Question

Find the area of the region enclosed between the graphs of y=ex and y=e−x and the line x=ln2.

Answer 1

The graphs of $y=e^x$ and $y=e^{-x}$ intersect if $x=0,$ therefore we should integrate $(e^x-e^{-x})$ on the interval of [0, ln 2]:

$\int\limits_0^{\ln 2} (e^x-e^{-x})\,dx = (e^x + e^{-x})\Big|_{0}^{\ln 2} = (2+0.5) - (1+1) = 0.5.$