What is the definite integral of #e^(2x)# from -2 to 2?

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Answer 1 · 2014-09-25T04:23:56

#int_(-2)^(2)e^(2x)dx#

Let #u=2x#

#du=2 dx#

#(du)/2=(2 dx)/2#

#(du)/2=dx#

#inte^(u)*(du)/2=1/2inte^udu#

Calculate the new boundaries

Upper Bound = #2#

#u=2x=2(2)=4#

Lower Bound = #-2#

#u=2x=2(-2)=-4#

#1/2int_(-4)^(4)e^udu#

#=1/2[e^u]_(-4)^(4)#

#=1/2[e^(4)-e^(-4)]#

#=1/2[e^(4)-1/e^(4)]#

#=[e^(4)/2-1/(2e^(4))]#

#=27.2899172#