Answer
$A = 4$
Work Step by Step
We can calculate the area $A$ of the region that lies under the curve:
$A = \int_{0}^{\pi}(2~sin~x-sin~2x)~dx$
$A = \int_{0}^{\pi}(2~sin~x-2~sin~x~cos~x)~dx$
$A = \int_{0}^{\pi}2~sin~x~(1-cos~x)~dx$
Let $u = 1-cos~x$
$\frac{du}{dx} = sin~x$
$dx = \frac{du}{sin~x}$
When $x = 0$, then $u = 0$
When $x = \pi$, then $u = 2$
$\int_{0}^{2}2~sin~x~(u)~\frac{du}{sin~x}$
$=\int_{0}^{2}2u~du$
$= u^2~\vert_{0}^{2}$
$= 2^2-0^2$
$= 4$
