The curve C has equation `y = ½(eˣ + e⁻ˣ)` for `0 ≤ x ≤ ln 5`. Find

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2011). The mark or mark type is shown in brackets.

1. [M1A1] Uses mean value formula and integrates. M.V. = (1/(ln 5 - 0)) ∫₀^(ln 5) (½(eˣ + e⁻ˣ)) dx = (1/(2 ln 5)) [eˣ - e⁻ˣ]₀^(ln 5)
2. [M1A1] Substitutes limits and evaluates. = (1/(2 ln 5)) [ (e^(ln 5) - e^(-ln 5)) - (e⁰ - e⁰) ] = (1/(2 ln 5)) [ (5 - 1/5) - (1 - 1) ] = (1/(2 ln 5)) [24/5] = 12/(5 ln 5) (≈ 1.49)
3. [M1A1] Differentiates and finds 1+(y')². y' = ½(eˣ - e⁻ˣ) ⇒ 1+(y')² = 1 + (¼(eˣ - e⁻ˣ)²) = 1 + ¼(e²ˣ - 2 + e⁻²ˣ) = ¼(4 + e²ˣ - 2 + e⁻²ˣ) = ¼(e²ˣ + 2 + e⁻²ˣ) = ¼(eˣ + e⁻ˣ)²
4. [M1A1] Integrates and obtains result. s = ∫₀^(ln 5) √(¼(eˣ + e⁻ˣ)²) dx = ∫₀^(ln 5) ½(eˣ + e⁻ˣ) dx = ½ [eˣ - e⁻ˣ]₀^(ln 5) = ½ [ (5 - 1/5) - (1-1) ] = ½ [24/5] = 12/5
5. [M1] Uses surface area formula. S = 2π ∫₀^(ln 5) y √(1+(y')²) dx = 2π ∫₀^(ln 5) ½(eˣ + e⁻ˣ) * ½(eˣ + e⁻ˣ) dx = (π/2) ∫₀^(ln 5) (eˣ + e⁻ˣ)² dx
6. [M1] Integrates. = (π/2) ∫₀^(ln 5) (e²ˣ + 2 + e⁻²ˣ) dx = (π/2) [e²ˣ/2 + 2x - e⁻²ˣ/2]₀^(ln 5)
7. [A1] Substitutes limits and evaluates. = (π/2) [ (e^(2 ln 5)/2 + 2 ln 5 - e^(-2 ln 5)/2) - (e⁰/2 + 0 - e⁰/2) ] = (π/2) [ (25/2 + 2 ln 5 - 1/50) - (1/2 - 1/2) ] = (π/2) [25/2 + 2 ln 5 - 1/50] = π [25/4 + ln 5 - 1/100]
8. [A1] evaluates. ≈ 24.7