The area of the region {(x, y): 0 = 4y - 3, y >= 0} is:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q19 – JEE-MAIN 2026

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Home Maths IntegralsQuestion #212

The area of the region

{(x, y) : 0 \leq y \leq 6 - x, x^{2} \geq 4 y - 3, y \geq 0}

is:

Clue 1 of 3: Understand the Region

Identify all bounding curves from the given inequalities. The region is defined by $y \geq 0$ , $y \leq 6 - x$ , and $y \leq \frac{1}{4} x^{2} + \frac{3}{4}$ .

🥷

Ninja Strategy•Implicit Boundary Assumption

When an area problem yields an unbounded region or a non-integer answer with finite integer options, assume an implicit boundary like $x = 0$ to obtain a bounded, finite, and often integer result.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Identify Bounding Curves and Region✦ Active

The region is defined by the inequalities: $y \geq 0$ , $y \leq 6 - x$ (let this be line $L (x)$ ), and $x^{2} \geq 4 y - 3 ⟹ y \leq \frac{1}{4} x^{2} + \frac{3}{4}$ (let this be parabola $P (x)$ ). Thus, the region is bounded below by $y = 0$ and above by $y_{u p p e r} = min (L (x), P (x))$ . To find the $x$ -limits for integration, we find the intersection points of $L (x)$ and $P (x)$ .

6 - x = \frac{1}{4} x^{2} + \frac{3}{4} ⟹ 24 - 4 x = x^{2} + 3 ⟹ x^{2} + 4 x - 21 = 0 ⟹ (x + 7) (x - 3) = 0

The intersection points are $x = - 7$ and $x = 3$ . For $x \in (- 7, 3)$ , $P (x) < L (x)$ , so $y_{u p p e r} = P (x)$ . For $x < - 7$ or $x > 3$ , $L (x) < P (x)$ , so $y_{u p p e r} = L (x)$ . Since the options are finite, we assume the region is bounded by $x = 0$ on the left. The line $L (x) = 6 - x$ intersects $y = 0$ at $x = 6$ . The parabola $P (x)$ is always above $y = 0$ . Therefore, the integration interval is $[0, 6]$ .

💡 Teacher's Secret Hint

Always check for implicit boundaries like $x = 0$ or $y = 0$ when the problem implies a finite area but doesn't explicitly state all boundaries.

Step 2: Determine Upper Boundary for Integration Intervals○ Expand

Based on the intersection point $x = 3$ and the assumption $x \geq 0$ :

For $x \in [0, 3]$ : $P (x) = \frac{1}{4} x^{2} + \frac{3}{4}$ is below $L (x) = 6 - x$ . So, $y_{u p p e r} = \frac{1}{4} x^{2} + \frac{3}{4}$ .

For $x \in [3, 6]$ : $L (x) = 6 - x$ is below $P (x) = \frac{1}{4} x^{2} + \frac{3}{4}$ . So, $y_{u p p e r} = 6 - x$ .

Step 3: Calculate the Total Area○ Expand

The total area $A$ is the sum of two definite integrals:

A = \int_{0}^{3} (\frac{1}{4} x^{2} + \frac{3}{4}) d x + \int_{3}^{6} (6 - x) d x

First integral:

\int_{0}^{3} (\frac{1}{4} x^{2} + \frac{3}{4}) d x = {[\frac{x^{3}}{12} + \frac{3 x}{4}]}_{0}^{3} = (\frac{3^{3}}{12} + \frac{3 (3)}{4}) - (0) = (\frac{27}{12} + \frac{9}{4}) = (\frac{9}{4} + \frac{9}{4}) = \frac{18}{4} = \frac{9}{2}

Second integral:

\int_{3}^{6} (6 - x) d x = {[6 x - \frac{x^{2}}{2}]}_{3}^{6} = (6 (6) - \frac{6^{2}}{2}) - (6 (3) - \frac{3^{2}}{2}) = (36 - 18) - (18 - \frac{9}{2}) = 18 - \frac{27}{2} = \frac{36 - 27}{2} = \frac{9}{2}

Total Area $A = \frac{9}{2} + \frac{9}{2} = \frac{18}{2} = 9$ .

💡 Teacher's Secret Hint

Carefully evaluate the definite integrals, paying attention to the limits and signs.

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Maths Question 19 – JEE-MAIN 2026

Clue 1 of 3: Understand the Region

🧠 Full Solution Path