CAIE P2 2012 November — Question 6 7 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2012
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeTrapezium rule with stated number of strips
DifficultyModerate -0.3 Part (a) is a straightforward trapezium rule application with only 2 intervals requiring basic calculator work and formula recall. Part (b) involves routine algebraic manipulation (expanding and simplifying before integrating) followed by standard integration of exponential terms. Both parts are mechanical applications of standard techniques with no problem-solving or insight required, making this slightly easier than average.
Spec1.06d Natural logarithm: ln(x) function and properties1.08b Integrate x^n: where n != -1 and sums1.09f Trapezium rule: numerical integration

6
  1. Use the trapezium rule with two intervals to estimate the value of $$\int _ { 0 } ^ { 1 } \frac { 1 } { 6 + 2 \mathrm { e } ^ { x } } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.
  2. Find \(\int \frac { \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } } { \mathrm { e } ^ { 2 x } } \mathrm {~d} x\).

AnswerMarks Guidance
(a) State or imply correct ordinates 0.125, 0.08743…, 0.21511…B1
Use correct formula, or equivalent, correctly with \(h = 0.5\) and three ordinatesM1
Obtain answer 0.11 with no errors seenA1 [3]
(b) Attempt to expand brackets and divide by \(e^{2x}\)M1
Integrate a term of form \(ke^{x}\) or \(ke^{-2x}\) correctlyA1
Obtain 2 correct termsA1
Fully correct integral \(x + 4e^{-x} - 2e^{-2x} + c\)A1 [4]
(a) State or imply correct ordinates 0.125, 0.08743…, 0.21511… | B1 |
Use correct formula, or equivalent, correctly with $h = 0.5$ and three ordinates | M1 |
Obtain answer 0.11 with no errors seen | A1 | [3]

(b) Attempt to expand brackets and divide by $e^{2x}$ | M1 |
Integrate a term of form $ke^{x}$ or $ke^{-2x}$ correctly | A1 |
Obtain 2 correct terms | A1 |
Fully correct integral $x + 4e^{-x} - 2e^{-2x} + c$ | A1 | [4]
6
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule with two intervals to estimate the value of

$$\int _ { 0 } ^ { 1 } \frac { 1 } { 6 + 2 \mathrm { e } ^ { x } } \mathrm {~d} x$$

giving your answer correct to 2 decimal places.
\item Find $\int \frac { \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } } { \mathrm { e } ^ { 2 x } } \mathrm {~d} x$.
\end{enumerate}

\hfill \mbox{\textit{CAIE P2 2012 Q6 [7]}}