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Q. The value of $\frac{6}{\pi} \int\limits_{\pi / 6}^{\pi / 3}[2 \sin x] d x$, where [.] represents greatest integer function is

Integrals

Solution:

$\frac{\pi}{6} < x < \frac{\pi}{3} \Rightarrow \frac{1}{2} < \sin x < \frac{\sqrt{3}}{2} $
$\Rightarrow 1 < 2 \sin x < \sqrt{3}$
$ \therefore {[2 \sin x]=1}$