Assertion (A): int_{frac{pi}{2}}^{frac{3 pi}{ | vedclass

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(D) Consider the integral $I = \int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} [2 \sin x] dx$. In the interval $\left[\frac{\pi}{2}, \pi\right]$,$0 \le \sin x

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$A$ is false,$R$ is true

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(D) Consider the integral $I = \int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} [2 \sin x] dx$. In the interval $\left[\frac{\pi}{2}, \piight]$,$0 \le \sin x \le 1$,so $0 \le 2 \sin x \le 2$. Specifically,for $x \in \left[\frac{\pi}{2}, \piight)$,$0 \le 2 \sin x More precisely,$2 \sin x = 1 \implies \sin x = \frac{1}{2} \implies x = \frac{5\pi}{6}$. Thus,$[2 \sin x] = 1$ for $x \in \left[\frac{\pi}{6}, \frac{5\pi}{6}ight]$ is not applicable here. For $x \in \left[\frac{\pi}{2}, \frac{5\pi}{6}ight)$,$1 \le 2 \sin x For $x \in \left[\frac{5\pi}{6}, \piight]$,$0 \le 2 \sin x For $x \in \left[\pi, \frac{7\pi}{6}ight]$,$-1 \le 2 \sin x For $x \in \left[\frac{7\pi}{6}, \frac{3\pi}{2}ight]$,$-2 \le 2 \sin x Calculating the integral: $\int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} 1 dx + \int_{\frac{5\pi}{6}}^{\pi} 0 dx + \int_{\pi}^{\frac{7\pi}{6}} (-1) dx + \int_{\frac{7\pi}{6}}^{\frac{3\pi}{2}} (-2) dx = \left(\frac{5\pi}{6} - \frac{\pi}{2}ight) + 0 - \left(\frac{7\pi}{6} - \piight) - 2\left(\frac{3\pi}{2} - \frac{7\pi}{6}ight) = \frac{\pi}{3} - \frac{\pi}{6} - 2\left(\frac{\pi}{3}ight) = \frac{\pi}{6} - \frac{2\pi}{3} = -\frac{\pi}{2} 
eq 0$. Thus,$A$ is false. The function $f(x) = 2 \sin x$ has derivative $f'(x) = 2 \cos x$. In $\left[\frac{\pi}{2}, \frac{3\pi}{2}ight]$,$\cos x \le 0$,so $f'(x) \le 0$,meaning $f(x)$ is decreasing. Thus,$R$ is true.

Assertion $(A)$: $\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} [2 \sin x] dx = 0$,where $[.]$ denotes the greatest integer function.
Reason $(R)$: $2 \sin x$ is a decreasing function in $\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$.

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Assertion $(A)$: $\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} [2 \sin x] dx = 0$,where $[.]$ denotes the greatest integer function. Reason $(R)$: $2 \sin x$ is a decreasing function in $\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$.