A) \[\frac{1}{2}\]
B) \[-2\]
C) \[1\]
D) \[-1\]
E) \[2\]
Correct Answer: B
Solution :
\[\int_{0}^{\pi }{|\cos x|}\,dx=\int_{0}^{\pi /2}{\cos x\,dx}-\int_{\pi /2}^{\pi }{\cos xdx}\] \[=[-\sin x]_{0}^{\pi /2}-[-\sin x]_{\pi /2}^{\pi }\] \[=-\sin \frac{\pi }{2}+\sin 0+\sin \pi -\sin \frac{\pi }{2}\] \[=-2\sin \frac{\pi }{2}=-2\]
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