An old IIT problem

$\begin{aligned} \int_\pi ^{2\pi} \lfloor 2 \sin x \rfloor dx & = \int_\pi ^\frac{7\pi}{6} (-1) \space dx + \int_\frac{7\pi}{6} ^\frac{11\pi}{6} (-2) \space dx + \int_\frac{11\pi}{6}^{2\pi} (-1) \space dx \\ & = -\left[ \frac{7\pi}{6} - \pi \right] -2 \left[ \frac{11\pi}{6} - \frac{7\pi}{6} \right] --\left[2\pi - \frac{11\pi}{6} \right] \\ & = -\frac{\pi}{6} -\frac{8\pi}{6} -\frac{\pi}{6} \\ & = -\frac{5\pi}{3} \end{aligned}$

$\Rightarrow |A|+|B| = 5+3 = \boxed{8}$