| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Moderate -0.8 This is a straightforward C2 trapezium rule question requiring only calculator work to complete a table and mechanical application of the formula with given values. No conceptual difficulty or problem-solving is involved—purely procedural computation that any student who has learned the trapezium rule can execute. |
| Spec | 1.09f Trapezium rule: numerical integration |
| \(x\) | 1 | 1.1 | 1.2 | 1.3 | 1.4 | 1.5 |
| \(y\) | 0.7071 | 0.7591 | 0.8090 | 0.9037 | 0.9487 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\{x=1.3\}\), \(y=0.8572\) (only) | B1 cao | Must be 0.8572 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2} \times 0.1\) | B1 | For using \(\frac{1}{2} \times 0.1\) or 0.05 or equivalent |
| \(\{0.7071+0.9487+2(0.7591+0.8090+\text{"0.8572"}+0.9037)\}\) | M1 | First bracket contains first \(y\) value plus last \(y\) value; second bracket is sum of remaining \(y\) values multiplied by 2; no additional values; omitting one value from 2nd bracket may be a slip |
| \(\ldots\{0.7071+0.9487+2(0.7591+0.8090+\text{"0.8572"}+0.9037)\}\) | A1ft | Correct bracket following through candidate's \(y\) value from part (a) |
| \(\{0.05(8.3138)\}=0.41569\) = awrt \(0.416\) | A1 | — |
## Question 2:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\{x=1.3\}$, $y=0.8572$ (only) | B1 cao | Must be 0.8572 cao |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2} \times 0.1$ | B1 | For using $\frac{1}{2} \times 0.1$ or 0.05 or equivalent |
| $\{0.7071+0.9487+2(0.7591+0.8090+\text{"0.8572"}+0.9037)\}$ | M1 | First bracket contains first $y$ value plus last $y$ value; second bracket is sum of remaining $y$ values multiplied by 2; no additional values; omitting one value from 2nd bracket may be a slip |
| $\ldots\{0.7071+0.9487+2(0.7591+0.8090+\text{"0.8572"}+0.9037)\}$ | A1ft | Correct bracket following through candidate's $y$ value from part (a) |
| $\{0.05(8.3138)\}=0.41569$ = awrt $0.416$ | A1 | — |
---2.
$$y = \frac { x } { \sqrt { ( 1 + x ) } }$$
\begin{enumerate}[label=(\alph*)]
\item Complete the table below with the value of $y$ corresponding to $x = 1.3$, giving your answer to 4 decimal places.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & 1 & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 \\
\hline
$y$ & 0.7071 & 0.7591 & 0.8090 & & 0.9037 & 0.9487 \\
\hline
\end{tabular}
\end{center}
\item Use the trapezium rule, with all the values of $y$ in the completed table, to obtain an approximate value for
$$\int _ { 1 } ^ { 1.5 } \frac { x } { \sqrt { } ( 1 + x ) } \mathrm { d } x$$
giving your answer to 3 decimal places.\\
You must show clearly each stage of your working.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2013 Q2 [5]}}