If for a real number y,[y] is the greatest in | vedclass

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(C) We need to evaluate $I = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} [2 \sin x] \,dx$.For $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$,$\sin x$ ranges from $

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$-\frac{\pi}{2}$

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(C) We need to evaluate $I = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} [2 \sin x] \,dx$.For $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$,$\sin x$ ranges from $1$ to $-1$,so $2 \sin x$ ranges from $2$ to $-2$.We break the integral based on the values of $[2 \sin x]$:$1) 	ext{ For } x \in [\frac{\pi}{2}, \frac{5\pi}{6}], \frac{1}{2} \leq \sin x \leq 1 \Rightarrow 1 \leq 2 \sin x \leq 2$. Thus $[2 \sin x] = 1$ for $x \in [\frac{\pi}{2}, \frac{5\pi}{6})$ and $2$ at $x = \frac{\pi}{2}$.$2) 	ext{ For } x \in [\frac{5\pi}{6}, \pi], 0 \leq \sin x \leq \frac{1}{2} \Rightarrow 0 \leq 2 \sin x \leq 1$. Thus $[2 \sin x] = 0$ for $x \in [\frac{5\pi}{6}, \pi)$.$3) 	ext{ For } x \in [\pi, \frac{7\pi}{6}], -\frac{1}{2} \leq \sin x \leq 0 \Rightarrow -1 \leq 2 \sin x \leq 0$. Thus $[2 \sin x] = -1$ for $x \in [\pi, \frac{7\pi}{6})$.$4) 	ext{ For } x \in [\frac{7\pi}{6}, \frac{3\pi}{2}], -1 \leq \sin x \leq -\frac{1}{2} \Rightarrow -2 \leq 2 \sin x \leq -1$. Thus $[2 \sin x] = -2$ for $x \in [\frac{7\pi}{6}, \frac{3\pi}{2})$.$I = \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} (1) \,dx + \int_{\frac{5\pi}{6}}^{\pi} (0) \,dx + \int_{\pi}^{\frac{7\pi}{6}} (-1) \,dx + \int_{\frac{7\pi}{6}}^{\frac{3\pi}{2}} (-2) \,dx$$I = (\frac{5\pi}{6} - \frac{\pi}{2}) + 0 - (\frac{7\pi}{6} - \pi) - 2(\frac{3\pi}{2} - \frac{7\pi}{6})$$I = \frac{\pi}{3} - \frac{\pi}{6} - 2(\frac{9\pi - 7\pi}{6}) = \frac{\pi}{6} - 2(\frac{2\pi}{6}) = \frac{\pi}{6} - \frac{2\pi}{3} = \frac{\pi - 4\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2}$.

If for a real number $y$,$[y]$ is the greatest integer less than or equal to $y$,then the value of the integral $\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} [2 \sin x] \,dx$ is

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