Answer
$A= {e^{\pi /2}} - 2$
Work Step by Step
$$\eqalign{
& {\text{Let the functions}} \cr
& y = \cos x,{\text{ }}y = {e^x},{\text{ }}x = \frac{\pi }{2} \cr
& {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr
& \cr
& {\text{We can find the area integrating with respect to }}x \cr
& {\text{Use }}A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx,{\text{ with }}f\left( x \right) \geqslant g\left( x \right) \cr
& {\text{Let }} \cr
& f\left( x \right) = {e^x}{\text{, }}g\left( x \right) = \cos x \cr
& x = a = 0{\text{ and }}x = b = \frac{\pi }{2} \cr
& {\text{Therefore}}{\text{,}} \cr
& \underbrace {A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx}_ \Downarrow \cr
& A = \int_0^{\pi /2} {\left( {{e^x} - \cos x} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {{e^x} - \sin x} \right]_0^{\pi /2} \cr
& A = \left[ {{e^{\pi /2}} - \sin \left( {\frac{\pi }{2}} \right)} \right] - \left[ {{e^0} - \sin \left( 0 \right)} \right] \cr
& {\text{Simplifying}} \cr
& A = {e^{\pi /2}} - 1 - 1 + 0 \cr
& A = {e^{\pi /2}} - 2 \cr} $$
