JEE Main 2025 Apr 8 Shift 2, Mathematics Q21: Area as Definite Integral

JEE Main2025Apr 8, Shift 2Mathematics

Let the area of the bounded region ${(x, y) : 0 \leq 9 x \leq y^{2}, y \geq 3 x - 6}$ be $A$ . Then $6 A$ is equal to ______

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Solution

The parabola $y^{2} = 9 x$ opens rightward, and the line $y = 3 x - 6$ meets it at points found by substituting $x = \frac{y ^{2}}{9}$ : $y = \frac{y ^{2}}{3} - 6$ , giving $y^{2} - 3 y - 18 = 0$ , so $y = 6$ or $y = - 3$ .

The bounded region is bounded above by the lower branch of the parabola $y = - 3 x$ from $x = 0$ down to the intersection, and below/right by the line, along with the constraint $x \geq 0$ .

Integrating in $x$ from $0$ to $1$ (where the lower parabola branch meets the line $y = 3 x - 6$ at $x = 1, y = - 3$ ): the region between $y = - 3 x$ (upper) and $y = 3 x - 6$ (lower) has area $\int_{0}^{1} [(- 3 x) - (3 x - 6)] d x$ .

Evaluating: $\int_{0}^{1} (- 3 x) d x = - 3 \cdot \frac{2}{3} = - 2$ , and $\int_{0}^{1} (3 x - 6) d x = \frac{3}{2} - 6 = - \frac{9}{2}$ .

So $A = - 2 - (- \frac{9}{2}) = \frac{5}{2}$ square units.

Therefore $6 A = 6 \times \frac{5}{2} = 15$ .

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JEE Main 2025 Apr 8 Shift 2, Mathematics Q21: Area as Definite Integral

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