JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 29

Region S. Between $y=x^3$ (below) and $y=\sqrt x$ (above) for $0\le x\le1$: area $\frac23-\frac14=\frac5{12}$.

The cut. For $x\ge0$, $y=2x$ meets $y=\sqrt x$ at $x=\frac14$ and stays above $y=x^3$ on $[0,1]$ (they meet only at $x=\sqrt2$). So it cuts off the piece between $y=2x$ and $y=\sqrt x$ on $[0,\frac14]$:
$$R_1=\int_0^{1/4}(\sqrt x-2x)\,dx=\frac1{12}-\frac1{16}=\frac1{48}.$$

Ratio. $R_2=\frac5{12}-\frac1{48}=\frac{19}{48}$, so $\frac{R_2}{R_1}=19$.

Answer: 19