A) \[\frac{-5\pi }{3}\]
B) \[-\pi \]
C) \[\frac{5\pi }{3}\]
D) \[-2\pi \]
Correct Answer: A
Solution :
| Let \[\begin{align} & I=\int_{\pi }^{2\pi }{\left[ 2\sin x \right]}dx \\ & \\ \end{align}\] |
| \[\pi \le x<7\pi /6\Rightarrow -1\le 2\sin x<0\] |
| \[\begin{align} & \Rightarrow \left[ 2\sin x \right]=-1 \\ & 7\pi /6\le x<11\pi /6\Rightarrow -2\le 2\sin x<-1 \\ \end{align}\] |
| \[\Rightarrow \left[ 2\sin x \right]=-1\] |
| \[\therefore I=\int\limits_{\pi }^{7\pi /6}{-dx+\int\limits_{7\pi /6}^{11\pi /6}{-2dx+\int\limits_{11\pi /6}^{2\pi }{-1dx}}}\] |
| =\[\left( -\frac{7\pi }{6}+\pi \right)+2\left( \frac{11\pi }{6}+\frac{7\pi }{6} \right)+\left( -2\pi +\frac{11\pi }{6} \right)\] |
| =\[-\frac{\pi }{6}-\frac{8\pi }{6}-\frac{\pi }{6}=-\frac{10\pi }{6}=\frac{-5\pi }{3}\] |
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