Question:

$$\int_{0}^{16\pi/3} |\sin x| dx = \int_{0}^{5\pi} |\sin x| dx + \int_{5\pi}^{5\pi + \pi/3} |\sin x| dx = 5 \int_{0}^{\pi} |\sin x| dx + \int_{0}^{\pi/3} |\sin x| dx$$<br>$$= 5[-\cos x]_0^{\pi} + [-\cos x]_0^{\pi/3} = 10 + \left( -\frac{1}{2} + 1 \right) = \frac{21}{2}$$