Date | May 2011 | Marks available | 6 | Reference code | 11M.1.hl.TZ1.7 |
Level | HL only | Paper | 1 | Time zone | TZ1 |
Command term | Find | Question number | 7 | Adapted from | N/A |
Question
Markscheme
METHOD 1
area=√3∫0arctanxdx A1
attempting to integrate by parts M1
=[xarctanx]√30−√3∫0x11+x2dx A1A1
=[xarctanx]√30−[12ln(1+x2)]√30 A1
Note: Award A1 even if limits are absent.
=π√3−12ln4 A1
(=π√33−ln2)
METHOD 2
area=π√33−π3∫0tanydy M1A1A1
=π√33+[ln|cosy|]π30 M1A1
=π√33+ln12 (=π√33−ln2) A1
[6 marks]
Examiners report
Many candidates were able to write down the correct expression for the required area, although in some cases with incorrect integration limits. However, very few managed to achieve any further marks due to a number of misconceptions, in particular arctanx=cotx=cosxsinx. Candidates who realised they should use integration by parts were in general very successful in answering this question. It was pleasing to see a few alternative correct approaches to this question.