How do you integrate $int e^(3-x)dx$ from $[3,4]$ ?

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Answer 1 · 2017-05-12T03:11:16

$int_3^4(e^(3-x))dx=1- 1/e$

$int_3^4(e^(3-x))dx$

Do a $u$ -substitution:
$u=3-x$
$(du)/dx=-1$
$dx=-du$

Remember to change the bounds of the integral according to the $u$ values
$int_(0)^(-1)(e^u)(-1)du$

$int_0^(-1)(-e^u)du$

$=[-e^u]_0^(-1)$

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

$=1- 1/e$

How do you integrate int e^(3-x)dxint e^(3-x)dx from [3,4][3,4]?