Question
If for a real number $$y,\,\left[ y \right]$$ is the greatest integer less than or
equal to $$y,$$ then the value of the integral $$\int\limits_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left[ {2\,\sin \,x} \right]dx} $$ is-
A.
$$ - \pi $$
B.
$$0$$
C.
$$\frac{{ - \pi }}{2}$$
D.
$$\frac{\pi }{2}$$
Answer :
$$\frac{{ - \pi }}{2}$$
Solution :
In the range $$\frac{\pi }{2}$$ to $$\frac{{3\pi }}{2},$$ we have to find the value of $${\left[ {2\,\sin \,x} \right]}$$
\[\left[ {2\,\sin \,x} \right] = \left\{ \begin{array}{l}
2\,\,\,\,\,\,\,\,\,\,{\rm{if }}x = \frac{\pi }{2}\\
1\,\,\,\,\,\,\,\,\,\,\,{\rm{if }}\frac{\pi }{2} < x \le \frac{{5\pi }}{6}\\
0\,\,\,\,\,\,\,\,\,\,{\rm{if }}\frac{{5\pi }}{6} < x \le \pi \\
- 1\,\,\,\,\,\,\,{\rm{if }}\pi < x \le \frac{{7\pi }}{6}\\
- 2\,\,\,\,\,\,\,{\rm{if }}\frac{{7\pi }}{6} < x \le \frac{{3\pi }}{2}
\end{array} \right.\]
$$\eqalign{
& {\text{Thus}} \cr
& I = \int_{\frac{\pi }{2}}^{\frac{{5\pi }}{6}} {1.dx} + \int_{\frac{{5\pi }}{6}}^\pi {0.dx} + \int_\pi ^{\frac{{7\pi }}{6}} {\left( { - 1} \right).dx} + \int_{\frac{{7\pi }}{6}}^{\frac{{3\pi }}{2}} {\left( { - 2} \right).dx} \cr
& {\text{or }}I = \left[ {\frac{{5\pi }}{6} - \frac{\pi }{2}} \right] + 0 - 1\left[ {\frac{{7\pi }}{6} - \pi } \right] - 2\left[ {\frac{{3\pi }}{2} - \frac{{7\pi }}{6}} \right] \cr
& = \frac{{2\pi }}{6} - \frac{\pi }{6} - \frac{{4\pi }}{6} \cr
& = \frac{{ - 3\pi }}{6} \cr
& = \frac{{ - \pi }}{2} \cr} $$