Chapter 14 - Review - Review Exercises - Page 1071: 12

$5.5243$

Here, we have: $y=e^x, y=e^{-x}, x=0$ to $x=2$ The area is given by $A=\int_0^{2} (e^x-e^{-x})] \ dx$ or, $= [e^x+e^{-x}]_0^2$ or, $=[e^2+e^{-2}]-[e^0+e^{-0}]$ or, $=e^2+\dfrac{1}{e^2}-2$ Therefore, the required area is: $Area=5.5243$