| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Moderate -0.8 This is a straightforward application of the trapezium rule requiring only substitution into the given function (y = e^(0.5x²)) and applying a standard numerical method formula. The calculations are routine with no conceptual challenges or problem-solving required, making it easier than average for A-level. |
| Spec | 1.09f Trapezium rule: numerical integration |
| \(x\) | 0 | 0.4 | 0.8 | 1.2 | 1.6 | 2 |
| \(y\) | \(\mathrm { e } ^ { 0 }\) | \(\mathrm { e } ^ { 0.08 }\) | \(\mathrm { e } ^ { 0.72 }\) | \(\mathrm { e } ^ { 2 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either \(e^{0.32}\) and \(e^{1.28}\) or awrt 1.38 and 3.60 (or a mixture of e's and decimals) | B1 | Complete correct table values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times 0.4\) or \(0.2\) outside brackets | B1 | Outside brackets |
| \(e^0 + 2(e^{0.08} + e^{0.32} + e^{0.72} + e^{1.28}) + e^2\) | M1\(\checkmark\) | For structure of trapezium rule |
| \(= 0.2 \times 24.61203164... = 4.922406... = 4.922\) (4sf) | A1 cao |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Either $e^{0.32}$ and $e^{1.28}$ or awrt 1.38 and 3.60 (or a mixture of e's and decimals) | B1 | Complete correct table values |
## Part (b) Way 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 0.4$ or $0.2$ outside brackets | B1 | Outside brackets |
| $e^0 + 2(e^{0.08} + e^{0.32} + e^{0.72} + e^{1.28}) + e^2$ | M1$\checkmark$ | For structure of trapezium rule |
| $= 0.2 \times 24.61203164... = 4.922406... = 4.922$ (4sf) | A1 cao | |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fb1924cc-9fa3-4fde-ba4d-6fb095f7f70b-02_519_451_210_749}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows part of the curve with equation $y = \mathrm { e } ^ { 0.5 x ^ { 2 } }$. The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $x$-axis, the $y$-axis and the line $x = 2$.
\begin{enumerate}[label=(\alph*)]
\item Complete the table with the values of $y$ corresponding to $x = 0.8$ and $x = 1.6$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & 0 & 0.4 & 0.8 & 1.2 & 1.6 & 2 \\
\hline
$y$ & $\mathrm { e } ^ { 0 }$ & $\mathrm { e } ^ { 0.08 }$ & & $\mathrm { e } ^ { 0.72 }$ & & $\mathrm { e } ^ { 2 }$ \\
\hline
\end{tabular}
\end{center}
\item Use the trapezium rule with all the values in the table to find an approximate value for the area of $R$, giving your answer to 4 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2008 Q1 [4]}}