Question

The value of $ \int \limits^{2 \pi} _ 0 [ 2 \, \sin \, x ] \, dx , $ where [.] represents the greatest integral functions, is

• A. $ - \frac{ 5 \pi}{ 3 } $
• B. $ - \pi $
• C. $ \frac { 5 \pi }{ 3 } $
• D. $ - 2 { \pi } $

Answer 1

Answer: (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 $[2 \sin x$ ]

We know that, $\sin \frac{\pi}{6}=\frac{1}{2}$
$\therefore \sin \left(\pi+\frac{\pi}{6}\right)=\sin \frac{7 \pi}{6}=-\frac{1}{2} $
$\Rightarrow \sin \frac{11 \pi}{6}=\sin \left(2 \pi-\frac{\pi}{6}\right)=-\sin \frac{\pi}{6}=-\frac{1}{2} $
$\Rightarrow \sin \frac{9 \pi}{6}=\sin \frac{3 \pi}{6}=-1$
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(-\frac{1}{2}, 0\right) $
$\Rightarrow 2 \sin x=(0,-1),(-2,-1),(-1,0)$
$\Rightarrow [2 \sin x] =-1 $
$=\int\limits_{\pi}^{7 \pi / 6}[2 \sin x] d x+\int\limits_{7 \pi / 6}^{11 \pi / 6}[2 \sin x] d x $
$+\int\limits_{11 \pi / 6}^{2 \pi}[2 \sin x] d x $
$=\int\limits_{\pi}^{7 \pi / 6}-1 d x+\int\limits_{7 \pi / 6}^{11 \pi / 6}-2 d x+\int\limits_{11 \pi / 6}^{2 \pi}-1 d x $
$=-\frac{\pi}{6}-2\left(\frac{4 \pi}{6}\right)-\frac{\pi}{6}=-\frac{10 \pi}{6}=-\frac{5 \pi}{3} $