# How do you find the area of the region bounded by the curves y=sin(x), y=e^x, x=0, and x=pi/2 ?

##### 1 Answer
Sep 22, 2014

Over the interval $\left[0 , \frac{\pi}{2}\right]$ the function $y = {e}^{x}$ is greater that the function $y = \sin \left(x\right)$. This is why $y = \sin \left(x\right)$ is subtracted from $y = {e}^{x}$.

${\int}_{0}^{\frac{\pi}{2}} {e}^{x} - \sin \left(x\right) \mathrm{dx}$

$= {\left[{e}^{x} - \left(- \cos \left(x\right)\right)\right]}_{0}^{\frac{\pi}{2}}$

$= {\left[{e}^{x} + \cos \left(x\right)\right]}_{0}^{\frac{\pi}{2}}$

$= \left[{e}^{\frac{\pi}{2}} + \cos \left(\frac{\pi}{2}\right) - \left({e}^{0} + \cos \left(0\right)\right)\right]$

$= \left[{e}^{\frac{\pi}{2}} + 0 - \left(1 + 1\right)\right]$

$= \left[{e}^{\frac{\pi}{2}} - \left(2\right)\right]$

$= \left[{e}^{\frac{\pi}{2}} - 2\right]$

$= 2.81048 \to$ Solution