| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| 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 followed by routine integration of a polynomial. Part (i) requires careful arithmetic with the trapezium rule formula and a simple volume calculation. Part (ii) involves substituting into a given polynomial and integrating term-by-term using standard power rule. While multi-part and requiring accuracy, it demands only direct application of standard C2 techniques with no problem-solving insight or conceptual challenges. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2} \times 0.2\ (0 + 0 + 2(0.5 + 0.7 + 0.75 + 0.7 + 0.5))\) | M3 | M2 if one error, M1 if two errors; condone omission of zeros; or M3 for \(0.05 + 0.12 + 0.145 + 0.145 + 0.12 + 0.05\); may be unsimplified, must be summed; basic shape of formula must be correct; must be 6 strips; M0 if brackets omitted but allow recovery; M0 if \(h = 1\) or \(1.2\); Area \(= 6.3\) and \(0.53\) imply M0 |
| \([= 0.63]\) | ||
| \((\text{their } 0.63) \times 50\) | M1 | |
| \(31.5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
|---|---|---|
| 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; NB \(F(0.9) = -0.496\ldots\) |
| \(50 \times \text{their}\ \pm F(0.9)\) | M1dep* | |
| \(24.8\) to \(24.9\) cao | A1 |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \times 0.2\ (0 + 0 + 2(0.5 + 0.7 + 0.75 + 0.7 + 0.5))$ | M3 | M2 if one error, M1 if two errors; condone omission of zeros; or M3 for $0.05 + 0.12 + 0.145 + 0.145 + 0.12 + 0.05$; may be unsimplified, must be summed; basic shape of formula must be correct; must be 6 strips; M0 if brackets omitted but allow recovery; M0 if $h = 1$ or $1.2$; Area $= 6.3$ and $0.53$ imply M0 |
| $[= 0.63]$ | | |
| $(\text{their } 0.63) \times 50$ | M1 | |
| $31.5$ | A1 | |
---
## Question 9(ii)(A):
| 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 $-0.01968$ |
---
## Question 9(ii)(B):
| 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; NB $F(0.9) = -0.496\ldots$ |
| $50 \times \text{their}\ \pm F(0.9)$ | M1dep* | |
| $24.8$ to $24.9$ cao | A1 | |
---9 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]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-4_437_640_470_715}
\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]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-4_574_808_1402_632}
\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 2012 Q9 [12]}}