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Date May 2018 Marks available 4 Reference code 18M.1.hl.TZ2.8
Level HL only Paper 1 Time zone TZ2
Command term Use and Find Question number 8 Adapted from N/A

Question

Use the substitution u=x12u=x12 to find dxx32+x12dxx32+x12.

[4]
a.

Hence find the value of 1291dxx32+x121291dxx32+x12, expressing your answer in the form arctan qq, where qQ.

[3]
b.

Markscheme

dudx=12x12 (accept du=12x12dx or equivalent)       A1

substitution, leading to an integrand in terms of u     M1

2uduu3+u or equivalent      A1

= 2 arctan (x)(+c)     A1

[4 marks]

 

a.

 

1291dxx32+x12 = arctan 3 − arctan 1     A1

tan(arctan 3 − arctan 1) = 311+3×1      (M1)

tan(arctan 3 − arctan 1) = 12

arctan 3 − arctan 1 = arctan 12     A1

[3 marks]

b.

Examiners report

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

Syllabus sections

Topic 6 - Core: Calculus » 6.7 » Integration by substitution.
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