Question

The value of $\underset{0}{\overset{2\pi }{\int }}[2\sin x]dx,$ 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
$\left(\pi ,\frac{7\pi }{6}\right),\left(\frac{7\pi }{6},\frac{11\pi }{6}\right),\left(\frac{11\pi }{6},2\pi \right)$
$\sin x=\left(0,-\frac{1}{2}\right),\left(-1,-\frac{1}{2}\right),\left(-\frac{1}{2},0\right)$
$\Rightarrow 2\sin x=(0,-1),(-2,-1),(-1,0)$
$\Rightarrow [2\sin x]=-1$
$=\underset{\pi }{\overset{7\pi /6}{\int }}[2\sin x]dx+\underset{7\pi /6}{\overset{11\pi /6}{\int }}[2\sin x]dx$
$+\underset{11\pi /6}{\overset{2\pi }{\int }}[2\sin x]dx$
$=\underset{\pi }{\overset{7\pi /6}{\int }}-1dx+\underset{7\pi /6}{\overset{11\pi /6}{\int }}-2dx+\underset{11\pi /6}{\overset{2\pi }{\int }}-1dx$
$=-\frac{\pi }{6}-2\left(\frac{4\pi }{6}\right)-\frac{\pi }{6}=-\frac{10\pi }{6}=-\frac{5\pi }{3}$