<p>Given \(f: \mathbb{R} \to \mathbb{R}\) such that \(f(2-x) = f(2+x)\) and \(f(4-x) = f(4+x)\) for all \(x \in \mathbb{R}\). If \(\int_0^2 f(x)\,dx = 5\), find \(\int_{10}^{50} f(x)\,dx\).</p>
Step-by-Step Solution
Key Concept: The two symmetry conditions f(2-x)=f(2+x) and f(4-x)=f(4+x) imply f has period 4. This is because the axes of symmetry at x=2 and x=4 are separated by 2 units, forcing the function to repeat every 4 units.
<p><strong>Step 1: Establish periodicity from symmetries</strong></p><p>From f(2-x)=f(2+x), the function is symmetric about x=2.</p><p>From f(4-x)=f(4+x), the function is symmetric about x=4.</p><p>Composing these symmetries: if f is symmetric about x=2 and x=4, then f(x)=f(4-x)=f(4+(x-4))=f(8-x). But also f(x)=f(2-(x-2))=f(4-x). Therefore f(x+4)=f(x), so <strong>f has period 4</strong>.</p><p><strong>Step 2: Compute the integral over one period</strong></p><p>∫₀⁴ f(x)dx = ∫₀² f(x)dx + ∫₂⁴ f(x)dx</p><p>Using symmetry about x=2: Let u=2-t in ∫₂⁴ f(x)dx, when x goes from 2 to 4, substitute x=2+t where t goes 0 to 2.</p><p>∫₂⁴ f(x)dx = ∫₀² f(2+t)dt = ∫₀² f(2-t)dt = ∫₀² f(x)dx = 5</p><p>Therefore: ∫₀⁴ f(x)dx = 5 + 5 = 10</p><p><strong>Step 3: Count periods in [10, 50]</strong></p><p>The interval [10, 50] has length 40 = 10×4, which is exactly 10 complete periods.</p><p>∫₁₀⁵⁰ f(x)dx = 10 × ∫₀⁴ f(x)dx = 10 × 10 = 100</p><p><strong>∴ Answer: 100</strong></p>
Correct Answer: A