| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | January |
| Marks | 14 |
| 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 requiring table completion, trapezium rule application, and substitution-based integration. While part (c) involves a non-trivial substitution, the substitution is provided and the steps are mechanical. This is slightly easier than average as it's a standard C4 numerical methods question with clear guidance. |
| Spec | 1.08h Integration by substitution1.09f Trapezium rule: numerical integration |
| \(x\) | 2 | 3 | 4 | 5 |
| \(y\) | 0.2 | 0.1745 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(x=3 \Rightarrow y=0.1847\) | B1 | awrt |
| \(x=5 \Rightarrow y=0.1667\) | B1 | awrt or \(\frac{1}{6}\), (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(I\approx\frac{1}{2}\left[0.2+0.1667+2(0.1847+0.1745)\right]\) | B1 M1 A1ft | |
| \(\approx 0.543\) | A1 | 0.542 or 0.543, (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{dx}{du}=2(u-4)\) | B1 | |
| \(\int\frac{1}{4+\sqrt{(x-1)}}\,dx=\int\frac{1}{u}\times2(u-4)\,du\) | M1 | |
| \(=\int\left(2-\frac{8}{u}\right)du\) | A1 | |
| \(=2u-8\ln u\) | M1 A1 | |
| \(x=2\Rightarrow u=5,\quad x=5\Rightarrow u=6\) | B1 | |
| \(\left[2u-8\ln u\right]_5^6=(12-8\ln6)-(10-8\ln5)\) | M1 | |
| \(=2+8\ln\left(\frac{5}{6}\right)\) | A1 | (8) [14] |
## Question 7:
### Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $x=3 \Rightarrow y=0.1847$ | B1 | awrt |
| $x=5 \Rightarrow y=0.1667$ | B1 | awrt or $\frac{1}{6}$, (2) |
### Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $I\approx\frac{1}{2}\left[0.2+0.1667+2(0.1847+0.1745)\right]$ | B1 M1 A1ft | |
| $\approx 0.543$ | A1 | 0.542 or 0.543, (4) |
### Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{dx}{du}=2(u-4)$ | B1 | |
| $\int\frac{1}{4+\sqrt{(x-1)}}\,dx=\int\frac{1}{u}\times2(u-4)\,du$ | M1 | |
| $=\int\left(2-\frac{8}{u}\right)du$ | A1 | |
| $=2u-8\ln u$ | M1 A1 | |
| $x=2\Rightarrow u=5,\quad x=5\Rightarrow u=6$ | B1 | |
| $\left[2u-8\ln u\right]_5^6=(12-8\ln6)-(10-8\ln5)$ | M1 | |
| $=2+8\ln\left(\frac{5}{6}\right)$ | A1 | (8) **[14]** |7.
$$I = \int _ { 2 } ^ { 5 } \frac { 1 } { 4 + \sqrt { } ( x - 1 ) } \mathrm { d } x$$
\begin{enumerate}[label=(\alph*)]
\item Given that $y = \frac { 1 } { 4 + \sqrt { } ( x - 1 ) }$, complete the table below with values of $y$ corresponding to $x = 3$ and $x = 5$. Give your values to 4 decimal places.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 2 & 3 & 4 & 5 \\
\hline
$y$ & 0.2 & & 0.1745 & \\
\hline
\end{tabular}
\end{center}
\item Use the trapezium rule, with all of the values of $y$ in the completed table, to obtain an estimate of $I$, giving your answer to 3 decimal places.
\item Using the substitution $x = ( u - 4 ) ^ { 2 } + 1$, or otherwise, and integrating, find the exact value of $I$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2011 Q7 [14]}}