JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Area Under Curves

The parabola $y^2=8x$ and the line $y=\sqrt{2}x$ meet where $2x^2=8x$, that is at $x=0$ and $x=4$, and between these the parabola lies above the line, so the region they enclose has area ${\int }_0^4(2\sqrt{2}\sqrt{x}-\sqrt{2}x)dx=\frac{32\sqrt{2}}{3}-8\sqrt{2}=\frac{8\sqrt{2}}{3}$. The triangle cut off by $y=\sqrt{2}x$, $x=1$ and $y=2\sqrt{2}$ has vertices $(1,\sqrt{2})$, $(1,2\sqrt{2})$ and $(2,2\sqrt{2})$; it is right-angled with legs $\sqrt{2}$ and $1$, so its area is $\frac{\sqrt{2}}{2}$. Since this triangle sits entirely inside the region between the curves, the part lying outside it is $\frac{8\sqrt{2}}{3}-\frac{\sqrt{2}}{2}=\frac{16\sqrt{2}-3\sqrt{2}}{6}=\frac{13\sqrt{2}}{6}$.