The value of ${\int }_0^{2\pi } [2\sin x]dx$ where [.] represents the greatest integral function is 

${\int }_0^{2\pi } [2\sin x]dx={\int }_0^{\frac{\pi }{6}} 0dx+{\int }_{\pi /6}^{5\pi /6} 1dx+{\int }_{5\pi /6}^{\pi } 0dx++{\int }_{\pi }^{7\pi /6} (-1)\mathrm{d}x+{\int }_{7\pi /6}^{11\pi /6} (-2)\mathrm{d}x+{\int }_{11\pi /6}^{2\pi } (-1)\mathrm{d}x$

$\begin{array}{l} \\ =\frac{4\pi }{6}-\frac{\pi }{6}-\frac{8\pi }{6}-\frac{\pi }{6}=-\pi \end{array}$