Area Under the Curve
Area between parabola and line
Grade 12
Question:
<p>The area (in sq. units) of the region described by \(\{(x, y) : y^2 \leq 2x\) and \(y \geq 4x - 1\}\) is</p>
<p>\(\dfrac{5}{64}\)</p>
<p>\(\dfrac{15}{64}\)</p>
<p>\(\dfrac{9}{32}\)</p>
<p>\(\dfrac{7}{32}\)</p>
Step-by-Step Solution
Key Concept: Find intersection points of the parabola y² = 2x and line y = 4x - 1, then integrate to find the enclosed area using the appropriate variable of integration.
<p><strong>Step 1:</strong> Find intersection points by substituting y = 4x - 1 into y² = 2x.</p><p>(4x - 1)² = 2x</p><p>16x² - 8x + 1 = 2x</p><p>16x² - 10x + 1 = 0</p><p>Using quadratic formula: x = (10 ± √(100 - 64))/32 = (10 ± 6)/32</p><p>So x = 1/2 or x = 1/8</p><p>At x = 1/2: y = 4(1/2) - 1 = 1</p><p>At x = 1/8: y = 4(1/8) - 1 = -1/2</p><p><strong>Step 2:</strong> Integrate with respect to y (more convenient). From y² = 2x, we get x = y²/2. From y = 4x - 1, we get x = (y + 1)/4.</p><p>The line is to the right of the parabola in the region bounded by intersection points.</p><p><strong>Step 3:</strong> Area = ∫₍₋₁/₂₎¹ [(y + 1)/4 - y²/2] dy</p><p>= ∫₍₋₁/₂₎¹ [y/4 + 1/4 - y²/2] dy</p><p>= [y²/8 + y/4 - y³/6]₍₋₁/₂₎¹</p><p>= [(1/8 + 1/4 - 1/6) - (1/32 - 1/8 + 1/48)]</p><p>= [(3/24 + 6/24 - 4/24) - (3/96 - 12/96 + 2/96)]</p><p>= [5/24 - (-7/96)]</p><p>= [20/96 + 7/96] = 27/96 = 9/32</p><p>∴ Answer: C (9/32 sq. units)</p>
Correct Answer: C