| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Session | Specimen |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Compare two trapezium rule estimates |
| Difficulty | Challenging +1.2 This is a multi-part question on numerical integration requiring understanding of trapezium/midpoint rules, error analysis, and extrapolation. While it involves several steps and concepts (ratio of differences, Richardson extrapolation, precision justification), the calculations are straightforward and follow standard Further Maths numerical methods procedures. The conceptual demand is moderate—students need to understand convergence behavior and error bounds, but no novel insight is required. |
| Spec | 1.09f Trapezium rule: numerical integration |
| A | A | B | C |
| 1 | \(h\) | Midpoint | Trapezium |
| 2 | 1 | 1.99851742 | 1.751283839 |
| 3 | 0.5 | 1.9638591 | 1.874900631 |
| 4 | 0.25 | 1.95135259 | 1.919379864 |
| 5 | 0.125 | 1.94682102 | 1.935366229 |
| N | O | P | Q |
| \(n\) | Simpson | differences | ratio |
| 1 | 1.91610623 | 0.01810005 | 0.3584931 |
| 2 | 1.93420628 | 0.00648874 | 0.3556525 |
| 4 | 1.94069502 | 0.00230774 | 0.3544828 |
| 8 | 1.94300275 | 0.00081805 | 0.3539885 |
| 16 | 1.94382081 | 0.00028958 | 0.3537638 |
| 32 | 1.94411039 | 0.00010244 | 0.3536568 |
| 64 | 1.94421283 | \(3.623 \mathrm { E } - 05\) | |
| 128 | 1.94424906 | ||
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (i) | 1.9353662285 (cid:100) I (cid:31) 1.946821025 |
| Answer | Marks |
|---|---|
| Midpoint Rule generates an over-estimate | B1 |
| Answer | Marks |
|---|---|
| [2] | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (ii) | (A) |
| Answer | Marks |
|---|---|
| Rule is a second order method | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 2.4 | N |
| Answer | Marks | Guidance |
|---|---|---|
| ratio | Condone minor arithmetic slips | |
| 7 | (ii) | (B) |
| Answer | Marks |
|---|---|
| 8 | M1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3.2b | A0 for incomplete extrapolation | |
| 7 | (iii) | P |
| Answer | Marks |
|---|---|
| has led to the values “swapping places”.) | B1 |
| Answer | Marks |
|---|---|
| [2] | 2.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (iv) | 2(cid:117)1.95135259(cid:14)1.91937986 |
| Answer | Marks |
|---|---|
| 3 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (v) | 0.354 |
| Answer | Marks |
|---|---|
| Sensible reasoning consistent with their answer. | M1 |
| Answer | Marks |
|---|---|
| [6] | 3.1a |
| Answer | Marks |
|---|---|
| 2.4 | Extrapolation using most accurate |
| Answer | Marks | Guidance |
|---|---|---|
| Question | AO1 | AO2 |
| 1i | 2 | 0 |
| 1ii | 2 | 1 |
| 2i | 0 | 1 |
| 2ii | 0 | 1 |
| 2iii | 2 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 3i | 1 | 1 |
| 3ii | 2 | 0 |
| 4i | 4 | 0 |
| 0 | 4 | |
| 4ii | 0 | 2 |
| 5i | 0 | 1 |
| 0 | 0 | 1 |
| 5ii | 2 | 0 |
| 5iii | 1 | 0 |
| 0 | 1 | 2 |
| 5iv | 1 | 0 |
| 5v | 0 | C |
| 0 | 0 | 1 |
| 6i | 2 | 1 |
| 6ii | 0 | 1 |
| 6iii | E | |
| 1 | 1 | 0 |
| 6ivA | 1 | 2 |
| 6ivB | P | |
| 0 | 2 | 0 |
| 6v | 0 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7i | 0 | 2 |
| 7iiA | 2 | 1 |
| 7iiB | 3 | 0 |
| 7iii | 0 | 2 |
| 7iv | 2 | 0 |
| 7v | 2 | 2 |
| Totals | 30 | 22 |
Question 7:
7 | (i) | 1.9353662285 (cid:100) I (cid:31) 1.946821025
The curve is concave down over this interval, so the
Trapezium Rule generates an underestimate, and the
Midpoint Rule generates an over-estimate | B1
E1
[2] | 2.2a
2.4
7 | (ii) | (A) | diffs ratio
0.12361679 0.359815
0.04447923 0.359412
0.01598637
r ≈ 0.36
Generally r should be close to 0.25, as the Trapezium
Rule is a second order method | M1
A1
C
E1
[3] | 1.1
I
1.1
2.4 | N
E
Differences and attempt to find
M
ratio | Condone minor arithmetic slips
7 | (ii) | (B) | E
0.36
1.93536623(cid:14)0.01598637(cid:117) (cid:32)1.9443(58563)
0.64
1.94
This agrees with T to two decimal places
8 | M1
A1
A1
E1
[4] | 1.1
1.1
1.1
3.2b | A0 for incomplete extrapolation
7 | (iii) | P
S
It is not appropriate.
Extrapolation is an approximate method (in this case it
has led to the values “swapping places”.) | B1
B1
[2] | 2.3
2.4
7 | (iv) | 2(cid:117)1.95135259(cid:14)1.91937986
(cid:32)1.940695
3 | M1
A1
[2] | 1.1
1.1
7 | (v) | 0.354
1.94424906(cid:14)0.00003623(cid:117)
1(cid:16)0.354
= 1.944268919 to 1.944268568
1.9443 looks secure, 1.94427 or 1.944269 is possible.
Sensible reasoning consistent with their answer. | M1
M1
M1
A1
A1
E1
[6] | 3.1a
1.1
2.1
1.1
3.2b
2.4 | Extrapolation using most accurate
Simpson value
Use of r = 0.354 or 0.35
ExtrapoNlation to infinity
E
Accept argument based on
comparison with 1.94424906 for
4 sf
M
or on further improvement due to
satisfactory demonstration of
convergence of r for 5 or 6 sf
Question | AO1 | AO2 | AO3(PS) | AO3(M) | Totals
1i | 2 | 0 | 0 | 0 | 2
1ii | 2 | 1 | 0 | 0 | 3
2i | 0 | 1 | 0 | 0 | 1
2ii | 0 | 1 | 0 | 0 | 1
2iii | 2 | 0 | 0 | 0 | N
2
3i | 1 | 1 | 0 | 0 | 2
3ii | 2 | 0 | 0 | 0 | 2
4i | 4 | 0 | 0 | E
0 | 4
4ii | 0 | 2 | 0 | 0 | 2
5i | 0 | 1 | M
0 | 0 | 1
5ii | 2 | 0 | 0 | 2 | 4
5iii | 1 | 0 | I
0 | 1 | 2
5iv | 1 | 0 | 0 | 1 | 2
5v | 0 | C
0 | 0 | 1 | 1
6i | 2 | 1 | 0 | 0 | 3
6ii | 0 | 1 | 0 | 0 | 1
6iii | E
1 | 1 | 0 | 0 | 2
6ivA | 1 | 2 | 0 | 0 | 3
6ivB | P
0 | 2 | 0 | 0 | 2
6v | 0 | 1 | 0 | 0 | 1
S
7i | 0 | 2 | 0 | 0 | 2
7iiA | 2 | 1 | 0 | 0 | 3
7iiB | 3 | 0 | 1 | 0 | 4
7iii | 0 | 2 | 0 | 0 | 2
7iv | 2 | 0 | 0 | 0 | 2
7v | 2 | 2 | 2 | 0 | 6
Totals | 30 | 22 | 3 | 5 | 60
PPMMTT
Y434 Mark Scheme June 20XX
BLANK PAGE
N
E
M
I
C
E
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12
7 Fig. 7 shows the graph of $y = \mathrm { f } ( x )$ for values of $x$ from 0 to 1 .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{662c2d48-228a-4b94-a4b4-cdd31634ef21-6_693_673_390_696}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
The following spreadsheet printout shows estimates of $\int _ { 0 } ^ { 1 } f ( x ) d x$ found using the midpoint and trapezium rules for different values of $h$, the strip width.
\begin{center}
\begin{tabular}{ | l | r | r | l | }
\hline
A & \multicolumn{1}{|c|}{A} & \multicolumn{1}{c|}{B} & \multicolumn{1}{c|}{C} \\
\hline
1 & $h$ & \multicolumn{1}{c|}{Midpoint} & Trapezium \\
\hline
2 & 1 & 1.99851742 & 1.751283839 \\
\hline
3 & 0.5 & 1.9638591 & 1.874900631 \\
\hline
4 & 0.25 & 1.95135259 & 1.919379864 \\
\hline
5 & 0.125 & 1.94682102 & 1.935366229 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item Without doing any further calculation, write down the smallest possible interval which contains the value of the integral. Justify your answer.
\item (A) - Calculate the ratio of differences, $r$, for the sequence of estimates calculated using the trapezium rule.
\begin{itemize}
\item Hence suggest a value for $r$ correct to 2 significant figures.
\item Comment on your suggested value for $r$.\\
(B) - Use extrapolation to find an improved approximation to the value of the integral.
\item State the value of the integral to two decimal places.
\item Explain why this precision is secure.
\end{itemize}
Using a similar approach with the sequence of estimates calculated using the midpoint rule, the approximation to the integral from extrapolation was found to be 1.94427 correct to 5 decimal places.
\item Andrea uses the extrapolated midpoint rule value and the value found in part (ii) ( $B$ ) to write down an interval which contains the value of the integral. Comment on the validity of Andrea's method.
\item Use the values from the spreadsheet output to calculate an approximation to the integral using Simpson's rule with $h = 0.125$. Give your answer to 5 decimal places.
Approximations to the integral using Simpson's rule are given in the spreadsheet output below. The number of applications of Simpson's rule is given in column N.
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
N & O & P & Q \\
\hline
$n$ & Simpson & differences & ratio \\
\hline
1 & 1.91610623 & 0.01810005 & 0.3584931 \\
\hline
2 & 1.93420628 & 0.00648874 & 0.3556525 \\
\hline
4 & 1.94069502 & 0.00230774 & 0.3544828 \\
\hline
8 & 1.94300275 & 0.00081805 & 0.3539885 \\
\hline
16 & 1.94382081 & 0.00028958 & 0.3537638 \\
\hline
32 & 1.94411039 & 0.00010244 & 0.3536568 \\
\hline
64 & 1.94421283 & $3.623 \mathrm { E } - 05$ & \\
\hline
128 & 1.94424906 & & \\
\hline
& & & \\
\hline
\end{tabular}
\end{center}
\item Use the spreadsheet output to find the value of the integral as accurately as possible. Justify the precision quoted.
\section*{END OF QUESTION PAPER}
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}\section*{}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Numerical Methods Q7 [19]}}