The value of $\int\limits_\pi ^{2\pi } {[2\sin x]dx} $ is equal to (where[.] is the G.I.F.)
A. $ - \pi $
B. $ - 2\pi $
C. $ - \dfrac{{5\pi }}{3}$
D. $\dfrac{{5\pi }}{3}$

Hint: Greatest Integer Function (G.I.F.) or floor function is a function that rounds off the real number(input) to the greatest integer less than or equal to the number. So here as well, we will first divide the interval over integrand to sub-intervals and then we will check the greatest integer of that interval using the graph of 2sinx i.e., over$\int\limits_\pi ^{2\pi } {[2\sin x]dx} $.

Complete step-by-step answer:

Here we are given that we need to find the integral of $\int\limits_\pi ^{2\pi } {[2\sin x]dx} $where [.] is the greatest integer function. So for this, first we will see the graph of the above-mentioned function.


 The graph on the right is the graph of 2sinx. We will consider the portion $\pi to2\pi $as it is the limit of our given integral.
Here, we can see that we have divided the graph in to intervals$\left( {\pi ,\dfrac{{7\pi }}{6}} \right),\left( {\dfrac{{7\pi }}{6},\dfrac{{3\pi }}{2}} \right),\left( {\dfrac{{3\pi }}{2},\dfrac{{11\pi }}{6}} \right),\left( {\dfrac{{11\pi }}{6},2\pi } \right)$
The greatest integer value of 2sinx in the intervals:
$\left( {\pi ,\dfrac{{7\pi }}{6}} \right)$ is -1
$\left( {\dfrac{{7\pi }}{6},\dfrac{{3\pi }}{2}} \right)$is -2
$\left( {\dfrac{{3\pi }}{2},\dfrac{{11\pi }}{6}} \right)$is -2
$\left( {\dfrac{{11\pi }}{6},2\pi } \right)$is -1
Therefore, we can write the given integral $\int\limits_\pi ^{2\pi } {[2\sin x]dx} $ as:
 $\int\limits_\pi ^{2\pi } {[2\sin x]dx} $=$\int\limits_\pi ^{\dfrac{{7\pi }}{6}} {[2\sin x]dx + \int\limits_{\dfrac{{7\pi }}{6}}^{\dfrac{{3\pi }}{2}} {[2\sin x]dx + \int\limits_{\dfrac{{3\pi }}{2}}^{\dfrac{{11\pi }}{6}} {[2\sin x]dx + \int\limits_{\dfrac{{11\pi }}{6}}^{2\pi } {[2\sin x]dx} } } } $
Substituting the values of [2sinx] in the respective integrals, we get
$
  \int\limits_\pi ^{2\pi } {[2\sin x]dx = \int\limits_\pi ^{\dfrac{{7\pi }}{6}} {( - 1)dx + \int\limits_{\dfrac{{7\pi }}{6}}^{\dfrac{{3\pi }}{2}} {( - 2)dx + \int\limits_{\dfrac{{3\pi }}{2}}^{\dfrac{{11\pi }}{6}} {( - 2)dx + \int\limits_{\dfrac{{11\pi }}{6}}^{2\pi } {( - 1)dx} } } } } \\
   \Rightarrow \int\limits_\pi ^{2\pi } {[2\sin x]dx = } [ - x]_\pi ^{\dfrac{{7\pi }}{6}} + [ - 2x]_{\dfrac{{7\pi }}{6}}^{\dfrac{{3\pi }}{2}} + [ - 2x]_{\dfrac{{3\pi }}{2}}^{\dfrac{{11\pi }}{6}} + [ - 1x]_{\dfrac{{11\pi }}{6}}^{2\pi } \\
   \Rightarrow \int\limits_\pi ^{2\pi } {[2\sin x]dx = - \left[ {\left( {\dfrac{{7\pi }}{6} - \pi } \right) + \left( {\dfrac{{3\pi }}{2} - \dfrac{{7\pi }}{6}} \right) + \left( {\dfrac{{11\pi }}{6} - \dfrac{{3\pi }}{2}} \right) + \left( {2\pi - \dfrac{{11\pi }}{6}} \right)} \right]} \\
   \Rightarrow \int\limits_\pi ^{2\pi } {[2\sin x]dx} = - \left[ {\left( {\dfrac{\pi }{6} + \dfrac{{4\pi }}{3} + \dfrac{\pi }{6}} \right)} \right] = - \left[ {\dfrac{{\pi + 8\pi + \pi }}{6}} \right] = - \dfrac{{10\pi }}{6} = - \dfrac{{5\pi }}{3} \\
 $
$ \Rightarrow \int\limits_\pi ^{2\pi } {[2\sin x]dx} = $$ - \dfrac{{5\pi }}{3}$
Hence, option(A) is correct.

Note: Students generally gets confused after seeing the question when G. I. F. of the function is mentioned in the integral. Take care of the graph while scaling it because we need a graph of function mentioned in the question. You are required to substitute only those values in the place of G. I. F. of the given equation which has been converted into the respective greatest integers.