| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area under piecewise-defined curve |
| Difficulty | Moderate -0.3 This is a straightforward piecewise function question requiring basic differentiation to find a maximum and integration of simple polynomials. The structure is clearly guided with two parts, and the techniques are standard C2 material with no conceptual challenges—slightly easier than average due to the simple functions involved. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = -x + 3\) | M1 | Differentiate |
| \(= 0\) when \(x = 3\) | M1 | Set \(= 0\) |
| \(\Rightarrow y = 1.5\) | A1, A1 | For \(x\); for \(y\) |
| Highest point is where \(x = 3\), \(\Rightarrow y = -\frac{9}{2} + 9 - 3 = 1.5\) | Alt. B2, M1, A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area \(= 1\times2 + \int_2^4\left(-\frac{1}{2}x^2 + 3x - 3\right)dx + 1\times2\) | M1 | |
| \(= 4 + \left[-\frac{x^3}{6} + \frac{3x^2}{2} - 3x\right]_2^4\) | A1, B1 | Adding on the 4 |
| \(= 4 + \left(-\frac{64}{6} + \frac{48}{2} - 12\right) - \left(-\frac{8}{6} + \frac{12}{2} - 6\right)\) | M1 | Definite integral |
| \(= 4 + 1\frac{1}{3} + 1\frac{1}{3} = 6\frac{2}{3}\) | A1, A1, A1 | \(2^2/_3\), c.a.o. [7] |
## Question 11:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = -x + 3$ | M1 | Differentiate |
| $= 0$ when $x = 3$ | M1 | Set $= 0$ |
| $\Rightarrow y = 1.5$ | A1, A1 | For $x$; for $y$ |
| Highest point is where $x = 3$, $\Rightarrow y = -\frac{9}{2} + 9 - 3 = 1.5$ | | Alt. B2, M1, A1 **[4]** |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area $= 1\times2 + \int_2^4\left(-\frac{1}{2}x^2 + 3x - 3\right)dx + 1\times2$ | M1 | |
| $= 4 + \left[-\frac{x^3}{6} + \frac{3x^2}{2} - 3x\right]_2^4$ | A1, B1 | Adding on the 4 |
| $= 4 + \left(-\frac{64}{6} + \frac{48}{2} - 12\right) - \left(-\frac{8}{6} + \frac{12}{2} - 6\right)$ | M1 | Definite integral |
| $= 4 + 1\frac{1}{3} + 1\frac{1}{3} = 6\frac{2}{3}$ | A1, A1, A1 | $2^2/_3$, c.a.o. **[7]** |11 The cross-section of a brick wall built on horizontal ground is given, for $0 \leq x \leq 6$, by the following function
$$\begin{array} { l l }
0 \leq x \leq 2 & y = 1 \\
2 \leq x \leq 4 & y = - \frac { 1 } { 2 } x ^ { 2 } + 3 x - 3 \\
4 \leq x \leq 6 & y = 1
\end{array}$$
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{13bfa97b-ec49-4f41-b3dd-d9a31a2c30e8-4_523_1327_633_413}
\end{center}
Units are metres.\\
(i) Show that the highest point on the wall is 1.5 metres above the ground.\\
(ii) Find the area of the cross-section of the wall.
\hfill \mbox{\textit{OCR MEI C2 Q11 [11]}}