Area Under the Curve
Area between two curves
Grade 12
Question:
<p>The parabolas \(y^2 = 4x\) and \(x^2 = 4y\) divide the square region bounded by the lines \(x = 4\), \(y = 4\) and the coordinate axes. If \(S_1, S_2, S_3\) are, respectively, the areas of these parts numbered from top to bottom; then \(S_1 : S_2 : S_3\) is</p>
<p>\(1 : 2 : 1\)</p>
<p>\(1 : 2 : 3\)</p>
<p>\(2 : 1 : 2\)</p>
<p>\(1 : 1 : 1\)</p>
Step-by-Step Solution
Key Concept: The two parabolas y² = 4x and x² = 4y intersect at (0,0) and (4,4), creating three distinct regions in the 4×4 square. Finding each area requires setting up integrals: S₁ (above both curves), S₂ (between curves), and S₃ (below both curves).
<p><strong>Step 1: Identify intersection points and regions.</strong></p><p>Parabolas y² = 4x and x² = 4y intersect at (0,0) and (4,4). For 0 ≤ x ≤ 4, we have y = 2√x (from y² = 4x) and y = x²/4 (from x² = 4y), with 2√x ≥ x²/4 in this interval.</p><p><strong>Step 2: Calculate S₃ (below both curves).</strong></p><p>S₃ = ∫₀⁴ (x²/4) dx = [x³/12]₀⁴ = 64/12 = 16/3</p><p><strong>Step 3: Calculate S₁ (above both curves).</strong></p><p>S₁ = ∫₀⁴ [4 - 2√x] dx = [4x - (4/3)x^(3/2)]₀⁴ = 16 - 32/3 = 16/3</p><p><strong>Step 4: Calculate S₂ (between curves).</strong></p><p>S₂ = ∫₀⁴ [2√x - x²/4] dx = [(4/3)x^(3/2) - x³/12]₀⁴ = 32/3 - 16/3 = 16/3</p><p><strong>Step 5: Find ratio.</strong></p><p>S₁ : S₂ : S₃ = (16/3) : (16/3) : (16/3) = <strong>1 : 1 : 1</strong></p><p>∴ Answer: D</p>
Correct Answer: D