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Date	November 2019	Marks available	6	Reference code	19N.1.AHL.TZ0.H_2
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Find	Question number	H_2	Adapted from	N/A

Question

Given that $\int_0^{{\text{ln}}\,k} {{{\text{e}}^{2x}}} {\text{d}}x = 12$ , find the value of $k$ .

Markscheme

$\frac{1}{2}{{\text{e}}^{2x}}$ seen (A1)

attempt at using limits in an integrated expression $\left( {\left[ {\frac{1}{2}{{\text{e}}^{2x}}} \right]_0^{{\text{ln}}\,k} = \frac{1}{2}{{\text{e}}^{2\,{\text{ln}}\,k}} - \frac{1}{2}{{\text{e}}^0}} \right)$ (M1)

$= \frac{1}{2}{{\text{e}}^{{\text{ln}}\,{k^2}}} - \frac{1}{2}{{\text{e}}^0}$ (A1)

Setting their equation $= 12$ M1

Note: their equation must be an integrated expression with limits substituted.

$\frac{1}{2}{k^2} - \frac{1}{2} = 12$ A1

$\left( {{k^2} = 25 \Rightarrow } \right)k = 5$ A1

Note: Do not award final A1 for $k = \pm 5$ .

[6 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.10—Indefinite integration, reverse chain, by substitution

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Topic 5 —Calculus » SL 5.11—Definite integrals, areas under curve onto x-axis and areas between curves

Topic 5 —Calculus