Definite Integration
Beta function / definite integrals
Grade None
Question:
<p>The value of \(\displaystyle\int_0^1 x(1-x)^{99}\,dx\) is ______ \(\times 10^{-5}\).</p>
<p>\(\dfrac{1}{10100}\)</p>
<p>\(\dfrac{11}{10100}\)</p>
<p>\(\dfrac{1}{10010}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Use integration by parts with u = x and dv = (1-x)^99 dx, or recognize the Beta function pattern: ∫₀¹ x^(m-1)(1-x)^(n-1) dx = B(m,n) = Γ(m)Γ(n)/Γ(m+n) = (m-1)!(n-1)!/(m+n-1)!
<p><strong>Step 1:</strong> Recognize the integral fits the Beta function form B(m,n) = ∫₀¹ x^(m-1)(1-x)^(n-1) dx</p><p>Here: ∫₀¹ x·(1-x)⁹⁹ dx = ∫₀¹ x¹(1-x)⁹⁹ dx, so m-1=1 → m=2, and n-1=99 → n=100</p><p><strong>Step 2:</strong> Apply Beta function formula: B(2,100) = Γ(2)Γ(100)/Γ(102) = (1!)·(99!)/(101!)</p><p><strong>Step 3:</strong> Simplify: = (1·99!)/(101·100·99!) = 1/(101·100) = 1/10100</p><p><strong>Step 4:</strong> Convert to required form: 1/10100 ≈ 0.0000990099... ≈ 9.901 × 10⁻⁵</p><p>∴ Answer: <strong>9.901</strong> (or approximately <strong>10</strong> when rounded, or <strong>1/10100</strong> as exact value)</p>
Correct Answer: A