| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | October |
| Marks | 6 |
| 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 clearly given values, followed by a simple volume calculation (area × speed × time) and a standard interpretation question. All steps are routine P2 techniques with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure and real-world context. |
| Spec | 1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identifies \(h = 3\) | B1 | May be stated or implied by sight of \(\dfrac{3}{2}\{\ldots+2(\ldots+\ldots+\ldots)\}\) |
| Area \(= \dfrac{3}{2}\{0+0+2(1.52+2.74+3.12+3.08)\}\) | M1 | Correct method using trapezium rule; condone slips on digits of heights e.g. 3.21 for 3.12 |
| \(= 31.38 \text{ m}^2\) | A1 | Accept 31.38 or awrt 31.4 (m²); exact fractional answer \(\dfrac{1569}{50}\) also acceptable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Calculates \("31.38" \times 1.5 \times 60\) | M1 | Or equivalent such as their (a) \(\times 90\) |
| \(= \text{awrt } 2800 \text{ m}^3\) | A1 | Units not necessary |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Underestimate — area of trapezia are smaller than cross sectional area | B1 | Must state underestimate AND give a valid reason; reason must reference 'trapezia' or 'area of a trapezium'; do not accept convex/concave without clarification |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identifies $h = 3$ | B1 | May be stated or implied by sight of $\dfrac{3}{2}\{\ldots+2(\ldots+\ldots+\ldots)\}$ |
| Area $= \dfrac{3}{2}\{0+0+2(1.52+2.74+3.12+3.08)\}$ | M1 | Correct method using trapezium rule; condone slips on digits of heights e.g. 3.21 for 3.12 |
| $= 31.38 \text{ m}^2$ | A1 | Accept 31.38 or awrt 31.4 (m²); exact fractional answer $\dfrac{1569}{50}$ also acceptable |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Calculates $"31.38" \times 1.5 \times 60$ | M1 | Or equivalent such as their (a) $\times 90$ |
| $= \text{awrt } 2800 \text{ m}^3$ | A1 | Units not necessary |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Underestimate — area of trapezia are smaller than cross sectional area | B1 | Must state **underestimate** AND give a **valid reason**; reason must reference 'trapezia' or 'area of a trapezium'; do not accept convex/concave without clarification |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{66abdef1-072e-41eb-a933-dd51a96330ff-14_488_1511_246_278}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A river is being studied.\\
At one particular place, the river is 15 m wide.\\
The depth, $y$ metres, of the river is measured at a point $x$ metres from one side of the river.
Figure 1 shows a plot of the cross-section of the river and the coordinate values $( x , y )$
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule with all the $y$ values given in Figure 1 to estimate the cross-sectional area of the river.
The water in the river is modelled as flowing at a constant speed of $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ across the whole of the cross-section.
\item Use the model and the answer to part (a) to estimate the volume of water flowing through this section of the river each minute, giving your answer in $\mathrm { m } ^ { 3 }$ to 2 significant figures.
Assuming the model,
\item state, giving a reason for your answer, whether your answer for part (b) is an overestimate or an underestimate of the true volume of water flowing through this section of the river each minute.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2023 Q6 [6]}}