A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\pi \]
D) 1
E) 0
Correct Answer: D
Solution :
\[\int_{0}^{\pi /2}{x\sin xdx=[-x\cos x]_{0}^{\pi /2}\cos xdx}\] \[=0+[\sin x]_{0}^{\pi /2}\] \[=1\]
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