The diagram shows parts of the graphs of y = 2+5eˣ and y = 4-3e²ˣ. Find the area of the shaded region. Give your answer in the form a+b ln3, where a and b are exact constants. [Figure 10.1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

1. [M1] Writes as a 3-term quadratic in eˣ in solvable form: 3e²ˣ +5eˣ -2[=0] soi
2. [M1 dep previous M1] Solve or factorises their 3-term quadratic in eˣ: eˣ = 1/3 or eˣ = 0.333[3...] [eˣ = -2]
3. [A1] x = ln(1/3) or -ln3 only solution, nfww
4. [M1 dep on use of 2 + 5eˣ=4 – 3e²ˣ to find the lower limit] Correct plan: ∫(from ln(1/3) to 0) (3e²ˣ + 5eˣ - 2) dx soi
5. [OR] ∫(from ln(1/3) to 0) (2+5eˣ)dx - ∫(from ln(1/3) to 0) (4-3e²ˣ)dx soi
6. [B2] -3/2 e²ˣ +5eˣ −2x OR (2x+5eˣ) and (4x-3/2e²ˣ)
7. [B1] for [∫3e²ˣdx=] 3/2 e²ˣ soi
8. [B1] for all other terms correct
9. [M1 FT correct substitution of limits 0 and k where k is exact into ae²ˣ + beˣ + cx or aex + bx or ae²ˣ + bx at least once, providing limits are consistent] 3/2 + 5 - 2ln(1/3) OR 5 - (2ln(1/3)+5e^(ln(1/3))) - 3/2 e^(2ln(1/3))
10. [OR] 2ln(1/3) - 4ln(1/3) - 3/2 e^(2ln(1/3)) oe
11. [M1 dep on correct plan for one correct term out of two or for a +2ln(1/3) with constant a] 14/3 - 2ln3 dep on all previous marks awarded
12. [A1] (implied by 10 marks total)