<p>The area bounded by the curve \(y = x(x-1)^2\), the \(y\)-axis and the line \(y = 2\) is \(a/b\), here \(a\) and \(b\) are co-prime then</p>
Step-by-Step Solution
Key Concept: Find intersection points of y = x(x-1)² with y = 2, then integrate the difference between the line and curve. The area requires splitting the region based on where the curve lies above or below the line.
<p><strong>Step 1:</strong> Find intersection of y = x(x-1)² with y = 2.</p><p>x(x-1)² = 2 gives x(x² - 2x + 1) = 2, or x³ - 2x² + x - 2 = 0</p><p>Factoring: (x-2)(x² + 1) = 0, so x = 2 (real solution)</p><p><strong>Step 2:</strong> Identify the bounded region. At x = 0: y = 0. At x = 2: y = 2. The line y = 2 is horizontal.</p><p><strong>Step 3:</strong> For 0 ≤ x ≤ 2, check which is above. At x = 1: curve gives y = 1(0)² = 0, line gives y = 2. So line is above.</p><p><strong>Step 4:</strong> Set up the integral:</p><p>Area = ∫₀² [2 - x(x-1)²] dx = ∫₀² [2 - x(x² - 2x + 1)] dx</p><p>= ∫₀² [2 - x³ + 2x² - x] dx</p><p>= ∫₀² [-x³ + 2x² - x + 2] dx</p><p><strong>Step 5:</strong> Integrate: [-x⁴/4 + 2x³/3 - x²/2 + 2x]₀²</p><p>= -16/4 + 16/3 - 4/2 + 4 = -4 + 16/3 - 2 + 4 = 16/3 - 2 = 16/3 - 6/3 = 10/3</p><p>∴ a = 10, b = 3 (coprime); Answer: a/b = 10/3</p>
Correct Answer: A