<p>The area (in square units) bounded by the curves \(y = \sqrt{x}\), \(2y - x + 3 = 0\), \(x\)-axis, and lying in the first quadrant is</p>
Step-by-Step Solution
Key Concept: Find intersection points of y = √x and the line 2y - x + 3 = 0, then integrate by breaking the region into parts bounded by different curves. The line intersects the x-axis at x = 3, creating a composite region.
<p><strong>Step 1:</strong> Find intersections. For y = √x and 2y - x + 3 = 0:</p><p>Substituting y = √x into 2y - x + 3 = 0:</p><p>2√x - x + 3 = 0 → 2√x = x - 3</p><p>Squaring: 4x = (x-3)² = x² - 6x + 9</p><p>x² - 10x + 9 = 0 → (x-1)(x-9) = 0</p><p>x = 1 (valid, since 2(1) - 1 + 3 = 4 > 0) or x = 9 (valid: 2(3) - 9 + 3 = 0)</p><p>Intersection points: (1, 1) and (9, 3)</p><p><strong>Step 2:</strong> The line 2y - x + 3 = 0 gives y = (x-3)/2, crossing x-axis at x = 3.</p><p>In first quadrant region:</p><ul><li>From x = 0 to x = 1: Area under y = √x</li><li>From x = 1 to x = 3: Area under y = √x</li><li>From x = 3 to x = 9: Area under the line y = (x-3)/2</li></ul><p><strong>Step 3:</strong> Calculate area:</p><p>A₁ = ∫₀¹ √x dx = [⅔x^(3/2)]₀¹ = ⅔</p><p>A₂ = ∫₁³ √x dx = [⅔x^(3/2)]₁³ = ⅔(3√3 - 1)</p><p>A₃ = ∫₃⁹ (x-3)/2 dx = [1/4(x-3)²]₃⁹ = 1/4(36) = 9</p><p><strong>Step 4:</strong> Total area = ⅔ + ⅔(3√3 - 1) + 9 = ⅔(1 + 3√3 - 1) + 9 = 2√3 + 9</p><p>∴ Answer: <strong>D (2√3 + 9 square units)</strong></p>
Correct Answer: D