JEE Main 2026 — Area Under Curves Question with Solution

The given region is defined by the inequalities:
$x\ge 0$
$0\le y\le 6-x\Rightarrow y\ge 0$ and $x\le 6-y$
$y^2\ge 4x-3\Rightarrow x\le \frac{y^2+3}{4}$

From these inequalities, for a given $y\ge 0$, the value of $x$ ranges from $0$ to $\min \left(6-y,\frac{y^2+3}{4}\right)$.

To find the point where the two bounding curves intersect, we equate them:
$6-y=\frac{y^2+3}{4}$

$24-4y=y^2+3$

$y^2+4y-21=0$

$(y+7)(y-3)=0$

Since $y\ge 0$, the intersection occurs at $y=3$.

For $0\le y\le 3$, the right boundary is the parabola $x=\frac{y^2+3}{4}$.
For $3\le y\le 6$, the right boundary is the line $x=6-y$.

The total area $A$ can be calculated by integrating with respect to $y$:
$A={\int }_0^3\frac{y^2+3}{4}dy+{\int }_3^6(6-y)dy$

Evaluating the first integral:
${\int }_0^3\frac{y^2+3}{4}dy=\frac{1}{4}{\left[\frac{y^3}{3}+3y\right]}_0^3=\frac{1}{4}(9+9)=\frac{18}{4}=\frac{9}{2}$

Evaluating the second integral:
${\int }_3^6(6-y)dy={\left[6y-\frac{y^2}{2}\right]}_3^6=(36-18)-\left(18-\frac{9}{2}\right)=18-13.5=\frac{9}{2}$

Total Area = $\frac{9}{2}+\frac{9}{2}=9$

Answer: $9$