A-LevelFurther MathematicsIntegration9231/12 · May/June 2015 · Q79 Marks
Let In = ∫₀^π xⁿ sin x dx, where n is a non-negative integer. Show that In = n(π)ⁿ⁻¹ – n(n − 1)In-2, for n ≥ 2. Find the exact value of I₄.
Mark scheme
Marking points from the official mark scheme (May/June 2015). The mark or mark type is shown in brackets.
- [M1A1] Iₙ = ∫ xⁿ sin x dx = [-xⁿ cos x]₀^π + ∫ n xⁿ⁻¹ cos x dx (LNR)
- [M1A1] = 0 + [n xⁿ⁻¹ sin x]₀^π - ∫ n(n-1)xⁿ⁻² sin x dx (LR)
- [A1] = n(π)ⁿ⁻¹ - n(n-1)Iₙ₋₂ (AG)
- [B1] I₀ = ∫₀^π sin x dx = [-cos x]₀^π = 1
- [M1A1] I₂ = π - 2, using reduction formula.
- [A1A1] I₄ = 4(π³) - 12(π-2) = π³ - 12π + 24 (If I₂ found by integration M1A1 (if correct) then M1A1 for I₄ from reduction formula.)
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