The value of $\underset{\pi }{\overset{2\pi }{\int }}\left[2\sin x\right]dx$ is equal to (where [.] represents the greatest integer function)

Let, $I={\displaystyle {\int }_{\pi }^{2\pi }[2\sin x]dx}$
 

For $\mathrm{\pi }\le x\le \frac{7\mathrm{\pi }}{6} \mathrm{or} \frac{11\mathrm{\pi }}{6}\le \mathrm{x}\le 2\mathrm{\pi }$

  $-1\le 2\hspace{0.17em}\sin x<0$

$\Rightarrow [2\sin x]=-1$

For $\frac{7\pi }{6}\le x<\frac{11\pi }{6}$

$-2\le 2\sin x<-1$

$\Rightarrow [2\sin x]=-2$

$\therefore I={\displaystyle \underset{\pi }{\overset{7\pi /6}{\int }}-1dx+}{\displaystyle \underset{7\pi /6}{\overset{11\pi /6}{\int }}-2dx+}{\displaystyle \underset{11\pi /6}{\overset{2\pi }{\int }}-1dx}$

$=\left(-\frac{7\pi }{6}+\pi \right)+2\left(-\frac{11\pi }{6}+\frac{7\pi }{6}\right)+\left(-2\pi +\frac{11\pi }{6}\right)$

$=-\frac{\pi }{6}-\frac{8\pi }{6}-\frac{\pi }{6}=-\frac{10\pi }{6}=\frac{-5\pi }{3}$