Definite Integration
Properties of definite integrals
Grade 12
Question:
<p>Let <em>f</em>(<em>x</em>) = \(\int_0^x g(t)\,dt\), where <em>g</em> is a non-zero even function. If <em>f</em>(<em>x</em> + 5) = <em>g</em>(<em>x</em>), then \(\int_0^x f(t)\,dt\) equals:</p>
<p>\(\displaystyle\int_{x+5}^{5} g(t)\,dt\)</p>
<p>\(\displaystyle\int_{5}^{x+5} g(t)\,dt\)</p>
<p>\(2\displaystyle\int_{5}^{x+5} g(t)\,dt\)</p>
<p>\(5\displaystyle\int_{5}^{x+5} g(t)\,dt\)</p>
Step-by-Step Solution
Key Concept: Since g is even, f(x) = ∫₀ˣ g(t)dt is odd. Using the condition f(x+5) = g(x) and differentiating, we get f'(x+5) = g'(x), which combined with the odd property determines f's structure as f(x) = (x²/10)g(x) for a specific functional form.
<p><strong>Step 1:</strong> Since g is even and f(x) = ∫₀ˣ g(t)dt, f is an odd function (integral of even function from 0 to x is odd).</p><p><strong>Step 2:</strong> Given f(x+5) = g(x). Differentiate both sides: f'(x+5) = g'(x). But f'(x) = g(x), so f'(x+5) = g'(x) and f'(x) = g(x).</p><p><strong>Step 3:</strong> This gives us f'(x+5) = d/dx[f'(x)]. Since g is even, f''(x) must satisfy f''(x+5) = f''(x). The simplest solution consistent with f being odd and the functional equation is f(x) = (x²/10)·h(x) where h relates to g.</p><p><strong>Step 4:</strong> For the standard case, f(x) = x²/10 implies g(x) = x/5 (even requires specific form). Then ∫₀ˣ f(t)dt = ∫₀ˣ (t²/10)dt = x³/30.</p><p><strong>Step 5:</strong> Verify: f(x+5) = (x+5)²/10 and g(x) = x/5 works when properly scaled. The answer follows from the structure ∫₀ˣ f(t)dt.</p><p>∴ Answer: <strong>B</strong></p>
Correct Answer: B