Definite Integration
Integral Equations
Grade 12
Question:
<p>Let <span>\( k = \displaystyle\int_0^1 f(t)\,dt \)</span> and <span>\( g(x) = x - k \)</span>. If <span>\( f(x) = \dfrac{x^3}{2} + 1 - x\displaystyle\int_0^x (t-k)\,dt \)</span>, then which of the following are correct?</p><p>(a) <span>\( k = \dfrac{3}{2} \)</span></p><p>(b) <span>\( f(x) = 1 + \dfrac{3x^2}{2} \)</span></p><p>(c) Some other option</p><p>(d) Some other option</p>
<p>\( k = \dfrac{3}{2} \)</p>
<p>\( f(x) = 1 + \dfrac{3x^2}{2} \)</p>
<p>\( g(x) = x - \dfrac{3}{2} \)</p>
<p>\( f(x) = 1 + kx^2 \) for some \( k \)</p>
Step-by-Step Solution
Key Concept: Use the self-referential nature of the definition: since f(x) contains an integral of g(t) = t-k, differentiate the f(x) equation and use the boundary condition that ∫₀¹ f(t)dt = k to find k, then verify f(x).
<p><strong>Step 1:</strong> Differentiate f(x) with respect to x using Leibniz rule.</p><p>Given: f(x) = x³/2 + 1 - x∫₀ˣ (t-k)dt</p><p>f'(x) = 3x²/2 - [∫₀ˣ (t-k)dt + x(x-k)]</p><p>f'(x) = 3x²/2 - ∫₀ˣ (t-k)dt - x(x-k)</p><p><strong>Step 2:</strong> Compute ∫₀ˣ (t-k)dt = x²/2 - kx</p><p>f'(x) = 3x²/2 - (x²/2 - kx) - x² + kx</p><p>f'(x) = 3x²/2 - x²/2 + kx - x² + kx = 2kx</p><p><strong>Step 3:</strong> Integrate to find f(x): f(x) = kx² + C</p><p>From f(x) = x³/2 + 1 - x∫₀ˣ (t-k)dt at x=0: f(0) = 1, so C = 1</p><p>Therefore: f(x) = kx² + 1</p><p><strong>Step 4:</strong> Use the constraint k = ∫₀¹ f(t)dt</p><p>k = ∫₀¹ (kt² + 1)dt = k/3 + 1</p><p>k - k/3 = 1 → 2k/3 = 1 → k = 3/2</p><p><strong>Step 5:</strong> Substitute k = 3/2: f(x) = (3/2)x² + 1 = 1 + 3x²/2</p><p>∴ <strong>Answer: A (k = 3/2) and C (f(x) = 1 + 3x²/2) are correct</strong></p>
Correct Answer: A, C