Let S be the region bounded by the curves y = x ^ 3 and y ^ 2 = x ....

Intersection point of ${y}^{2} = x \text{ and } y = 2\left|x\right|$ in the first quadrant will be \left(\dfrac{1}{4}, \dfrac{1}{2}\right)$</p><p>Also, intersection point of$y = {x}^{3} \text{ and } {y}^{2} = x$in first quadrant will be \left(1, 1\right)$

Area of the region $S = {∫}_{0}^{1}( \sqrt{x} - {x}^{3} )dx$ $= {\left[\dfrac{2{x}^{\dfrac{3}{2}}}{3} - \dfrac{{x}^{4}}{4}\right]}_{1}^{0}$ $= \dfrac{5}{12}$

Area of region ${R}_{1} = {∫}_{0}^{\dfrac{1}{4}}\left(\sqrt{x} - 2x\right)dx$ $= {\left[\dfrac{2{x}^{\dfrac{3}{2}}}{3} - {x}^{2}\right]}_{0}^{\dfrac{1}{4}} = \dfrac{1}{48}$

$∴ {R}_{2} = S - {R}_{1} = \dfrac{19}{48}$

So, $\dfrac{{R}_{2}}{{R}_{1}} = 19$