OCR MEI Further Numerical Methods 2023 June — Question 9 9 marks

Exam BoardOCR MEI
ModuleFurther Numerical Methods (Further Numerical Methods)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeSimpson's rule application
DifficultyChallenging +1.2 This is a structured Further Maths numerical methods question requiring application of standard techniques (Simpson's rule from trapezium rule results, Richardson extrapolation) with some analysis of convergence rates. While it involves multiple parts and requires understanding of error analysis, each step follows well-established procedures taught in Further Numerical Methods courses. The conceptual demand is moderate—students must recognize the ratio pattern indicates order of convergence and apply extrapolation formulas—but no novel insight is required.
Spec1.09f Trapezium rule: numerical integration
9 The trapezium rule is used to calculate 3 approximations to \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x\) with 1,2 and 4 strips respectively. The results are shown in Table 9.1. \begin{table}[h]
\captionsetup{labelformat=empty} \caption{Table 9.1}
\(n\)\(\mathrm {~T} _ { n }\)
10.52764369
20.66617652
40.72534275
\end{table}
  1. Use these results to determine two approximations to \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x\) using Simpson's rule.
  2. Use your answers to part (a) to state the value of \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x\) as accurately as you can, justifying the precision quoted. Table 9.2 shows some further approximations found using the trapezium rule, together with some analysis of these approximations. \begin{table}[h]
    \captionsetup{labelformat=empty} \caption{Table 9.2}
    \(n\)\(\mathrm { T } _ { n }\)differenceratio
    10.5276437
    20.66617650.138533
    40.72534270.0591660.42709
    80.74988210.0245390.41475
    160.75988580.0100040.40766
    320.76392210.0040360.40348
    640.76554040.0016180.40095
    \end{table}
  3. Explain what can be deduced about the order of the method in this case.
  4. Use extrapolation to obtain the value of \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x\) as accurately as you can, justifying the precision quoted.
Question 9:
AnswerMarks Guidance
9(a) oe or
4×0.66617652−0.52764369
3
oe
4×0.72534275−0.66617652
[S =] 0.712354 to 0.71235413 and
2
3
[S =] 0.7450648 to 0.745065
AnswerMarks
4M1
A11.1
1.1allow A1 if not attributed, but A0 if wrongly attributed
NB 0.745064826667
allow SC1 for both answers correct unsupported
[2]
AnswerMarks Guidance
9(b) 0.7 because the two Simpson’s estimates
only agree to this precisionB1 2.2b
difference between S and S is large enough to suggest value
4 2
closer to 0.8 than 0.7
[1]
B1
B1
AnswerMarks Guidance
9(c) ratio of differences seems to be converging
[to 0.4]
0.25 < r < 0.5 hence order of (convergence
of) method is between first and second order
AnswerMarks
(in this case)2.2b
2.2bDo not allow eg order of convergence is between 1 and 2
[2]
AnswerMarks Guidance
9(d) 0.4
0.7655404+0.001618×1−0.4
0.766619 to 0.766624
0.767 or 0.7666 since extrapolation much
AnswerMarks
more accurate oeM1
A1
A1
AnswerMarks
A13.1a
1.1
1.1
AnswerMarks
3.2aextrapolation with values of T , associated difference and r
n
deduced from table; allow for partial extrapolation
allow 0.4 to 0.401 for r; allow for partial extrapolation
NB 0.7661876 with r = 0.4
NB 0.7661891371 with r = 0.40095
NB 0.766619066667 and 0.76662334316 from extrapolation
to infinity with 0.4 and 0.40095 respectively
allow 0.77 is secure by comparison of T with extrapolated
64
value oe
if M0, allow SC2 for extrapolation with r = 0.25 to obtain
0.766079…
[4]
AnswerMarks
ResponseMark
PMT
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OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
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Question 9:
9 | (a) | oe or
4×0.66617652−0.52764369
3
oe
4×0.72534275−0.66617652
[S =] 0.712354 to 0.71235413 and
2
3
[S =] 0.7450648 to 0.745065
4 | M1
A1 | 1.1
1.1 | allow A1 if not attributed, but A0 if wrongly attributed
NB 0.745064826667
allow SC1 for both answers correct unsupported
[2]
9 | (b) | 0.7 because the two Simpson’s estimates
only agree to this precision | B1 | 2.2b | or 0.8 because the approximations are increasing and the
difference between S and S is large enough to suggest value
4 2
closer to 0.8 than 0.7
[1]
B1
B1
9 | (c) | ratio of differences seems to be converging
[to 0.4]
0.25 < r < 0.5 hence order of (convergence
of) method is between first and second order
(in this case) | 2.2b
2.2b | Do not allow eg order of convergence is between 1 and 2
[2]
9 | (d) | 0.4
0.7655404+0.001618×1−0.4
0.766619 to 0.766624
0.767 or 0.7666 since extrapolation much
more accurate oe | M1
A1
A1
A1 | 3.1a
1.1
1.1
3.2a | extrapolation with values of T , associated difference and r
n
deduced from table; allow for partial extrapolation
allow 0.4 to 0.401 for r; allow for partial extrapolation
NB 0.7661876 with r = 0.4
NB 0.7661891371 with r = 0.40095
NB 0.766619066667 and 0.76662334316 from extrapolation
to infinity with 0.4 and 0.40095 respectively
allow 0.77 is secure by comparison of T with extrapolated
64
value oe
if M0, allow SC2 for extrapolation with r = 0.25 to obtain
0.766079…
[4]
Response | Mark
PMT
Need to get in touch?
If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in
touch with our customer support centre.
Call us on
01223 553998
Alternatively, you can email us on
support@ocr.org.uk
For more information visit
ocr.org.uk/qualifications/resource-finder
ocr.org.uk
Twitter/ocrexams
/ocrexams
/company/ocr
/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
between the specification and a resource, please contact us.
Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more
information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.
9 The trapezium rule is used to calculate 3 approximations to $\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x$ with 1,2 and 4 strips respectively. The results are shown in Table 9.1.

\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 9.1}
\begin{tabular}{ | c | c | }
\hline
$n$ & $\mathrm {~T} _ { n }$ \\
\hline
1 & 0.52764369 \\
\hline
2 & 0.66617652 \\
\hline
4 & 0.72534275 \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Use these results to determine two approximations to $\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x$ using Simpson's rule.
\item Use your answers to part (a) to state the value of $\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x$ as accurately as you can, justifying the precision quoted.

Table 9.2 shows some further approximations found using the trapezium rule, together with some analysis of these approximations.

\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 9.2}
\begin{tabular}{|l|l|l|l|}
\hline
$n$ & $\mathrm { T } _ { n }$ & difference & ratio \\
\hline
1 & 0.5276437 &  &  \\
\hline
2 & 0.6661765 & 0.138533 &  \\
\hline
4 & 0.7253427 & 0.059166 & 0.42709 \\
\hline
8 & 0.7498821 & 0.024539 & 0.41475 \\
\hline
16 & 0.7598858 & 0.010004 & 0.40766 \\
\hline
32 & 0.7639221 & 0.004036 & 0.40348 \\
\hline
64 & 0.7655404 & 0.001618 & 0.40095 \\
\hline
\end{tabular}
\end{center}
\end{table}
\item Explain what can be deduced about the order of the method in this case.
\item Use extrapolation to obtain the value of $\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x$ as accurately as you can, justifying the precision quoted.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2023 Q9 [9]}}
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