JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Area Under Curves

The inequality $|x-y|\le y$ means $-y\le x-y\le y$, i.e. $x\ge 0$ and $x\le 2y$, so $y\ge \frac{x}{2}$. Together with $y\le 4\sqrt{x}$, the region lies between the line $y=\frac{x}{2}$ below and the curve $y=4\sqrt{x}$ above for $x\ge 0$. These meet where $4\sqrt{x}=\frac{x}{2}$, giving $x=64$. The area is ${\int }_0^{64}\left(4\sqrt{x}-\frac{x}{2}\right)dx={\left[\frac{8x^{3/2}}{3}-\frac{x^2}{4}\right]}_0^{64}=\frac{8\cdot 512}{3}-1024=\frac{4096}{3}-\frac{3072}{3}$, so the area of the region is $\frac{1024}{3}$.