# What is the net area between f(x) = x-4cos^2x  and the x-axis over x in [0, 3pi ]?

Nov 19, 2016

$= 25.564$ areal units, nearly

#### Explanation:

In $\left[0 , 3 \pi\right]$, the periodic curve extends,

from $\left(0 , - 4\right) \to \left(9.425 , 5.425\right)$. 3pi=9.425 units, nearly.

The graph has been zoomed to include the part of the curve, under

reference.

The area $= \int \left(x - 4 {\cos}^{2} x\right) \mathrm{dx}$, from $x = 0 \to x = 3 \pi$

$= \int \left(x - 2 \left(1 + \cos 2 x\right)\right) \mathrm{dx}$, for the limits

= [x^2/2-2x-2(sin 2x)/2], between the limits

$= \left[\frac{9}{2} {\left(\pi\right)}^{2} - 6 \pi - 0\right]$

$= 25.564$ areal units, nearly.

Note that there are two negative parts for the area, between x-axis

and the curve, that add up to the total
graph{x-4cos(x)cos(x) [-40, 40, -20, 20]}