_ ,0 ^ , /2 x - [ x] ,dx is equal to - Vidyadip

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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