A) \[0\]
B) \[1\]
C) \[-1\]
D) \[2\]
Correct Answer: B
Solution :
\[\int_{0}^{\pi /2}{x\,\sin \,x\,dx=[x(-\cos x)]_{0}^{\pi /2}}\] \[-\int_{0}^{\pi /2}{1\,(-\cos \,x)dx}\] \[=\left[ -\frac{\pi }{2}\,\cos \frac{\pi }{2}-0 \right]+[\operatorname{six}\,x]_{0}^{\pi /2}\] \[=0+1-0=1\]You need to login to perform this action.
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