Definite Integration
Inequalities involving integrals
Grade 12
Question:
<p>Consider <span>\(0 \leq \int_0^1 (f'(x)-1)^2\, dx\)</span>. If <span>\(f(0)=0\)</span> and <span>\(f(1)=1\)</span>, then find the minimum value of <span>\(\int_0^1 (f'(x))^2\, dx\)</span>.</p>
<p>A) 0</p>
<p>B) 1</p>
<p>C) 2</p>
<p>D) 1/2</p>
Step-by-Step Solution
Key Concept: Expand (f'(x)-1)² and use the constraint that ∫₀¹(f'(x)-1)²dx ≥ 0 to establish a relationship between ∫₀¹(f'(x))²dx and ∫₀¹f'(x)dx. The boundary conditions f(0)=0, f(1)=1 give ∫₀¹f'(x)dx = 1.
<p><strong>Step 1:</strong> Expand the given non-negative integral:</p><p>∫₀¹(f'(x)-1)²dx ≥ 0</p><p>∫₀¹[(f'(x))² - 2f'(x) + 1]dx ≥ 0</p><p><strong>Step 2:</strong> Separate the integral:</p><p>∫₀¹(f'(x))²dx - 2∫₀¹f'(x)dx + ∫₀¹1·dx ≥ 0</p><p><strong>Step 3:</strong> Use boundary conditions. From f(0)=0 and f(1)=1:</p><p>∫₀¹f'(x)dx = f(1) - f(0) = 1</p><p><strong>Step 4:</strong> Substitute into the inequality:</p><p>∫₀¹(f'(x))²dx - 2(1) + 1 ≥ 0</p><p>∫₀¹(f'(x))²dx ≥ 1</p><p><strong>Step 5:</strong> Equality holds when f'(x) - 1 = 0, i.e., f'(x) = 1 (constant), which gives f(x) = x satisfying both boundary conditions.</p><p>∴ Minimum value = <strong>1</strong></p>
Correct Answer: B