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Date	May 2017	Marks available	3	Reference code	17M.1.SL.TZ1.S_5
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Find	Question number	S_5	Adapted from	N/A

Question

Find $\int {x{{\text{e}}^{{x^2} - 1}}{\text{d}}x}$ .

[4]

Find $f(x)$ , given that $f’(x) = x{{\text{e}}^{{x^2} - 1}}$ and $f( - 1) = 3$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach to set up integration by substitution/inspection (M1)

eg $\,\,\,\,\,$ $u = {x^2} - 1,{\text{ d}}u = 2x,{\text{ }}\int {2x{{\text{e}}^{{x^2} - 1}}{\text{d}}x}$

correct expression (A1)

eg $\,\,\,\,\,$ $\frac{1}{2}\int {2x{{\text{e}}^{{x^2} - 1}}{\text{d}}x,{\text{ }}\frac{1}{2}\int {{{\text{e}}^u}{\text{d}}u} }$

$\frac{1}{2}{{\text{e}}^{{x^2} - 1}} + c$ A2 N4

Notes: Award A1 if missing “ $+ c$ ”.

[4 marks]

substituting $x = - 1$ into their answer from (a) (M1)

eg $\,\,\,\,\,$ $\frac{1}{2}{{\text{e}}^0},{\text{ }}\frac{1}{2}{{\text{e}}^{1 - 1}} = 3$

correct working (A1)

eg $\,\,\,\,\,$ $\frac{1}{2} + c = 3,{\text{ }}c = 2.5$

$f(x) = \frac{1}{2}{{\text{e}}^{{x^2} - 1}} + 2.5$ A1 N2

[3 marks]

Examiners report

[N/A]

Syllabus sections

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

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