The area bounded by the curves $y = \sqrt{x}, 2y - x + 3 = 0$, and the $x$-axis in the first quadrant is:

1. Identify the boundaries: Curve 1: $y = \sqrt{x}$ (or $x = y^2$). Curve 2: $x = 2y + 3$. Boundary 3: $y = 0$ ($x$-axis). 2. Find intersection points: Solve $x = y^2$ and $x = 2y + 3$. $y^2 = 2y + 3 \implies y^2 - 2y - 3 = 0 \implies (y-3)(y+1) = 0$. Since we are in the first quadrant ($y>0$), $y = 3$. Intersection is at $y=3, x=9$. Point $(9,3)$. 3. Set up the integral with respect to $y$ (Horizontal Strips): This avoids splitting the area. $y$ goes from 0 to 3. Right boundary: line $x = 2y + 3$. Left boundary: curve $x = y^2$. Area $= \int_0^3 (x_{right} - x_{left}) dy = \int_0^3 ((2y + 3) - y^2) dy$. 4. Evaluate: $= [y^2 + 3y - y^3/3]_0^3$ $= (9 + 9 - 9) - 0 = 9$.

Area Under Curves

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The area bounded by the curves $y = x, 2 y - x + 3 = 0$ , and the $x$ -axis in the first quadrant is:

Region bounded by curve and line

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$9$

$3$

$6$

$9/2$

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About This Question

Subject: mathematics
Chapter: integral calculus
Topic: area under curves
Difficulty: Medium
Year: 2025

Solution

Correct Answer:

$9$

Identify the boundaries: Curve 1: $y = x$ (or $x = y^{2}$ ). Curve 2: $x = 2 y + 3$ . Boundary 3: $y = 0$ ( $x$ -axis).
Find intersection points: Solve $x = y^{2}$ and $x = 2 y + 3$ . $y^{2} = 2 y + 3 ⟹ y^{2} - 2 y - 3 = 0 ⟹ (y - 3) (y + 1) = 0$ . Since we are in the first quadrant ( $y > 0$ ), $y = 3$ . Intersection is at $y = 3, x = 9$ . Point $(9, 3)$ .
Set up the integral with respect to $y$ (Horizontal Strips): This avoids splitting the area. $y$ goes from 0 to 3. Right boundary: line $x = 2 y + 3$ . Left boundary: curve $x = y^{2}$ . Area $= \int_{0}^{3} (x_{r i g h t} - x_{l e f t}) d y = \int_{0}^{3} ((2 y + 3) - y^{2}) d y$ .
Evaluate: $= [y^{2} + 3 y - y^{3} /3]_{0}^{3}$ $= (9 + 9 - 9) - 0 = 9$ .

This medium difficulty mathematics question is from the chapter integral calculus, covering the topic of area under curves. It appeared in the 2025 exam.

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Area Under Curves

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Area Under Curves

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Solution

The area bounded by the parabolas $y^{2} = 4 x$ and $x^{2} = 4 y$ is:

Area bounded by the curves $y = ln x, y = 0$ , and $x = e$ is: