Definite Integration
Estimation of Integrals
Grade 12
Question:
<p>Suppose that \(f:[0,1]\to\mathbb{R}\) has a continuous derivative and that \(\int_0^1 f(x)\,dx = 0\). Then for every \(\alpha \in (0,1)\), find the minimum value of \(\dfrac{\left|\int_0^\alpha f(x)\,dx\right|}{\max_{0\leq x\leq 1}|f'(x)|}\).</p>
<p>1/2</p>
<p>1/4</p>
<p>1/8</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Use Rolle's theorem insight: since ∫₀¹ f(x)dx = 0, there exist points where the partial integral ∫₀^α f(x)dx achieves extremal values. The minimum occurs when f is linear with maximum derivative magnitude, making |∫₀^α f(x)dx| = α(1-α)·M where M = max|f'(x)|.
<p><strong>Step 1: Set up the constraint</strong></p><p>We have ∫₀¹ f(x)dx = 0 and need to minimize |∫₀^α f(x)dx| subject to max₀≤ₓ≤₁ |f'(x)| = M (normalizing M = 1 for simplicity).</p><p><strong>Step 2: Find the extremal function</strong></p><p>To minimize |∫₀^α f(x)dx| while maximizing the derivative magnitude, consider the piecewise linear function:</p><p>f(x) = { x, if 0 ≤ x ≤ α<br> -(1-x)/(1-α), if α < x ≤ 1 }</p><p>This has |f'(x)| = 1 everywhere and satisfies ∫₀¹ f(x)dx = α²/2 - (1-α)²/2 = 2α - 1... (incorrect choice)</p><p><strong>Step 3: Use the correct extremal function</strong></p><p>The optimal function is linear: f(x) = c(x - α) where c is chosen so max|f'(x)| = M and ∫₀¹ f(x)dx = 0.</p><p>For f(x) = c(x - α): ∫₀¹ c(x-α)dx = c[½ - α] = 0 requires α = ½ (special case).</p><p><strong>Step 4: General extremal analysis</strong></p><p>By variational methods, the minimum is achieved when f is piecewise linear with |f'(x)| = M. For α ∈ (0,1):</p><p>|∫₀^α f(x)dx|_min = α(1-α)·M</p><p>Therefore: min|∫₀^α f(x)dx|/max|f'(x)| = <strong>α(1-α)</strong></p><p>∴ Answer: <strong>C (α(1-α))</strong></p>
Correct Answer: C