| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule accuracy improvement explanation |
| Difficulty | Moderate -0.3 This is a standard C4 trapezium rule question with straightforward integration by parts. Part (a) requires routine application of the trapezium rule formula, part (b) tests understanding of accuracy (increase number of strips), and part (c) involves standard integration by parts of xe^(2x) form. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08i Integration by parts1.09f Trapezium rule: numerical integration |
| \(x\) | 0 | 0.5 | 1 | 1.5 | 2 |
| \(y\) | 2 | 4.077 | 7.389 | 10.043 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(\approx \frac{1}{2} \times 0.5 \times \left[2 + 2(4.077 + 7.389 + 10.043) + 0\right]\) | B1; M1 | B1: outside brackets \(\frac{1}{2}\times 0.5\) or \(\frac{0.5}{2}\) or 0.25 or \(\frac{1}{4}\). M1: for structure of trapezium rule \([\ldots]\), condone missing 0 |
| \(= \frac{1}{4} \times 45.018 = 11.2545 = 11.25\) (2 dp) | A1 cao | No errors allowed. Working must be seen to demonstrate trapezium rule. Actual area is 12.39953751… |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Any one of: increase the number of strips / use more trapezia / make \(h\) smaller / increase the number of \(x\) and/or \(y\) values used / shorter/smaller intervals for \(x\) / more values of \(y\) / more intervals of \(x\) / increase \(n\) | B1 | B0 for: smaller values of \(x\) and/or \(y\); use more decimal places |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = 2-x \Rightarrow \frac{du}{dx} = -1\); \(\frac{dv}{dx} = e^{2x} \Rightarrow v = \frac{1}{2}e^{2x}\) | ||
| \((2-x)e^{2x} \rightarrow \pm\lambda(2-x)e^{2x} \pm \int \mu e^{2x}\{dx\}\) or \(\pm xe^{2x} \rightarrow \pm\lambda xe^{2x} \pm \int \mu e^{2x}\{dx\}\) | M1 | |
| \(= \frac{1}{2}(2-x)e^{2x} - \int -\frac{1}{2}e^{2x}\{dx\}\) | A1 | \((2-x)e^{2x} \rightarrow \frac{1}{2}(2-x)e^{2x} - \int -\frac{1}{2}e^{2x}\{dx\}\) either unsimplified or simplified |
| \(= \frac{1}{2}(2-x)e^{2x} + \frac{1}{4}e^{2x}\) | A1 oe | Correct expression: \(\frac{1}{2}(2-x)e^{2x} + \frac{1}{4}e^{2x}\) or \(\frac{5}{4}e^{2x} - xe^{2x}\) or equivalent |
| Area \(= \left[\frac{1}{2}(2-x)e^{2x} + \frac{1}{4}e^{2x}\right]_0^2\) | ||
| \(= \left(0 + \frac{1}{4}e^4\right) - \left(\frac{1}{2}(2)e^0 + \frac{1}{4}e^0\right)\) | dM1 | Applies limits of 2 and 0 to all terms and subtracts the correct way round. Evidence of proper consideration of limit of 0 needed; just subtracting zero is M0 |
| \(= \frac{1}{4}e^4 - \frac{5}{4}\) | A1 oe | \(\frac{1}{4}e^4 - \frac{5}{4}\) or \(\frac{e^4-5}{4}\) cao. Do not allow \(\frac{1}{4}e^4 - \frac{5}{4}e^0\) unless simplified. 12.39953751… without seeing \(\frac{1}{4}e^4 - \frac{5}{4}\) is A0 |
## Question 2:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $\approx \frac{1}{2} \times 0.5 \times \left[2 + 2(4.077 + 7.389 + 10.043) + 0\right]$ | B1; M1 | B1: outside brackets $\frac{1}{2}\times 0.5$ or $\frac{0.5}{2}$ or 0.25 or $\frac{1}{4}$. M1: for structure of trapezium rule $[\ldots]$, condone missing 0 |
| $= \frac{1}{4} \times 45.018 = 11.2545 = 11.25$ (2 dp) | A1 cao | No errors allowed. Working must be seen to demonstrate trapezium rule. Actual area is 12.39953751… |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Any one of: increase the number of strips / use more trapezia / make $h$ smaller / increase the number of $x$ and/or $y$ values used / shorter/smaller intervals for $x$ / more values of $y$ / more intervals of $x$ / increase $n$ | B1 | B0 for: smaller values of $x$ and/or $y$; use more decimal places |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = 2-x \Rightarrow \frac{du}{dx} = -1$; $\frac{dv}{dx} = e^{2x} \Rightarrow v = \frac{1}{2}e^{2x}$ | | |
| $(2-x)e^{2x} \rightarrow \pm\lambda(2-x)e^{2x} \pm \int \mu e^{2x}\{dx\}$ or $\pm xe^{2x} \rightarrow \pm\lambda xe^{2x} \pm \int \mu e^{2x}\{dx\}$ | M1 | |
| $= \frac{1}{2}(2-x)e^{2x} - \int -\frac{1}{2}e^{2x}\{dx\}$ | A1 | $(2-x)e^{2x} \rightarrow \frac{1}{2}(2-x)e^{2x} - \int -\frac{1}{2}e^{2x}\{dx\}$ either unsimplified or simplified |
| $= \frac{1}{2}(2-x)e^{2x} + \frac{1}{4}e^{2x}$ | A1 oe | Correct expression: $\frac{1}{2}(2-x)e^{2x} + \frac{1}{4}e^{2x}$ or $\frac{5}{4}e^{2x} - xe^{2x}$ or equivalent |
| Area $= \left[\frac{1}{2}(2-x)e^{2x} + \frac{1}{4}e^{2x}\right]_0^2$ | | |
| $= \left(0 + \frac{1}{4}e^4\right) - \left(\frac{1}{2}(2)e^0 + \frac{1}{4}e^0\right)$ | dM1 | Applies limits of 2 and 0 **to all terms** and subtracts the correct way round. Evidence of proper consideration of limit of 0 needed; just subtracting zero is M0 |
| $= \frac{1}{4}e^4 - \frac{5}{4}$ | A1 oe | $\frac{1}{4}e^4 - \frac{5}{4}$ or $\frac{e^4-5}{4}$ cao. Do not allow $\frac{1}{4}e^4 - \frac{5}{4}e^0$ unless simplified. 12.39953751… without seeing $\frac{1}{4}e^4 - \frac{5}{4}$ is A0 |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e14881c1-5ba5-4868-92ee-8bc58d4884dc-03_606_1070_251_445}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation
$$y = ( 2 - x ) \mathrm { e } ^ { 2 x } , \quad x \in \mathbb { R }$$
The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $x$-axis and the $y$-axis.
The table below shows corresponding values of $x$ and $y$ for $y = ( 2 - x ) \mathrm { e } ^ { 2 x }$
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & 0.5 & 1 & 1.5 & 2 \\
\hline
$y$ & 2 & 4.077 & 7.389 & 10.043 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule with all the values of $y$ in the table, to obtain an approximation for the area of $R$, giving your answer to 2 decimal places.
\item Explain how the trapezium rule can be used to give a more accurate approximation for the area of $R$.
\item Use calculus, showing each step in your working, to obtain an exact value for the area of $R$. Give your answer in its simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2014 Q2 [9]}}