| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2020 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Area under curve requiring parts |
| Difficulty | Standard +0.3 This is a straightforward application of integration by parts for part (a), followed by splitting the integral and using the result in part (b). The integration by parts setup for ln(x)/x² is standard, and part (b) is routine algebraic manipulation. Slightly easier than average due to the scaffolding provided. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{\ln x}{x^2}dx = \int x^{-2}\ln x\,dx = -x^{-1}\ln x + \int x^{-1}\times\frac{1}{x}dx\) | M1 | Attempts by parts to reach \(\pm ax^{-1}\ln x \pm b\int x^{-1}\times\frac{1}{x}dx\), \(a,b\neq 0\); formula must be correct if stated |
| \(= -x^{-1}\ln x - x^{-1}(+c)\) | dM1 A1 | o.e. such as \(-\frac{\ln x}{x} - \frac{1}{x} + c\); condone omission of \(c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{3+2x+\ln x}{x^2} = 3x^{-2}+2x^{-1}+x^{-2}\ln x\) | B1 | o.e.; \(\frac{2}{x}\) may be implied by \(\int\frac{2x}{x^2}dx = 2\ln x\) or \(\ln x^2\) |
| Area \(= \left[-\frac{3}{x}+2\ln x - \frac{\ln x}{x} - \frac{1}{x}\right]_2^4\) | M1 | Following through on part (a) |
| Substitutes 2 and 4 into expression of correct form | M1 | Must attempt to simplify using one log law correctly |
| \(= 1 + 2\ln 2\) | A1 | Exact; likely \(1+2\ln 2\) or \(1+\ln 4\) |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{\ln x}{x^2}dx = \int x^{-2}\ln x\,dx = -x^{-1}\ln x + \int x^{-1}\times\frac{1}{x}dx$ | M1 | Attempts by parts to reach $\pm ax^{-1}\ln x \pm b\int x^{-1}\times\frac{1}{x}dx$, $a,b\neq 0$; formula must be correct if stated |
| $= -x^{-1}\ln x - x^{-1}(+c)$ | dM1 A1 | o.e. such as $-\frac{\ln x}{x} - \frac{1}{x} + c$; condone omission of $c$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3+2x+\ln x}{x^2} = 3x^{-2}+2x^{-1}+x^{-2}\ln x$ | B1 | o.e.; $\frac{2}{x}$ may be implied by $\int\frac{2x}{x^2}dx = 2\ln x$ or $\ln x^2$ |
| Area $= \left[-\frac{3}{x}+2\ln x - \frac{\ln x}{x} - \frac{1}{x}\right]_2^4$ | M1 | Following through on part (a) |
| Substitutes 2 and 4 into expression of correct form | M1 | Must attempt to simplify using one log law correctly |
| $= 1 + 2\ln 2$ | A1 | Exact; likely $1+2\ln 2$ or $1+\ln 4$ |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{79ac81c3-cd05-4f28-8840-3c8a6960e7b7-14_600_1022_255_461}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Find $\int \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x$
Figure 3 shows a sketch of part of the curve with equation
$$y = \frac { 3 + 2 x + \ln x } { x ^ { 2 } } \quad x > 0.5$$
The finite region $R$, shown shaded in Figure 3, is bounded by the curve, the line with equation $x = 2$, the $x$-axis and the line with equation $x = 4$
\item Use the answer to part (a) to find the exact area of $R$, writing your answer in simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2020 Q5 [7]}}