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Date May 2017 Marks available 4 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  x e x 2 1 d x .

[4]
a.

Find f ( x ) , given that f ( x ) = x e x 2 1 and f ( 1 ) = 3 .

[3]
b.

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 ,  d u = 2 x ,   2 x e x 2 1 d x

correct expression     (A1)

eg 1 2 2 x e x 2 1 d x ,   1 2 e u d u

1 2 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 1 2 e 0 ,   1 2 e 1 1 = 3

correct working     (A1)

eg 1 2 + c = 3 ,   c = 2.5

f ( x ) = 1 2 e x 2 1 + 2.5     A1     N2

[3 marks]

b.

Examiners report

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

Syllabus sections

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

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