| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| 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. The algebra is straightforward and the method is routine textbook material, making it slightly easier than average but still requiring multiple steps and careful execution. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Puts \(10-x = 10x - x^2 - 8\) and rearranges to three-term quadratic | M1 | Or puts \(y = 10(10-y)-(10-y)^2-8\) and rearranges |
| Solves "\(x^2 - 11x + 18 = 0\)" using acceptable method | M1 | Or solves "\(y^2 - 9y + 8 = 0\)" |
| \(x = 2\), \(x = 9\) (may be on diagram or in part (b) limits) | A1 | Or obtains \(y=8\), \(y=1\) |
| Substitutes their \(x\) into a given equation to give \(y=\) | M1 | Or substitutes their \(y\) to give \(x=\) |
| \(y = 8\), \(y = 1\) | A1 | Or \(x=2\), \(x=9\). Just one pair of correct coordinates with no working is M0M0A0M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int(10x - x^2 - 8)\,dx = \frac{10x^2}{2} - \frac{x^3}{3} - 8x\ \{+c\}\) | M1 A1 A1 | M1: \(x^n \to x^{n+1}\) for any one term. 1st A1: at least two of three terms correct. 2nd A1: all three correct |
| \(\left[\frac{10x^2}{2} - \frac{x^3}{3} - 8x\right]_2^9 = (\ldots) - (\ldots)\) | dM1 | Substitutes 9 and 2 and subtracts |
| \(= 90 - \frac{4}{3} = 88\frac{2}{3}\) or \(\frac{266}{3}\) | ||
| Area of trapezium \(= \frac{1}{2}(8+1)(9-2) = 31.5\) | B1 | Using any correct method (could be integration or triangle minus rectangle plus triangle) |
| Area of \(R\) is \(88\frac{2}{3} - 31.5 = 57\frac{1}{6}\) or \(\frac{343}{6}\) | M1 A1 cao | M1: Their Area under curve \(-\) Their Area under line. A1: Accept \(57.1\overline{6}\) but not \(57.16\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_2^9 (10x-x^2-8)-(10-x)\,dx\) | M1 | 3rd M1: Uses difference between two functions in integral |
| \(\int_2^9 -x^2+11x-18\,dx\) | ||
| \(= -\frac{x^3}{3} + \frac{11x^2}{2} - 18x\ \{+c\}\) | A1 A1 | A1: at least two of three simplified terms correct. A1: Correct integration |
| \(\left[-\frac{x^3}{3}+\frac{11x^2}{2}-18x\right]_2^9 = (\ldots)-(\ldots)\) | dM1 | Substitutes 9 and 2 and subtracts |
| B1 implied by final answer rounding to 57.2 | B1 | |
| \(40.5 - (-16\frac{2}{3}) = 57\frac{1}{6}\) cao | M1 A1 |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Puts $10-x = 10x - x^2 - 8$ and rearranges to three-term quadratic | M1 | Or puts $y = 10(10-y)-(10-y)^2-8$ and rearranges |
| Solves "$x^2 - 11x + 18 = 0$" using acceptable method | M1 | Or solves "$y^2 - 9y + 8 = 0$" |
| $x = 2$, $x = 9$ (may be on diagram or in part (b) limits) | A1 | Or obtains $y=8$, $y=1$ |
| Substitutes their $x$ into a given equation to give $y=$ | M1 | Or substitutes their $y$ to give $x=$ |
| $y = 8$, $y = 1$ | A1 | Or $x=2$, $x=9$. Just one pair of correct coordinates with no working is M0M0A0M1A0 |
## Part (b) Method 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int(10x - x^2 - 8)\,dx = \frac{10x^2}{2} - \frac{x^3}{3} - 8x\ \{+c\}$ | M1 A1 A1 | M1: $x^n \to x^{n+1}$ for any one term. 1st A1: at least two of three terms correct. 2nd A1: all three correct |
| $\left[\frac{10x^2}{2} - \frac{x^3}{3} - 8x\right]_2^9 = (\ldots) - (\ldots)$ | dM1 | Substitutes 9 and 2 and subtracts |
| $= 90 - \frac{4}{3} = 88\frac{2}{3}$ or $\frac{266}{3}$ | | |
| Area of trapezium $= \frac{1}{2}(8+1)(9-2) = 31.5$ | B1 | Using any correct method (could be integration or triangle minus rectangle plus triangle) |
| Area of $R$ is $88\frac{2}{3} - 31.5 = 57\frac{1}{6}$ or $\frac{343}{6}$ | M1 A1 cao | M1: Their Area under curve $-$ Their Area under line. A1: Accept $57.1\overline{6}$ but not $57.16$ |
## Part (b) Method 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_2^9 (10x-x^2-8)-(10-x)\,dx$ | M1 | 3rd M1: Uses difference between two functions in integral |
| $\int_2^9 -x^2+11x-18\,dx$ | | |
| $= -\frac{x^3}{3} + \frac{11x^2}{2} - 18x\ \{+c\}$ | A1 A1 | A1: at least two of three simplified terms correct. A1: Correct integration |
| $\left[-\frac{x^3}{3}+\frac{11x^2}{2}-18x\right]_2^9 = (\ldots)-(\ldots)$ | dM1 | Substitutes 9 and 2 and subtracts |
| B1 implied by final answer rounding to 57.2 | B1 | |
| $40.5 - (-16\frac{2}{3}) = 57\frac{1}{6}$ cao | M1 A1 | |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f1ef99f0-4ad4-49d8-bee7-d5bb9cc84660-07_823_1081_267_404}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the line with equation $y = 10 - x$ and the curve with equation $y = 10 x - x ^ { 2 } - 8$
The line and the curve intersect at the points $A$ and $B$, and $O$ is the origin.
\begin{enumerate}[label=(\alph*)]
\item Calculate the coordinates of $A$ and the coordinates of $B$.
The shaded area $R$ is bounded by the line and the curve, as shown in Figure 2.
\item Calculate the exact area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2012 Q5 [12]}}