Indefinite Integration
Integration using functional equations
Grade 12
Question:
<p>If \(f(x) = f(x) + xf'(x)\) then \(\int g(x)\,dx\) is equal to:</p>
<p>(a) \((x-1)f(x) + k\)</p>
<p>(b) \((x-1)f(x) + k\)</p>
<p>(c) \(f(x) + k\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: The condition g(x) = f(x) + xf'(x) represents the derivative of xf(x), since d/dx[xf(x)] = f(x) + xf'(x). Therefore, integrating g(x) yields xf(x) plus a constant.
<p><strong>Step 1:</strong> Recognize the structure of g(x) = f(x) + xf'(x)</p><p>Notice that if we differentiate xf(x) using the product rule:</p><p>d/dx[xf(x)] = 1·f(x) + x·f'(x) = f(x) + xf'(x) = g(x)</p><p><strong>Step 2:</strong> Apply the fundamental theorem of calculus</p><p>Since g(x) is the derivative of xf(x), we integrate:</p><p>∫g(x)dx = ∫[d/dx(xf(x))]dx = xf(x) + C</p><p><strong>Step 3:</strong> Verify by differentiating the result</p><p>d/dx[xf(x) + C] = f(x) + xf'(x) = g(x) ✓</p><p>∴ Answer: <strong>B</strong> (∫g(x)dx = xf(x) + C)</p>
Correct Answer: B