A-LevelFurther MathematicsIntegration9231/12 · May/June 2013 · Q48 Marks
Let I_n = ∫₀¹ 1/(1 + x²)^n dx. Prove that, for every positive integer n, 2nI_{n+1} = 2⁻ⁿ + (2n − 1)I_n. Given that I₁ = π/4, find the exact value of I₃.
- (a)Let I_n = ∫₀¹ 1/(1 + x²)^n dx. Prove that, for every positive integer n, 2nI_{n+1} = 2⁻ⁿ + (2n − 1)I_n.[5]
- (b)Given that I₁ = π/4, find the exact value of I₃.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2013). The mark or mark type is shown in brackets.
- M1A1 Integrates by parts. I_n = [x/(1+x²)^n]₀¹ + ∫₀¹ n(1+x²)^(-n-1).2x² dx
- M1A1 Rearranges. = 2⁻ⁿ + 2n ∫₀¹ (1+x²)^(-n-1) (1+x²-1) dx
- A1 Obtains result. ∴ 2nI_{n+1} = 2⁻ⁿ + (2n-1)I_n. (AG)
- M1A1 Uses reduction formula to find I₂. 2I₂ = 2⁻¹ + (2-1)I₁ ⇒ I₂ = 1/4 + 1/8π
- A1 Uses reduction formula to find I₃. 4I₃ = 2⁻² + (4-1)I₂ ⇒ I₃ = 1/4 + 3/32π
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