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Date May 2017 Marks available 3 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 xex21dx.

[4]
a.

Find f(x), given that f(x)=xex21 and f(1)=3.

[3]
b.

Markscheme

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

egu=x21, du=2x, 2xex21dx

correct expression     (A1)

eg122xex21dx, 12eudu

12ex21+c     A2     N4

 

Notes: Award A1 if missing “+c”.

 

[4 marks]

a.

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

eg12e0, 12e11=3

correct working     (A1)

eg12+c=3, c=2.5

f(x)=12ex21+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 f(g(x))g(x)dx .

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