<p>The area of the region \(A = \{(x, y) \in R \times R \mid 0 \le x \le 3, 0 \le y \le 4, y \le x^2 + 3x\}\) is equal to:</p>
Step-by-Step Solution
Key Concept: The region is bounded by y = 0 (bottom), y = 4 (top), x = 0 (left), x = 3 (right), and y = x² + 3x (curve). You must integrate the minimum of {4, x² + 3x} minus 0 over [0,3], finding where x² + 3x = 4 to split the integral.
<p><strong>Step 1:</strong> Find where y = x² + 3x intersects y = 4.</p><p>x² + 3x = 4 → x² + 3x - 4 = 0 → (x + 4)(x - 1) = 0</p><p>Since 0 ≤ x ≤ 3, the intersection is at x = 1.</p><p><strong>Step 2:</strong> For 0 ≤ x ≤ 1: x² + 3x ≤ 4, so the region extends from y = 0 to y = x² + 3x.</p><p>Area₁ = ∫₀¹ (x² + 3x) dx = [x³/3 + 3x²/2]₀¹ = 1/3 + 3/2 = 11/6</p><p><strong>Step 3:</strong> For 1 ≤ x ≤ 3: x² + 3x ≥ 4, so the region extends from y = 0 to y = 4.</p><p>Area₂ = ∫₁³ 4 dx = 4(3 - 1) = 8</p><p><strong>Step 4:</strong> Total area = 11/6 + 8 = 11/6 + 48/6 = 59/6 ≈ 9.833</p><p>∴ Answer: 9.833</p>
Correct Answer: 9.833