| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Easy -1.2 This is a straightforward C2 question requiring only basic substitution into a given function and direct application of the trapezium rule formula with provided values. Both parts are routine procedural tasks with no problem-solving or conceptual challenge beyond recall of standard methods. |
| Spec | 1.02b Surds: manipulation and rationalising denominators1.09f Trapezium rule: numerical integration |
| \(x\) | 1 | 1.25 | 1.5 | 1.75 | 2 |
| \(y\) | 1.414 | 1.803 | 2.016 | 2.236 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At \(x = 1.25\), \(y = 1.601\) (only) | B1 cao | 1.601 may not be in the table and can score if seen as part of working in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2} \times 0.25\) or \(\frac{1}{8}\) | B1 | For using \(\frac{1}{2} \times 0.25\) or \(\frac{1}{8}\) or equivalent |
| Structure \(\{\ldots\}\) correct | M1 | Must contain first \(y\) value plus last \(y\) value; second bracket multiplied by 2 containing remaining \(y\) values. M0 if any \(x\) values used instead of \(y\) values |
| \(\frac{1}{2} \times 0.25 \times \{1.414 + 2.236 + 2(\text{their } 1.601 + 1.803 + 2.016)\}\) | A1ft | Correct expression following through candidate's \(y\) value found in part (a) |
| \(\left\{= \frac{1}{8}(14.49)\right\} = 1.81125\); answer \(1.81\) or awrt \(1.81\) | A1 |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| At $x = 1.25$, $y = 1.601$ (only) | B1 cao | 1.601 may not be in the table and can score if seen as part of working in (b) |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2} \times 0.25$ or $\frac{1}{8}$ | B1 | For using $\frac{1}{2} \times 0.25$ or $\frac{1}{8}$ or equivalent |
| Structure $\{\ldots\}$ correct | M1 | Must contain first $y$ value **plus** last $y$ value; second bracket multiplied by 2 containing remaining $y$ values. M0 if any $x$ values used instead of $y$ values |
| $\frac{1}{2} \times 0.25 \times \{1.414 + 2.236 + 2(\text{their } 1.601 + 1.803 + 2.016)\}$ | A1ft | Correct expression following through candidate's $y$ value found in part (a) |
| $\left\{= \frac{1}{8}(14.49)\right\} = 1.81125$; answer $1.81$ or awrt $1.81$ | A1 | |
**Note:** Correct answer only in (b) scores no marks. If required accuracy not seen in (a), full marks can still be scored in (b).
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e6b490c0-80c4-4e15-b587-ac052ee27db7-02_738_1257_274_340}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation $y = \sqrt { } \left( x ^ { 2 } + 1 \right) , x \geqslant 0$
The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $x$-axis and the lines $x = 1$ and $x = 2$
The table below shows corresponding values for $x$ and $y$ for $y = \sqrt { } \left( x ^ { 2 } + 1 \right)$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 1 & 1.25 & 1.5 & 1.75 & 2 \\
\hline
$y$ & 1.414 & & 1.803 & 2.016 & 2.236 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Complete the table above, giving the missing value of $y$ to 3 decimal places.
\item Use the trapezium rule, with all the values of $y$ in the completed table, to find an approximate value for the area of $R$, giving your answer to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2014 Q1 [5]}}