Question
$\int_{\,0}^{\,\pi /2} {\{ x - [\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} ]$.
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$ f(x)=\left\{\begin{array}{ll} \frac{\sin (p+1) x+\sin x}{x}, & x<0 \\ q & , x=0 \\ \frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{3 / 2}} & , x>0 \end{array}\right.$
$R$ में $x$ के सभी मानों के लिए सतत् है: