OCR MEI Further Numerical Methods Specimen — Question 7 19 marks

Exam BoardOCR MEI
ModuleFurther Numerical Methods (Further Numerical Methods)
SessionSpecimen
Marks19
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeCompare two trapezium rule estimates
DifficultyChallenging +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.
Spec1.09f Trapezium rule: numerical integration

7 Fig. 7 shows the graph of \(y = \mathrm { f } ( x )\) for values of \(x\) from 0 to 1 . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{662c2d48-228a-4b94-a4b4-cdd31634ef21-6_693_673_390_696} \captionsetup{labelformat=empty} \caption{Fig. 7}
\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.
AABC
1\(h\)MidpointTrapezium
211.998517421.751283839
30.51.96385911.874900631
40.251.951352591.919379864
50.1251.946821021.935366229
  1. Without doing any further calculation, write down the smallest possible interval which contains the value of the integral. Justify your answer.
  2. (A) - Calculate the ratio of differences, \(r\), for the sequence of estimates calculated using the trapezium rule.
    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.
  3. 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.
  4. 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.
    NOPQ
    \(n\)Simpsondifferencesratio
    11.916106230.018100050.3584931
    21.934206280.006488740.3556525
    41.940695020.002307740.3544828
    81.943002750.000818050.3539885
    161.943820810.000289580.3537638
    321.944110390.000102440.3536568
    641.94421283\(3.623 \mathrm { E } - 05\)
    1281.94424906
  5. Use the spreadsheet output to find the value of the integral as accurately as possible. Justify the precision quoted. \section*{END OF QUESTION PAPER} }{www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.
    OCR is part of the }\section*{}

Question 7:
AnswerMarks Guidance
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
AnswerMarks
Midpoint Rule generates an over-estimateB1
E1
AnswerMarks
[2]2.2a
2.4
AnswerMarks Guidance
7(ii) (A)
0.12361679 0.359815
0.04447923 0.359412
0.01598637
r ≈ 0.36
Generally r should be close to 0.25, as the Trapezium
AnswerMarks
Rule is a second order methodM1
A1
C
E1
AnswerMarks
[3]1.1
I
1.1
AnswerMarks
2.4N
E
Differences and attempt to find
M
AnswerMarks Guidance
ratioCondone minor arithmetic slips
7(ii) (B)
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
AnswerMarks
8M1
A1
A1
E1
AnswerMarks
[4]1.1
1.1
1.1
AnswerMarks Guidance
3.2bA0 for incomplete extrapolation
7(iii) P
S
It is not appropriate.
Extrapolation is an approximate method (in this case it
AnswerMarks
has led to the values “swapping places”.)B1
B1
AnswerMarks
[2]2.3
2.4
AnswerMarks Guidance
7(iv) 2(cid:117)1.95135259(cid:14)1.91937986
(cid:32)1.940695
AnswerMarks
3M1
A1
AnswerMarks
[2]1.1
1.1
AnswerMarks Guidance
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.
AnswerMarks
Sensible reasoning consistent with their answer.M1
M1
M1
A1
A1
E1
AnswerMarks
[6]3.1a
1.1
2.1
1.1
3.2b
AnswerMarks
2.4Extrapolation 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
AnswerMarks Guidance
QuestionAO1 AO2
1i2 0
1ii2 1
2i0 1
2ii0 1
2iii2 0
2
AnswerMarks Guidance
3i1 1
3ii2 0
4i4 0
04
4ii0 2
5i0 1
00 1
5ii2 0
5iii1 0
01 2
5iv1 0
5v0 C
00 1
6i2 1
6ii0 1
6iiiE
11 0
6ivA1 2
6ivBP
02 0
6v0 1
S
AnswerMarks Guidance
7i0 2
7iiA2 1
7iiB3 0
7iii0 2
7iv2 0
7v2 2
Totals30 22
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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
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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}

}{www.ocr.org.uk}) after the live examination series.

If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.

For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.\\
OCR is part of the 
}\section*{}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods  Q7 [19]}}