| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2011 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Curve with minimum point |
| Difficulty | Standard +0.3 Part (i) requires standard differentiation using the product rule and solving for a stationary point—routine A-level calculus. Part (ii) involves integration by parts applied twice, which is a standard P3/C4 technique. Both parts follow predictable methods with no novel insight required, making this slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Obtain correct derivative in any form | A1 | |
| Equate derivative to zero and solve for \(x\) | M1 | |
| Obtain answer \(x = e^{-\frac{1}{2}}\), or equivalent | A1 | |
| Obtain answer \(y = -\frac{1}{2}e^{-1}\), or equivalent | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Obtain \(\frac{1}{3}x^3 \ln x - \frac{1}{3}\int x^2 dx\), or equivalent | M1* A1 | |
| Integrate again and obtain \(\frac{1}{3}x^3 \ln x - \frac{1}{9}x^3\), or equivalent | A1 | |
| Use limits \(x = 1\) and \(x = e\), having integrated twice | M1(dep*) | |
| Obtain answer \(\frac{1}{9}(2e^3 + 1)\), or exact equivalent | A1 | [5] |
**(i) Use product rule**
Obtain correct derivative in any form | A1 |
Equate derivative to zero and solve for $x$ | M1 |
Obtain answer $x = e^{-\frac{1}{2}}$, or equivalent | A1 |
Obtain answer $y = -\frac{1}{2}e^{-1}$, or equivalent | A1 | [5]
**(ii) Attempt integration by parts reaching $kx^3 \ln x \pm k \int x^3 \cdot \frac{1}{x}dx$**
Obtain $\frac{1}{3}x^3 \ln x - \frac{1}{3}\int x^2 dx$, or equivalent | M1* A1 |
Integrate again and obtain $\frac{1}{3}x^3 \ln x - \frac{1}{9}x^3$, or equivalent | A1 |
Use limits $x = 1$ and $x = e$, having integrated twice | M1(dep*) |
Obtain answer $\frac{1}{9}(2e^3 + 1)$, or exact equivalent | A1 | [5]
[SR: An attempt reaching $ax^2(x\ln x - x) + b \int 2x(x\ln x - x)dx$ scores M1. Then give the first A1 for $I = x^2(x\ln x - x) - 2I + \int 2x^2 dx$, or equivalent.]9\\
\includegraphics[max width=\textwidth, alt={}, center]{f421f03c-57c9-4feb-91b9-a7f9b12f96ce-3_666_956_1231_593}
The diagram shows the curve $y = x ^ { 2 } \ln x$ and its minimum point $M$.\\
(i) Find the exact values of the coordinates of $M$.\\
(ii) Find the exact value of the area of the shaded region bounded by the curve, the $x$-axis and the line $x = \mathrm { e }$.
\hfill \mbox{\textit{CAIE P3 2011 Q9 [10]}}