Question 834948
{{{ graph( 300, 300, -2, 5, -2, 5, x, e^(1-x)) }}} 
From the graph, you can find the area by breaking up the area into two parts,
{{{A=int(x,dx,0,1)+int(e^(1-x),dx,1,infinity)}}}
{{{int(x,dx,0,1)=(1/2)x^2+C=(1/2)(1^2-0^2)=1/2}}}
{{{int(e^(1-x),dx,1,infinity)=-e^(1-x)+C=-e^(1-infinity)+e^(1-1)=0+1=1}}}
{{{A=1/2+1=3/2}}}