Chapter 14 - Review - Review Exercises - Page 1071: 14

$1+\dfrac{3}{e^2}$

Here, we have: $y=x, y=xe^{-x}$ and $x=0$ to $x=2$ The area is given by $A=\int_{0}^2 (x-xe^{-x}) \ dx$ or, $= [x-(-x e^{-x} -e^{-x})]_0^2$ or, $=[x+xe^{-x}+e^{-x}]_0^2$ or, $=[2+2e^{-2}+e^{-2}]-[0+0e^{-0}+e^{-0}]$ Therefore, the required area is: $Area=1+\dfrac{3}{e^2}$