A) \[\frac{{{\pi }^{2}}}{8}\]
B) \[\frac{{{\pi }^{2}}}{8}-1\]
C) \[\frac{{{\pi }^{2}}}{8}-2\]
D) None of these
Correct Answer: A
Solution :
\[\int_{0}^{\pi /2}{\{x-[\sin x]\}\,dx=\int_{0}^{\pi /2}{xdx-\int_{\,0}^{\,\pi /2}{[\sin x]\,\,dx}}}\] \[=\left( \frac{{{x}^{2}}}{2} \right)_{0}^{\pi /2}\]\[=\frac{{{\pi }^{2}}}{8}\], \[[\because \int_{\,0}^{\,\pi /2}{[\sin x]\,dx=0}]\].
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