| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Simpson's rule application |
| Difficulty | Standard +0.3 This is a straightforward application of standard numerical integration formulas. Part (a) requires recall of the trapezium rule formula relating T_n to M_n, part (b) involves simple arithmetic calculations using Simpson's rule weighting (1/3 of 2M_n + T_n), and part (c) asks for basic interpretation of convergence. All techniques are standard textbook exercises with no novel problem-solving required. |
| Spec | 1.09f Trapezium rule: numerical integration |
| F | G | H | I | ||
| 3 | \(n\) | \(\mathrm { M } _ { \mathrm { n } }\) | \(\mathrm { T } _ { \mathrm { n } }\) | \(\mathrm { S } _ { 2 \mathrm { n } }\) | |
| 4 | 1 | 0.2436699 | 0.1479020 | ||
| 5 | 2 | 0.2306967 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | π1+π1 |
| Answer | Marks |
|---|---|
| =(G4+H4)/2 oe | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (b) | 2π1+π1 2π2+π2 4π2βπ1 |
| Answer | Marks |
|---|---|
| 0.2306967 0.1957860 0.2190598 | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | use of appropriate formula, may be implied by one correct |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | 16Γπ‘βπππ 0.2190598βπ‘βπππ 0.2117472 |
| Answer | Marks |
|---|---|
| from Richardsonβs extrapolation | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | 0.2190598β0.2117473 |
| Answer | Marks |
|---|---|
| n | T |
| n | S |
| Answer | Marks | Guidance |
|---|---|---|
| 0.2436699 | 0.1479020 | 0.2117473 |
| 0.2306967 | 0.1957860 | 0.2190598 |
Question 5:
5 | (a) | π1+π1
used
2
=(G4+H4)/2 oe | M1
A1
[2] | 1.1a
1.1
5 | (b) | 2π1+π1 2π2+π2 4π2βπ1
π = π¨π« π = or or
2 4
3 3 3
π1+π1
π = used soi
2
2
M T S
n n 2n
0.2436699 0.1479020 0.2117473
0.2306967 0.1957860 0.2190598 | M1
A1
A1
A1
[4] | 3.1a
1.1
1.1
1.1 | use of appropriate formula, may be implied by one correct
answer
if all correct values given to greater precision allow
M1A1A1A0
if table incorrect or incomplete, allow up to M1A1A1 for
work seen in the space below
5 | (c) | 16Γπ‘βπππ 0.2190598βπ‘βπππ 0.2117472
oe
15
0.2195473 β 0.21954731
0.220 is possible due to increased accuracy
from Richardsonβs extrapolation | M1
A1
A1
[3] | 3.1a
1.1
1.1 | 0.2190598β0.2117473
or 0.2190598+ ; may be implied by
16
0.2195168β¦
allow 0.22 is secure by comparison with S
4
M
n | T
n | S
2n
0.2436699 | 0.1479020 | 0.2117473
0.2306967 | 0.1957860 | 0.2190598
5 Kai uses the midpoint rule, trapezium rule and Simpson's rule to find approximations to $\int _ { \mathrm { a } } ^ { \mathrm { b } } \mathrm { f } ( \mathrm { x } ) \mathrm { dx }$, where $a$ and $b$ are constants. The associated spreadsheet output is shown in the table. Some of the values are missing.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
& F & G & H & I & \\
\hline
3 & $n$ & $\mathrm { M } _ { \mathrm { n } }$ & $\mathrm { T } _ { \mathrm { n } }$ & $\mathrm { S } _ { 2 \mathrm { n } }$ & \\
\hline
4 & 1 & 0.2436699 & 0.1479020 & & \\
\hline
5 & 2 & 0.2306967 & & & \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Write down a suitable spreadsheet formula for cell H 5 .
\item Complete the copy of the table in the Printed Answer Booklet, giving the values correct to 7 decimal places.
\item Use your answers to part (b) to determine the value of $\int _ { a } ^ { b } f ( x ) d x$ as accurately as you can, justifying the precision quoted.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2022 Q5 [9]}}