Definite Integration
Properties of definite integrals
Grade 12
Question:
<p>Let <i>f</i> be a differentiable function from <i>R</i> to <i>R</i> such that \(|f(x) - f(y)| \leq 2|x - y|^{3/2}\), for all \(x, y \in R\). If \(f(0) = 1\) then \[\int_0^1 f^2(x)\,dx\] is equal to:</p>
<p>1</p>
<p>2</p>
<p>\(\dfrac{1}{2}\)</p>
<p>0</p>
Step-by-Step Solution
Key Concept: The Lipschitz-type condition |f(x) - f(y)| ≤ 2|x - y|^(3/2) forces f'(x) = 0 everywhere (since the derivative must satisfy |f'(x)| ≤ lim[h→0] 2|h|^(1/2)/|h| = 0), making f constant. Combined with f(0) = 1, we get f(x) = 1 for all x.
<p><strong>Step 1:</strong> Analyze the given condition. For differentiable f, we have:</p><p>|f(x+h) - f(x)| ≤ 2|h|^(3/2) for all h</p><p><strong>Step 2:</strong> Divide by |h| (h ≠ 0):</p><p>|f(x+h) - f(x)|/|h| ≤ 2|h|^(1/2)</p><p><strong>Step 3:</strong> Take the limit as h → 0:</p><p>|f'(x)| = lim[h→0] |f(x+h) - f(x)|/|h| ≤ lim[h→0] 2|h|^(1/2) = 0</p><p><strong>Step 4:</strong> Therefore f'(x) = 0 for all x ∈ ℝ, which means f is constant on ℝ.</p><p><strong>Step 5:</strong> Since f(0) = 1 and f is constant, we have f(x) = 1 for all x ∈ ℝ.</p><p><strong>Step 6:</strong> Calculate the integral:</p><p>∫₀¹ f²(x)dx = ∫₀¹ 1² dx = ∫₀¹ 1 dx = x|₀¹ = 1</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: A