How do you find the area under the curve #y=1/2(e^x + e^-x)# from #x=-2# to #x=2#?

1 Answer
sjc
Nov 14, 2017

#e^(-2)(e^4-1)#

Explanation:

area#=int_(-2)^(2)1/2(e^x+e^(-x))dx#

#=1/2[e^x-e^(-x)]_(-2)^2#

#=1/2{[e^x-e^(-x)]^2-[e^x-e^(-x)]_(-2)}#

#1/2{e^2-e^(-2)-e^(-2)+e^(- -2)}#

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

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

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

#=e^(-2)(e^4-1)#

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