Answer
$\frac{9}{2}$
Work Step by Step
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
$I= F(\pi)-F(0)$
$= -9(cos\frac{\pi}{3})-[-9(cos\frac{0}{3})]$
$= (-9\times\frac{1}{2})-(-9\times1)$
$= -\frac{9}{2}+9 = \frac{9}{2}$
