It is given that I_n = ∫_0^1 x^n / √(1 + 2x) dx. Show that, for n ≥ 1, (2n + 1)I_n = √3 - nI_{n-1}. Show that I_3 = 2/35 (√3 + 1).

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2013). The mark or mark type is shown in brackets.

1. [M1A1] (a) Integrates by parts. I_n = [x^n(1+2x)^(1/2)]_0^1 - ∫_0^1 n x^(n-1)(1+2x)^(1/2)dx = √3 - n∫_0^1 x^(n-1)/√(1+2x) dx - 2n∫_0^1 x^n/(1+2x)^(3/2) dx
2. [A1] Obtains result. ⇒(2n+1)I_n = √3 - nI_{n-1} (AG)
3. [M1] Alternatively: d/dx {x^n(1+2x)^(-1/2)} = nx^(n-1)(1+2x)^(-1/2) - x^n(1+2x)^(-3/2)
4. [A1] ⇒ x^n/(1+2x)^(1/2) = (1+2x)x^(n-1)/(1+2x)^(1/2) - (1+2x)x^n/(1+2x)^(3/2)
5. [A1] ∫_0^1 [x^n(1+2x)^(-1/2)] dx = ∫_0^1 nx^(n-1)(1+2x)^(-1/2) dx - ∫_0^1 x^n(1+2x)^(-3/2) dx
6. [B1] (b) Finds I_0 (or I_1). I_0 = [√(1+2x)]_0^1 = √3 − 1
7. [M1] Uses Red. Form. 3I_1 = √3 − I_0 ⇒ I_1 = 1/3
8. [A1A1] Finds I_2 and I_3. ⇒ I_2 = (√3 - 2/3)/5 = (3√3-2)/15; I_3 = (√3 - 3(3√3-2)/15)/7 = (5√3 - 3√3 + 2)/35 = (2√3 + 2)/35 = 2/35(√3+1) (AG)