Definite Integration
Beta function / integration by parts
Grade 12
Question:
<p>The value of the integral \(I = \int_0^1 x(1-x)^n\, dx\) is</p>
<p>\(\dfrac{1}{n+1}\)</p>
<p>\(\dfrac{1}{n+2}\)</p>
<p>\(\dfrac{1}{n+1} - \dfrac{1}{n+2}\)</p>
<p>\(\dfrac{1}{n+1} + \dfrac{1}{n+2}\)</p>
Step-by-Step Solution
Key Concept: Use the Beta function property or integrate by parts twice: recognizing that ∫₀¹ x(1-x)ⁿ dx = B(2,n+1) = 1/(n+1)(n+2), where the Beta function relates to factorials.
<p><strong>Step 1:</strong> Use integration by parts with u = x, dv = (1-x)ⁿ dx</p><p>Then du = dx, v = -(1-x)ⁿ⁺¹/(n+1)</p><p><strong>Step 2:</strong> Apply integration by parts formula:</p><p>I = [-x(1-x)ⁿ⁺¹/(n+1)]₀¹ + ∫₀¹ (1-x)ⁿ⁺¹/(n+1) dx</p><p>The first term vanishes at both limits (at x=1: (1-1)ⁿ⁺¹=0; at x=0: 0·1=0)</p><p><strong>Step 3:</strong> Evaluate the remaining integral:</p><p>I = (1/(n+1)) ∫₀¹ (1-x)ⁿ⁺¹ dx = (1/(n+1)) · [-(1-x)ⁿ⁺²/(n+2)]₀¹</p><p>= (1/(n+1)) · (1/(n+2))</p><p><strong>Step 4:</strong> Simplify:</p><p>I = 1/[(n+1)(n+2)]</p><p>∴ Answer: <strong>1/[(n+1)(n+2)]</strong> or equivalently <strong>1/(n+1)(n+2)</strong></p>
Correct Answer: C