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#