| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Combined region areas |
| Difficulty | Moderate -0.3 This is a standard C2 integration question with routine factorisation, finding roots, differentiation for gradient/turning points, and calculating areas under curves. All parts follow textbook procedures with no novel problem-solving required, though the multi-part structure and area calculation between curve and line require careful setup, making it slightly easier than average overall. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(x(x^2 - 6x + 5)\) | M1 | |
| \(= x(x-1)(x-5)\) | M1 A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| 1 and 5 | B1 ft | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 3x^2 - 12x + 5\) | M1 A1 | |
| At \(x = 1\): \(\frac{dy}{dx} = 3 - 12 + 5 = -4\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int(x^3 - 6x^2 + 5x)dx = \frac{x^4}{4} - \frac{6x^3}{3} + \frac{5x^2}{2}\) | M1 A1 | |
| \([\ldots]_0^1 = \frac{1}{4} - 2 + \frac{5}{2} = \frac{3}{4}\) (Region \(R\)) | M1 A1 ft | |
| Evaluating at 5: \(\frac{625}{4} - 250 + \frac{125}{2} = -31\frac{1}{4}\) | A1 | |
| To find \(S\): \(-31\frac{1}{4} - \frac{3}{4} = -32\) | M1 | |
| Total Area \(= 32 + \frac{3}{4} = 32\frac{3}{4}\) | A1 | (7 marks) |
# Question 7:
## Part (a):
$x(x^2 - 6x + 5)$ | M1 |
$= x(x-1)(x-5)$ | M1 A1 | (3 marks)
## Part (b):
1 and 5 | B1 ft | (1 mark)
## Part (c):
$\frac{dy}{dx} = 3x^2 - 12x + 5$ | M1 A1 |
At $x = 1$: $\frac{dy}{dx} = 3 - 12 + 5 = -4$ | A1 | (3 marks)
## Part (d):
$\int(x^3 - 6x^2 + 5x)dx = \frac{x^4}{4} - \frac{6x^3}{3} + \frac{5x^2}{2}$ | M1 A1 |
$[\ldots]_0^1 = \frac{1}{4} - 2 + \frac{5}{2} = \frac{3}{4}$ (Region $R$) | M1 A1 ft |
Evaluating at 5: $\frac{625}{4} - 250 + \frac{125}{2} = -31\frac{1}{4}$ | A1 |
To find $S$: $-31\frac{1}{4} - \frac{3}{4} = -32$ | M1 |
Total Area $= 32 + \frac{3}{4} = 32\frac{3}{4}$ | A1 | (7 marks)
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7.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{e4d38009-aa70-4c55-9765-c54044ffaa31-3_771_972_1322_557}
\end{center}
\end{figure}
Fig. 2 shows part of the curve $C$ with equation $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = x ^ { 3 } - 6 x ^ { 2 } + 5 x$.\\
The curve crosses the $x$-axis at the origin $O$ and at the points $A$ and $B$.
\begin{enumerate}[label=(\alph*)]
\item Factorise $\mathrm { f } ( x )$ completely.
\item Write down the $x$-coordinates of the points $A$ and $B$.
\item Find the gradient of $C$ at $A$.
The region $R$ is bounded by $C$ and the line $O A$, and the region $S$ is bounded by $C$ and the line $A B$.
\item Use integration to find the area of the combined regions $R$ and $S$, shown shaded in Fig.2.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{e4d38009-aa70-4c55-9765-c54044ffaa31-4_736_727_338_402}
\end{center}
\end{figure}
Fig. 3 shows a sketch of part of the curve $C$ with equation $y = x ^ { 3 } - 7 x ^ { 2 } + 15 x + 3 , x \geq 0$. The point $P$, on $C$, has $x$-coordinate 1 and the point $Q$ is the minimum turning point of $C$.\\
(a) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(b) Find the coordinates of $Q$.\\
(c) Show that $P Q$ is parallel to the $x$-axis.\\
(d) Calculate the area, shown shaded in Fig. 3, bounded by $C$ and the line $P Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q7 [17]}}