Area Under the Curve
Area between parabola and line
Grade 12
Question:
<p>Let \(A = \{(x, y) : y^2 \leq 4x,\ y - 2x \geq -4\}\). The area (in square units) of the region \(A\) is</p>
<p>8</p>
<p>9</p>
<p>10</p>
<p>11</p>
Step-by-Step Solution
Key Concept: The region A is bounded by a parabola y² = 4x and a line y - 2x = -4. You must find their intersection points and integrate with respect to the appropriate variable (y-integration is cleaner here since the parabola is naturally expressed as x = y²/4).
<p><strong>Step 1: Find intersection points</strong></p><p>Parabola: y² = 4x → x = y²/4</p><p>Line: y - 2x = -4 → x = (y+4)/2</p><p>At intersection: y²/4 = (y+4)/2</p><p>y² = 2(y+4) → y² - 2y - 8 = 0</p><p>(y-4)(y+2) = 0 → y = 4 or y = -2</p><p></p><p><strong>Step 2: Determine which curve is on the right</strong></p><p>Check at y = 0: Parabola gives x = 0; Line gives x = 2</p><p>So the line x = (y+4)/2 is to the right of the parabola x = y²/4 in the region between y = -2 and y = 4.</p><p></p><p><strong>Step 3: Set up the integral with respect to y</strong></p><p>Area = ∫₋₂⁴ [(y+4)/2 - y²/4] dy</p><p></p><p><strong>Step 4: Evaluate</strong></p><p>= ∫₋₂⁴ [(y+4)/2 - y²/4] dy</p><p>= [y²/4 + 2y - y³/12]₋₂⁴</p><p>= [16/4 + 8 - 64/12] - [4/4 - 4 + 8/12]</p><p>= [4 + 8 - 16/3] - [1 - 4 + 2/3]</p><p>= [12 - 16/3] - [-3 + 2/3]</p><p>= [36/3 - 16/3] - [-9/3 + 2/3]</p><p>= 20/3 - (-7/3)</p><p>= 20/3 + 7/3 = 27/3 = <strong>9</strong></p><p>∴ Answer: B</p>
Correct Answer: B