| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule applied to real-world data |
| Difficulty | Moderate -0.3 This is a straightforward application of the trapezium rule and basic integration to a practical context. Part (i) requires routine trapezium rule calculation with given data points, then simple multiplication for volume. Part (ii) involves substituting into a polynomial, basic polynomial integration, and applying limits. All techniques are standard C2 level with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2} \times 0.2(0 + 0 + 2(0.5 + 0.7 + 0.75 + 0.7 + 0.5))\) \([= 0.63]\) | M3 | M2 if one error, M1 if two errors; condone omission of zeros; or M3 for \(0.05 + 0.12 + 0.145 + 0.145 + 0.12 + 0.05\); must be 6 strips; M0 if \(h=1\) or \(1.2\); Area \(= 6.3\) and \(0.53\) imply M0 |
| (their \(0.63\)) \(\times 50\) | M1 | |
| \(31.5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(3.8 \times 0.2^4 - 6.8 \times 0.2^3 + 7.7 \times 0.2^2 - 4.2 \times 0.2\) | M1 | \(\pm 0.58032\) implies M1; condone one sign error |
| \(0.01968\) cao isw | A1 | Or B2 if unsupported; allow \(-0.01968\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{3.8x^5}{5} - \frac{6.8x^4}{4} + \frac{7.7x^3}{3} - \frac{4.2x^2}{2} + c\) | M2 | M1 for two terms correct excluding \(c\); condone omission of \(c\); accept \(2.56\) to \(2.57\) for coefficient of \(x^3\); allow M1 if all signs reversed |
| \(F(0.9) - F(0)\) | M1* | As long as at least M1 awarded |
| \(50 \times\) their \(\pm F(0.9)\) | M1dep* | NB \(F(0.9) = -0.496...\) |
| \(24.8\) to \(24.9\) cao | A1 |
## Question 3(i):
$\frac{1}{2} \times 0.2(0 + 0 + 2(0.5 + 0.7 + 0.75 + 0.7 + 0.5))$ $[= 0.63]$ | M3 | M2 if one error, M1 if two errors; condone omission of zeros; or M3 for $0.05 + 0.12 + 0.145 + 0.145 + 0.12 + 0.05$; must be 6 strips; M0 if $h=1$ or $1.2$; Area $= 6.3$ and $0.53$ imply M0
(their $0.63$) $\times 50$ | M1 |
$31.5$ | A1 |
**[5 marks]**
---
## Question 3(ii)(A):
$3.8 \times 0.2^4 - 6.8 \times 0.2^3 + 7.7 \times 0.2^2 - 4.2 \times 0.2$ | M1 | $\pm 0.58032$ implies M1; condone one sign error
$0.01968$ cao isw | A1 | Or B2 if unsupported; allow $-0.01968$
**[2 marks]**
---
## Question 3(ii)(B):
$\frac{3.8x^5}{5} - \frac{6.8x^4}{4} + \frac{7.7x^3}{3} - \frac{4.2x^2}{2} + c$ | M2 | M1 for two terms correct excluding $c$; condone omission of $c$; accept $2.56$ to $2.57$ for coefficient of $x^3$; allow M1 if all signs reversed
$F(0.9) - F(0)$ | M1* | As long as at least M1 awarded
$50 \times$ their $\pm F(0.9)$ | M1dep* | NB $F(0.9) = -0.496...$
$24.8$ to $24.9$ cao | A1 |
**[5 marks]**
---3 A farmer digs ditches for flood relief. He experiments with different cross-sections. Assume that the surface of the ground is horizontal.
\begin{enumerate}[label=(\roman*)]
\item \begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-3_432_640_410_745}
\captionsetup{labelformat=empty}
\caption{Fig. 9.1}
\end{center}
\end{figure}
Fig. 9.1 shows the cross-section of one ditch, with measurements in metres. The width of the ditch is 1.2 m and Fig. 9.1 shows the depth every 0.2 m across the ditch.
Use the trapezium rule with six intervals to estimate the area of cross-section. Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.
\item Another ditch is 0.9 m wide, with cross-section as shown in Fig. 9.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-3_579_813_1336_656}
\captionsetup{labelformat=empty}
\caption{Fig. 9.2}
\end{center}
\end{figure}
With $x$ - and $y$-axes as shown in Fig. 9.2, the curve of the ditch may be modelled closely by $y = 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x$.\\
(A) The actual ditch is 0.6 m deep when $x = 0.2$. Calculate the difference between the depth given by the model and the true depth for this value of $x$.\\
(B) Find $\int \left( 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x \right) \mathrm { d } x$. Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 Q3 [12]}}