| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| 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 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 | 1.02q 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 | Mark | Guidance |
| \(-x^2 + 2x + 24 = x + 4\) (eliminating \(y\)) | B1 | Implied by resulting quadratic |
| \(x^2 - x - 20 = 0 \Rightarrow (x-5)(x+4) = 0\) | M1 | Attempt to solve resulting quadratic |
| \(x = 5,\ x = -4\) | A1 | Both values |
| \(y = 9\) and \(y = 0\) | B1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int(-x^2 + 2x + 24)\,dx = -\frac{x^3}{3} + \frac{2x^2}{2} + 24x\) | M1A1A1 | M1: \(x^n \to x^{n+1}\) for any one term; 1st A1 two of three terms correct; 2nd A1 correct answer |
| \(\left[-\frac{x^3}{3} + \frac{2x^2}{2} + 24x\right]_{-4}^{5}\) substituting limits and subtracting | dM1 | Substitutes \(5\) and \(-4\), subtracts either way |
| \(\left(103\frac{1}{3}\right) - \left(-58\frac{2}{3}\right) = 162\) | ||
| Area of \(\Delta = \frac{1}{2}(9)(9) = 40.5\) | M1 | Correct method for triangle area |
| Area of \(R = 162 - 40.5 = 121.5\) | M1, A1 oe cao | Area under curve minus area of triangle |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\{\)Curve \(=\) Line\(\} \Rightarrow y = -(y-4)^2 + 2(y-4) + 24\) | B1 | Eliminating \(x\) correctly |
| \(y^2 - 9y \{=0\} \Rightarrow y(y-9)\{=0\} \Rightarrow y = \ldots\) | M1 | Attempt to solve resulting quadratic to give \(y\) values |
| So \(y = 0, 9\) | A1 | Both \(y=0\) and \(y=9\) |
| So corresponding \(x\)-values are \(x=-4\) and \(x=5\) | B1ft | Correctly substituting \(y\) values into line or parabola to give both correct \(x\)-values |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Area of \(R = \displaystyle\int_{-4}^{5}(-x^2+2x+24)-(x+4)\,dx\) | M1 (3rd), M1 (4th) | Uses integral of \((x+4)\) with correct ft limits; uses "curve" \(-\) "line" with correct ft limits |
| \(= -\dfrac{x^3}{3} + \dfrac{x^2}{2} + 20x \{+c\}\) | M1 | \(x^n \to x^{n+1}\) for any one term |
| A1ft | At least two out of three terms correct | |
| A1 | Correct answer (ignore \(+c\)) | |
| \(\left[-\dfrac{x^3}{3}+\dfrac{x^2}{2}+20x\right]_{-4}^{5} = (\ldots\ldots)-(\ldots\ldots)\) | dM1 | Substitutes 5 and \(-4\) (or their limits) into integrated function and subtracts either way round |
| \(\left\{\left(-\dfrac{125}{3}+\dfrac{25}{2}+100\right)-\left(\dfrac{64}{3}+8-80\right)\right\} = \left(70\dfrac{5}{6}\right)-\left(-50\dfrac{2}{3}\right)\) | M1, M1 | See above working for 3rd and 4th M1 |
| Area of \(R = 121.5\) | A1 cao |
# Question 9(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-x^2 + 2x + 24 = x + 4$ (eliminating $y$) | B1 | Implied by resulting quadratic |
| $x^2 - x - 20 = 0 \Rightarrow (x-5)(x+4) = 0$ | M1 | Attempt to solve resulting quadratic |
| $x = 5,\ x = -4$ | A1 | Both values |
| $y = 9$ and $y = 0$ | B1ft | |
---
# Question 9(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int(-x^2 + 2x + 24)\,dx = -\frac{x^3}{3} + \frac{2x^2}{2} + 24x$ | M1A1A1 | M1: $x^n \to x^{n+1}$ for any one term; 1st A1 two of three terms correct; 2nd A1 correct answer |
| $\left[-\frac{x^3}{3} + \frac{2x^2}{2} + 24x\right]_{-4}^{5}$ substituting limits and subtracting | dM1 | Substitutes $5$ and $-4$, subtracts either way |
| $\left(103\frac{1}{3}\right) - \left(-58\frac{2}{3}\right) = 162$ | | |
| Area of $\Delta = \frac{1}{2}(9)(9) = 40.5$ | M1 | Correct method for triangle area |
| Area of $R = 162 - 40.5 = 121.5$ | M1, A1 oe cao | Area under curve minus area of triangle |
## Question 9(a) — Aliter Way 2:
| Working | Mark | Guidance |
|---|---|---|
| $\{$Curve $=$ Line$\} \Rightarrow y = -(y-4)^2 + 2(y-4) + 24$ | B1 | Eliminating $x$ correctly |
| $y^2 - 9y \{=0\} \Rightarrow y(y-9)\{=0\} \Rightarrow y = \ldots$ | M1 | Attempt to solve resulting quadratic to give $y$ values |
| So $y = 0, 9$ | A1 | Both $y=0$ and $y=9$ |
| So corresponding $x$-values are $x=-4$ and $x=5$ | B1ft | Correctly substituting $y$ values into line or parabola to give **both correct** $x$-values |
**[4 marks]**
---
## Question 9(b) — Aliter Way 2:
| Working | Mark | Guidance |
|---|---|---|
| Area of $R = \displaystyle\int_{-4}^{5}(-x^2+2x+24)-(x+4)\,dx$ | M1 (3rd), M1 (4th) | Uses integral of $(x+4)$ with correct ft limits; uses "curve" $-$ "line" with correct ft limits |
| $= -\dfrac{x^3}{3} + \dfrac{x^2}{2} + 20x \{+c\}$ | M1 | $x^n \to x^{n+1}$ for any one term |
| | A1ft | At least two out of three terms correct |
| | A1 | Correct answer (ignore $+c$) |
| $\left[-\dfrac{x^3}{3}+\dfrac{x^2}{2}+20x\right]_{-4}^{5} = (\ldots\ldots)-(\ldots\ldots)$ | dM1 | Substitutes 5 and $-4$ (or **their limits**) into integrated function and subtracts either way round |
| $\left\{\left(-\dfrac{125}{3}+\dfrac{25}{2}+100\right)-\left(\dfrac{64}{3}+8-80\right)\right\} = \left(70\dfrac{5}{6}\right)-\left(-50\dfrac{2}{3}\right)$ | M1, M1 | See above working for 3rd and 4th M1 |
| Area of $R = 121.5$ | A1 cao | |
**[7 marks] [11 total]**
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c9758792-ca4c-4837-bd7c-e695fe0c0cdf-14_360_956_278_504}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The straight line with equation $y = x + 4$ cuts the curve with equation $y = - x ^ { 2 } + 2 x + 24$ at the points $A$ and $B$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Use algebra to find the coordinates of the points $A$ and $B$.
The finite region $R$ is bounded by the straight line and the curve and is shown shaded in Figure 3.
\item Use calculus to find the exact area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2011 Q9 [11]}}