What is the net area between f(x) = cos2x-xsinx  and the x-axis over x in [0, 3pi ]?

Apr 13, 2018

The net area between $f \left(x\right)$ and the $x$-axis from $0$ to $3 \pi$ is $- 3 \pi$.

Explanation:

color(white)=int_0^(3pi)(cos2x−xsinx)dx

$= {\int}_{0}^{3 \pi} \cos 2 x$ dx−int_0^(3pi)xsinx $\mathrm{dx}$

=1/2sin2x|_0^(3pi)−int_0^(3pi)xsinx $\mathrm{dx}$

=1/2sin(3pi)-1/2sin(0)−int_0^(3pi)xsinx $\mathrm{dx}$

=−int_0^(3pi)xsinx $\mathrm{dx}$

Using the DI method (an easier way to understand integration by parts):

Multiply the diagonals with the appropriate sign, and add them (and don't forget the negative sign in front of the integral from before):

$= - \left(\left(x \cdot - \cos x - 1 \cdot - \sin x\right) {|}_{0}^{3 \pi}\right)$

$= - \left(\left(- x \cos x + \sin x\right) {|}_{0}^{3 \pi}\right)$

$= - \left(\left(\sin x - x \cos x\right) {|}_{0}^{3 \pi}\right)$

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

$= - \left(\left(0 + 3 \pi\right) - \left(0 - 0\right)\right)$

$= - 3 \pi$

That is the net area under the curve. Hope this helped!