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:

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

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

$=1-e^(-1)$

$~= 0.63212$

How do you evaluate int_1^2 e^(1-x)dxint_1^2 e^(1-x)dx?