Definite Integration
Properties of Definite Integrals
Grade 12
Question:
<p>Suppose that <em>f</em> is a function on the interval [1, 3] such that \(-1 \leq f(x) \leq 1\) for all <em>x</em> and \(\int_1^3 f(x)\,dx = 0\). Find the maximum value of \(\int_1^3 \dfrac{f(x)}{x}\,dx\) be?</p>
<p>\(\log\dfrac{16}{3}\)</p>
<p>\(\log\dfrac{4}{3}\)</p>
<p>\(1+\log\dfrac{4}{3}\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: To maximize ∫₁³ [f(x)/x]dx subject to constraints -1 ≤ f(x) ≤ 1 and ∫₁³ f(x)dx = 0, recognize that the coefficient 1/x is decreasing on [1,3], so we should assign f(x) = 1 where 1/x is largest (near x=1) and f(x) = -1 where 1/x is smaller (near x=3), while respecting the zero integral constraint.
<p><strong>Step 1: Set up the optimization problem.</strong> We want to maximize I = ∫₁³ [f(x)/x]dx subject to:<br/>• -1 ≤ f(x) ≤ 1 for all x ∈ [1,3]<br/>• ∫₁³ f(x)dx = 0</p><p><strong>Step 2: Apply the extremal principle.</strong> Since we're maximizing a linear functional with linear constraints, the maximum occurs at an extreme point. The optimal f(x) should be bang-bang: f(x) ∈ {-1, +1} almost everywhere. Set f(x) = 1 on [1, a] and f(x) = -1 on [a, 3].</p><p><strong>Step 3: Apply the integral constraint.</strong><br/>∫₁³ f(x)dx = ∫₁ᵃ 1·dx + ∫ₐ³ (-1)·dx = (a-1) - (3-a) = 2a - 4 = 0<br/>∴ a = 2</p><p><strong>Step 4: Calculate the maximum value.</strong><br/>I = ∫₁² (1/x)dx + ∫₂³ (-1/x)dx<br/>= [ln x]₁² - [ln x]₂³<br/>= ln 2 - (ln 3 - ln 2)<br/>= 2ln 2 - ln 3<br/>= ln(4/3)</p><p><strong>Step 5: Verify this is achievable.</strong> The function f(x) = 1 for x ∈ [1,2) and f(x) = -1 for x ∈ [2,3] satisfies all constraints.</p><p>∴ Maximum value = <strong>ln(4/3)</strong> or equivalently <strong>2ln 2 - ln 3</strong></p>
Correct Answer: A