<p>The area (in sq. units) of the region \(A = \{(x, y) : x^2 \leq y \leq x + 2\}\) is ______.</p>
Step-by-Step Solution
Key Concept: The region A is bounded between the parabola y = x² (below) and the line y = x + 2 (above); find intersection points to set integration limits, then integrate the difference of upper and lower curves.
<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 = -1 or x = 2</p><p><strong>Step 2: Verify which curve is above</strong></p><p>At x = 0: parabola gives y = 0, line gives y = 2. So y = x + 2 is above y = x² 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>= [4/2 + 4 - 8/3] - [1/2 - 2 + 1/3]</p><p>= [2 + 4 - 8/3] - [1/2 - 2 + 1/3]</p><p>= [6 - 8/3] - [-3/2 + 1/3]</p><p>= [18/3 - 8/3] - [-9/6 + 2/6]</p><p>= 10/3 - (-7/6)</p><p>= 10/3 + 7/6</p><p>= 20/6 + 7/6 = 27/6</p><p>∴ Answer: <strong>9/2 sq. units</strong></p>
Correct Answer: 9