Mathematics - Area Under Curves Question with Solution | TestHub

Intersection point of $y^2=x \& y=2\left|x\right|$ in the first quadrant will be $\left(\frac{1}{4},\frac{1}{2}\right)$

Also, intersection point of $y=x^3 \& y^2=x$ in first quadrant will be $\left(1,1\right)$

Area of the region $\mathrm{S}={\int }_0^1(\sqrt{x}-x^3)dx$ $={\left[\frac{2x^{{\displaystyle \frac{3}{2}}}}{3}-\frac{x^4}{4}\right]}_1^0$ $=\frac{5}{12}$

Area of region $R_1={\int }_0^{\frac{1}{4}}\left(\sqrt{x}-2x\right)dx$ $={\left[\frac{2x^{{\displaystyle \frac{3}{2}}}}{3}-x^2\right]}_0^{\frac{1}{4}}=\frac{1}{48}$

$\therefore R_2=S-R_1=\frac{19}{48}$

So, $\frac{R_2}{R_1}=19$