<p>Given the region \(x^2 \le y \le x + 2\), find the area of the region (in square units).</p>
Step-by-Step Solution
Key Concept: Find intersection points of the parabola y = x² and line y = x + 2, then integrate the difference (upper curve minus lower curve) between these bounds.
<p><strong>Step 1: Find intersection points</strong></p><p>Set x² = x + 2</p><p>x² - x - 2 = 0</p><p>(x - 2)(x + 1) = 0</p><p>x = 2 or x = -1</p><p><strong>Step 2: Determine which curve is above</strong></p><p>Test x = 0: line gives y = 2, parabola gives y = 0</p><p>So y = x + 2 is the upper curve on [-1, 2]</p><p><strong>Step 3: Set up the integral</strong></p><p>Area = ∫₋₁² [(x + 2) - x²] dx</p><p><strong>Step 4: Evaluate</strong></p><p>= ∫₋₁² (x + 2 - x²) dx</p><p>= [x²/2 + 2x - x³/3]₋₁²</p><p>At x = 2: 4/2 + 4 - 8/3 = 2 + 4 - 8/3 = 18/3 - 8/3 = 10/3</p><p>At x = -1: 1/2 - 2 + 1/3 = 3/6 - 12/6 + 2/6 = -7/6</p><p>Area = 10/3 - (-7/6) = 20/6 + 7/6 = 27/6 = 9/2 = 4.5 square units</p><p>∴ <strong>Answer: 4.5</strong></p>
Correct Answer: 4.5