Let I_n denote ∫ (from 0 to ∞) xⁿe⁻²ˣ dx.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2012). The mark or mark type is shown in brackets.

1. [M1] Integrates by parts to obtain reduction formula: ∫ (from 0 to ∞) xⁿe⁻²ˣ dx = [-xⁿe⁻²ˣ/2] (from 0 to ∞) - ∫ (from 0 to ∞) n xⁿ⁻¹ (-e⁻²ˣ/2) dx
2. [A1] = (n/2)I_(n-1) (AG)
3. [B1] States proposition: P_n: I_n = n! / 2^(n+1)
4. [B1] Proves base case: I₀ = ∫ (from 0 to ∞) e⁻²ˣ dx = [-e⁻²ˣ/2] (from 0 to ∞) = 1/2. For n=1, I₁ = ½ * 1 * I₀ = 1/4. Also 1!/2^(1+1) = 1/4. So P₁ is true.
5. [B1] Shows P_k ⇒ P_(k+1): P_k: I_k = k! / 2^(k+1) for some integer k.
6. [M1A1] :. I_(k+1) = (k+1)/2 * I_k = (k+1)/2 * k! / 2^(k+1) = (k+1)! / 2^(k+2)
7. [A1] States conclusion: Hence by PMI P_n is true for all positive integers n.