MCQ
$\int_{\,0}^{\,\pi /2} {\{ x - [\sin x]\} \,dx} $ is equal to
  • $\frac{{{\pi ^2}}}{8}$
  • B
    $\frac{{{\pi ^2}}}{8} - 1$
  • C
    $\frac{{{\pi ^2}}}{8} - 2$
  • D
    None of these

Answer

Correct option: A.
$\frac{{{\pi ^2}}}{8}$
a
(a) $\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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