Question

Area bounded by the curve $y=\sqrt{(\sin \left[x\right]+\left[\sin x\right])},$ where $[.]$ denotes the greatest integer function, lines $x=1$ and $x=\frac{\pi }{2}$ and the $x-$axis is

• A. $(\frac{\pi }{2}-1)$ sq.unit
• B. $\sqrt{\sin 1}(\frac{\pi }{2}-1)$ sq.unit
• C. $\sqrt{\cos 1}(\frac{\pi }{2}-1)$ sq.unit
• D. $\sqrt{\frac{\pi }{2}}(\frac{\pi }{2}-1)$ sq.unit

Answer 1

The correct option is B $\sqrt{\sin 1}(\frac{\pi }{2}-1)$ sq.unit

$\because 1\le x<\frac{\pi }{2}$
$\therefore \left[x\right]=1$ and $\sin 1\le \sin x<1$
and $\left[\sin x\right]=0$
$\therefore$ Required Area$={\int }_1^{\frac{\pi }{2}}y.dx$
$={\int }_1^{\frac{\pi }{2}}\sqrt{sin1}dx$
$=\sqrt{\sin 1}{\left[x\right]}_1^{\frac{\pi }{2}}$
$=\sqrt{\sin 1}(\frac{\pi }{2}-1)$.sq.unit