JEE Main 2026 — 6 April, Morning Shift — Question 39

The region is bounded by $y = 6 - x$, $y^2 = 4x - 3$, and $x \geq 0$. Solve for intersection: $y = 6 - x \Rightarrow x = 6 - y$. Substitute into $y^2 = 4x - 3$: $y^2 = 4(6 - y) - 3 = 24 - 4y - 3 = 21 - 4y$. Thus $y^2 + 4y - 21 = 0 \Rightarrow (y + 7)(y - 3) = 0 \Rightarrow y = 3$ (since $y \geq 0$). For $y = 0$, from $y^2 = 4x - 3$ we get $x = \frac{3}{4}$. The line $y = 6 - x$ gives $x = 6$ when $y = 0$. The region is bounded on the left by $x = \frac{y^2 + 3}{4}$ and on the right by $x = 6 - y$, for $y$ from 0 to 3. Area $A = \int_{0}^{3} \left[ (6 - y) - \frac{y^2 + 3}{4} \right] dy = \int_{0}^{3} \left( 6 - y - \frac{y^2}{4} - \frac{3}{4} \right) dy = \int_{0}^{3} \left( \frac{21}{4} - y - \frac{y^2}{4} \right) dy$. Evaluate: $\left[ \frac{21}{4}y - \frac{y^2}{2} - \frac{y^3}{12} \right]_{0}^{3} = \frac{63}{4} - \frac{9}{2} - \frac{27}{12} = \frac{63}{4} - \frac{18}{4} - \frac{9}{4} = \frac{36}{4} = 9$.