| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2003 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Area under curve using integration |
| Difficulty | Moderate -0.8 This is a straightforward integration question requiring only the standard integral of e^(2x) and solving a simple logarithmic equation. Part (i) is routine application of integration with substitution or recognition of the reverse chain rule, and part (ii) involves basic algebraic manipulation of exponentials and logarithms. Both parts are below average difficulty for A-level, being standard textbook exercises with no problem-solving insight required. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply indefinite integral of \(e^{2x}\) is \(\frac{1}{2}e^{2x}\), or equivalent | B1 | |
| Substitute correct limits correctly | M1 | |
| Obtain answer \(R = \frac{1}{2}e^{2p} - \frac{1}{2}\), or equivalent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(R = 5\) and use logarithmic method to obtain an equation in \(2p\) | M1* | |
| Solve for \(p\) | M1 (dep*) | |
| Obtain answer \(p = 1.2\) \((1.1989...)\) | A1 |
# Question 3(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply indefinite integral of $e^{2x}$ is $\frac{1}{2}e^{2x}$, or equivalent | B1 | |
| Substitute correct limits correctly | M1 | |
| Obtain answer $R = \frac{1}{2}e^{2p} - \frac{1}{2}$, or equivalent | A1 | |
**Total: [3]**
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# Question 3(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $R = 5$ and use logarithmic method to obtain an equation in $2p$ | M1* | |
| Solve for $p$ | M1 (dep*) | |
| Obtain answer $p = 1.2$ $(1.1989...)$ | A1 | |
**Total: [3]**
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{a31a4b4e-83a6-47d9-9679-3471b3da1b6e-2_488_664_863_737}
The diagram shows the curve $y = \mathrm { e } ^ { 2 x }$. The shaded region $R$ is bounded by the curve and by the lines $x = 0 , y = 0$ and $x = p$.\\
(i) Find, in terms of $p$, the area of $R$.\\
(ii) Hence calculate the value of $p$ for which the area of $R$ is equal to 5 . Give your answer correct to 2 significant figures.
\hfill \mbox{\textit{CAIE P2 2003 Q3 [6]}}