A) \[-\pi \]
B) \[-2\pi \]
C) \[-\frac{5\pi }{3}\]
D) \[\frac{5\pi }{3}\]
Correct Answer: C
Solution :
\[\int_{\pi }^{2\pi }{[2\sin x]dx=\int_{\pi }^{\pi +(\pi /6)}{(-1)dx+\int_{\pi +(\pi /6)}^{\pi +(\pi /2)}{\,(-2)dx}}}\] \[+\int_{\pi +(\pi /2)}^{\pi +(\pi /2)+(\pi /3)}{\,(-2)dx+\int_{\pi +(\pi /2)+(\pi /3)}^{2\pi }{\,(-1)dx}}\] \[=-\frac{\pi }{6}-2\left[ \frac{\pi }{2}-\frac{\pi }{6} \right]-2\left[ \frac{\pi }{3} \right]-1\left[ \frac{\pi }{2}-\frac{\pi }{3} \right]\] \[=-\frac{\pi }{6}-\frac{2\pi }{3}-\frac{2\pi }{3}-\frac{\pi }{6}\]\[=-\frac{\pi }{6}-\frac{8\pi }{6}-\frac{\pi }{6}=-\frac{10\pi }{6}=-\frac{5\pi }{3}\].
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