A) \[\pi \]
B) 0
C) 1
D) \[{{\pi }^{2}}\]
Correct Answer: A
Solution :
\[I=\int_{0}^{\pi }{x\sin xdx=\int_{0}^{\pi }{(\pi -x)\sin x\,dx}}\] Þ \[2I=\pi \int_{0}^{\pi }{\sin xdx=\pi [-\cos x]_{0}^{\pi }\Rightarrow I=\pi }\].
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