Definite Integration
Properties of Definite Integrals
Grade 12
Question:
<p>If \(f(2-x) = f(2+x)\) and \(f(4-x) = f(4+x)\) and \(f(x)\) is a function for which \(\int_0^2 f(x)\,dx = 5\), then \(\int_0^{50} f(x)\,dx\) is equal to</p>
<p>(a) 125</p>
<p>(b) \(\int_{-4}^{46} f(x)\,dx\)</p>
<p>(c) \(\int_1^{51} f(x)\,dx\)</p>
<p>(d) \(\int_2^{52} f(x)\,dx\)</p>
Step-by-Step Solution
Key Concept: The function has two axes of symmetry (at x=2 and x=4), which means f is periodic with period 4. Once you identify the period, you can express the integral over [0,50] as a multiple of the integral over one period [0,4].
<p><strong>Step 1:</strong> Analyze the symmetry conditions.</p><p>Given: f(2-x) = f(2+x) and f(4-x) = f(4+x)</p><p>This means f is symmetric about x=2 AND symmetric about x=4.</p><p><strong>Step 2:</strong> Determine the period.</p><p>If f is symmetric about x=2: f(x) = f(4-x)</p><p>If f is symmetric about x=4: f(x) = f(8-x)</p><p>Combining: f(4-x) = f(8-x), which gives f(y) = f(y+4) by substitution y=4-x</p><p>Therefore, f has period T = 4.</p><p><strong>Step 3:</strong> Find ∫₀⁴ f(x)dx.</p><p>By symmetry about x=2: ∫₀⁴ f(x)dx = 2∫₀² f(x)dx = 2(5) = 10</p><p><strong>Step 4:</strong> Calculate ∫₀⁵⁰ f(x)dx.</p><p>Since 50 = 12(4) + 2:</p><p>∫₀⁵⁰ f(x)dx = 12∫₀⁴ f(x)dx + ∫₀² f(x)dx = 12(10) + 5 = 125</p><p>∴ Answer: <strong>125</strong></p>
Correct Answer: A