| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Moderate -0.8 This is a standard C2 integration question requiring students to find intersection points (routine algebra) and calculate area between a curve and line using definite integration. The setup is straightforward with simple polynomial functions, and the method is a textbook application of ∫(upper - lower)dx. Easier than average A-level questions due to its routine nature and clear structure. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks |
|---|---|
| (a) Either solving \(0 = x(6-x)\) and showing \(x = 6\) (and \(x = 0\)) | B1 (1) |
| Answer | Marks |
|---|---|
| (b) Solving \(2x = 6x - x^2\) (i.e. \(x^2 = 4x\)) to \(x = ...\) | M1 |
| \(x = 4\) (and \(x = 0\)) | A1 |
| Conclusion: when \(x = 4\), \(y = 8\) and when \(x = 0\), \(y = 0\) | A1 (3) |
| (c) \[({\text{Area}} = \int_0^{(4)} (6x - x^2) \, dx\] Limits not required | M1 |
| Correct integration: \(3x^2 - \frac{x^3}{3}\) (+ c) | A1 |
| Correct use of correct limits on their result above (see notes on limits) | M1 |
| \[\left[ 3x^2 - \frac{x^3}{3} \right]_4^1 - \left[ 3x^2 - \frac{x^3}{3} \right]_0 \text{ with limits substituted } [= 48 - 21\frac{1}{3} = 26\frac{2}{3}]\] | |
| Area of triangle = \(2 \times 8 = 16\) (Can be awarded even if no M scored, i.e. B1) | A1 |
| Shaded area = \(±\)(area under curve \(-\) area of triangle) applied correctly | M1 |
| \[(= 26\frac{2}{3} - 16) = 10\frac{2}{3}\] (awrt 10.7) | A1 (6) [10] |
| Answer | Marks |
|---|---|
| - Verifying \((4,8)\) satisfies both the line and the curve | M1 (attempt at both), |
| - Both shown successfully | A1 |
| Answer | Marks |
|---|---|
| \[= (\pm)[2x^2 - \frac{x^3}{3} (+ c)]\] | A1, |
| Answer | Marks |
|---|---|
| Totally correct, unsimplified \(±\) expression (may be implied by correct ans.) | A1 |
**(a)** Either solving $0 = x(6-x)$ and showing $x = 6$ (and $x = 0$) | B1 (1) |
or showing $(6,0)$ (and $x = 0$) satisfies $y = 6x - x^2$ [allow for showing $x = 6$]
**(b)** Solving $2x = 6x - x^2$ (i.e. $x^2 = 4x$) to $x = ...$ | M1 |
$x = 4$ (and $x = 0$) | A1 |
Conclusion: when $x = 4$, $y = 8$ and when $x = 0$, $y = 0$ | A1 (3) |
**(c)** $$({\text{Area}} = \int_0^{(4)} (6x - x^2) \, dx$$ Limits not required | M1 |
Correct integration: $3x^2 - \frac{x^3}{3}$ (+ c) | A1 |
Correct use of correct limits on their result above (see notes on limits) | M1 |
$$\left[ 3x^2 - \frac{x^3}{3} \right]_4^1 - \left[ 3x^2 - \frac{x^3}{3} \right]_0 \text{ with limits substituted } [= 48 - 21\frac{1}{3} = 26\frac{2}{3}]$$ | |
Area of triangle = $2 \times 8 = 16$ (Can be awarded even if no M scored, i.e. B1) | A1 |
Shaded area = $±$(area under curve $-$ area of triangle) applied correctly | M1 |
$$(= 26\frac{2}{3} - 16) = 10\frac{2}{3}$$ (awrt 10.7) | A1 (6) [10] |
**Notes:**
**(b)** In scheme first A1: need only give $x = 4$
If verifying approach used:
- Verifying $(4,8)$ satisfies both the line and the curve | M1 (attempt at both),
- Both shown successfully | A1
- For final A1, $(0,0)$ needs to be mentioned; accept "clear from diagram"
**(c) Alternative** Using Area = $± \int_0^{(4)} [(6x - x^2) - 2x] \, dx$ approach
(i) If candidate integrates separately can be marked as main scheme
If combine to work with = $± \int_0^{(4)} (4x - x^2) \, dx$, first M mark and third M mark
$$= (\pm)[2x^2 - \frac{x^3}{3} (+ c)]$$ | A1,
Correct use of correct limits on their result second M1,
Totally correct, unsimplified $±$ expression (may be implied by correct ans.) | A1
$10^{2/3}$ A1 [Allow this if, having given -10⅔, they correct it]
M1 for correct use of correct limits: Must substitute correct limits for their strategy into a changed expression and subtract, either way round, e.g. $± [[ ]_4^0 - [ ]_0]$
If a long method is used, e.g. finding three areas, this mark only gained for correct strategy and all limits need to be correct for this strategy.
Use of trapezium rule: M0A0MA0, possible A1 for triangle
M1 (if correct application of trap. rule from $x = 0$ to $x = 4$) A0
---\includegraphics{figure_2}
In Figure 2 the curve $C$ has equation $y = 6x - x^2$ and the line $L$ has equation $y = 2x$.
\begin{enumerate}[label=(\alph*)]
\item Show that the curve $C$ intersects the $x$-axis at $x = 0$ and $x = 6$. [1]
\item Show that the line $L$ intersects the curve $C$ at the points $(0, 0)$ and $(4, 8)$. [3]
\end{enumerate}
The region $R$, bounded by the curve $C$ and the line $L$, is shown shaded in Figure 2.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Use calculus to find the area of $R$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2008 Q7 [10]}}