## Calculus: Early Transcendentals 8th Edition

$A = 4$
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 $u = 2$ $\int_{0}^{2}2~sin~x~(u)~\frac{du}{sin~x}$ $=\int_{0}^{2}2u~du$ $= u^2~\vert_{0}^{2}$ $= 2^2-0^2$ $= 4$