| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Sequential multi-part (building on previous) |
| Difficulty | Challenging +1.2 This is a structured multi-part integration question requiring finding an x-intercept, performing integration by parts with a non-standard substitution (u=1/x), and calculating an area. While it requires careful algebraic manipulation and the integration by parts technique is slightly disguised, the question provides clear scaffolding through its parts and follows a predictable pattern for C4 questions. The main challenge is recognizing the substitution needed for part (b), but this is a standard C4 technique that students would have practiced. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts |
| Answer | Marks |
|---|---|
| \(y = 4x - xe^{\frac{1}{x}}, x \geq 0\) | |
| \(y = 0 \Rightarrow 4x - xe^{\frac{1}{x}} = 0 \Rightarrow x(4 - e^{\frac{1}{x}}) = 0\) |
| Answer | Marks |
|---|---|
| Attempts to solve \(e^{\frac{1}{x}} = 4\) giving \(x = \ldots\) in terms of \(\pm \lambda \ln \mu\) where \(\mu > 0\) | M1 |
| \(\ln 2\), cao. (Ignore \(x = 0\).) | A1 |
| [2] |
| Answer | Marks |
|---|---|
| \(\left\{\int xe^{\frac{1}{x}}dx\right\} = 2xe^{\frac{1}{x}} - \int 2e^{\frac{1}{x}}\{dx\}\) | M1 |
| \(\alpha xe^{\frac{1}{x}} - \beta\int e^{\frac{1}{x}}\{dx\}, \alpha > 0, \beta > 0\) | M1 |
| \(2xe^{\frac{1}{x}} - \int 2e^{\frac{1}{x}}\{dx\}\), with or without \(dx\) | A1 (ht on ePEN) |
| \(2xe^{\frac{1}{x}} - 4e^{\frac{1}{x}} + c\) | A1 |
| [3] |
| Answer | Marks |
|---|---|
| \(\int 4x dx = 2x^2\) | B1 |
| \(4x \rightarrow 2x^2\) or \(\frac{4x^2}{2}\) o.e. | |
| \(\int\left[(4x - xe^{\frac{1}{x}})dx\right] = \left[2x^2 - \left\{2xe^{\frac{1}{x}} - 4e^{\frac{1}{x}}\right\}\right]\) | M1 (ht2 or lft to their limits) |
| \(= 2(4\ln 2)^2 - 2(4\ln 2)e^{\frac{1}{4\ln 2}} + 4e^{\frac{1}{4\ln 2}} - \left\{2(0)^2 - 2(0)e^{\frac{0}{0}} + 4e^{\frac{0}{0}}\right\}\) | |
| \(= (32(\ln 2)^2 - 32(\ln 2) + 16) - (-4)\) | |
| \(= 32(\ln 2)^2 - 32(\ln 2) + 12\) | A1 |
| [3] |
## Part (a)
$y = 4x - xe^{\frac{1}{x}}, x \geq 0$ | |
$y = 0 \Rightarrow 4x - xe^{\frac{1}{x}} = 0 \Rightarrow x(4 - e^{\frac{1}{x}}) = 0$ | |
| |
Attempts to solve $e^{\frac{1}{x}} = 4$ giving $x = \ldots$ in terms of $\pm \lambda \ln \mu$ where $\mu > 0$ | M1 |
$\ln 2$, cao. (Ignore $x = 0$.) | A1 |
| [2] |
## Part (b)
$\left\{\int xe^{\frac{1}{x}}dx\right\} = 2xe^{\frac{1}{x}} - \int 2e^{\frac{1}{x}}\{dx\}$ | M1 |
$\alpha xe^{\frac{1}{x}} - \beta\int e^{\frac{1}{x}}\{dx\}, \alpha > 0, \beta > 0$ | M1 |
$2xe^{\frac{1}{x}} - \int 2e^{\frac{1}{x}}\{dx\}$, with or without $dx$ | A1 (ht on ePEN) |
$2xe^{\frac{1}{x}} - 4e^{\frac{1}{x}} + c$ | A1 |
| [3] |
## Part (c)
$\int 4x dx = 2x^2$ | B1 |
$4x \rightarrow 2x^2$ or $\frac{4x^2}{2}$ o.e. | |
$\int\left[(4x - xe^{\frac{1}{x}})dx\right] = \left[2x^2 - \left\{2xe^{\frac{1}{x}} - 4e^{\frac{1}{x}}\right\}\right]$ | M1 (ht2 or lft to their limits) |
$= 2(4\ln 2)^2 - 2(4\ln 2)e^{\frac{1}{4\ln 2}} + 4e^{\frac{1}{4\ln 2}} - \left\{2(0)^2 - 2(0)e^{\frac{0}{0}} + 4e^{\frac{0}{0}}\right\}$ | |
$= (32(\ln 2)^2 - 32(\ln 2) + 16) - (-4)$ | |
$= 32(\ln 2)^2 - 32(\ln 2) + 12$ | A1 |
| [3] |
---\includegraphics{figure_1}
Figure 1 shows a sketch of part of the curve with equation $y = 4x - xe^{\frac{1}{x}}, x \geqslant 0$
The curve meets the $x$-axis at the origin $O$ and cuts the $x$-axis at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $\ln 2$, the $x$ coordinate of the point $A$.
[2]
\item Find
$$\int xe^{\frac{1}{x}} dx$$
[3]
\item Find, by integration, the exact value for the area of $R$.
Give your answer in terms of $\ln 2$
[3]
\end{enumerate}
The finite region $R$, shown shaded in Figure 1, is bounded by the $x$-axis and the curve with equation
$$y = 4x - xe^{\frac{1}{x}}, x \geqslant 0$$
\hfill \mbox{\textit{Edexcel C4 2015 Q3 [8]}}