Explanation

(a)
It is a question of greatest integer function. We have, subdivide the interval $\pi\ to\ 2\pi$ as under keeping in view that we have to evaluate $\lbrack 2\sin x\rbrack$
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We know that, ${sin\ }{\frac{\pi}{6} = \frac{1}{2}}$
$\sin{\left( \pi + \frac{\pi}{6} \right) = \sin{\frac{7\pi}{6} = - \frac{1}{2}}}$
$\sin{\frac{11\pi}{6} = \sin{\left( 2\pi - \frac{\pi}{6} \right) = - \sin{\frac{\pi}{6} = - \frac{1}{2}}}}$
$\sin{\frac{9\pi}{6} = \sin{\frac{3\pi}{6} = - 1}}$
Hence, we divide the interval $\pi\ to\ 2\pi\ as\ \left( \pi,\frac{7\pi}{6} \right),\left( \frac{7\pi}{6},\frac{11\pi}{6} \right),\left( \frac{11\pi}{6},\ 2\pi \right)$
$\sin{x =}\left( 0,\ - \frac{1}{2} \right),\left( - 1,\ - \frac{1}{2} \right),\left( 0,\ - \frac{1}{2} \right)$
$2\sin{x = (0, - 1),( - 2, - 1),(0, - 1)}$
$\lbrack 2\sin{x\rbrack = - 1}$
$\therefore\ I = I_{1} + I_{2} + I_{3}$
$= \int_{}^{}{- 1\ dx + \int_{}^{}{- 2\ dx}\int_{}^{}{- 1\ dx}}$
Between proper limits
$= - \frac{\pi}{6} - 2\left( \frac{4\pi}{6} \right) - \frac{\pi}{6} = - \frac{10\pi}{6} = - \frac{5\pi}{3}$