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

Apr 7, 2017

We start by checking for x-intercepts within the given interval.

$f \left(x\right) = \cos x \sin x \to f \left(x\right) = \frac{1}{2} \sin 2 x$

If we set $f \left(x\right)$ to $0$ and solve for $x$.

$0 = \frac{1}{2} \sin 2 x$

$0 = \sin 2 x$

$2 x = 2 \pi \mathmr{and} 2 x = \pi$

$x = \pi \mathmr{and} x = \frac{\pi}{2}$

Since trig functions repeat, and the period of this function is $\pi$, we will have more than two crossings. If we analyze the graph, we realize that in $\left[0 , 3 \pi\right]$ there are $6$ intercepts.

graph{sinxcosx [-4.5, 15.5, -5.16, 4.84]}

This means that there are $3$ regions of equal area above the x-axis, and three areas of equal area below the x-axis. Since we're talking about net area, we should subtract the negative area (below the x-axis) from the positive area (above the x-axis).

Our expression for area will be

$A = 3 \left({\int}_{0}^{\frac{\pi}{2}} \frac{1}{2} \sin 2 x \mathrm{dx} - {\int}_{\frac{\pi}{2}}^{\pi} \frac{1}{2} \sin 2 x \mathrm{dx}\right)$

$A = 3 \left({\left[- \cos 2 x\right]}_{0}^{\frac{\pi}{2}} - {\left[- \cos 2 x\right]}_{\frac{\pi}{2}}^{\pi}\right)$

$A = 3 \left(- \cos \pi - \cos 0 - \left(- \cos 2 \pi - \cos \pi\right)\right)$

$A = 3 \left(1 - 1 + 1 - 1\right)$

$A = 0$

A geometric view of this gives sense to this result: whenever a positive area occurs, it is cancelled out by the negative area, giving a final result of $0$ (because there are an equal number of positive and negative areas).

Hopefully this helps!