<p>The area (in sq. units) of the region bounded by the curve x² = 4y and the straight line x = 4y − 2 is:</p>
Step-by-Step Solution
Key Concept: Find intersection points of the parabola and line, then integrate the difference between the line (as x-function) and parabola to get the bounded area.
<p><strong>Step 1:</strong> Find intersection points by solving x² = 4y and x = 4y - 2 simultaneously.</p><p>From x = 4y - 2: y = (x+2)/4</p><p>Substitute into x² = 4y: x² = 4·(x+2)/4 = x + 2</p><p>x² - x - 2 = 0 → (x-2)(x+1) = 0 → x = 2 or x = -1</p><p>When x = 2: y = 1; When x = -1: y = 1/4</p><p>Points: (-1, 1/4) and (2, 1)</p><p><strong>Step 2:</strong> Express curves as functions of x.</p><p>Parabola: x² = 4y → y = x²/4</p><p>Line: x = 4y - 2 → y = (x+2)/4</p><p><strong>Step 3:</strong> Determine which curve is above. At x = 0: line gives y = 1/2, parabola gives y = 0. Line is above.</p><p><strong>Step 4:</strong> Calculate area:</p><p>A = ∫₋₁² [(x+2)/4 - x²/4] dx = (1/4)∫₋₁² (x + 2 - x²) dx</p><p>= (1/4)[x²/2 + 2x - x³/3]₋₁²</p><p>= (1/4)[(2 + 4 - 8/3) - (1/2 - 2 + 1/3)]</p><p>= (1/4)[6 - 8/3 - (-3/2 + 1/3)]</p><p>= (1/4)[6 - 8/3 + 3/2 - 1/3] = (1/4)[6 + 3/2 - 3]</p><p>= (1/4)·9/2 = <strong>9/8 sq. units</strong></p><p>∴ Answer: A</p>
Correct Answer: A