JEE Main 2026 Apr 6 Shift 1, Mathematics Q19: Area as Definite Integral

The boundary curves are the line $x=6-y$ and the parabola $x=\frac{y^2+3}{4}$ (from $y^2=4x-3$). They meet where $6-y=\frac{y^2+3}{4}$, i.e. $y^2+4y-21=0$, giving $y=3$ (rejecting $y=-7$).

For $0\le y\le 3$, the region runs from $x=0$ on the left to $x=\min (6-y,(y^2+3)/4)$, but the constraint $y^2\ge 4x-3$ means $x\le (y^2+3)/4$, while $x\le 6-y$ from the line. On $[0,3]$, $(y^2+3)/4\le 6-y$, so the right boundary is the parabola.

Wait — the region requires $y\le 6-x$ (i.e. $x\le 6-y$) AND $y^2\ge 4x-3$ (i.e. $x\le (y^2+3)/4$). So the right boundary at each $y$ is the smaller of the two, which is the parabola on $[0,3]$ and the line on $[3,6]$.

Hence

$A={\int }_0^3\frac{y^2+3}{4}dy+{\int }_3^6(6-y)dy={\left[\frac{y^3}{12}+\frac{3y}{4}\right]}_0^3+{\left[6y-\frac{y^2}{2}\right]}_3^6$

$=\left(\frac{27}{12}+\frac{9}{4}\right)+\left(18-\frac{27}{2}\right)=\frac{9}{2}+\frac{9}{2}=9$.