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(C) We know that for $x \in [\pi/2, 3\pi/2]$,$\sin x$ ranges from $-1$ to $1$,so $2\sin x$ ranges from $-2$ to $2$.We break the integral at points whe

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

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(C) We know that for $x \in [\pi/2, 3\pi/2]$,$\sin x$ ranges from $-1$ to $1$,so $2\sin x$ ranges from $-2$ to $2$.We break the integral at points where $[2\sin x]$ changes its value:$I = \int_{\pi/2}^{3\pi/2} [2\sin x] \, dx$$= \int_{\pi/2}^{5\pi/6} [2\sin x] \, dx + \int_{5\pi/6}^{\pi} [2\sin x] \, dx + \int_{\pi}^{7\pi/6} [2\sin x] \, dx + \int_{7\pi/6}^{3\pi/2} [2\sin x] \, dx$In the interval $[\pi/2, 5\pi/6]$,$1 \le 2\sin x In the interval $[5\pi/6, \pi]$,$0 \le 2\sin x In the interval $[\pi, 7\pi/6]$,$-1 \le 2\sin x In the interval $[7\pi/6, 3\pi/2]$,$-2 \le 2\sin x Substituting these values:$I = \int_{\pi/2}^{5\pi/6} (1) \, dx + \int_{5\pi/6}^{\pi} (0) \, dx + \int_{\pi}^{7\pi/6} (-1) \, dx + \int_{7\pi/6}^{3\pi/2} (-2) \, dx$$I = (5\pi/6 - \pi/2) + 0 - (7\pi/6 - \pi) - 2(3\pi/2 - 7\pi/6)$$I = (2\pi/6) - (\pi/6) - 2(2\pi/6) = \pi/6 - 4\pi/6 = -3\pi/6 = -\pi/2$.

List $I$	List $II$
$P.$ The number of polynomials $f(x)$ with non-negative integer coefficients of degree $\leq 2$,satisfying $f(0)=0$ and $\int_0^1 f(x) dx=1$,is	$1.$ $8$
$Q.$ The number of points in the interval $(-\sqrt{13}, \sqrt{13})$ at which $f(x)=\sin(x^2)+\cos(x^2)$ attains its maximum value,is	$2.$ $2$
$R.$ $\int_{-2}^2 \frac{3x^2}{1+e^x} dx$ equals	$3.$ $4$
$S.$ $\frac{\int_{-1/2}^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx}{\int_0^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx}$ equals	$4.$ $0$

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

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