Question 7 - A-Level Maths

OCR MEI Further Numerical Methods 2024 June — Question 7 14 marks

Exam Board	OCR MEI
Module	Further Numerical Methods (Further Numerical Methods)
Year	2024
Session	June
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Numerical integration
Type	Simpson's rule application
Difficulty	Standard +0.3 This is a straightforward numerical methods question requiring standard Simpson's rule calculations and understanding of convergence ratios. Part (a) uses the relationship S_{2n} = (2M_n + T_n)/3, part (b) requires basic precision justification, part (c) tests understanding of accumulated rounding errors, and part (d) asks for recognition of quartic convergence (ratio → 1/16). All techniques are standard for Further Maths numerical methods with no novel problem-solving required.
Spec	1.09f Trapezium rule: numerical integration

7 A student is using a spreadsheet to find approximations to \(\int _ { 0 } ^ { 1 } f ( x ) d x\) using the midpoint rule, the trapezium rule and Simpson's rule. Some of the associated spreadsheet output with \(n = 1\) and \(n = 2\), is shown in Table 7.1. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Table 7.1}

\(n\)	\(\mathrm { M } _ { n }\)	\(\mathrm {~T} _ { n }\)	\(\mathrm {~S} _ { 2 n }\)
1	0.612547	1
2	0.639735

\end{table}

Complete the copy of Table 7.1 in the Printed Answer Booklet. Give your answers correct to 5 decimal places.
State the value of \(\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x\) as accurately as possible. You must justify the precision quoted. The student calculates some more approximations using Simpson's rule. These approximations are shown in the associated spreadsheet output, together with some further analysis, in Table 7.2. The values of \(S _ { 2 }\) and \(S _ { 4 }\) have been blacked out, together with the associated difference and ratio. \begin{table}[h]
\captionsetup{labelformat=empty} \caption{Table 7.2}
n \(\mathrm { S } _ { 2 n }\) difference ratio
1
2
4 0.674353 -0.0209
8 0.665199 -0.00915 0.438059
16 0.661297 -0.0039 0.426286
32 0.659675 -0.00162 0.415762
64 0.659015 -0.00066 0.406785
\end{table}
The student checks some of her values with a calculator. She does not obtain 0.406785 when she calculates \(- 0.00066 \div ( - 0.00162 )\). Explain whether the value in the spreadsheet, or her value, is a more precise approximation to the ratio of differences in this case.
1. State the order of convergence of the values in the ratio column. You must justify your answer.
2. Explain what the values in the ratio column tell you about the order of the method in this case.
3. Comment on whether this is unusual.
Determine the value of \(\int _ { 0 } ^ { 1 } f ( x ) d x\) as accurately as you can. You must justify the precision quoted.

Show mark scheme Show mark scheme source

Question 7:

Answer	Marks	Guidance
7	(a)	n M T S

𝑛 𝑛 2𝑛

1 0.612547 1 0.74170

Answer	Marks
2 0.639735 0.80627 0.69525	B1

B1

Answer	Marks
B1	1.1

1.1

Answer	Marks
1.1	award B1 for each correct value given to 5 dp in the table.

if only one B mark awarded, allow SC1 for other 2 values given to

higher precision

if B0B0B0 allow SC1 for two values given to higher precision or

SC2 for all three values given to higher precision.

NB T = 0.806274 or 0.8062735

2 NB S = 0.741698

2 NB S = 0.69524783

4 [3]

Answer	Marks	Guidance
7	(b)	0.7 because S and S agree to this precision
4 2	B1	2.2b

4 2

ignore further comments unless contradictory

[1]

Answer	Marks	Guidance
7	(c)	spreadsheet stores values to greater accuracy

than it displays

student is working with the displayed values,

which are less accurate, so the student’s value is

Answer	Marks
less accurate oe	B1
B1	1.2
2.4	or spreadsheet works with the stored values so the answer in the

spreadsheet is more accurate

[2]

Answer	Marks	Guidance
7	(d)	(i)
first order convergence	B1	2.2b

[1]

Answer	Marks
n	M
𝑛	T
𝑛	S

2𝑛

Answer	Marks	Guidance
7	(d)	(ii)

0.25 < 0.4 < 0.5

Answer	Marks
so order of method is between 1 and 2	B1
B1	2.2b
2.2a	allow

eg order is higher than 1 and lower than 2

eg order of convergence of method is between 1 and 2

eg 1 < order < 2

do not allow

eg order is faster than 1 and slower than 2

eg convergence is between 1st and 2nd order

eg convergence is higher than 1st order and lower than 2nd order

eg convergence is faster than 1st order and slower than 2nd order

eg convergence is between 1st and 2nd order

[2]

Answer	Marks	Guidance
7	(d)	(iii)
(a) fourth order (method)	B1	2.5

convergence

[1]

Answer	Marks	Guidance
7	(e)	𝑟

𝑆+𝐷×

1−𝑟

𝑟

0.659015−0.00066×

1−𝑟

0.6585624 to 0.65858

accept 0.65856 or 0.6586 since extrapolation

Answer	Marks
greatly improves accuracy isw oe	M1

M1

A1

Answer	Marks
A1	3.1a

2.1

1.1

Answer	Marks
3.2a	𝑟

allow 𝑆±𝐷×

1−𝑟

S is value of S from the table, D is the associated difference and r

2n

= 0.4 or one of the (possibly rounded) ratios from the table

𝑟

allow 0.659015±0.00066×

1−𝑟

1 or 0.659675±0.00066×

1−𝑟

1 NB 0.659675−0.00066× earns M1M1

1−𝑟

0.4 ≤ r ≤ 0.406785

M1M1 may be implied by sight of correct value in range with

evidence of use of correct formula

allow 0.659 is certain since extrapolated value and

S (= 0.659015) agree to this precision; last (or best) Simpson’s

128 estimate

16×0.659015−0.659675

if M0M0 allow SC1 for = 0.658971

15 then SC1 for 0.659 seems secure by comparison with best

Simpson’s estimate oe

if M0M0 allow SC1 for

0.659015−0.00066×𝑟 = 0.65874 to 0.658751

then SC1 for 0.659 is secure by comparison with last Simpson’s

estimate oe

[4]

Answer	Marks
Response	Mark

PMT

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Though we make every effort to check our resources, there may be contradictions between published support and the

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Question 7:
7 | (a) | n M T S
𝑛 𝑛 2𝑛
1 0.612547 1 0.74170
2 0.639735 0.80627 0.69525 | B1
B1
B1 | 1.1
1.1
1.1 | award B1 for each correct value given to 5 dp in the table.
if only one B mark awarded, allow SC1 for other 2 values given to
higher precision
if B0B0B0 allow SC1 for two values given to higher precision or
SC2 for all three values given to higher precision.
NB T = 0.806274 or 0.8062735
2
NB S = 0.741698
2
NB S = 0.69524783
4
[3]
7 | (b) | 0.7 because S and S agree to this precision
4 2 | B1 | 2.2b | allow eg S ≈ S ≈ 0.7
4 2
ignore further comments unless contradictory
[1]
7 | (c) | spreadsheet stores values to greater accuracy
than it displays
student is working with the displayed values,
which are less accurate, so the student’s value is
less accurate oe | B1
B1 | 1.2
2.4 | or spreadsheet works with the stored values so the answer in the
spreadsheet is more accurate
[2]
7 | (d) | (i) | appear to be converging to a constant [0.4] so
first order convergence | B1 | 2.2b | or slowly decreasing, so convergence (slightly) faster than 1st order
[1]
n | M
𝑛 | T
𝑛 | S
2𝑛
7 | (d) | (ii) | (ratio of differences is converging to 0.4 and)
0.25 < 0.4 < 0.5
so order of method is between 1 and 2 | B1
B1 | 2.2b
2.2a | allow
eg order is higher than 1 and lower than 2
eg order of convergence of method is between 1 and 2
eg 1 < order < 2
do not allow
eg order is faster than 1 and slower than 2
eg convergence is between 1st and 2nd order
eg convergence is higher than 1st order and lower than 2nd order
eg convergence is faster than 1st order and slower than 2nd order
eg convergence is between 1st and 2nd order
[2]
7 | (d) | (iii) | this is unusual because Simpson’s rule is usually
(a) fourth order (method) | B1 | 2.5 | allow eg unusual because Simpson’s rule usually has fourth order
convergence
[1]
7 | (e) | 𝑟
𝑆+𝐷×
1−𝑟
𝑟
0.659015−0.00066×
1−𝑟
0.6585624 to 0.65858
accept 0.65856 or 0.6586 since extrapolation
greatly improves accuracy isw oe | M1
M1
A1
A1 | 3.1a
2.1
1.1
3.2a | 𝑟
allow 𝑆±𝐷×
1−𝑟
S is value of S from the table, D is the associated difference and r
2n
= 0.4 or one of the (possibly rounded) ratios from the table
𝑟
allow 0.659015±0.00066×
1−𝑟
1
or 0.659675±0.00066×
1−𝑟
1
NB 0.659675−0.00066× earns M1M1
1−𝑟
0.4 ≤ r ≤ 0.406785
M1M1 may be implied by sight of correct value in range with
evidence of use of correct formula
allow 0.659 is certain since extrapolated value and
S (= 0.659015) agree to this precision; last (or best) Simpson’s
128
estimate
16×0.659015−0.659675
if M0M0 allow SC1 for = 0.658971
15
then SC1 for 0.659 seems secure by comparison with best
Simpson’s estimate oe
if M0M0 allow SC1 for
0.659015−0.00066×𝑟 = 0.65874 to 0.658751
then SC1 for 0.659 is secure by comparison with last Simpson’s
estimate oe
[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
2024 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.

Show LaTeX source

7 A student is using a spreadsheet to find approximations to $\int _ { 0 } ^ { 1 } f ( x ) d x$ using the midpoint rule, the trapezium rule and Simpson's rule. Some of the associated spreadsheet output with $n = 1$ and $n = 2$, is shown in Table 7.1.

\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 7.1}
\begin{tabular}{ | c | c | c | c | }
\hline
$n$ & $\mathrm { M } _ { n }$ & $\mathrm {~T} _ { n }$ & $\mathrm {~S} _ { 2 n }$ \\
\hline
1 & 0.612547 & 1 &  \\
\hline
2 & 0.639735 &  &  \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Complete the copy of Table 7.1 in the Printed Answer Booklet. Give your answers correct to 5 decimal places.
\item State the value of $\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x$ as accurately as possible. You must justify the precision quoted.

The student calculates some more approximations using Simpson's rule. These approximations are shown in the associated spreadsheet output, together with some further analysis, in Table 7.2. The values of $S _ { 2 }$ and $S _ { 4 }$ have been blacked out, together with the associated difference and ratio.

\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 7.2}
\begin{tabular}{|l|l|l|l|}
\hline
n & $\mathrm { S } _ { 2 n }$ & difference & ratio \\
\hline
1 &  &  &  \\
\hline
2 & \multicolumn{2}{|c|}{} &  \\
\hline
4 & 0.674353 & -0.0209 &  \\
\hline
8 & 0.665199 & -0.00915 & 0.438059 \\
\hline
16 & 0.661297 & -0.0039 & 0.426286 \\
\hline
32 & 0.659675 & -0.00162 & 0.415762 \\
\hline
64 & 0.659015 & -0.00066 & 0.406785 \\
\hline
\end{tabular}
\end{center}
\end{table}
\item The student checks some of her values with a calculator. She does not obtain 0.406785 when she calculates $- 0.00066 \div ( - 0.00162 )$. Explain whether the value in the spreadsheet, or her value, is a more precise approximation to the ratio of differences in this case.
\item \begin{enumerate}[label=(\roman*)]
\item State the order of convergence of the values in the ratio column. You must justify your answer.
\item Explain what the values in the ratio column tell you about the order of the method in this case.
\item Comment on whether this is unusual.
\end{enumerate}\item Determine the value of $\int _ { 0 } ^ { 1 } f ( x ) d x$ as accurately as you can. You must justify the precision quoted.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2024 Q7 [14]}}

This paper (7 questions)

View full paper

Q1 4 Q2 6 Q3 6 Q4 10 Q5 10 Q6 10 Q7 14

n	\(\mathrm { S } _ { 2 n }\)	difference	ratio
1
2
4	0.674353	-0.0209
8	0.665199	-0.00915	0.438059
16	0.661297	-0.0039	0.426286
32	0.659675	-0.00162	0.415762
64	0.659015	-0.00066	0.406785