## University Calculus: Early Transcendentals (3rd Edition)

$\frac{9}{2}$
Let $I= \int^{\pi}_{0}3sin\frac{x}{3}dx$ $\int 3sin\frac{x}{3}dx= 3\int sin\frac{x}{3}dx$ $= 3\times(\frac{-cos\frac{x}{3}}{\frac{1}{3}})+C$ $= -9(cos\frac{x}{3})+C= F(x)$ By the second fundamental theorem of calculus, we have $I= F(\pi)-F(0)$ $= -9(cos\frac{\pi}{3})-[-9(cos\frac{0}{3})]$ $= (-9\times\frac{1}{2})-(-9\times1)$ $= -\frac{9}{2}+9 = \frac{9}{2}$