A-LevelFurther MathematicsVectors9231/12 · May/June 2012 · Q1114 Marks

Answer only one of the following two alternatives. EITHER Show that ∫ (from 0 to π) eˣ sin x dx = (1 + e^π) / 2. Given that Iₙ = ∫ (from 0 to π) eˣ sinⁿ x dx, show that, for n ≥ 2, Iₙ = n(n − 1) ∫ (from 0 to π) eˣ cos²x sinⁿ⁻²x dx – nIₙ and deduce that (n² + 1)Iₙ = n(n − 1)Iₙ₋₂. A curve has equation y = eˣ sin⁵ x. Find, in an exact form, the mean value of y over the interval 0 < x < π.

Show that ∫ (from 0 to π) eˣ sin x dx = (1 + e^π) / 2.[4]
Given that Iₙ = ∫ (from 0 to π) eˣ sinⁿ x dx, show that, for n ≥ 2, Iₙ = n(n − 1) ∫ (from 0 to π) eˣ cos²x sinⁿ⁻²x dx – nIₙ and deduce that (n² + 1)Iₙ = n(n − 1)Iₙ₋₂.[6]
A curve has equation y = eˣ sin⁵ x. Find, in an exact form, the mean value of y over the interval 0 < x < π.[4]

Mark scheme

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

[M1] I = ∫eˣ sin x dx = -eˣ cos x + ∫eˣ cos x dx
[A1] = -eˣ cos x + eˣ sin x - ∫eˣ sin x dx
[M1A1] ∴ 2I = eˣ(sin x – cos x) | from 0 to π = (e^π(0 – (-1))) – (e^0(0 – 1)) = (e^π + 1) ⇒ I = (1 + e^π)/2 (AG)
[M1] Iₙ = [sinⁿ x eˣ] from 0 to π - ∫eˣ (n sinⁿ⁻¹ x cos x)dx
[A1] = 0 - [n sinⁿ⁻¹ x cos x eˣ] from 0 to π + n ∫eˣ (cos²x(n-1)sinⁿ⁻² x – sinⁿ⁻¹ x sin x)dx
[A1] = 0 + n(n-1) ∫eˣ cos²x sinⁿ⁻²x dx - nIₙ (AG)
[M1A1] ∴ (n+1)Iₙ = n(n −1)Iₙ₋₂ – n(n −1)Iₙ
[A1] ∴ (n² +1)Iₙ = n(n −1)Iₙ₋₂ (AG)
[M1] I₅ = 20/26 I₃ = 6/13 I₁
[A1] I₅ = 3(1+e^π)/13
[M1A1] Mean value = (1/π) ∫eˣ sin⁵x dx = 3(1+e^π)/(13π)

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