Definite Integration
Integral equations
Grade None
Question:
<p><strong>Paragraph for Question nos. 580 to 582</strong><br>Let \(f(x)\) and \(g(x)\) are two continuous functions defined for \(0 \leq x \leq 1\), \(f(x) = \int_0^1 e^{x+t} f(t)\, dt\), \(g(x) = x + \int_0^1 e^{x+t} g(t)\, dt\).</p><p>The value of \(g(0)\) is:</p>
<p>(a) \(\dfrac{2}{3 - e^2}\)</p>
<p>(b) \(\dfrac{2}{e^2 - 2}\)</p>
<p>(c) \(\dfrac{2}{e^2 - 1}\)</p>
<p>(d) 0</p>
Step-by-Step Solution
Key Concept: Recognize that the integral ∫₀¹ e^(x+t) f(t)dt = e^x · ∫₀¹ e^t f(t)dt is a constant with respect to integration, so f(x) must be an exponential function. Use this structure to find g(0) by evaluating at x=0.
<p><strong>Step 1:</strong> For g(x), rewrite the integral term:</p><p>g(x) = x + ∫₀¹ e^(x+t) g(t)dt = x + e^x ∫₀¹ e^t g(t)dt</p><p>Let k = ∫₀¹ e^t g(t)dt (a constant)</p><p>So: g(x) = x + ke^x</p><p><strong>Step 2:</strong> Find k by substituting back into the integral condition:</p><p>∫₀¹ e^t g(t)dt = ∫₀¹ e^t[t + ke^t]dt = ∫₀¹ te^t dt + k∫₀¹ e^(2t)dt</p><p><strong>Step 3:</strong> Calculate ∫₀¹ te^t dt using integration by parts:</p><p>∫₀¹ te^t dt = [te^t]₀¹ - ∫₀¹ e^t dt = e - (e-1) = 1</p><p><strong>Step 4:</strong> Calculate ∫₀¹ e^(2t)dt = [e^(2t)/2]₀¹ = (e² - 1)/2</p><p><strong>Step 5:</strong> Set up the equation:</p><p>k = 1 + k·(e² - 1)/2</p><p>k - k(e² - 1)/2 = 1</p><p>k[1 - (e² - 1)/2] = 1</p><p>k[(2 - e² + 1)/2] = 1</p><p>k = 2/(3 - e²)</p><p><strong>Step 6:</strong> Find g(0):</p><p>g(0) = 0 + ke^0 = k = 2/(3 - e²)</p><p>∴ Answer: g(0) = 2/(3 - e²)</p>
Correct Answer: A