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+x12∫dxx32+x12.
[4]
a.
Hence find the value of 129∫1dxx32+x12129∫1dxx32+x12, expressing your answer in the form arctan qq, where q∈Q.
[3]
b.
Markscheme
dudx=12x−12 (accept du=12x−12dx 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.
129∫1dxx32+x12 = arctan 3 − arctan 1 A1
tan(arctan 3 − arctan 1) = 3−11+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.