Indefinite Integration
Integration and properties
Grade 12
Question:
<p>Let \(h(x) = \int\left(\int\left(\int g'''(x)\,dx\right)dx\right)dx\) with \(h(3) = g(3)\), \(h(1) = g(1)\) and \(h(0) - g(0) = 6\).</p><p>If \(f(x) = h(x) - g(x)\), then:</p>
<p>\(f(x)\) decreases in the interval \((1, 3)\)</p>
<p>\(f(x)\) decreases in the interval \((-\infty, 2)\)</p>
<p>\(f(4) = 6\)</p>
<p>\(f(2) = 6\)</p>
Step-by-Step Solution
Key Concept: Recognize that h(x) involves three successive integrations of g'''(x), which means h(x) = g(x) + cubic polynomial. Use the three boundary conditions to determine the polynomial coefficients uniquely.
<p><strong>Step 1:</strong> Let ∫g'''(x)dx = g''(x) + C₁</p><p>Then ∫(g''(x) + C₁)dx = g'(x) + C₁x + C₂</p><p>Finally ∫(g'(x) + C₁x + C₂)dx = g(x) + C₁·x²/2 + C₂x + C₃</p><p><strong>Step 2:</strong> So h(x) = g(x) + (C₁/2)x² + C₂x + C₃</p><p>Therefore f(x) = h(x) - g(x) = (C₁/2)x² + C₂x + C₃</p><p><strong>Step 3:</strong> Apply boundary conditions:</p><p>• h(3) = g(3) ⟹ (C₁/2)·9 + 3C₂ + C₃ = 0</p><p>• h(1) = g(1) ⟹ (C₁/2)·1 + C₂ + C₃ = 0</p><p>• h(0) - g(0) = 6 ⟹ C₃ = 6</p><p><strong>Step 4:</strong> From conditions 1 and 2:</p><p>(9C₁/2 + 3C₂ + 6) - (C₁/2 + C₂ + 6) = 0</p><p>4C₁ + 2C₂ = 0 ⟹ C₂ = -2C₁</p><p>From condition 2: (C₁/2) - 2C₁ + 6 = 0</p><p>-3C₁/2 = -6 ⟹ C₁ = 4, C₂ = -8</p><p><strong>Step 5:</strong> Thus f(x) = 2x² - 8x + 6 = 2(x² - 4x + 3) = 2(x-1)(x-3)</p><p>∴ f(1) = 0, f(3) = 0, f(0) = 6, and f(x) has zeros at x = 1 and x = 3</p>
Correct Answer: A,D