| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Substitution then integration by parts |
| Difficulty | Standard +0.8 This question requires two sophisticated techniques in sequence: first executing a non-trivial substitution with careful handling of limits and the differential, then applying integration by parts to a logarithmic integrand. While each technique is standard for P3/C4 level, the combination and the need to simplify (2x-2)ln x before integrating by parts makes this moderately challenging, above average difficulty. |
| Spec | 1.08h Integration by substitution1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply \(dx = 2t\,dt\) or equivalent | B1 | |
| Express the integral in terms of \(x\) and \(dx\) | M1 | |
| Obtain given answer \(\int_1^5 (2x-2)\ln x\,dx\), including change of limits | A1 | [3] AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt integration by parts obtaining \((ax^2 + bx)\ln x \pm \int(ax^2+bx)\frac{1}{x}\,dx\) or equivalent | M1 | |
| Obtain \((x^2 - 2x)\ln x - \int(x^2-2x)\frac{1}{x}\,dx\) or equivalent | A1 | |
| Obtain \((x^2-2x)\ln x - \frac{1}{2}x^2 + 2x\) | A1 | |
| Use limits correctly having integrated twice | M1 | |
| Obtain \(15\ln 5 - 4\) or exact equivalent | A1 | [5] |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply $dx = 2t\,dt$ or equivalent | B1 | |
| Express the integral in terms of $x$ and $dx$ | M1 | |
| Obtain given answer $\int_1^5 (2x-2)\ln x\,dx$, including change of limits | A1 | [3] **AG** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt integration by parts obtaining $(ax^2 + bx)\ln x \pm \int(ax^2+bx)\frac{1}{x}\,dx$ or equivalent | M1 | |
| Obtain $(x^2 - 2x)\ln x - \int(x^2-2x)\frac{1}{x}\,dx$ or equivalent | A1 | |
| Obtain $(x^2-2x)\ln x - \frac{1}{2}x^2 + 2x$ | A1 | |
| Use limits correctly having integrated twice | M1 | |
| Obtain $15\ln 5 - 4$ or exact equivalent | A1 | [5] |
**Note:** Equivalent for M1 is $(2x-2)(ax\ln x + bx) - \int(ax\ln x + bx)\,2dx$
---7 The integral $I$ is defined by $I = \int _ { 0 } ^ { 2 } 4 t ^ { 3 } \ln \left( t ^ { 2 } + 1 \right) \mathrm { d } t$.\\
(i) Use the substitution $x = t ^ { 2 } + 1$ to show that $I = \int _ { 1 } ^ { 5 } ( 2 x - 2 ) \ln x \mathrm {~d} x$.\\
(ii) Hence find the exact value of $I$.
\hfill \mbox{\textit{CAIE P3 2011 Q7 [8]}}