| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Integration of x^n·ln(x) |
| Difficulty | Moderate -0.3 This is a straightforward integration by parts question with a standard form (ln x with a power of x). The choice of u and dv is clear, the integration is routine, and evaluating at the limits e and 1 is simple since ln(1)=0 and ln(e)=1. Slightly easier than average due to the clean limits and standard technique. |
| Spec | 1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \frac{\ln x}{x^2}dx = \int x^{-2}\ln x\, dx = \frac{x^{-1}}{-1}\ln x - \int \frac{x^{-1}}{-1} \times \frac{1}{x}\,dx\) | M1A1 | Integration by parts correct way around; must see \(Ax^{-1}\ln x \pm B\int x^{-1} \times \frac{1}{x}\,dx\) |
| \(= \frac{x^{-1}}{-1}\ln x + \frac{x^{-1}}{-1}(+c)\) | M1A1 | Combining two \(x\) terms correctly; completely correct integral |
| \(\int_1^e \frac{\ln x}{x^2}dx = \left[\frac{-1}{x}\ln x - \frac{1}{x}\right]_1^e = \left(\frac{-1}{e}\ln e - \frac{1}{e}\right) - \left(\frac{-1}{1}\ln 1 - \frac{1}{1}\right)\) | M1 | Substituting limits 1 and e and subtracting |
| \(= 1 - \frac{2}{e}\) | A1 | cso and cao; allow \(-\frac{2}{e}+1\); accept \(1 + \frac{-2}{e}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int ue^{-u}\,du = -ue^{-u} - \int -e^{-u}\,du\) | M1A1 | Must see \(Aue^{-u} \pm \int e^{-u}\,du\) |
| \(\int ue^{-u}\,du = -ue^{-u} - e^{-u}(+c)\) | M1A1 | \(\int e^{-u}\,du \rightarrow e^{-u}\); completely correct integral |
| \(\int_1^e \frac{\ln x}{x^2}dx = \left[-ue^{-u} - e^{-u}\right]_0^1 = \left(-\frac{1}{e} - \frac{1}{e}\right) - (0-1)\) | M1 | Substituting limits 0 and 1 and subtracting |
| \(= 1 - \frac{2}{e}\) | A1 | cso and cao |
## Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{\ln x}{x^2}dx = \int x^{-2}\ln x\, dx = \frac{x^{-1}}{-1}\ln x - \int \frac{x^{-1}}{-1} \times \frac{1}{x}\,dx$ | M1A1 | Integration by parts correct way around; must see $Ax^{-1}\ln x \pm B\int x^{-1} \times \frac{1}{x}\,dx$ |
| $= \frac{x^{-1}}{-1}\ln x + \frac{x^{-1}}{-1}(+c)$ | M1A1 | Combining two $x$ terms correctly; completely correct integral |
| $\int_1^e \frac{\ln x}{x^2}dx = \left[\frac{-1}{x}\ln x - \frac{1}{x}\right]_1^e = \left(\frac{-1}{e}\ln e - \frac{1}{e}\right) - \left(\frac{-1}{1}\ln 1 - \frac{1}{1}\right)$ | M1 | Substituting limits 1 and e and subtracting |
| $= 1 - \frac{2}{e}$ | A1 | cso and cao; allow $-\frac{2}{e}+1$; accept $1 + \frac{-2}{e}$ |
**Alternative by substitution** $u = \ln x$:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int ue^{-u}\,du = -ue^{-u} - \int -e^{-u}\,du$ | M1A1 | Must see $Aue^{-u} \pm \int e^{-u}\,du$ |
| $\int ue^{-u}\,du = -ue^{-u} - e^{-u}(+c)$ | M1A1 | $\int e^{-u}\,du \rightarrow e^{-u}$; completely correct integral |
| $\int_1^e \frac{\ln x}{x^2}dx = \left[-ue^{-u} - e^{-u}\right]_0^1 = \left(-\frac{1}{e} - \frac{1}{e}\right) - (0-1)$ | M1 | Substituting limits 0 and 1 and subtracting |
| $= 1 - \frac{2}{e}$ | A1 | cso and cao |
---\begin{enumerate}
\item Use integration by parts to find the exact value of $\int _ { 1 } ^ { \mathrm { e } } \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x$
\end{enumerate}
Write your answer in the form $a + \frac { b } { \mathrm { e } }$, where $a$ and $b$ are integers.\\
\hfill \mbox{\textit{Edexcel C34 2017 Q2 [6]}}