Definite Integration
Integration involving greatest integer function
Grade 12
Question:
<p>If \(a > 1\), then \(\int_1^a [x]\,f'(x)\,dx\) equals (where \([x]\) denotes greatest integer function):</p>
<p>\([a]f(a) - \{f(1) + f(2) + \cdots + f([a])\}\)</p>
<p>\([a]f([a]) - \{f(1) + f(2) + \cdots + f([a]-1)\}\)</p>
<p>\([a]f(a) + \{f(1) + f(2) + \cdots + f([a])\}\)</p>
<p>\([a]f([a]) + \{f(1) + f(2) + \cdots + f([a]-1)\}\)</p>
Step-by-Step Solution
Key Concept: Use integration by parts with u = [x] (which is piecewise constant on intervals) and dv = f'(x)dx. Recognize that [x] jumps at integer values, so split the integral at these points where [x] changes value.
<p><strong>Step 1:</strong> For a > 1, let n = [a], so n ≤ a < n+1. Split the integral at integer points:</p><p>∫₁ᵃ [x]f'(x)dx = ∫₁² 1·f'(x)dx + ∫₂³ 2·f'(x)dx + ... + ∫ₙ₋₁ⁿ (n-1)f'(x)dx + ∫ₙᵃ n·f'(x)dx</p><p><strong>Step 2:</strong> On each interval [k, k+1), [x] = k (constant). Apply integration by parts on each piece:</p><p>∫ₖᵏ⁺¹ k·f'(x)dx = k[f(x)]ₖᵏ⁺¹ = k[f(k+1) - f(k)]</p><p><strong>Step 3:</strong> Sum all pieces:</p><p>= 1[f(2)-f(1)] + 2[f(3)-f(2)] + 3[f(4)-f(3)] + ... + (n-1)[f(n)-f(n-1)] + n[f(a)-f(n)]</p><p><strong>Step 4:</strong> Rearrange using telescoping (grouping by f values):</p><p>= -f(1) + f(2) + f(3) + ... + f(n) + n·f(a) - n·f(n)</p><p>= -f(1) + Σₖ₌₂ⁿ f(k) + n[f(a) - f(n)]</p><p><strong>Step 5:</strong> This simplifies to: <strong>n·f(a) - Σₖ₌₁ⁿ f(k)</strong> or equivalently <strong>[a]·f(a) - ∫₁ᵃ f(x)dx</strong></p><p>∴ Answer: A</p>
Correct Answer: A