Definite Integration
Minimization of integral using Cauchy-Schwarz
Grade 12
Question:
<p>If \(f\) is continuous and differentiable on \([0,1]\) with \(f(0)=0\) and \(f(1)=1\), then the minimum value of \(\displaystyle\int_0^1 (f'(x))^2\,dx\) is equal to</p>
Step-by-Step Solution
Key Concept: Use Cauchy-Schwarz inequality on the integral ∫₀¹ f'(x)dx = 1 to establish a lower bound for ∫₀¹ (f'(x))² dx. The minimum occurs when f'(x) is constant.
<p><strong>Step 1:</strong> Use the constraint from boundary conditions. Since f is continuous and differentiable on [0,1] with f(0)=0 and f(1)=1, we have:</p><p>∫₀¹ f'(x)dx = f(1) - f(0) = 1 - 0 = 1</p><p><strong>Step 2:</strong> Apply Cauchy-Schwarz inequality: [∫₀¹ f'(x)·1 dx]² ≤ [∫₀¹ (f'(x))² dx][∫₀¹ 1² dx]</p><p>This gives: 1² ≤ [∫₀¹ (f'(x))² dx]·1</p><p>Therefore: ∫₀¹ (f'(x))² dx ≥ 1</p><p><strong>Step 3:</strong> Equality in Cauchy-Schwarz holds when f'(x) is proportional to the constant function 1, meaning f'(x) = c (constant).</p><p>From ∫₀¹ f'(x)dx = 1, we get c·1 = 1, so f'(x) = 1.</p><p>This gives f(x) = x, which satisfies f(0)=0 and f(1)=1.</p><p>Verification: ∫₀¹ (1)² dx = 1</p><p><strong>∴ Answer: 1</strong></p>
Correct Answer: 1