The value of int_{pi}^{2pi} [2sin x] , dx,whe | vedclass

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(C) Let $I = \int_{\pi}^{2\pi} [2\sin x] \, dx$. In the interval $[\pi, 2\pi]$,$\sin x$ ranges from $0$ to $-1$ and back to $0$. Thus,$2\sin x$ ranges

Vedclass · Accepted Answer

$-\frac{5\pi}{3}$

Vedclass · Answer

(C) Let $I = \int_{\pi}^{2\pi} [2\sin x] \, dx$. In the interval $[\pi, 2\pi]$,$\sin x$ ranges from $0$ to $-1$ and back to $0$. Thus,$2\sin x$ ranges from $0$ to $-2$ and back to $0$.We split the integral based on the values of $[2\sin x]$:$1$. For $x \in [\pi, 7\pi/6]$,$2\sin x \in [-1, 0]$,so $[2\sin x] = -1$.$2$. For $x \in [7\pi/6, 3\pi/2]$,$2\sin x \in [-2, -1]$,so $[2\sin x] = -2$.$3$. For $x \in [3\pi/2, 11\pi/6]$,$2\sin x \in [-2, -1]$,so $[2\sin x] = -2$.$4$. For $x \in [11\pi/6, 2\pi]$,$2\sin x \in [-1, 0]$,so $[2\sin x] = -1$.$I = \int_{\pi}^{7\pi/6} (-1) \, dx + \int_{7\pi/6}^{3\pi/2} (-2) \, dx + \int_{3\pi/2}^{11\pi/6} (-2) \, dx + \int_{11\pi/6}^{2\pi} (-1) \, dx$$I = -1(\frac{7\pi}{6} - \pi) - 2(\frac{3\pi}{2} - \frac{7\pi}{6}) - 2(\frac{11\pi}{6} - \frac{3\pi}{2}) - 1(2\pi - \frac{11\pi}{6})$$I = -(\frac{\pi}{6}) - 2(\frac{2\pi}{6}) - 2(\frac{2\pi}{6}) - (\frac{\pi}{6})$$I = -\frac{\pi}{6} - \frac{4\pi}{6} - \frac{4\pi}{6} - \frac{\pi}{6} = -\frac{10\pi}{6} = -\frac{5\pi}{3}$.

The value of $\int_{\pi}^{2\pi} [2\sin x] \, dx$,where $[\cdot]$ represents the greatest integer function,is

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