The diagram shows the curve with equation y = 8e⁻¹⁄₂ˣ - 1. The curve meets the axes at the points A and B. The shaded region is bounded by the curve and the line segment AB. [Figure 5.1]

Mark scheme

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

1. [M1] Attempt solution of 8e⁻¹⁄₂ˣ −1=0 with use of a relevant logarithm property E.g. -1/2x = ln 8
2. [A1] Confirm x=6ln2 AG - necessary detail needed. Need to see ln8=3ln2 or 64 = 2^6 OE used.
3. [M1] Substitute x=6ln2 in 8e⁻¹⁄₂ˣ −1 and show use of a relevant logarithm property
4. [A1] Obtain 8e⁻³ˡⁿ² –1 and hence 8eˡⁿ(¹⁄₈) –1 or equivalent and verify answer 0 AG - necessary detail needed.
5. [M1] Integrate 8e⁻¹⁄₂ˣ -1 to obtain the form ke⁻¹⁄₂ˣ − x
6. [A1] Obtain correct -16e⁻¹⁄₂ˣ - x
7. [A1] Apply limits 0 and 6ln2 to obtain -16e⁻¹⁄₂ˣ |⁰_⁶ˡⁿ² −6ln2+16 OE 14-6ln2
8. [M1] Attempt area of triangle minus area under curve Allow decimals for this mark (14.56).
9. [A1] Obtain ½×6ln2×7-(14-6ln2) or equivalent, and hence 27ln2-14