| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Moderate -0.3 This is a standard C2 integration question involving finding intersection points by solving a quadratic equation, setting up an area integral between curve and line, and evaluating a polynomial integral. All techniques are routine for this level, though the multi-step nature and 11 total marks make it slightly more substantial than the most basic exercises. The working is methodical rather than requiring insight, placing it just below average difficulty. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks |
|---|---|
| \(x = -3, 1\) | M1 M1 A1 |
| \(\therefore (-3, 7), (1, 15)\) | A1 |
| Answer | Marks |
|---|---|
| \(\text{area} = \int_{-3}^{1} [(2x + 13) - (2x^2 + 6x + 7)] \, dx\) | M1 |
| \(= \int_{-3}^{1} (6 - 4x - 2x^2) \, dx\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= [6x - 2x^2 - \frac{2}{3}x^3]_{-3}^{1}\) | M1 A2 | |
| \(= (6 - 2 - \frac{2}{3}) - (-18 - 18 + 18) = 21\frac{1}{3}\) | M1 A1 | (11) |
| Answer | Marks |
|---|---|
| Total | (72) |
## (i)
$2x^2 + 6x + 7 = 2x + 13$
$x^2 + 2x - 3 = 0$
$(x + 3)(x - 1) = 0$
$x = -3, 1$ | M1 M1 A1 |
$\therefore (-3, 7), (1, 15)$ | A1 |
## (ii)
$\text{area} = \int_{-3}^{1} [(2x + 13) - (2x^2 + 6x + 7)] \, dx$ | M1 |
$= \int_{-3}^{1} (6 - 4x - 2x^2) \, dx$ | A1 |
## (iii)
$= [6x - 2x^2 - \frac{2}{3}x^3]_{-3}^{1}$ | M1 A2 |
$= (6 - 2 - \frac{2}{3}) - (-18 - 18 + 18) = 21\frac{1}{3}$ | M1 A1 | (11)
---
**Total** | (72)\includegraphics{figure_9}
The diagram shows the curve $y = 2x^2 + 6x + 7$ and the straight line $y = 2x + 13$.
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of the points where the curve and line intersect. [4]
\item Show that the area of the shaded region bounded by the curve and line is given by
$$\int_{-3}^{1} (6 - 4x - 2x^2) dx.$$ [2]
\item Hence find the area of the shaded region. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR C2 Q9 [11]}}