A-LevelMathematicsIntegration9709/22 · Oct/Nov 2022 · Q69 Marks

The diagram shows the curves y = 6 / (3x + 2) and y = 3e⁻ˣ – 3 for values of x between 0 and 4. The shaded region is bounded by the two curves and the lines x = 0 and x = 4. [Figure 6.1] Find the exact area of the shaded region, giving your answer in the form ln a + b + ceᵈ.

(a)The diagram shows the curves y = 6 / (3x + 2) and y = 3e⁻ˣ – 3 for values of x between 0 and 4. The shaded region is bounded by the two curves and the lines x = 0 and x = 4. Find the exact area of the shaded region, giving your answer in the form ln a + b + ceᵈ.[9]

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Mark scheme

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

[*M1] Integrate 6/(3x + 2) to obtain form k₁ ln(3x + 2) for any constant k₁.
[A1] Obtain correct 2ln(3x + 2)
[DM1] Apply limits correctly
[A1] Obtain 2ln14 - 2ln2 and hence ln 49 at this stage or later.
[M1] Integrate 3e⁻ˣ - 3 to obtain form k₂e⁻ˣ + k₃x for any non-zero constants k₂, k₃.
[A1] Obtain correct -3e⁻ˣ - 3x
[A1] Apply limits to obtain -3e⁻⁴ - 12 + 3 OE; implied if ∫(y₁ - y₂)dx approach used.
[M1] Use correct procedure to find exact total area
[A1] Obtain ln 49 + 9 + 3e⁻⁴

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