| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2003 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Stationary points then area/volume |
| Difficulty | Standard +0.3 This question requires finding a stationary point using the product rule and differentiation (standard technique), then integration by parts to find an area. Both are routine P3/C4 techniques with no novel insight required. The integration by parts for (3-x)e^(-2x) is straightforward, and the question clearly signposts what to do at each stage. Slightly above average difficulty due to the two-part nature and requiring careful execution of integration by parts, but well within standard A-level pure maths expectations. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use product or quotient rule to find derivative | M1 | |
| Obtain derivative in any correct form | A1 | |
| Equate derivative to zero and solve a linear equation in \(x\) | M1 | |
| Obtain answer \(3\frac{1}{2}\) only | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State first step of the form \(\pm\frac{1}{2}(3-x)e^{-2x} \pm \frac{1}{2}\int e^{-2x}dx\), with or without 3 | M1 | |
| State correct first step e.g. \(-\frac{1}{2}(3-x)e^{-2x} - \frac{1}{2}\int e^{-2x}dx\), or equivalent, with or without 3 | A1 | |
| Complete the integration correctly obtaining \(-\frac{1}{2}(3-x)e^{-2x} + \frac{1}{4}e^{-2x}\), or equivalent | A1 | |
| Substitute limits \(x = 0\) and \(x = 3\) correctly in the complete integral | M1 | |
| Obtain answer \(\frac{1}{4}(5 + e^{-6})\), or exact equivalent (allow \(e^0\) in place of 1) | A1 |
## Question 6(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use product or quotient rule to find derivative | M1 | |
| Obtain derivative in any correct form | A1 | |
| Equate derivative to zero and solve a linear equation in $x$ | M1 | |
| Obtain answer $3\frac{1}{2}$ only | A1 | |
**Total: [4]**
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## Question 6(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State first step of the form $\pm\frac{1}{2}(3-x)e^{-2x} \pm \frac{1}{2}\int e^{-2x}dx$, with or without 3 | M1 | |
| State correct first step e.g. $-\frac{1}{2}(3-x)e^{-2x} - \frac{1}{2}\int e^{-2x}dx$, or equivalent, with or without 3 | A1 | |
| Complete the integration correctly obtaining $-\frac{1}{2}(3-x)e^{-2x} + \frac{1}{4}e^{-2x}$, or equivalent | A1 | |
| Substitute limits $x = 0$ and $x = 3$ correctly in the complete integral | M1 | |
| Obtain answer $\frac{1}{4}(5 + e^{-6})$, or exact equivalent (allow $e^0$ in place of 1) | A1 | |
**Total: [5]**
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{79efa364-da5a-4888-85a9-dc4de1e0908e-3_543_825_287_660}
The diagram shows the curve $y = ( 3 - x ) \mathrm { e } ^ { - 2 x }$ and its minimum point $M$. The curve intersects the $x$-axis at $A$ and the $y$-axis at $B$.\\
(i) Calculate the $x$-coordinate of $M$.\\
(ii) Find the area of the region bounded by $O A , O B$ and the curve, giving your answer in terms of e.
\hfill \mbox{\textit{CAIE P3 2003 Q6 [9]}}