Question

The area bounded by the curve $(y-{\text{sin}}^{-1}x{)}^2=x-x^2$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is A

$\frac{\pi }{4}$

$(y-{\text{sin}}^{-1}x{)}^2=x-x^2$

$y-{\text{sin}}^{-1}x=\pm \sqrt{x-x^2}$

$y={\text{sin}}^{-1}x\pm \sqrt{x-x^2}$

The given curves are $y={\text{sin}}^{-1}x+\left(\sqrt{x-x^2}\right)$ and $y={\text{sin}}^{-1}x-\left(\sqrt{x-x^2}\right)$

Domain of the curve is $(0,1)$

${\sin }^{-1}x$ and $\sqrt{x-x^2}$ are positive in $(0,1)$

$\therefore y_2={\sin }^{-1}x+\sqrt{x-x^2}\ge y_1={\sin }^{-1}x-\sqrt{x-x^2}$

Required area is ${\int }_0^1(y_2-y_1)dx$

$={\int }_0^{1}2\sqrt{x-x^2}dx$

$=2{\int }_0^1\sqrt{{\left(\frac{1}{2}\right)}^2-{(x-\frac{1}{2})}^2}dx$

$=2{[\left(\frac{x-\frac{1}{2}}{2}\right)\sqrt{{\left(\frac{1}{2}\right)}^2-{(x-\frac{1}{2})}^2}+\left(\frac{{\left(\frac{1}{2}\right)}^2}{2}\right){\sin }^{-1}\left(\frac{x-\frac{1}{2}}{\frac{1}{2}}\right)]}_0^1$

$=\frac{\pi }{4}$