| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Curve-Line-Axis Bounded Region |
| Difficulty | Moderate -0.3 This is a standard C2 integration question requiring factorization to find x-intercepts, verification of a point, basic integration, and calculating area between a curve and line. All techniques are routine for this level, though part (d) requires recognizing to subtract the triangle area from the integral, which adds mild problem-solving beyond pure recall. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Puts \(y=0\), solves \((x-4)(x-1)=0\); points are \((1,0)\) and \((4,0)\) | M1, A1 (2) | M1 for attempt to find \(L\) and \(M\); accept \(x=1\) and \(x=4\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x=5\) gives \(y = 25-25+4=4\), so \((5,4)\) lies on curve | B1cso (1) | Working must be shown; usually \(0=0\) or \(4=4\) is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int(x^2-5x+4)\,dx = \frac{1}{3}x^3 - \frac{5}{2}x^2 + 4x \quad (+c)\) | M1A1 (2) | M1 for \(x^2 \to kx^3\), \(x \to kx^2\) or \(4 \to 4x\); A1 for correct integration of all three terms |
| Answer | Marks | Guidance |
|---|---|---|
| Area of triangle \(= \frac{1}{2}\times 4\times 4 = 8\) or \(\int(x-1)\,dx = \frac{1}{2}x^2-x\) with limits 1 to 5 giving 8 | B1 | B1 for this triangle only |
| Area under curve \(= \int_4^5\): \(\frac{1}{3}\times 5^3 - \frac{5}{2}\times 5^2+4\times 5\left[=-\frac{5}{6}\right]\) | M1 | 1st M1 for substituting 5 into their integrated function |
| \(\frac{1}{3}\times 4^3 - \frac{5}{2}\times 4^2+4\times 4\left[=-\frac{8}{3}\right]\) | M1 | 2nd M1 for substituting 4 into their integrated function |
| \(\int_4^5 = -\frac{5}{6} - \left(-\frac{8}{3}\right) = \frac{11}{6}\) (allow 1.83 or 1.8) | A1cao | |
| Area of \(R = 8 - \frac{11}{6} = 6\frac{1}{6}\) or \(\frac{37}{6}\) or \(6.1\overline{6}\) (not 6.17) | A1cao (5) |
## Question 7:
### Part (a):
| Puts $y=0$, solves $(x-4)(x-1)=0$; points are $(1,0)$ and $(4,0)$ | M1, A1 (2) | M1 for attempt to find $L$ and $M$; accept $x=1$ and $x=4$ |
### Part (b):
| $x=5$ gives $y = 25-25+4=4$, so $(5,4)$ lies on curve | B1cso (1) | Working must be shown; usually $0=0$ or $4=4$ is B0 |
### Part (c):
| $\int(x^2-5x+4)\,dx = \frac{1}{3}x^3 - \frac{5}{2}x^2 + 4x \quad (+c)$ | M1A1 (2) | M1 for $x^2 \to kx^3$, $x \to kx^2$ or $4 \to 4x$; A1 for correct integration of all three terms |
### Part (d):
| Area of triangle $= \frac{1}{2}\times 4\times 4 = 8$ or $\int(x-1)\,dx = \frac{1}{2}x^2-x$ with limits 1 to 5 giving 8 | B1 | B1 for this triangle only |
| Area under curve $= \int_4^5$: $\frac{1}{3}\times 5^3 - \frac{5}{2}\times 5^2+4\times 5\left[=-\frac{5}{6}\right]$ | M1 | 1st M1 for substituting 5 into their integrated function |
| $\frac{1}{3}\times 4^3 - \frac{5}{2}\times 4^2+4\times 4\left[=-\frac{8}{3}\right]$ | M1 | 2nd M1 for substituting 4 into their integrated function |
| $\int_4^5 = -\frac{5}{6} - \left(-\frac{8}{3}\right) = \frac{11}{6}$ (allow 1.83 or 1.8) | A1cao | |
| Area of $R = 8 - \frac{11}{6} = 6\frac{1}{6}$ or $\frac{37}{6}$ or $6.1\overline{6}$ (not 6.17) | A1cao (5) | |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e3faf018-37a8-48ef-b100-81402a8ec87f-09_696_821_205_516}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The curve $C$ has equation $y = x ^ { 2 } - 5 x + 4$. It cuts the $x$-axis at the points $L$ and $M$ as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the point $L$ and the point $M$.
\item Show that the point $N ( 5,4 )$ lies on $C$.
\item Find $\int \left( x ^ { 2 } - 5 x + 4 \right) \mathrm { d } x$.
The finite region $R$ is bounded by $L N , L M$ and the curve $C$ as shown in Figure 2.
\item Use your answer to part (c) to find the exact value of the area of $R$.\\
\section*{LU}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2010 Q7 [10]}}