| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Multiple Region or Composite Area |
| Difficulty | Standard +0.3 This is a straightforward area-between-curves question with clearly given equations and intersection points. Part (a) requires a single definite integral, part (b) adds a second region, and part (c) is simple division. All steps are routine C2 integration techniques with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
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| END |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempts integration \(\int(4x^2+6)\,dx = \frac{4x^3}{3}+6x\) | M1 A1 | Correct integration method – increase power by one |
| Uses limits 4 and \(-2\), finds area under curve \(\left[\frac{4x^3}{3}+6x\right]_{-2}^{4} = \left(109\frac{1}{3}\right)-\left(-22\frac{2}{3}\right)=132\) | dM1 A1 | Uses limits 4 and \(-2\) within integrated function; A1 for achieving 132 |
| Full method: Area of trapezium \(\frac{1}{2}\times(4+2)(22+70)-\text{"132"}\) or \(\left[4x^2+38x\right]_{-2}^{4}-\text{"132"}\) | M1 | Full attempt at area of \(R_1\); look for trapezium area or area under line minus 132 |
| So area \(= 276-132=144\) | A1 | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempts area of \(R_2\): \(\left[\frac{4x^3}{3}+6x\right]_{4}^{5}-\frac{(5-4)(70+78)}{2}\) | M1 | Uses limits 5 and 4; subtracts trapezium area. The 78 must be from correct method |
| Area of \(R_2 = 196\frac{2}{3}-(109\frac{1}{3})-74=13\frac{1}{3}\) | A1 | Allow \(13.\dot{3}\) and \(13.\overline{3}\); may be implied by final answer |
| Total Area shaded \(= 144+13\frac{1}{3}=157\frac{1}{3}\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((k)=10.8\) oe | B1 | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Subtracts and integrates: \(\pm\int\{(8x+38)-(4x^2+6)\}\,dx = \pm\left\{8\frac{x^2}{2}+38x-\frac{4x^3}{3}-6x\right\}\) | M1 A1 | Attempts to combine and integrate; correct method – increase power by one; condone bracketing error |
| Uses correct limits: \(\pm\left[4x^2+32x-\frac{4x^3}{3}\right]_{-2}^{4}=\left(106\frac{2}{3}\right)-\left(-37\frac{1}{3}\right)\) | dM1 A1 | Uses limits 4 and \(-2\); correct values seen |
| Full method (awarded on line 1): Area \(R_1=144\) | M1 A1 | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempts area of \(R_2\) using correct limits: \(\pm\left[-4x^2-32x+\frac{4x^3}{3}\right]_{4}^{5}=\left(106\frac{2}{3}\right)-\left(93\frac{1}{3}\right)\) | M1 | Uses limits 5 and 4 in their subtracted/integrated functions |
| Area of \(R_2=13\frac{1}{3}\) | A1 | May be implied by final answer |
| Total Area shaded \(=144+13\frac{1}{3}=157\frac{1}{3}\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((k)=10.8\) oe | B1 | [1] |
# Question 15:
## Part (a):
**Way 1: Integrates separately**
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts integration $\int(4x^2+6)\,dx = \frac{4x^3}{3}+6x$ | M1 A1 | Correct integration method – increase power by one |
| Uses limits 4 and $-2$, finds area under curve $\left[\frac{4x^3}{3}+6x\right]_{-2}^{4} = \left(109\frac{1}{3}\right)-\left(-22\frac{2}{3}\right)=132$ | dM1 A1 | Uses limits 4 and $-2$ within integrated function; A1 for achieving 132 |
| Full method: Area of trapezium $\frac{1}{2}\times(4+2)(22+70)-\text{"132"}$ or $\left[4x^2+38x\right]_{-2}^{4}-\text{"132"}$ | M1 | Full attempt at area of $R_1$; look for trapezium area or area under line minus 132 |
| So area $= 276-132=144$ | A1 | **[6]** |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts area of $R_2$: $\left[\frac{4x^3}{3}+6x\right]_{4}^{5}-\frac{(5-4)(70+78)}{2}$ | M1 | Uses limits 5 and 4; subtracts trapezium area. The 78 must be from correct method |
| Area of $R_2 = 196\frac{2}{3}-(109\frac{1}{3})-74=13\frac{1}{3}$ | A1 | Allow $13.\dot{3}$ and $13.\overline{3}$; may be implied by final answer |
| Total Area shaded $= 144+13\frac{1}{3}=157\frac{1}{3}$ | A1 | **[3]** |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $(k)=10.8$ oe | B1 | **[1]** |
---
**Way 2: Integrates line $-$ curve**
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Subtracts and integrates: $\pm\int\{(8x+38)-(4x^2+6)\}\,dx = \pm\left\{8\frac{x^2}{2}+38x-\frac{4x^3}{3}-6x\right\}$ | M1 A1 | Attempts to combine and integrate; correct method – increase power by one; condone bracketing error |
| Uses correct limits: $\pm\left[4x^2+32x-\frac{4x^3}{3}\right]_{-2}^{4}=\left(106\frac{2}{3}\right)-\left(-37\frac{1}{3}\right)$ | dM1 A1 | Uses limits 4 and $-2$; correct values seen |
| Full method (awarded on line 1): Area $R_1=144$ | M1 A1 | **[6]** |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts area of $R_2$ using correct limits: $\pm\left[-4x^2-32x+\frac{4x^3}{3}\right]_{4}^{5}=\left(106\frac{2}{3}\right)-\left(93\frac{1}{3}\right)$ | M1 | Uses limits 5 and 4 in their subtracted/integrated functions |
| Area of $R_2=13\frac{1}{3}$ | A1 | May be implied by final answer |
| Total Area shaded $=144+13\frac{1}{3}=157\frac{1}{3}$ | A1 | **[3]** |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $(k)=10.8$ oe | B1 | **[1]** |15.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{ce06b37a-aa57-4256-bec8-7277c8a9fc65-44_851_1506_212_260}
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\caption{Figure 3}
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A design for a logo consists of two finite regions $R _ { 1 }$ and $R _ { 2 }$, shown shaded in Figure 3 .\\
The region $R _ { 1 }$ is bounded by the straight line $l$ and the curve $C$.\\
The region $R _ { 2 }$ is bounded by the straight line $l$, the curve $C$ and the line with equation $x = 5$\\
The line $l$ has equation $y = 8 x + 38$\\
The curve $C$ has equation $y = 4 x ^ { 2 } + 6$\\
Given that the line $l$ meets the curve $C$ at the points $( - 2,22 )$ and $( 4,70 )$, use integration to find
\begin{enumerate}[label=(\alph*)]
\item the area of the larger lower region, labelled $R _ { 1 }$
\item the exact value of the total area of the two shaded regions.
Given that
$$\frac { \text { Area of } R _ { 1 } } { \text { Area of } R _ { 2 } } = k$$
\item find the value of $k$.
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\hfill \mbox{\textit{Edexcel C12 2018 Q15 [10]}}