| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| 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 (algebra then calculus), both techniques are routine and commonly practiced at this level, 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 integrals |
| Answer | Marks | Guidance |
|---|---|---|
| sub for \(y\) in \(y = 3x + 20\), \(y = 5\) or \(26\) | M1, M1, A1, M1, A1 | (5 marks) |
| Answer | Marks |
|---|---|
| \(\int(10-3x-x^2)dx = 10x - \frac{3}{2}x^2 - \frac{x^3}{3}\) | M1, A1, M1, A2/1/0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 11\frac{1}{3} - 45\frac{5}{6} = 57\frac{1}{6}\) | M1, A1 | (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int(x^2+6x+10)dx = \frac{x^3}{3} + 3x^2 + 10x\) | M1 A2 | |
| use of limits \(= \left(\frac{8}{3} + 12 + 20\right) - \left(-\frac{125}{3} + 75 - 50\right) = (108\frac{1}{2})\) | M1 | |
| Area of Trapezium \(= \frac{1}{2}(5+26)(2--5) = (51\frac{1}{4})\) | B1 \(\checkmark\) | |
| Shaded area \(=\) Trapezium \(- \int = 108\frac{1}{2} - 51\frac{1}{4} = 57\frac{1}{4}\) | M1 A1 | (7 marks) |
**(a)** $x^2 + 6x + 10 = 3x + 20$
$\Rightarrow x^2 + 3x - 10 = 0$
$(x+5)(x-2) = 0$ so $x = -5$ or $2$
sub for $y$ in $y = 3x + 20$, $y = 5$ or $26$ | M1, M1, A1, M1, A1 | (5 marks)
**(b)** line – curve, $= 10 - 3x - x^2$
$\int(10-3x-x^2)dx = 10x - \frac{3}{2}x^2 - \frac{x^3}{3}$ | M1, A1, M1, A2/1/0 |
$\left[10x - \frac{3}{2}x^2 - \frac{x^3}{3}\right]_{-5}^2 = \left(20 - \frac{3}{2} \times 4 - \frac{8}{3}\right) - \left(-50 - \frac{3}{2} \times 25 + \frac{125}{3}\right)$
$= 11\frac{1}{3} - 45\frac{5}{6} = 57\frac{1}{6}$ | M1, A1 | (7 marks)
**(Subtotal: 12 marks)**
Guidance: (a) 1st M1: putting curve = line. 3rd M1: obtaining at least one $y$ value. Don't need A and B identified.
(b) 1st M1: for $\pm(10-3x-x^2)$. 3rd M1 (2nd on ALT): using their limits. $\checkmark$ their $x$ values from (a).
**ALT (b):**
$\int(x^2+6x+10)dx = \frac{x^3}{3} + 3x^2 + 10x$ | M1 A2 |
use of limits $= \left(\frac{8}{3} + 12 + 20\right) - \left(-\frac{125}{3} + 75 - 50\right) = (108\frac{1}{2})$ | M1 |
Area of Trapezium $= \frac{1}{2}(5+26)(2--5) = (51\frac{1}{4})$ | B1 $\checkmark$ |
Shaded area $=$ Trapezium $- \int = 108\frac{1}{2} - 51\frac{1}{4} = 57\frac{1}{4}$ | M1 A1 | (7 marks)8.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{13bca882-27da-40f2-99d8-4fdeb6629c4e-14_1102_1317_308_340}
\end{center}
\end{figure}
The line with equation $y = 3 x + 20$ cuts the curve with equation $y = x ^ { 2 } + 6 x + 10$ at the points $A$ and $B$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Use algebra to find the coordinates of $A$ and the coordinates of $B$.
The shaded region $S$ is bounded by the line and the curve, as shown in Figure 2.
\item Use calculus to find the exact area of $S$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2005 Q8 [12]}}