The diagram shows the curves y = √2π – 2x and y = sin²x for 0 ≤ x ≤ π. The shaded region is bounded by the two curves and the line x = 0. Find the exact area of the shaded region. [Figure 7.1]

Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

1. [M1] Integrate √(2π-2x) to obtain form k₁(2π – 2x)^3/2 any constant k₁
2. [A1] Obtain correct -(2π−2x)^3/2 / 3 OE
3. [A1] Apply limits to obtain (2π)^(3/2) / 3 or exact equivalent at this stage or later
4. [B1] Use identity -sin² x = −½+½cos2x OE
5. [M1] Integrate to obtain form k₂x + k₃ sin 2x any constants k₂, k₃
6. [A1] Obtain correct -½x+¼sin 2x OE if area taken as positive
7. [M1] Substituting correct limits into their -½x-¼sin 2x Can be SOI by ±½π Allow one slip
8. [A1] Obtain (2π)^(3/2) / 3 - ½π or exact equivalent CWO