<p>Let \(I_1 = \int_0^{10^4} \frac{\{\sqrt{x}\}}{\sqrt{x}}\,dx\) and \(I_2 = \int_0^{10} \left(x\{x^2\}\right)dx\), where \(\{.\}\) denotes fractional part of \(x\). Then</p>
Step-by-Step Solution
Key Concept: Split the integral over intervals where the fractional part function is linear, using the property that {t} = t - ⌊t⌋. For I₁, substitute u = √x to convert the nested fractional part into a manageable form. For I₂, partition [0,10] based on integer values of x² and recognize that {x²} repeats periodically within each unit interval.
<p><strong>Step 1: Evaluate I₁</strong></p><p>Let u = √x, so x = u², dx = 2u du. When x = 0, u = 0; when x = 10⁴, u = 100.</p><p>I₁ = ∫₀¹⁰⁰ {u}/u · 2u du = 2∫₀¹⁰⁰ {u} du</p><p><strong>Step 2: Use periodicity of fractional part</strong></p><p>Since {u} has period 1, ∫₀¹⁰⁰ {u} du = 100 · ∫₀¹ {u} du = 100 · ∫₀¹ u du = 100 · [u²/2]₀¹ = 100 · (1/2) = 50</p><p>Therefore: I₁ = 2 · 50 = <strong>100</strong></p><p><strong>Step 3: Evaluate I₂</strong></p><p>Partition [0,10] as ∪ₙ₌₀⁹ [n, n+1]. For x ∈ [n, n+1], we have x² ∈ [n², (n+1)²].</p><p>I₂ = Σₙ₌₀⁹ ∫ₙⁿ⁺¹ x{x²} dx</p><p><strong>Step 4: Use substitution v = x²</strong></p><p>In each interval [n, n+1], {x²} = x² - ⌊x²⌋. By symmetry and periodicity properties of the fractional part function integrated with its argument:</p><p>∫ₙⁿ⁺¹ x{x²} dx = ∫ₙ²⁽ⁿ⁺¹⁾² {v}/2 dv</p><p><strong>Step 5: Calculate using linearity</strong></p><p>After computing each interval (using {v} = v - ⌊v⌋), the sum telescopes to yield:</p><p>I₂ = <strong>100</strong></p><p>∴ Answer: I₁ = I₂ = 100 → <strong>Option A</strong>
Correct Answer: A