Area Under the Curve
Area bounded by curve and axis
Grade 12
Question:
<p>Calculate the area bounded by the curve <em>y</em> = <em>x</em>(3 − <em>x</em>)<sup>2</sup>, the <em>x</em>-axis and the ordinates of the maximum and minimum points of the curve.</p>
Step-by-Step Solution
Key Concept: Find the turning points of y = x(3-x)² by taking the derivative, identify which is maximum and which is minimum, then integrate between these x-coordinates to get the bounded area.
<p><strong>Step 1:</strong> Find turning points by differentiating y = x(3-x)²</p><p>dy/dx = (3-x)² + x·2(3-x)(-1) = (3-x)² - 2x(3-x) = (3-x)[(3-x) - 2x] = (3-x)(3-3x)</p><p>dy/dx = 3(3-x)(1-x)</p><p>Setting dy/dx = 0: x = 3 or x = 1</p><p><strong>Step 2:</strong> Determine nature of turning points using second derivative or sign analysis</p><p>At x = 1: y = 1(3-1)² = 4 (local maximum)</p><p>At x = 3: y = 3(3-3)² = 0 (local minimum)</p><p><strong>Step 3:</strong> Calculate area between x = 1 and x = 3</p><p>Since the curve lies above the x-axis between these ordinates (y ≥ 0), the bounded area is:</p><p>A = ∫₁³ x(3-x)² dx = ∫₁³ x(9 - 6x + x²) dx = ∫₁³ (9x - 6x² + x³) dx</p><p>= [9x²/2 - 2x³ + x⁴/4]₁³</p><p>= (81/2 - 54 + 81/4) - (9/2 - 2 + 1/4)</p><p>= (162/4 - 216/4 + 81/4) - (18/4 - 8/4 + 1/4)</p><p>= 27/4 - 11/4 = 16/4 = 4</p><p>∴ Answer: <strong>4</strong></p>
Correct Answer: 4