CAIE P2 2008 November — Question 5 6 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2008
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeDefinite integral with logarithmic form
DifficultyModerate -0.8 This is a straightforward definite integral requiring recognition of two standard logarithmic forms: ∫(1/x)dx = ln|x| and ∫(1/(2x+1))dx = (1/2)ln|2x+1|. The main steps are applying these standard results, substituting limits, and simplifying using log laws. While it requires careful algebraic manipulation at the end, there's no problem-solving or novel insight needed—just direct application of memorized techniques.
Spec1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)

5 Show that \(\int _ { 1 } ^ { 2 } \left( \frac { 1 } { x } - \frac { 4 } { 2 x + 1 } \right) \mathrm { d } x = \ln \frac { 18 } { 25 }\).

AnswerMarks Guidance
Integrate and state term \(\ln x\)B1
Obtain term of the form \(k\ln(2x + 1)\)M1
State correct term \(-2\ln(2x + 1)\)A1
Substitute limits correctlyM1
Use law for the logarithm of a product, quotient or powerM1
Obtain given answer correctlyA1 [6]
Integrate and state term $\ln x$ | B1 |
Obtain term of the form $k\ln(2x + 1)$ | M1 |
State correct term $-2\ln(2x + 1)$ | A1 |
Substitute limits correctly | M1 |
Use law for the logarithm of a product, quotient or power | M1 |
Obtain given answer correctly | A1 | [6]
5 Show that $\int _ { 1 } ^ { 2 } \left( \frac { 1 } { x } - \frac { 4 } { 2 x + 1 } \right) \mathrm { d } x = \ln \frac { 18 } { 25 }$.

\hfill \mbox{\textit{CAIE P2 2008 Q5 [6]}}