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

Nov 9, 2016

The area is $= \frac{3 \pi}{2} \left(3 \pi - 1\right)$

#### Explanation:

The area is ${\int}_{0}^{3 \pi} \left(x - {\sin}^{2} x\right) \mathrm{dx}$
$\cos 2 x = 1 - 2 {\sin}^{2} x$
$\therefore {\sin}^{2} x = \frac{1 - \cos 2 x}{2}$
The area is ${\int}_{0}^{3 \pi} \left(x - {\sin}^{2} x\right) \mathrm{dx} = {\int}_{0}^{3 \pi} \left(x - \frac{1}{2} + \cos \frac{2 x}{2}\right) \mathrm{dx}$

$= {\left[{x}^{2} / 2 - \frac{x}{2} + \sin \frac{2 x}{4}\right]}_{0}^{3 \pi}$
$= \left(\frac{9 {\pi}^{2}}{2} - \frac{3 \pi}{2} + \sin \frac{6 \pi}{4}\right) - \left(0\right)$

$= \frac{9 {\pi}^{2}}{2} - \frac{3 \pi}{2} = \frac{3 \pi}{2} \left(3 \pi - 1\right)$