Area Under the Curve
Area between curves
Grade 12
Question:
<p>Area of region <br> \(\{x \in R; x \geq 0, y \geq 0, y \geq x - 2 \text{ and } y \leq \sqrt{x}\}\) is equal to:</p>
<p>\(\dfrac{8}{3}\) sq. units</p>
<p>\(\dfrac{10}{3}\) sq. units</p>
<p>\(\dfrac{10}{3}\) sq. units</p>
<p>\(\dfrac{16}{3}\) sq. units</p>
Step-by-Step Solution
Key Concept: The region is bounded by y = √x (upper curve) and y = x - 2 (lower line) in the first quadrant. Find intersection points to determine integration limits, then integrate the difference of functions.
<p><strong>Step 1:</strong> Identify the region. We need x ≥ 0, y ≥ 0, y ≥ x - 2, and y ≤ √x.</p><p><strong>Step 2:</strong> Find intersection of y = √x and y = x - 2: √x = x - 2. Squaring: x = x² - 4x + 4, so x² - 5x + 4 = 0, giving (x - 1)(x - 4) = 0. Thus x = 1 or x = 4. Since √x = x - 2 requires x ≥ 2, only x = 4 is valid (where y = 2).</p><p><strong>Step 3:</strong> For 0 ≤ x ≤ 2: y ranges from 0 to √x (since x - 2 ≤ 0 here, constraint y ≥ x - 2 is redundant with y ≥ 0).</p><p><strong>Step 4:</strong> For 2 ≤ x ≤ 4: y ranges from (x - 2) to √x.</p><p><strong>Step 5:</strong> Calculate area:</p><p>A = ∫₀² √x dx + ∫₂⁴ (√x - (x - 2)) dx</p><p>= [⅔x^(3/2)]₀² + [⅔x^(3/2) - x²/2 + 2x]₂⁴</p><p>= ⅔(2√2) + [(⅔·8 - 8 + 8) - (⅔·2√2 - 2 + 4)]</p><p>= 4√2/3 + [16/3 - 4√2/3 - 2]</p><p>= 16/3 - 2 = 10/3</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B