| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Curve-Line Intersection Area |
| Difficulty | Moderate -0.3 This is a standard C2 integration question requiring finding intersection points by solving a quadratic equation, then computing a definite integral of the difference between two functions. While it involves multiple steps, the techniques are routine and well-practiced at this level, making it slightly easier than average. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(2x^2 + 6x + 7 = 2x + 13\) | M1 | |
| \(x^2 + 2x - 3 = 0\) | M1 | |
| \((x+3)(x-1) = 0\) | A1 | |
| \(x = -3, 1\) | A1 | |
| \(\therefore (-3, 7), (1, 15)\) | A1 | |
| (b) \(\text{area under curve} = \int_{-3}^{1} (2x^2 + 6x + 7) \, dx\) | M1 | |
| \(= [\frac{2}{3}x^3 + 3x^2 + 7x]_{-3}^{1}\) | M1 A2 | |
| \(= (\frac{2}{3} + 3 + 7) - (-18 + 27 - 21) = 22\frac{2}{3}\) | M1 | |
| \(\text{area of trapezium} = \frac{1}{2} \times (7 + 15) \times 4 = 44\) | B1 | |
| \(\text{shaded area} = 44 - 22\frac{2}{3} = 21\frac{1}{3}\) | M1 A1 | (11) |
(a) $2x^2 + 6x + 7 = 2x + 13$ | M1 |
$x^2 + 2x - 3 = 0$ | M1 |
$(x+3)(x-1) = 0$ | A1 |
$x = -3, 1$ | A1 |
$\therefore (-3, 7), (1, 15)$ | A1 |
(b) $\text{area under curve} = \int_{-3}^{1} (2x^2 + 6x + 7) \, dx$ | M1 |
$= [\frac{2}{3}x^3 + 3x^2 + 7x]_{-3}^{1}$ | M1 A2 |
$= (\frac{2}{3} + 3 + 7) - (-18 + 27 - 21) = 22\frac{2}{3}$ | M1 |
$\text{area of trapezium} = \frac{1}{2} \times (7 + 15) \times 4 = 44$ | B1 |
$\text{shaded area} = 44 - 22\frac{2}{3} = 21\frac{1}{3}$ | M1 A1 | (11)\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the points where the curve and line intersect.
\item Find the area of the shaded region bounded by the curve and line.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q7 [11]}}