| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with logarithmic form |
| Difficulty | Moderate -0.3 This is a multi-part question testing standard integration techniques. Part (a) requires algebraic manipulation then basic integration of exponential functions; part (b) is a straightforward logarithmic integration; part (c) applies the trapezium rule mechanically. All parts are routine applications of standard methods with no problem-solving insight required, making it slightly easier than average but not trivial due to the algebraic manipulation needed in part (a). |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Obtain integrand \(2e^{-2x}+\frac{1}{2}e^{-x}\) | B1 | |
| Obtain integral of form \(k_1e^{-2x}+k_2e^{-x}\) | M1 | |
| Obtain answer \(-e^{-2x}-\frac{1}{2}e^{-x}\), condoning absence of \(+c\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Integrate to obtain \(\frac{1}{2}\ln(2x+5)\) | B1 | |
| Show correct use of \(p\ln k=\ln k^p\) law at least once | M1 | |
| Show correct use of \(\ln m - \ln n = \ln\frac{m}{n}\) law | M1 | |
| Obtain \(\ln\frac{5}{3}\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply correct ordinates \(\log 2\), \(\log 5\), \(\log 8\) or decimal equivalents | B1 | |
| Use correct formula, or equivalent, correctly with \(h=3\) and 3 ordinates | M1 | |
| Obtain answer 3.9 with no errors seen | A1 | [3] |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtain integrand $2e^{-2x}+\frac{1}{2}e^{-x}$ | B1 | |
| Obtain integral of form $k_1e^{-2x}+k_2e^{-x}$ | M1 | |
| Obtain answer $-e^{-2x}-\frac{1}{2}e^{-x}$, condoning absence of $+c$ | A1 | [3] |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrate to obtain $\frac{1}{2}\ln(2x+5)$ | B1 | |
| Show correct use of $p\ln k=\ln k^p$ law at least once | M1 | |
| Show correct use of $\ln m - \ln n = \ln\frac{m}{n}$ law | M1 | |
| Obtain $\ln\frac{5}{3}$ | A1 | [4] |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply correct ordinates $\log 2$, $\log 5$, $\log 8$ or decimal equivalents | B1 | |
| Use correct formula, or equivalent, correctly with $h=3$ and 3 ordinates | M1 | |
| Obtain answer 3.9 with no errors seen | A1 | [3] |
---6
\begin{enumerate}[label=(\alph*)]
\item Find $\int \frac { 4 + \mathrm { e } ^ { x } } { 2 \mathrm { e } ^ { 2 x } } \mathrm {~d} x$.
\item Without using a calculator, find $\int _ { 2 } ^ { 10 } \frac { 1 } { 2 x + 5 } \mathrm {~d} x$, giving your answer in the form $\ln k$.
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{f85c4010-17b1-441c-ae8a-e77573d1b0c3-3_446_755_580_735}
The diagram shows the curve $y = \log _ { 10 } ( x + 2 )$ for $0 \leqslant x \leqslant 6$. The region bounded by the curve and the lines $x = 0 , x = 6$ and $y = 0$ is denoted by $R$. Use the trapezium rule with 2 strips to find an estimate of the area of $R$, giving your answer correct to 1 decimal place.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2016 Q6 [10]}}