Date | May 2008 | Marks available | 9 | Reference code | 08M.3ca.hl.TZ2.2 |
Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ2 |
Command term | Find | Question number | 2 | Adapted from | N/A |
Question
Find the exact value of \(\int_0^\infty {\frac{{{\text{d}}x}}{{(x + 2)(2x + 1)}}} \).
Markscheme
Let \(\frac{1}{{(x + 2)(2x + 1)}} = \frac{A}{{x + 2}} + \frac{B}{{2x + 1}} = \frac{{A(2x + 1) + B(x + 2)}}{{(x + 2)(2x + 1)}}\) M1A1
\(x = - 2 \to A = - \frac{1}{3}\) A1
\(x = - \frac{1}{2} \to B = \frac{2}{3}\) A1 N3
\(I = \frac{1}{3}\int_0^h {\left[ {\frac{2}{{(2x + 1)}} - \frac{1}{{(x + 2)}}} \right]{\text{d}}x} \) M1
\( = \frac{1}{3}\left[ {\ln (2x + 1) - \ln (x + 2)} \right]_0^h\) A1
\( = \frac{1}{3}\left[ {\mathop {\lim }\limits_{h \to \infty } \left( {\ln \left( {\frac{{2h + 1}}{{h + 2}}} \right)} \right) - \ln \frac{1}{2}} \right]\) A1
\( = \frac{1}{3}\left( {\ln 2 - \ln \frac{1}{2}} \right)\) A1
\( = \frac{2}{3}\ln 2\) A1
Note: If the logarithms are not combined in the third from last line the last three A1 marks cannot be awarded.
Total [9 marks]
Examiners report
Not a difficult question but combination of the logarithms obtained by integration was often replaced by a spurious argument with infinities to get an answer. \(\log (\infty + 1)\) was often seen.