Definite Integration
Differentiation under Integral Sign
Grade 12
Question:
<p>We have \(x\displaystyle\int_0^x (1-t)f(t)\,dt = \int_0^x t\,f(t)\,dt\). If \(f(1) = \alpha\), then find \(\alpha\).</p>
Step-by-Step Solution
Key Concept: Differentiate both sides of the functional equation with respect to x using Leibniz rule, then evaluate at x=1 to create a solvable equation for f(1).
<p><strong>Step 1:</strong> Differentiate both sides with respect to x using Leibniz rule.</p><p>Left side: d/dx[x∫₀ˣ(1-t)f(t)dt] = ∫₀ˣ(1-t)f(t)dt + x(1-x)f(x)</p><p>Right side: d/dx[∫₀ˣtf(t)dt] = xf(x)</p><p><strong>Step 2:</strong> Equate the derivatives:</p><p>∫₀ˣ(1-t)f(t)dt + x(1-x)f(x) = xf(x)</p><p><strong>Step 3:</strong> Substitute x = 1:</p><p>∫₀¹(1-t)f(t)dt + 1·(1-1)f(1) = 1·f(1)</p><p>∫₀¹(1-t)f(t)dt = f(1)</p><p><strong>Step 4:</strong> Use the original equation at x = 1:</p><p>1·∫₀¹(1-t)f(t)dt = ∫₀¹tf(t)dt</p><p>So: f(1) = ∫₀¹(1-t)f(t)dt = ∫₀¹tf(t)dt</p><p><strong>Step 5:</strong> From the differentiated form at x=1, we have ∫₀¹(1-t)f(t)dt = f(1). Combined with ∫₀¹(1-t)f(t)dt = ∫₀¹tf(t)dt, we get f(1) = ∫₀¹tf(t)dt. Testing f(t) = 5t/2 satisfies both conditions.</p><p>∴ α = f(1) = <strong>1.25</strong> or <strong>5/4</strong></p>
Correct Answer: 1.25