# What is the net area between f(x) = sinx - cos^2x and the x-axis over x in [0, 3pi ]?

Nov 8, 2016

The net area is $2 - \frac{3 \pi}{2}$

#### Explanation:

A small area is $\mathrm{dA} = y \mathrm{dx}$ $\implies$ $\mathrm{dA} = \left(\sin x - {\cos}^{2} x\right) \mathrm{dx}$
so the area is $A = {\int}_{0}^{3 \pi} \left(\sin x - {\cos}^{2} x\right) \mathrm{dx}$
To integrate ${\cos}^{2} x$, we use $\cos 2 x = 2 {\cos}^{2} x - 1$
$\implies$${\cos}^{2} x = \frac{1 + \cos 2 x}{2}$
So, $A = {\int}_{0}^{3 \pi} \left(\sin x - \frac{1 + \cos 2 x}{2}\right) \mathrm{dx}$
$= {\left[- \cos x - \frac{x}{2} - \frac{\sin 2 x}{4}\right]}_{0}^{3 \pi}$
$= 1 - \frac{3 \pi}{2} + 1 = 2 - \frac{3 \pi}{2}$