Question

Find the area enclosed between the curves $y^2-2y{e}^{si{n}^{-1}x}+x^2-1+\left[x\right]+e^{2si{n}^{-1}x}=0$

and line x = 0 and $x=\frac{1}{2}$ is (where [.] denotes greatest integer function)

• A. $\frac{\sqrt{3}}{4}+\frac{\pi }{6}$
• B. $\frac{\sqrt{3}}{2}+\frac{\pi }{6}$
• C. $\frac{\sqrt{3}}{4}-\frac{\pi }{6}$
• D. $\frac{\sqrt{3}}{2}-\frac{\pi }{6}$

Answer 1

Answer: (A)

$\frac{\sqrt{3}}{4}+\frac{\pi }{6}$

$(y-e^{si{n}^{-}1x})+x^2+\left[x\right]-1=0$
$\therefore x\in (0,1/2)$;[x] =0
$\Rightarrow y-e^{si{n}^{-}1x}=\pm \sqrt{1-x^2}$
$\Rightarrow y=e^{si{n}^{-}1x}\pm \sqrt{1-x^2}$
$\Rightarrow \stackrel{\frac{1}{2}}{\underset{0}{\int }}ydx=\stackrel{\frac{1}{2}}{\underset{0}{\int }}|e^{si{n}^{-}1x}+\sqrt{1+x^2}-(e^{si{n}^{-}1x}-\sqrt{1-x^2})|dx$
$\Rightarrow A=2\stackrel{\frac{1}{2}}{\underset{0}{\int }}\sqrt{1-x^2}dx$ ; Let $x=sin\theta$ so $dx=cos\theta d\theta$
$\Rightarrow A=\stackrel{\frac{\pi }{6}}{\underset{0}{\int }}2co{s}^2\theta d\theta ={[\theta +\frac{sin2\theta }{2}]}_0^{\pi /6}$
$\Rightarrow A=\pi /6+\frac{\sqrt{3}}{4}$