Question

The area of the region, bounded by the curves $y={\sin }^{-1}x+x(1-x)$ and $y={\sin }^{-1}x-x(1-x)$ in the first quadrant, is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{1}{3}$

${\sin }^{-1}x$ is defined, if $-1\le x\le 1$
In first quadrant $0\le x\le 1$ and $x(1-x)\ge 0$
$\therefore y={\sin }^{-1}x+x(1-x)$ ...... (i)
Lies above $y={\sin }^{-1}x-x(1-x)$ ...... (ii)
On solving, we get
$2x(1-x)=0$
$\Rightarrow x=0,1$
$\therefore$ Required area $={\int }_0^1(y_1-y_2)dx$
$={\int }_0^1[\{{\sin }^{-1}x+x(1-x\left)\right\}-\{{\sin }^{-1}x-x(1-x\left)\right\}]dx$
$=2{\int }_0^1(x-x^2)dx$
$=2{[\frac{x^2}{2}-\frac{x^3}{3}]}_0^1=2(\frac{1}{2}-\frac{1}{3})=\frac{1}{3}$