| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| 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 C2 trapezium rule application with standard follow-up parts. Part (i) is routine trapezium rule calculation with given data, (ii) tests understanding of over/underestimates using basic reasoning, (iii) is simple percentage calculation, and (iv) is standard polynomial integration. All parts are textbook-standard with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration |
| Time \(( t\) seconds \()\) | 0 | 10 | 20 | 30 | 40 | 50 | 60 |
| Speed \(\left( v \mathrm {~m} \mathrm {~s} ^ { - 1 } \right)\) | 28 | 19 | 14 | 11 | 12 | 16 | 22 |
| Answer | Marks |
|---|---|
| (i) \(\frac{1}{2}(10)[28 + 2(19 + 14 + 11 + 12 + 16) + 22]\) | M1 A1 |
| \(= 5[28 + 124 + 22] = 5 \times 174 = 870\) m | A1 A1 |
| (ii) Overestimate because curve is concave (bends upward); underestimate using right-hand rectangles (or left-hand for decreasing portion): | B1 |
| \(10(19 + 14 + 11 + 12 + 16 + 22) = 10 \times 94 = 940\)... using lower rectangles: \(10(19+14+11+11+12+16) = 830\) m | M1 A1 |
| (iii) Model gives \(28 - 10 + 0.015(100) = 28 - 10 + 1.5 = 19.5\); difference \(= 0.5\); \(\frac{0.5}{19} \approx 2.6\% < 3\%\) | M1 A1 |
| (iv) \(\int_0^{60}(28 - t + 0.015t^2)\,\mathrm{d}t = \left[28t - \frac{t^2}{2} + 0.005t^3\right]_0^{60}\) | M1 A1 |
| \(= 1680 - 1800 + 1080 = 960\) m | A1 |
## Question 10:
**(i)** $\frac{1}{2}(10)[28 + 2(19 + 14 + 11 + 12 + 16) + 22]$ | M1 A1 |
$= 5[28 + 124 + 22] = 5 \times 174 = 870$ m | A1 A1 |
**(ii)** Overestimate because curve is concave (bends upward); underestimate using right-hand rectangles (or left-hand for decreasing portion): | B1 |
$10(19 + 14 + 11 + 12 + 16 + 22) = 10 \times 94 = 940$... using lower rectangles: $10(19+14+11+11+12+16) = 830$ m | M1 A1 |
**(iii)** Model gives $28 - 10 + 0.015(100) = 28 - 10 + 1.5 = 19.5$; difference $= 0.5$; $\frac{0.5}{19} \approx 2.6\% < 3\%$ | M1 A1 |
**(iv)** $\int_0^{60}(28 - t + 0.015t^2)\,\mathrm{d}t = \left[28t - \frac{t^2}{2} + 0.005t^3\right]_0^{60}$ | M1 A1 |
$= 1680 - 1800 + 1080 = 960$ m | A1 |
---10 Fig. 10 shows the speed of a car, in metres per second, during one minute, measured at 10-second intervals.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2bdf241f-4538-4227-ba00-fe843d1b3aca-4_732_748_379_657}
\captionsetup{labelformat=empty}
\caption{Fig. 10}
\end{center}
\end{figure}
The measured speeds are shown below.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
Time $( t$ seconds $)$ & 0 & 10 & 20 & 30 & 40 & 50 & 60 \\
\hline
Speed $\left( v \mathrm {~m} \mathrm {~s} ^ { - 1 } \right)$ & 28 & 19 & 14 & 11 & 12 & 16 & 22 \\
\hline
\end{tabular}
\end{center}
(i) Use the trapezium rule with 6 strips to find an estimate of the area of the region bounded by the curve, the line $t = 60$ and the axes. [This area represents the distance travelled by the car.]\\
(ii) Explain why your calculation in part (i) gives an overestimate for this area. Use appropriate rectangles to calculate an underestimate for this area.
The speed of the car may be modelled by $v = 28 - t + 0.015 t ^ { 2 }$.\\
(iii) Show that the difference between the value given by the model when $t = 10$ and the measured value is less than $3 \%$ of the measured value.\\
(iv) According to this model, the distance travelled by the car is
$$\int _ { 0 } ^ { 60 } \left( 28 - t + 0.015 t ^ { 2 } \right) \mathrm { d } t$$
Find this distance.
\hfill \mbox{\textit{OCR MEI C2 2007 Q10 [12]}}