# How do you find the area under the graph of f(x)=cos(x) on the interval [-pi/2,pi/2] ?

This is an integration problem. We will find the area under the curve $\cos \left(x\right)$ over the interval $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$. This is inclusive because of the square brackets.
On the unit circle remember that the positive side of y-axis corresponds to $\frac{\pi}{2}$ and a coordinate of $\left(0 , 1\right) .$ The y-coordinate corresponds to $1.$
On the unit circle remember that the negative side of y-axis corresponds to $- \frac{\pi}{2}$ and a coordinate of $\left(0 , - 1\right) .$ The y-coordinate corresponds to $- 1.$
${\int}_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \cos \left(x\right) \mathrm{dx}$
$= {\left[\sin \left(x\right)\right]}_{- \frac{\pi}{2}}^{\frac{\pi}{2}} = \left[\sin \left(\frac{\pi}{2}\right) - \sin \left(- \frac{\pi}{2}\right)\right] = 1 - \left(- 1\right) = 2$