| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2018 |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Stationary points then area/volume |
| Difficulty | Standard +0.3 This is a standard P4/Further Pure integration question with routine steps: solving for x-intercept requires basic algebraic manipulation, integration by parts follows the standard formula with straightforward substitution, and finding area is direct application of definite integration. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 0 \Rightarrow 4x - xe^{\frac{1}{2}x} = 0 \Rightarrow x(4 - e^{\frac{1}{2}x}) = 0\) | — | Setting up equation |
| \(e^{\frac{1}{2}x} = 4 \Rightarrow x_A = 4\ln 2\) | M1 | Attempts to solve \(e^{\frac{1}{2}x} = 4\) giving \(x = ...\) in terms of \(\pm\lambda\ln\mu\) where \(\mu > 0\) |
| \(4\ln 2\) | A1 | cao (Ignore \(x = 0\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left\{\int xe^{\frac{1}{2}x}\,dx\right\} = 2xe^{\frac{1}{2}x} - \int 2e^{\frac{1}{2}x}\,\{dx\}\) | M1 | Integration by parts in the form \(\alpha xe^{\frac{1}{2}x} - \beta\int e^{\frac{1}{2}x}\,\{dx\}\), \(\alpha > 0\), \(\beta > 0\) |
| \(2xe^{\frac{1}{2}x} - \int 2e^{\frac{1}{2}x}\,\{dx\}\), with or without \(dx\) | A1 | Can be un-simplified |
| \(= 2xe^{\frac{1}{2}x} - 4e^{\frac{1}{2}x}\,\{+c\}\) | A1 | Can be un-simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left\{\int 4x\,dx\right\} = 2x^2\) | B1 | \(4x \to 2x^2\) or \(\frac{4x^2}{2}\) oe |
| \(\left\{\int_0^{4\ln 2}(4x - xe^{\frac{1}{2}x})\,dx\right\} = \left[2x^2 - \left(2xe^{\frac{1}{2}x} - 4e^{\frac{1}{2}x}\right)\right]_0^{4\ln 2}\) | — | 4ln2 or ln16 or their limits |
| \(= \left(2(4\ln 2)^2 - 2(4\ln 2)e^{\frac{1}{2}(4\ln 2)} + 4e^{\frac{1}{2}(4\ln 2)}\right) - \left(2(0)^2 - 2(0)e^{\frac{1}{2}(0)} + 4e^{\frac{1}{2}(0)}\right)\) | M1 | Complete method applying limits of \(x_A\) and 0 to all terms of form \(\pm Ax^2 \pm Bxe^{\frac{1}{2}x} \pm Ce^{\frac{1}{2}x}\), subtracting correct way round |
| \(= (32(\ln 2)^2 - 32(\ln 2) + 16) - (4)\) | A1 | Correct three term exact quadratic expression in \(\ln 2\), e.g. \(32(\ln 2)^2 - 32(\ln 2) + 12\) |
| \(= 32(\ln 2)^2 - 32(\ln 2) + 12\) | A1 | Also allow \(32\ln^2 2 - 32(\ln 2) + 12\) |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 0 \Rightarrow 4x - xe^{\frac{1}{2}x} = 0 \Rightarrow x(4 - e^{\frac{1}{2}x}) = 0$ | — | Setting up equation |
| $e^{\frac{1}{2}x} = 4 \Rightarrow x_A = 4\ln 2$ | M1 | Attempts to solve $e^{\frac{1}{2}x} = 4$ giving $x = ...$ in terms of $\pm\lambda\ln\mu$ where $\mu > 0$ |
| $4\ln 2$ | A1 | cao (Ignore $x = 0$) |
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left\{\int xe^{\frac{1}{2}x}\,dx\right\} = 2xe^{\frac{1}{2}x} - \int 2e^{\frac{1}{2}x}\,\{dx\}$ | M1 | Integration by parts in the form $\alpha xe^{\frac{1}{2}x} - \beta\int e^{\frac{1}{2}x}\,\{dx\}$, $\alpha > 0$, $\beta > 0$ |
| $2xe^{\frac{1}{2}x} - \int 2e^{\frac{1}{2}x}\,\{dx\}$, with or without $dx$ | A1 | Can be un-simplified |
| $= 2xe^{\frac{1}{2}x} - 4e^{\frac{1}{2}x}\,\{+c\}$ | A1 | Can be un-simplified |
## Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left\{\int 4x\,dx\right\} = 2x^2$ | B1 | $4x \to 2x^2$ or $\frac{4x^2}{2}$ oe |
| $\left\{\int_0^{4\ln 2}(4x - xe^{\frac{1}{2}x})\,dx\right\} = \left[2x^2 - \left(2xe^{\frac{1}{2}x} - 4e^{\frac{1}{2}x}\right)\right]_0^{4\ln 2}$ | — | 4ln2 or ln16 or their limits |
| $= \left(2(4\ln 2)^2 - 2(4\ln 2)e^{\frac{1}{2}(4\ln 2)} + 4e^{\frac{1}{2}(4\ln 2)}\right) - \left(2(0)^2 - 2(0)e^{\frac{1}{2}(0)} + 4e^{\frac{1}{2}(0)}\right)$ | M1 | Complete method applying limits of $x_A$ and 0 to all terms of form $\pm Ax^2 \pm Bxe^{\frac{1}{2}x} \pm Ce^{\frac{1}{2}x}$, subtracting correct way round |
| $= (32(\ln 2)^2 - 32(\ln 2) + 16) - (4)$ | A1 | Correct three term exact quadratic expression in $\ln 2$, e.g. $32(\ln 2)^2 - 32(\ln 2) + 12$ |
| $= 32(\ln 2)^2 - 32(\ln 2) + 12$ | A1 | Also allow $32\ln^2 2 - 32(\ln 2) + 12$ |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4de08317-5fb9-4789-8d57-ccf463224c78-14_614_858_303_552}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve with equation $y = 4 x - x \mathrm { e } ^ { \frac { 1 } { 2 } x } , x \geqslant 0$
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 $A$ .
\item Find $\int x \mathrm { e } ^ { \frac { 1 } { 2 } x } \mathrm {~d} x$
The finite region $R$ ,shown shaded in Figure 2,is bounded by the $x$-axis and the curve with equation $y = 4 x - x \mathrm { e } ^ { \frac { 1 } { 2 } x } , x \geqslant 0$
\item Find,by integration,the exact value for the area of $R$ .
Give your answer in terms of $\ln 2$\\
\includegraphics[max width=\textwidth, alt={}, center]{4de08317-5fb9-4789-8d57-ccf463224c78-18_2655_1943_114_118}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2018 Q5 [8]}}