<p>The area (in sq. units) of the region \(\{x \in \mathbb{R} : x \geq 0,\ y \geq 0,\ y \geq x - 2 \text{ and } y \leq \sqrt{x}\}\) is:</p>
Step-by-Step Solution
Key Concept: Identify the bounded region by finding intersection points of y = √x and y = x - 2, then determine which curve forms the upper/lower boundary in different x-intervals. The constraint y ≥ x - 2 is automatically satisfied for x ∈ [0,1] since √x ≥ x - 2 in this range.
<p><strong>Step 1:</strong> Find intersection of y = √x and y = x - 2.</p><p>√x = x - 2 ⟹ x = (x-2)² ⟹ x = x² - 4x + 4 ⟹ x² - 5x + 4 = 0 ⟹ (x-1)(x-4) = 0</p><p>So x = 1 or x = 4. At x = 1: y = 1 - 2 = -1 (invalid since y ≥ 0). At x = 4: y = 2. Intersection point: (4, 2).</p><p><strong>Step 2:</strong> Determine the region boundaries. For x ∈ [0, 1]: y = x - 2 < 0, so the lower bound is y = 0 (from y ≥ 0). For x ∈ [1, 4]: y = x - 2 ≥ 0, so the lower bound is y = x - 2.</p><p><strong>Step 3:</strong> Set up the integral:</p><p>Area = ∫₀¹ √x dx + ∫₁⁴ (√x - (x-2)) dx</p><p>= ∫₀¹ √x dx + ∫₁⁴ (√x - x + 2) dx</p><p><strong>Step 4:</strong> Evaluate:</p><p>∫₀¹ √x dx = [⅔x^(3/2)]₀¹ = ⅔</p><p>∫₁⁴ (√x - x + 2) dx = [⅔x^(3/2) - x²/2 + 2x]₁⁴</p><p>= (⅔·8 - 8 + 8) - (⅔ - ½ + 2) = 16/3 - (⅔ + 3/2) = 16/3 - 13/6 = 32/6 - 13/6 = 19/6</p><p>Total Area = ⅔ + 19/6 = 4/6 + 19/6 = 23/6 sq. units</p><p>∴ Answer: C</p>
Correct Answer: C