| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Session | Specimen |
| Marks | 12 |
| 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 integrating the difference of two functions. While it involves multiple steps (solving quadratic, setting up integral, integrating polynomial), these are all routine techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 - 3x + 8 = x + 5\) | M1 | Line = curve |
| \(x^2 - 4x + 3 = 0\) | M1 | 3TQ = 0 |
| \(0 = (x-3)(x-1)\) | M1 | Solving |
| \(A\) is \((1, 6)\); \(B\) is \((3, 8)\) | A1; A1 | |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int(x^2 - 3x + 8)\,dx = \left[\frac{x^3}{3} - \frac{3x^2}{2} + 8x\right]\) | M1 A2(1,0) | Integration |
| Area below curve \(= (9 - \frac{27}{2} + 24) - (\frac{1}{3} - \frac{3}{2} + 8) = 12\frac{2}{3}\) | M1 | Use of Limits |
| Trapezium \(= \frac{1}{2} \times 2 \times (6+8) = 14\) | B1 | |
| Area = Trapezium \(-\) Integral \(= 14 - 12\frac{2}{3} = 1\frac{1}{3}\) | M1, A1 | |
| (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-x^2 + 4x - 3\) | M1 | Line \(-\) curve |
| \(\int(-x^2 + 4x - 3)\,dx = \left[-\frac{x^3}{3} + 2x^2 - 3x\right]\) | M1 A2(1,0) | Integration |
| Area \(= \int_1^3(\ldots)\,dx = (-9+18-9) - (-\frac{1}{3}+2-3)\) | M1 | Use of limits |
| \(= 1\frac{1}{3}\) | A2 | |
| (7) |
# Question 8:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 - 3x + 8 = x + 5$ | M1 | Line = curve |
| $x^2 - 4x + 3 = 0$ | M1 | 3TQ = 0 |
| $0 = (x-3)(x-1)$ | M1 | Solving |
| $A$ is $(1, 6)$; $B$ is $(3, 8)$ | A1; A1 | |
| | **(5)** | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int(x^2 - 3x + 8)\,dx = \left[\frac{x^3}{3} - \frac{3x^2}{2} + 8x\right]$ | M1 A2(1,0) | Integration |
| Area below curve $= (9 - \frac{27}{2} + 24) - (\frac{1}{3} - \frac{3}{2} + 8) = 12\frac{2}{3}$ | M1 | Use of Limits |
| Trapezium $= \frac{1}{2} \times 2 \times (6+8) = 14$ | B1 | |
| Area = Trapezium $-$ Integral $= 14 - 12\frac{2}{3} = 1\frac{1}{3}$ | M1, A1 | |
| | **(7)** | |
### ALT Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-x^2 + 4x - 3$ | M1 | Line $-$ curve |
| $\int(-x^2 + 4x - 3)\,dx = \left[-\frac{x^3}{3} + 2x^2 - 3x\right]$ | M1 A2(1,0) | Integration |
| Area $= \int_1^3(\ldots)\,dx = (-9+18-9) - (-\frac{1}{3}+2-3)$ | M1 | Use of limits |
| $= 1\frac{1}{3}$ | A2 | |
| | **(7)** | |
---8.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{afaf76d8-2a1f-4239-8275-70fad4f418c1-2_616_712_1658_713}
\end{center}
\end{figure}
The line with equation $y = x + 5$ cuts the curve with equation $y = x ^ { 2 } - 3 x + 8$ at the points $A$ and $B$, as shown in Fig. 2.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the points $A$ and $B$.
\item Find the area of the shaded region between the curve and the line, as shown in Fig. 2.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q8 [12]}}