JEE Main 2025 — 7 April, Evening Shift — Question 26

Finding Intersection Points First, we find the intersection of the parabola $y = 1+x^2$ with the boundary lines.

Intersection of $y = 1+x^2$ and $y = x+7$:

x^2 + 1 &= x + 7 \\ x^2 - x - 6 &= 0 \\ (x-3)(x+2) &= 0 \end{aligned}$$ Points: $x = -2$ and $x = 3$. Intersection of $y = 1+x^2$ and $y = 11-3x$: $$\begin{aligned} x^2 + 1 &= 11 - 3x \\ x^2 + 3x - 10 &= 0 \\ (x+5)(x-2) &= 0 \end{aligned}$$ Points: $x = -5$ and $x = 2$. Intersection of the two lines: $$\begin{aligned} x + 7 &= 11 - 3x \\ 4x &= 4 \\ x &= 1 \end{aligned}$$ Setting up the Definite Integral The upper boundary is defined by $y = x+7$ for $x \in [-2, 1]$ and $y = 11-3x$ for $x \in [1, 2]$. The lower boundary is the parabola $y = 1+x^2$.

A = \int_{-2}^{1} [(x+7) - (1+x^2)] , dx + \int_{1}^{2} [(11-3x) - (1+x^2)] , dx

Evaluating the Area Part 1 ($A_1$): $$\begin{aligned} A_1 &= \int_{-2}^{1} (6 + x - x^2) \, dx \\ &= \left[ 6x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{1} \\ &= \left( 6 + \frac{1}{2} - \frac{1}{3} \right) - \left( -12 + 2 + \frac{8}{3} \right) \\ &= \frac{37}{6} - \left( -\frac{22}{3} \right) = \frac{81}{6} = \frac{27}{2} \end{aligned}$$ Part 2 ($A_2$): $$\begin{aligned} A_2 &= \int_{1}^{2} (10 - 3x - x^2) \, dx \\ &= \left[ 10x - \frac{3x^2}{2} - \frac{x^3}{3} \right]_{1}^{2} \\ &= \left( 20 - 6 - \frac{8}{3} \right) - \left( 10 - \frac{3}{2} - \frac{1}{3} \right) \\ &= \frac{34}{3} - \frac{49}{6} = \frac{19}{6} \end{aligned}$$ Final Calculation

A = A_1 + A_2 = \frac{27}{2} + \frac{19}{6} = \frac{81 + 19}{6} = \frac{100}{6} = \frac{50}{3}

$$Therefore,$$

3A = 3 \times \frac{50}{3} = 50

$$$$