JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Area Under Curves

The region requires $y^2\le 4x$ with $0\le x<4$ (so $-2\sqrt{x}\le y\le 2\sqrt{x}$), together with


For $y>0$, this reduces to $g(x)=\frac{x(x-1)(x-2)}{(x-3)(x-4)}>0$; for $y<0$ it reduces to $g(x)<0$ (the extra factor $y$ flips the sign).

Sign of $g(x)$ on $(0,4)$, using that $(x-4)<0$ throughout this range:

• $x\in (0,1)$: $(+)(-)(-)/[(-)(-)]=(+)/(+)>0$
• $x\in (1,2)$: $(+)(+)(-)/[(-)(-)]=(-)/(+)<0$
• $x\in (2,3)$: $(+)(+)(+)/[(-)(-)]=(+)/(+)>0$
• $x\in (3,4)$: $(+)(+)(+)/[(+)(-)]=(+)/(-)<0$

So: for $x\in (0,1)\cup (2,3)$ (where $g>0$), only $y>0$ is allowed, giving strip height $2\sqrt{x}-0=2\sqrt{x}$; for $x\in (1,2)\cup (3,4)$ (where $g<0$), only $y<0$ is allowed, giving strip height $0-(-2\sqrt{x})=2\sqrt{x}$.

Either way, every $x\in (0,4)$ contributes a strip of height exactly $2\sqrt{x}$ (just on one side of the axis or the other), so the total area is simply