Definite Integration
Integral Equations
Grade 12
Question:
<p>If \(\displaystyle\int_0^x f(t)\,dt = e^x - ae^{2x}\int_0^1 f(t)e^{-t}\,dt\), then \(f(1) + 2f(2)\) is equal to:</p>
<p>(a) \(e - 4e^4\)</p>
<p>(b) \(e - 2e^4\)</p>
<p>(c) \(e - 2e^2\)</p>
<p>(d) \(2e^2 - e^4\)</p>
Step-by-Step Solution
Key Concept: Differentiate both sides of the functional equation to extract f(x), then use the boundary condition at x=0 to find the unknown constant a.
<p><strong>Step 1:</strong> Differentiate both sides with respect to x:</p><p>f(x) = e^x - 2ae^(2x)∫₀¹ f(t)e^(-t)dt</p><p>Let k = ∫₀¹ f(t)e^(-t)dt (a constant), so f(x) = e^x - 2ake^(2x)</p><p><strong>Step 2:</strong> Use the boundary condition. At x=0, the left side is ∫₀⁰ f(t)dt = 0.</p><p>Right side: e⁰ - ae⁰·k = 1 - ak</p><p>Therefore: 0 = 1 - ak, so ak = 1</p><p><strong>Step 3:</strong> Substitute back: f(x) = e^x - 2e^(2x)</p><p><strong>Step 4:</strong> Calculate f(1) + 2f(2):</p><p>f(1) = e - 2e² = e(1 - 2e)</p><p>f(2) = e² - 2e⁴</p><p>2f(2) = 2e² - 4e⁴</p><p>f(1) + 2f(2) = e - 2e² + 2e² - 4e⁴ = e - 4e⁴</p><p>∴ Answer: B</p>
Correct Answer: B