Mathematics - Area Under Curves Question with Solution | TestHub

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