undo substitution:

#=-e^(1-x)]_1^2#

#= -e^(1-2)+e^(1-1) #

#=1-e^(-1)#

#~= 0.63212#