| 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 multi-part question testing standard C2 skills: plotting points, applying trapezium rule with given ordinates, integrating a simple polynomial, and basic optimization. All techniques are routine with no novel problem-solving required, making it easier than average for A-level. |
| 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 |
|---|---|---|
| Graph: points and curve | B1, B1 | Points; Curve |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = \frac{h}{2}\left(y_0 + 2(y_2+y_4+y_6+y_8)+y_{10}\right)\) | M1 | Method |
| \(= \frac{2}{2}(0+2(9+15+18+15)+10) = 124\) | A1, A1 | Substitution; c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = \int_0^{10}(6t-0.5t^2)\,dt = \left[3t^2 - \frac{0.5t^3}{3}\right]_0^{10}\) | M1, A2,1 | Integration; Each term |
| \(= 300 - 167 = 133\) | M1, A1 | Substitute; c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Graph sketch | B1 | |
| John's estimate is under because all the trapezia are under. The area under the curve looks to be over. | B1 | Total: 2 |
## Question 11(i):
Graph: points and curve | B1, B1 | Points; Curve | **Total: 2**
## Question 11(ii):
$A = \frac{h}{2}\left(y_0 + 2(y_2+y_4+y_6+y_8)+y_{10}\right)$ | M1 | Method
$= \frac{2}{2}(0+2(9+15+18+15)+10) = 124$ | A1, A1 | Substitution; c.a.o. | **Total: 3**
## Question 11(iii):
$A = \int_0^{10}(6t-0.5t^2)\,dt = \left[3t^2 - \frac{0.5t^3}{3}\right]_0^{10}$ | M1, A2,1 | Integration; Each term
$= 300 - 167 = 133$ | M1, A1 | Substitute; c.a.o. | **Total: 5**
## Question 11(iv):
Graph sketch | B1 |
John's estimate is under because all the trapezia are under. The area under the curve looks to be over. | B1 | **Total: 2**
---(i) The speed-time graph on the insert sheet provides the axes and the first two points plotted. Plot the remainder of these points and join them with a smooth curve.
The area between this curve and the $t$-axis represents the distance travelled by the car in this time.\\
(ii) Using the trapezium rule with 6 values of $t$ estimate the area under the curve to give the distance travelled. Illustrate on your graph the area found.\\
(iii) John's teacher suggests that the equation of the curve could be $v = 6 t - \frac { 1 } { 2 } t ^ { 2 }$. Find, by calculus, the area between this curve and the $t$ axis.\\
(iv) Plot this curve on your graph. Comment on whether the estimates obtained in parts (ii) and (iii) are overestimates or underestimates.
12 Fig. 12 shows a window. The base and sides are parts of a rectangle with dimensions $2 x$ metres horizontally by $y$ metres vertically. The top is a semicircle of radius $x$ metres. The perimeter of the window is 10 metres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{73d1c02b-1b7b-426d-a171-c762597cfed4-4_428_433_1638_766}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{center}
\end{figure}
(i) Express $y$ as a function of $x$.\\
(ii) Find the total area, $A \mathrm {~m} ^ { 2 }$, in terms of $x$ and $y$. Use your answer to part (i) to show that this simplifies to
$$A = 10 x - 2 x ^ { 2 } - \frac { 1 } { 2 } \pi x ^ { 2 }$$
(iii) Prove that for the maximum value of $A$, $y = x$ exactly.\\
\section*{MEI STRUCTURED MATHEMATICS }
\section*{CONCEPTS FOR ADVANCED MATHEMATICS, C2}
\section*{Practice Paper C2-B \\
Insert sheet for question 11}
\hfill \mbox{\textit{OCR MEI C2 Q11 [12]}}