| 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 -0.8 This is a straightforward application of the trapezium rule and rectangular approximation with given data points, followed by routine integration of a polynomial. All techniques are standard C2 procedures requiring only careful arithmetic and basic calculus recall, with no problem-solving insight needed. The comparison in part (iv) is a simple substitution and comment. |
| Spec | 1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(47.625 \text{ [m}^3\text{]}\) to 3 sf or more, with correct method shown | 4 | M3 for \(\frac{1.5}{2} \times (2.3 + 2 + 2[2.7 + 3.3 + 4 + 4.8 + 5.2 + 5.2 + 4.4])\) |
| (ii) \(43.05\) | 2 | M1 for \(1.5 \times (2.3+2.7+3.3+4+4.8+5.2+4.4+2)\) |
| (iii) \(-0.013x^4/4 + 0.16x^3/3 - 0.082x^2/2 + 2.4x\) o.e. | M2 | M1 for three terms correct |
| Their integral evaluated at \(x = 12\) (and \(0\)) only | M1 | Dep on integration attempted |
| \(47.6\) to \(47.7\) | A1 | |
| (iv) \(5.30...\) found; compared with \(5.2\) s.o.i. | 1, D1 |
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $47.625 \text{ [m}^3\text{]}$ to 3 sf or more, with correct method shown | 4 | M3 for $\frac{1.5}{2} \times (2.3 + 2 + 2[2.7 + 3.3 + 4 + 4.8 + 5.2 + 5.2 + 4.4])$ |
| **(ii)** $43.05$ | 2 | M1 for $1.5 \times (2.3+2.7+3.3+4+4.8+5.2+4.4+2)$ |
| **(iii)** $-0.013x^4/4 + 0.16x^3/3 - 0.082x^2/2 + 2.4x$ o.e. | M2 | M1 for three terms correct |
| Their integral evaluated at $x = 12$ (and $0$) only | M1 | Dep on integration attempted |
| $47.6$ to $47.7$ | A1 | |
| **(iv)** $5.30...$ found; compared with $5.2$ s.o.i. | 1, D1 | |
---3 Fig. 11 shows the cross-section of a school hall, with measurements of the height in metres taken at 1.5 m intervals from O .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f56da008-e7f5-45b9-8db8-e2ba09ab0161-2_579_1385_1035_424}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
(i) Use the trapezium rule with 8 strips to calculate an estimate of the area of the cross-section.\\
(ii) Use 8 rectangles to calculate a lower bound for the area of the cross-section.
The curve of the roof may be modelled by $y = - 0.013 x ^ { 3 } + 0.16 x ^ { 2 } - 0.082 x + 2.4$, where $x$ metres is the horizontal distance from O across the hall, and $y$ metres is the height.\\
(iii) Use integration to find the area of the cross-section according to this model.\\
(iv) Comment on the accuracy of this model for the height of the hall when $x = 7.5$.
\hfill \mbox{\textit{OCR MEI C2 Q3 [12]}}