| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| 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 integration by parts question with standard setup (xe^{2x}), followed by routine trapezium rule application. Part (a) requires one application of integration by parts with clear choice of u and dv, part (b) is calculator work, and part (c) is direct formula application. Slightly easier than average due to the predictable structure and minimal problem-solving required. |
| Spec | 1.08i Integration by parts1.09f Trapezium rule: numerical integration |
| \(x\) | 0 | 0.2 | 0.4 | 0.6 | 0.8 | 1 |
| \(y = x \mathrm { e } ^ { 2 x }\) | 0 | 0.29836 | 1.99207 | 7.38906 |
| Answer | Marks |
|---|---|
| (a) \(\int xe^{2x}dx = \frac{1}{2}xe^{2x} - \frac{1}{2}\int e^{2x}dx\) | M1 A1 |
| Attempting parts in the right direction | |
| \(= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}\) | A1 |
| \(\left[\frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}\right]_0^1 = \frac{1}{4} + \frac{1}{4}e^2\) | M1 A1 |
| Answer | Marks |
|---|---|
| (b) \(x = 0.4 \Rightarrow y \approx 0.89022\) | B1 |
| \(x = 0.8 \Rightarrow y \approx 3.96243\) | B1 |
| Both are required to 5 d.p. |
| Answer | Marks |
|---|---|
| (c) \(I = \frac{1}{2} \times 0.2 \times [\ldots]\) | B1 |
| \(\approx \ldots \times [0 + 7.38906 + 2(0.29836 + .89022 + 1.99207 + 3.96243)]\) | M1 A1ft |
| ft their answers to (b) | |
| \(\approx 0.1 \times 21.67522\) | |
| \(\approx 2.168\) | cao A1 |
| Note: \(\frac{1}{4} + \frac{1}{4}e^2 \approx 2.097\ldots\) |
**(a)** $\int xe^{2x}dx = \frac{1}{2}xe^{2x} - \frac{1}{2}\int e^{2x}dx$ | M1 A1 |
Attempting parts in the right direction | |
$= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}$ | A1 |
$\left[\frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}\right]_0^1 = \frac{1}{4} + \frac{1}{4}e^2$ | M1 A1 |
**Total for (a): [5]**
**(b)** $x = 0.4 \Rightarrow y \approx 0.89022$ | B1 |
$x = 0.8 \Rightarrow y \approx 3.96243$ | B1 |
Both are required to 5 d.p. | |
**Total for (b): [1]**
**(c)** $I = \frac{1}{2} \times 0.2 \times [\ldots]$ | B1 |
$\approx \ldots \times [0 + 7.38906 + 2(0.29836 + .89022 + 1.99207 + 3.96243)]$ | M1 A1ft |
ft their answers to (b) | |
$\approx 0.1 \times 21.67522$ | |
$\approx 2.168$ | cao A1 |
Note: $\frac{1}{4} + \frac{1}{4}e^2 \approx 2.097\ldots$ | |
**Total for (c): [4]**
**Total: [10]**
---5.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{7fa2c564-d1e5-4fd0-a690-e3189daea332-06_586_1079_260_427}
\end{center}
\end{figure}
Figure 1 shows the graph of the curve with equation
$$y = x \mathrm { e } ^ { 2 x } , \quad x \geqslant 0$$
The finite region $R$ bounded by the lines $x = 1$, the $x$-axis and the curve is shown shaded in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Use integration to find the exact value for the area of $R$.
\item Complete the table with the values of $y$ corresponding to $x = 0.4$ and 0.8 .
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | }
\hline
$x$ & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1 \\
\hline
$y = x \mathrm { e } ^ { 2 x }$ & 0 & 0.29836 & & 1.99207 & & 7.38906 \\
\hline
\end{tabular}
\end{center}
\item Use the trapezium rule with all the values in the table to find an approximate value for this area, giving your answer to 4 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2005 Q5 [10]}}