A) \[\frac{\pi }{2}\]
B) \[-\frac{\pi }{2}\]
C) \[0\]
D) \[\pi \]
Correct Answer: B
Solution :
\[\int_{\pi /2}^{3\pi /2}{[\sin x]}dx=\int_{\pi /2}^{\pi }{[\sin x]}\,\,dx+\] \[\int_{\pi }^{3\pi /2}{[\sin x]}\,\,dx\] \[=\int_{\pi /2}^{\pi }{0\,\,dx}+\int_{\pi }^{3\pi /2}{(-1)\,\,dx}=-[x]_{x}^{3\pi /2}\] \[=-\left( \frac{3\pi }{2}-\pi \right)=-\frac{\pi }{2}\]
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