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Date May 2017 Marks available 4 Reference code 17M.1.sl.TZ1.5
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 5 Adapted from N/A

Question

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

[4]
a.

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

[3]
b.

Markscheme

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]

a.

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]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 6 - Calculus » 6.4 » Integration by inspection, or substitution of the form \(\mathop \int \nolimits f\left( {g\left( x \right)} \right)g'\left( x \right){\text{d}}x\) .

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