Area Under the Curve
Area bounded by curves
Grade 12
Question:
<p>The area (in sq. units) of the region \(A = \{(x, y) \in \mathbb{R} \times \mathbb{R} \mid 0 \leq x \leq 3, 0 \leq y \leq 4, y \leq x^2 + 3x\}\) is ______ (up to three decimal places).</p>
Step-by-Step Solution
Key Concept: The region is bounded by y = x² + 3x from below and y = 4 from above within 0 ≤ x ≤ 3. Find where y = 4 intersects the parabola to determine integration limits, then integrate the difference of upper and lower bounds.
<p><strong>Step 1:</strong> Find intersection of y = x² + 3x and y = 4: x² + 3x = 4 → x² + 3x - 4 = 0 → (x+4)(x-1) = 0 → x = 1 (since 0 ≤ x ≤ 3)</p><p><strong>Step 2:</strong> For 0 ≤ x ≤ 1: y is bounded by 0 ≤ y ≤ x² + 3x (parabola is below y = 4)</p><p><strong>Step 3:</strong> For 1 ≤ x ≤ 3: y is bounded by 0 ≤ y ≤ 4 (the line y = 4 is below the parabola)</p><p><strong>Step 4:</strong> Area = ∫₀¹ (x² + 3x)dx + ∫₁³ 4dx</p><p><strong>Step 5:</strong> ∫₀¹ (x² + 3x)dx = [x³/3 + 3x²/2]₀¹ = 1/3 + 3/2 = 11/6</p><p><strong>Step 6:</strong> ∫₁³ 4dx = 4(3-1) = 8</p><p><strong>Step 7:</strong> Total Area = 11/6 + 8 = 11/6 + 48/6 = 59/6 ≈ 9.833</p><p>∴ Answer: <strong>9.833</strong></p>
Correct Answer: 9