| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Standard +0.3 This is a straightforward multi-part integration question with standard techniques. Part (a) is simple substitution, part (b) is routine trapezium rule application, and part (c) uses a given substitution to integrate—all standard C4 procedures requiring no novel insight. The parametric curve setup suggests follow-up parts, but the shown work is below average difficulty. |
| Spec | 1.08h Integration by substitution1.09f Trapezium rule: numerical integration |
| \(x\) | 1 | 2 | 3 | 4 |
| \(y\) | 0.5 | 0.8284 | 1.3333 |
| Answer | Marks | Guidance |
|---|---|---|
| \(1.0981\) | B1 cao | Correct answer only |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(\approx \frac{1}{2} \times 1 \times [0.5 + 2(0.8284 + \text{their } 1.0981) + 1.3333]\) | B1; M1 | Outside brackets \(\frac{1}{2}\times1\) or \(\frac{1}{2}\); for structure of trapezium rule |
| \(= \frac{1}{2} \times 5.6863 = 2.843\) (3 dp) | A1 | Anything that rounds to 2.843 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{u = 1+\sqrt{x}\} \Rightarrow \frac{du}{dx} = \frac{1}{2}x^{-\frac{1}{2}}\) or \(\frac{dx}{du} = 2(u-1)\) | B1 | |
| \(\left\{\int \frac{x}{1+\sqrt{x}}dx =\right\} \int \frac{(u-1)^2}{u} \cdot 2(u-1)\, du\) | M1 | \(\frac{x}{1+\sqrt{x}}\) becoming \(\frac{(u-1)^2}{u}\) (ignore integral sign) |
| \(\frac{(u-1)^2}{u} \cdot 2(u-1)\) | A1 | \(\frac{x}{1+\sqrt{x}}dx\) becoming \(\frac{(u-1)^2}{u}\cdot 2(u-1)\{du\}\) |
| \(= 2\int \frac{(u-1)^3}{u}\,du = \{2\}\int \frac{(u^3-3u^2+3u-1)}{u}\,du\) | M1 | Expands to give a "four term" cubic in \(u\), e.g. \(\pm Au^3 \pm Bu^2 \pm Cu \pm D\) |
| \(= \{2\}\int\left(u^2-3u+3-\frac{1}{u}\right)du\) | M1 | An attempt to divide at least three terms in *their cubic* by \(u\) |
| \(= \{2\}\left(\frac{u^3}{3} - \frac{3u^2}{2} + 3u - \ln u\right)\) | A1 | \(\int\frac{(u-1)^3}{u}\to\left(\frac{u^3}{3}-\frac{3u^2}{2}+3u-\ln u\right)\) |
| \(\text{Area}(R) = \left[\frac{2u^3}{3}-3u^2+6u-2\ln u\right]_2^3\) | ||
| \(= \left(\frac{2(3)^3}{3}-3(3)^2+6(3)-2\ln 3\right) - \left(\frac{2(2)^3}{3}-3(2)^2+6(2)-2\ln 2\right)\) | M1 | Applies limits of 3 and 2 in \(u\) or 4 and 1 in \(x\); subtracts either way round |
| \(= \frac{11}{3}+2\ln 2-2\ln 3\) or \(\frac{11}{3}+2\ln\left(\frac{2}{3}\right)\) or \(\frac{11}{3}-\ln\left(\frac{9}{4}\right)\) etc | A1 | Correct exact answer or equivalent |
# Question 4:
## Part (a):
| $1.0981$ | B1 cao | Correct answer only |
## Part (b):
| Area $\approx \frac{1}{2} \times 1 \times [0.5 + 2(0.8284 + \text{their } 1.0981) + 1.3333]$ | B1; M1 | Outside brackets $\frac{1}{2}\times1$ or $\frac{1}{2}$; for structure of trapezium rule |
| $= \frac{1}{2} \times 5.6863 = 2.843$ (3 dp) | A1 | Anything that rounds to 2.843 |
## Part (c):
| $\{u = 1+\sqrt{x}\} \Rightarrow \frac{du}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$ or $\frac{dx}{du} = 2(u-1)$ | B1 | |
| $\left\{\int \frac{x}{1+\sqrt{x}}dx =\right\} \int \frac{(u-1)^2}{u} \cdot 2(u-1)\, du$ | M1 | $\frac{x}{1+\sqrt{x}}$ becoming $\frac{(u-1)^2}{u}$ (ignore integral sign) |
| $\frac{(u-1)^2}{u} \cdot 2(u-1)$ | A1 | $\frac{x}{1+\sqrt{x}}dx$ becoming $\frac{(u-1)^2}{u}\cdot 2(u-1)\{du\}$ |
| $= 2\int \frac{(u-1)^3}{u}\,du = \{2\}\int \frac{(u^3-3u^2+3u-1)}{u}\,du$ | M1 | Expands to give a "four term" cubic in $u$, e.g. $\pm Au^3 \pm Bu^2 \pm Cu \pm D$ |
| $= \{2\}\int\left(u^2-3u+3-\frac{1}{u}\right)du$ | M1 | An attempt to divide at least three terms in *their cubic* by $u$ |
| $= \{2\}\left(\frac{u^3}{3} - \frac{3u^2}{2} + 3u - \ln u\right)$ | A1 | $\int\frac{(u-1)^3}{u}\to\left(\frac{u^3}{3}-\frac{3u^2}{2}+3u-\ln u\right)$ |
| $\text{Area}(R) = \left[\frac{2u^3}{3}-3u^2+6u-2\ln u\right]_2^3$ | | |
| $= \left(\frac{2(3)^3}{3}-3(3)^2+6(3)-2\ln 3\right) - \left(\frac{2(2)^3}{3}-3(2)^2+6(2)-2\ln 2\right)$ | M1 | Applies limits of 3 and 2 in $u$ or 4 and 1 in $x$; subtracts either way round |
| $= \frac{11}{3}+2\ln 2-2\ln 3$ or $\frac{11}{3}+2\ln\left(\frac{2}{3}\right)$ or $\frac{11}{3}-\ln\left(\frac{9}{4}\right)$ etc | A1 | Correct exact answer or equivalent |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a98d4a7f-1e6d-4294-9b5c-c945e8fbe83e-05_650_1143_223_427}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation $y = \frac { x } { 1 + \sqrt { } x }$. The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $x$-axis, the line with equation $x = 1$ and the line with equation $x = 4$.
\begin{enumerate}[label=(\alph*)]
\item Complete the table with the value of $y$ corresponding to $x = 3$, giving your answer to 4 decimal places.\\
(1)
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 \\
\hline
$y$ & 0.5 & 0.8284 & & 1.3333 \\
\hline
\end{tabular}
\end{center}
\item Use the trapezium rule, with all the values of $y$ in the completed table, to obtain an estimate of the area of the region $R$, giving your answer to 3 decimal places.
\item Use the substitution $u = 1 + \sqrt { } x$, to find, by integrating, the exact area of $R$.\\
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a98d4a7f-1e6d-4294-9b5c-c945e8fbe83e-07_743_1568_219_182}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve $C$ with parametric equations
$$x = 1 - \frac { 1 } { 2 } t , \quad y = 2 ^ { t } - 1$$
The curve crosses the $y$-axis at the point $A$ and crosses the $x$-axis at the point $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2013 Q4 [12]}}