| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule applied to real-world data |
| Difficulty | Moderate -0.8 This is a straightforward application of the trapezium rule with equally-spaced ordinates (h=4) and a simple formula for substitution. Part (a) requires basic calculator work, part (b) is direct application of the trapezium rule formula, and part (c) is a simple multiplication (area × speed × time). All steps are routine with no problem-solving or conceptual challenges beyond standard C2 level. |
| Spec | 1.09f Trapezium rule: numerical integration |
| \(x\) | 0 | 4 | 8 | 12 | 16 | 20 |
| \(y\) | 0 | 2.771 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Missing \(y\) values: \(1.6(00)\) and \(3.2(00)\) | B1 | |
| \(3.394\) | B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = \frac{1}{2} \times 4 \times \{(0+0) + 2(1.6 + 2.771 + 3.394 + 3.2)\}\) | B1, M1, A1ft | |
| \(= 43.86\) (or more accurate value) (or 43.9, or 44) | A1 | (4 marks) Answer only: No marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Volume \(= A \times 2 \times 60\) | M1 | M mark can be implied |
| \(= 5260 \text{ m}^3\) (or 5270, or 5280) | A1 | (2 marks) Answer only: Allow |
## Question 6:
**(a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Missing $y$ values: $1.6(00)$ and $3.2(00)$ | B1 | |
| $3.394$ | B1 | (2 marks) |
**(b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = \frac{1}{2} \times 4 \times \{(0+0) + 2(1.6 + 2.771 + 3.394 + 3.2)\}$ | B1, M1, A1ft | |
| $= 43.86$ (or more accurate value) (or 43.9, or 44) | A1 | (4 marks) Answer only: No marks |
**(c)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Volume $= A \times 2 \times 60$ | M1 | M mark can be implied |
| $= 5260 \text{ m}^3$ (or 5270, or 5280) | A1 | (2 marks) Answer only: Allow |
---6. A river, running between parallel banks, is 20 m wide. The depth, $y$ metres, of the river measured at a point $x$ metres from one bank is given by the formula
$$y = \frac { 1 } { 10 } x \sqrt { } ( 20 - x ) , \quad 0 \leqslant x \leqslant 20$$
\begin{enumerate}[label=(\alph*)]
\item Complete the table below, giving values of $y$ to 3 decimal places.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & 0 & 4 & 8 & 12 & 16 & 20 \\
\hline
$y$ & 0 & & 2.771 & & & 0 \\
\hline
\end{tabular}
\end{center}
\item Use the trapezium rule with all the values in the table to estimate the cross-sectional area of the river.
Given that the cross-sectional area is constant and that the river is flowing uniformly at $2 \mathrm {~ms} ^ { - 1 }$,
\item estimate, in $\mathrm { m } ^ { 3 }$, the volume of water flowing per minute, giving your answer to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2005 Q6 [8]}}