| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area involving fractional powers |
| Difficulty | Moderate -0.8 This is a straightforward C2 integration question requiring basic power rule application and finding area under a curve. Part (a) is routine integration of polynomial terms, and part (b) requires finding the x-intercept (solving 3x - x^(3/2) = 0) and evaluating a definite integral. All steps are standard textbook exercises with no problem-solving insight needed, making it easier than average. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(\int\left(3x-x^2\right)dx\right) = \frac{3x^2}{2} - \frac{x^3}{5} \{+c\}\) | M1 | Either \(3x \to \pm 2x^2\) or \(x^2 \to \pm \mu x^3\), \(\lambda, \mu \neq 0\) |
| A1 | At least one term correctly integrated. Can be simplified or un-simplified but power must be simplified. Then isw. | |
| Both terms correctly integrated | ||
| A1 | ||
| [3] | ||
| (b) \(0 = 3x - x^2 \Rightarrow 0 = 3 - x^2\) or \(0 = x\left(3-x^2\right) \Rightarrow x = ...\) | M1 | Sets \(y=0\), in order to find the correct \(x^2 = 3\) or \(x = 9\) (isw if \(x^2 = 3\) is followed by \(x = \sqrt{3}\)) |
| dM1 | Just seeing \(x = \sqrt{3}\) without the correct \(x^2 = 3\) gains M0. May just see \(x = 9\). | |
| \(\left\{\text{Area}(S) = \left[\frac{3x^2}{2} - \frac{2}{5}x^3\right]_0^9\right\}\) | ||
| \(= \left(\frac{3(9)}{2} - \left(\frac{2}{5}\right)(9)^2\right) - \{0\}\) | Applies the limit 9 on an integrated function with no wrong lower limit | |
| \(= \frac{243}{2} - \frac{486}{5} - \{0\} = \frac{243}{10}\) or 24.3 | A1 oe | |
| [3] |
| Answer | Marks |
|---|---|
| M1 | Either \(3x \to \pm 2x^2\) or \(x^2 \to \pm \mu x^3\), \(\lambda, \mu \neq 0\) |
| 1st A1 | At least one term correctly integrated. Can be simplified or un-simplified but power must be simplified. Then isw. |
| 2nd A1 | Both terms correctly integrated. Can be un-simplified (as in the scheme) but the \(n+1\) in each denominator and power should be a single number. (e.g. \(2 - \text{ not } 1+1\)) Ignore subsequent work if there are errors simplifying. Ignore integral signs in their answer. |
| Common Error | Common Error: \(0 = 3x - x^2 \Rightarrow x^2 = 3\) so \(x = \sqrt{3}\) Then uses limit \(\sqrt{3}\) etc gains M1 M0 A0 so 1/3 |
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\int\left(3x-x^2\right)dx\right) = \frac{3x^2}{2} - \frac{x^3}{5} \{+c\}$ | M1 | Either $3x \to \pm 2x^2$ or $x^2 \to \pm \mu x^3$, $\lambda, \mu \neq 0$ |
| | A1 | At least one term correctly integrated. Can be simplified or un-simplified but power must be simplified. Then isw. |
| | | Both terms correctly integrated |
| | | A1 |
| | [3] | |
| **(b)** $0 = 3x - x^2 \Rightarrow 0 = 3 - x^2$ or $0 = x\left(3-x^2\right) \Rightarrow x = ...$ | M1 | Sets $y=0$, in order to find the correct $x^2 = 3$ or $x = 9$ (isw if $x^2 = 3$ is followed by $x = \sqrt{3}$) |
| | dM1 | Just seeing $x = \sqrt{3}$ without the correct $x^2 = 3$ gains M0. May just see $x = 9$. |
| $\left\{\text{Area}(S) = \left[\frac{3x^2}{2} - \frac{2}{5}x^3\right]_0^9\right\}$ | | |
| $= \left(\frac{3(9)}{2} - \left(\frac{2}{5}\right)(9)^2\right) - \{0\}$ | | Applies the limit 9 on an integrated function with no wrong lower limit |
| $= \frac{243}{2} - \frac{486}{5} - \{0\} = \frac{243}{10}$ or 24.3 | A1 oe | |
| | [3] | |
**Question 7 Notes:**
| M1 | Either $3x \to \pm 2x^2$ or $x^2 \to \pm \mu x^3$, $\lambda, \mu \neq 0$ |
|---|---|
| 1st A1 | At least one term correctly integrated. Can be simplified or un-simplified but power must be simplified. Then isw. |
| 2nd A1 | Both terms correctly integrated. Can be un-simplified (as in the scheme) but the $n+1$ in each denominator and power should be a single number. (e.g. $2 - \text{ not } 1+1$) Ignore subsequent work if there are errors simplifying. Ignore integral signs in their answer. |
| Common Error | Common Error: $0 = 3x - x^2 \Rightarrow x^2 = 3$ so $x = \sqrt{3}$ Then uses limit $\sqrt{3}$ etc gains M1 M0 A0 so 1/3 |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{582cda45-80fc-43a8-90e6-1cae08cb1534-12_563_812_244_630}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve with equation
$$y = 3 x - x ^ { \frac { 3 } { 2 } } , \quad x \geqslant 0$$
The finite region $S$, bounded by the $x$-axis and the curve, is shown shaded in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Find
$$\int \left( 3 x - x ^ { \frac { 3 } { 2 } } \right) \mathrm { d } x$$
\item Hence find the area of $S$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2016 Q7 [6]}}