| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Combined region areas |
| Difficulty | Standard +0.3 This is a standard C2 integration question requiring students to recognize that the curve crosses the x-axis at x=1, necessitating splitting the integral into two parts and taking absolute values. While it requires understanding of signed areas and involves polynomial expansion before integration, it follows a well-practiced procedure with no novel insight needed—slightly easier than average for A-level. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = x(x^2 - 6x + 5) = x^3 - 6x^2 + 5x\) | M1, A1 | Attempt to multiply out, must be a cubic. Award A1 for their final version of expansion |
| \(\int(x^3 - 6x^2 + 5x)\,dx = \frac{x^4}{4} - \frac{6x^3}{3} + \frac{5x^2}{2}\) | M1, A1ft | Attempt to integrate \(x^n \rightarrow x^{n+1}\). A1ft on their expansion provided in form \(ax^p + bx^q + \ldots\), at least two terms, all integrated correctly |
| \(\left[\frac{x^4}{4} - 2x^3 + \frac{5x^2}{2}\right]_0^1 = \left(\frac{1}{4} - 2 + \frac{5}{2}\right) - 0 = \frac{3}{4}\) | M1 | Substitutes and subtracts for one integral. Integral must be 'changed'. For \([\,]_0^1\): \(-0\) for bottom limit can be implied |
| \(\left[\frac{x^4}{4} - 2x^3 + \frac{5x^2}{2}\right]_1^2 = (4 - 16 + 10) - \frac{3}{4} = -\frac{11}{4}\) | M1, A1(both) | Substitutes and subtracts for two integrals. Must have limits 1 and 0, and 2 and 1. \(\frac{3}{4}\) and \(-\frac{11}{4}\) o.e. \(-\frac{11}{4}\) may not be seen explicitly; can be implied by subsequent working |
| \(\therefore\) total area \(= \frac{3}{4} + \frac{11}{4}\) | M1 | \(\lvert\) their value for \([\,]_0^1\rvert + \lvert\) their value for \([\,]_1^2\rvert\). Dependent on at least one value coming from integration |
| \(= \frac{7}{2}\) o.e. | A1cso (9) | N.B. c.s.o. |
## Question 7:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = x(x^2 - 6x + 5) = x^3 - 6x^2 + 5x$ | M1, A1 | Attempt to multiply out, must be a cubic. Award A1 for their final version of expansion |
| $\int(x^3 - 6x^2 + 5x)\,dx = \frac{x^4}{4} - \frac{6x^3}{3} + \frac{5x^2}{2}$ | M1, A1ft | Attempt to integrate $x^n \rightarrow x^{n+1}$. A1ft on their expansion provided in form $ax^p + bx^q + \ldots$, at least two terms, all integrated correctly |
| $\left[\frac{x^4}{4} - 2x^3 + \frac{5x^2}{2}\right]_0^1 = \left(\frac{1}{4} - 2 + \frac{5}{2}\right) - 0 = \frac{3}{4}$ | M1 | Substitutes and subtracts for one integral. Integral must be 'changed'. For $[\,]_0^1$: $-0$ for bottom limit can be implied |
| $\left[\frac{x^4}{4} - 2x^3 + \frac{5x^2}{2}\right]_1^2 = (4 - 16 + 10) - \frac{3}{4} = -\frac{11}{4}$ | M1, A1(both) | Substitutes and subtracts for two integrals. Must have limits 1 and 0, and 2 and 1. $\frac{3}{4}$ and $-\frac{11}{4}$ o.e. $-\frac{11}{4}$ may not be seen explicitly; can be implied by subsequent working |
| $\therefore$ total area $= \frac{3}{4} + \frac{11}{4}$ | M1 | $\lvert$ their value for $[\,]_0^1\rvert + \lvert$ their value for $[\,]_1^2\rvert$. Dependent on at least one value coming from integration |
| $= \frac{7}{2}$ o.e. | A1cso **(9)** | N.B. c.s.o. |7.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{872356ab-68d3-43ee-8b76-650a2697d80e-08_1052_1116_351_413}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve $C$ with equation
$$y = x ( x - 1 ) ( x - 5 )$$
Use calculus to find the total area of the finite region, shown shaded in Figure 1, that is between $x = 0$ and $x = 2$ and is bounded by $C$, the $x$-axis and the line $x = 2$.\\
(9)\\
\hfill \mbox{\textit{Edexcel C2 2007 Q7 [9]}}