JEE Main 2026 — Mathematics Calculus

The given region is defined by the inequalities:

$$x \geq 0$$x≥0

$0 \leq y \leq 6 - x \Rightarrow y \geq 0$0≤y≤6−x⇒y≥0 and $x \leq 6 - y$x≤6−y

$$y^2 \geq 4x - 3 \Rightarrow x \leq \dfrac{y^2 + 3}{4}$$y2≥4x−3⇒x≤4y2+3​

From these inequalities, for a given $y \geq 0$y≥0, the value of $x$x ranges from $0$0 to $\min\left(6 - y, \dfrac{y^2 + 3}{4}\right)$min(6−y,4y2+3​).

To find the point where the two bounding curves intersect, we equate them:

$$6 - y = \dfrac{y^2 + 3}{4}$$6−y=4y2+3​

$$24 - 4y = y^2 + 3$$24−4y=y2+3

$$y^2 + 4y - 21 = 0$$y2+4y−21=0

$$(y + 7)(y - 3) = 0$$(y+7)(y−3)=0

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

For $0 \leq y \leq 3$0≤y≤3, the right boundary is the parabola $x = \dfrac{y^2 + 3}{4}$x=4y2+3​.

For $3 \leq y \leq 6$3≤y≤6, the right boundary is the line $x = 6 - y$x=6−y.

The total area $A$A can be calculated by integrating with respect to $y$y:

$$A = \int_{0}^{3} \dfrac{y^2 + 3}{4} dy + \int_{3}^{6} (6 - y) dy$$A=∫03​4y2+3​dy+∫36​(6−y)dy

Evaluating the first integral:

$$\int_{0}^{3} \dfrac{y^2 + 3}{4} dy = \dfrac{1}{4} \left[ \dfrac{y^3}{3} + 3y \right]_{0}^{3} = \dfrac{1}{4} (9 + 9) = \dfrac{18}{4} = \dfrac{9}{2}$$∫03​4y2+3​dy=41​[3y3​+3y]03​=41​(9+9)=418​=29​

Evaluating the second integral:

$$\int_{3}^{6} (6 - y) dy = \left[ 6y - \dfrac{y^2}{2} \right]_{3}^{6} = (36 - 18) - \left(18 - \dfrac{9}{2}\right) = 18 - 13.5 = \dfrac{9}{2}$$∫36​(6−y)dy=[6y−2y2​]36​=(36−18)−(18−29​)=18−13.5=29​

Total Area = $\dfrac{9}{2} + \dfrac{9}{2} = 9$29​+29​=9

Answer: $9$9