| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Tangent or Normal Bounded Area |
| Difficulty | Standard +0.3 This is a standard C2 area-between-curves question with all key information provided (coordinates, that the line is tangent at A). Students must find the tangent equation, set up the integral of (line - curve), and evaluate—routine techniques with no novel insight required, though slightly above average due to multiple steps and cubic integration. |
| Spec | 1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int\left(\frac{1}{8}x^3 + \frac{3}{4}x^2\right)dx = \frac{x^4}{32} + \frac{x^3}{4} \{+c\}\) | M1A1 | M1: \(x^n \to x^{n+1}\) on either term. A1: \(\frac{x^4}{32} + \frac{x^3}{4}\), any correct simplified or unsimplified form (+ c not required) |
| \(\left[\frac{x^4}{32}+\frac{x^3}{4}\right]_{-4}^{2} = \left(\frac{16}{32}+\frac{8}{4}\right)-\left(\frac{256}{32}+\frac{(-64)}{4}\right)\) | dM1 | Substitutes limits of 2 and \(-4\) into integrated function and subtracts either way round |
| \(= \frac{21}{2}\) | A1 | \(\frac{21}{2}\) or 10.5 |
| At \(x=-4\), \(y = -8+12=4\) or at \(x=2\), \(y=1+3=4\) | ||
| Area of Rectangle \(= 6 \times 4 = 24\), or Area of Rectangles \(= 4\times4=16\) and \(2\times4=8\) | M1 | Evidence of \((4--2)\times\) their \(y_{-4}\) or \((4--2)\times\) their \(y_2\), or Evidence of \(4\times\) their \(y_{-4}\) and \(2\times\) their \(y_2\) |
| So, \(\text{area}(R) = 24 - \frac{21}{2} = \frac{27}{2}\) | dddM1A1 | Area rectangle \(-\) integrated answer. Dependent on all previous method marks and requires: Rectangle > integration > 0. A1: \(\frac{27}{2}\) or 13.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\pm\int \left(\text{"their 4"} - \left(\frac{1}{8}x^3+\frac{3}{4}x^2\right)\right)dx\) | 4th M1 | Line \(-\) curve. Condone missing brackets and allow either way round |
| \(= 4x - \frac{x^4}{32} - \frac{x^3}{4}\{+c\}\) | 1st M1, 1st A1ft | M1: \(x^n\to x^{n+1}\) on either curve term. A1ft: \(-\frac{x^4}{32}-\frac{x^3}{4}\), any correct simplified or unsimplified form of curve terms, follow through sign errors |
| \(\left[\ \right]_{-4}^{2} = \left(8-\frac{16}{32}-\frac{8}{4}\right)-\left(-16-\frac{256}{32}-\frac{(-64)}{4}\right)\) | 2nd M1, 3rd M1, 2nd A1 | 2nd M1: substitutes limits of 2 and \(-4\). 3rd M1 for \(\pm(\text{"8"}-\text{"}-16\text{"})\): substitutes limits into line part and subtracts. 2nd A1 for correct \(\pm\) underlined expression |
| \(= \frac{27}{2}\) | 3rd A1 | \(\frac{27}{2}\) or 13.5 |
# Question 6:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int\left(\frac{1}{8}x^3 + \frac{3}{4}x^2\right)dx = \frac{x^4}{32} + \frac{x^3}{4} \{+c\}$ | M1A1 | M1: $x^n \to x^{n+1}$ on either term. A1: $\frac{x^4}{32} + \frac{x^3}{4}$, any correct simplified or unsimplified form (+ c not required) |
| $\left[\frac{x^4}{32}+\frac{x^3}{4}\right]_{-4}^{2} = \left(\frac{16}{32}+\frac{8}{4}\right)-\left(\frac{256}{32}+\frac{(-64)}{4}\right)$ | dM1 | Substitutes limits of 2 and $-4$ into integrated function and subtracts either way round |
| $= \frac{21}{2}$ | A1 | $\frac{21}{2}$ or 10.5 |
| At $x=-4$, $y = -8+12=4$ or at $x=2$, $y=1+3=4$ | | |
| Area of Rectangle $= 6 \times 4 = 24$, or Area of Rectangles $= 4\times4=16$ and $2\times4=8$ | M1 | Evidence of $(4--2)\times$ their $y_{-4}$ **or** $(4--2)\times$ their $y_2$, or Evidence of $4\times$ their $y_{-4}$ **and** $2\times$ their $y_2$ |
| So, $\text{area}(R) = 24 - \frac{21}{2} = \frac{27}{2}$ | dddM1A1 | Area rectangle $-$ integrated answer. Dependent on all previous method marks and requires: **Rectangle > integration > 0**. A1: $\frac{27}{2}$ or 13.5 |
### Alternative method:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\pm\int \left(\text{"their 4"} - \left(\frac{1}{8}x^3+\frac{3}{4}x^2\right)\right)dx$ | 4th M1 | Line $-$ curve. Condone missing brackets and allow either way round |
| $= 4x - \frac{x^4}{32} - \frac{x^3}{4}\{+c\}$ | 1st M1, 1st A1ft | M1: $x^n\to x^{n+1}$ on either curve term. A1ft: $-\frac{x^4}{32}-\frac{x^3}{4}$, any correct simplified or unsimplified form of curve terms, follow through sign errors |
| $\left[\ \right]_{-4}^{2} = \left(8-\frac{16}{32}-\frac{8}{4}\right)-\left(-16-\frac{256}{32}-\frac{(-64)}{4}\right)$ | 2nd M1, 3rd M1, 2nd A1 | 2nd M1: substitutes limits of 2 and $-4$. 3rd M1 for $\pm(\text{"8"}-\text{"}-16\text{"})$: substitutes limits into line part and subtracts. 2nd A1 for correct $\pm$ underlined expression |
| $= \frac{27}{2}$ | 3rd A1 | $\frac{27}{2}$ or 13.5 |
> If the final answer is $-13.5$ you can withhold the final A1. If $-13.5$ then "becomes" $+13.5$ allow the A1.
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f07cc9ed-a820-46c8-a3a3-3c780cf20fa7-09_796_1132_121_397}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve $C$ with equation
$$y = \frac { 1 } { 8 } x ^ { 3 } + \frac { 3 } { 4 } x ^ { 2 } , \quad x \in \mathbb { R }$$
The curve $C$ has a maximum turning point at the point $A$ and a minimum turning point at the origin $O$.
The line $l$ touches the curve $C$ at the point $A$ and cuts the curve $C$ at the point $B$.
The $x$ coordinate of $A$ is - 4 and the $x$ coordinate of $B$ is 2 .
The finite region $R$, shown shaded in Figure 3, is bounded by the curve $C$ and the line $l$.\\
Use integration to find the area of the finite region $R$.\\
\hfill \mbox{\textit{Edexcel C2 2014 Q6 [7]}}