| 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 C2 integration and trapezium rule question. Part (i) involves routine differentiation to find maximum height and standard integration of a quadratic. Part (ii) applies the trapezium rule formula with clearly given ordinates—a textbook exercise requiring only method recall and careful arithmetic, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(6.25\) | B2 | M1 for \(x = 5\) used to find \(y\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((V =)\) area of cross-section \(\times\) length | E1 | |
| \(\left(\frac{100}{4}\right)\left[\frac{10}{2}x^2 - \frac{1}{3}x^3\right]\) o.e. | M1 | |
| \([\text{val at } x = 10] - [\text{val at } x = 0]\) | M1 | Subs of correct limits into their integrand |
| \(4166\) to \(4167\) or \(4170\) | A2 | A1 for \(166.6...\) or \(16666.6...\) or \(41.6...\)rot to 3 sf or more |
| Answer | Marks | Guidance |
|---|---|---|
| \(52.62\) | B4 | M3 for \(\frac{2}{2}\times[2.15x2 + 2(5.64x2 + 6.44x2)]\) o.e.; Or M2 if one slip; Or M1 if 2 slips or one trap evaluated |
| Their\((5262)\) \(-\) their \((4167)\) | M1 | Must be \(> 0\) |
## Question 3:
**Part iA:**
$6.25$ | B2 | M1 for $x = 5$ used to find $y$ | **[2]**
**Part iB:**
$(V =)$ area of cross-section $\times$ length | E1 |
$\left(\frac{100}{4}\right)\left[\frac{10}{2}x^2 - \frac{1}{3}x^3\right]$ o.e. | M1 |
$[\text{val at } x = 10] - [\text{val at } x = 0]$ | M1 | Subs of correct limits into their integrand |
$4166$ to $4167$ or $4170$ | A2 | A1 for $166.6...$ or $16666.6...$ or $41.6...$rot to 3 sf or more | **[5]**
**Part ii:**
$52.62$ | B4 | M3 for $\frac{2}{2}\times[2.15x2 + 2(5.64x2 + 6.44x2)]$ o.e.; Or M2 if one slip; Or M1 if 2 slips or one trap evaluated |
Their$(5262)$ $-$ their $(4167)$ | M1 | Must be $> 0$ | **[5][12]**3
\begin{enumerate}[label=(\roman*)]
\item A tunnel is 100 m long. Its cross-section, shown in Fig. 9.1, is modelled by the curve
$$y = \frac { 1 } { 4 } \left( 10 x - x ^ { 2 } \right) ,$$
where $x$ and $y$ are horizontal and vertical distances in metres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-3_512_819_493_700}
\captionsetup{labelformat=empty}
\caption{Figure 9.1}
\end{center}
\end{figure}
Using this model,\\
(A) find the greatest height of the tunnel,\\
(B) explain why $100 \int _ { 0 } ^ { 10 } y \mathrm {~d} x$ gives the volume, in cubic metres, of earth removed to make the tunnel. Calculate this volume.\\[0pt]
[5]
\item The roof of the tunnel is re-shaped to allow for larger vehicles. Fig. 9.2 shows the new crosssection.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-3_506_942_1703_629}
\captionsetup{labelformat=empty}
\caption{Not to scale}
\end{center}
\end{figure}
Fig. 9.2
Use the trapezium rule with 5 strips to estimate the new cross-sectional area.\\
Hence estimate the volume of earth removed when the tunnel is re-shaped.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 Q3 [12]}}