Question 1032464
As shown by the graph, the curves shift at {{{x=0}}} so break up the integral accordingly.
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*[illustration fd1.JPG].
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{{{A=int((g(x)-f(x)),dx,-3,0)+int((f(x)-g(x)),dx,0,2)}}}
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{{{A[1]=int((3e^(4x)+1-4e^(4x)),dx,-3,0)}}}
{{{A[1]=int((-e^(4x)+1),dx,-3,0)}}}
{{{A[1]=-e^(4x)/4+x+C}}}
{{{A[1]=-(1-e^(-12))/4+(0+3)}}}
{{{A[1]=11/4-1/(4e^12)}}}

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{{{A[2]=int((4e^(4x)-(3e^(4x)+1)),dx,0,2)}}}
{{{A[2]=int((e^(4x)-1),dx,0,2)}}}
{{{A[2]=e^(4x)/4-x+C}}}
{{{A[2]=(e^(8)-1)/4-(2-0)}}}
{{{A[2]=e^(8)/4-1/4-2}}}
{{{A[2]=e^(8)/4-9/4}}}
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So then,
{{{A=A[1]+A[2]}}}
{{{A=11/4-1/(4e^12)+e^(8)/4-9/4}}}
{{{A=1/2+(1/4)(e^8-1/e^12)}}}