Further numerical methods (Further Pure 1)

FP1-6 Further numerical methods (Further Pure 1) — revision question pack

2 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section FP1-6. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

FP1-6.1 · Numerical solution of first order and second order differential equations.

Explanation

Three finite-difference approximations are quoted in the specification and must be known. The forward difference $\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_n\approx\dfrac{y_{n+1}-y_n}{h}$ steps forward from a single known value.
The central difference $\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_n\approx\dfrac{y_{n+1}-y_{n-1}}{2h}$ is more accurate but needs two consecutive starting values.
The second-derivative formula is $\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_n\approx\dfrac{y_{n+1}-2y_n+y_{n-1}}{h^2}$ .
In every case, rearrange to make $y_{n+1}$ the subject and then step forward, keeping more decimal places in the working than the question asks for in the answer.
The forward difference has error of order $h$ , whereas both central formulae have error of order $h^2$ , which is why halving the step size reduces their error by about a factor of four.

Worked example

Given $\dfrac{\mathrm{d}y}{\mathrm{d}x}=x+y$ and $y=1$ at $x=0$ , use the approximation $\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_n\approx\dfrac{y_{n+1}-y_n}{h}$ with $h=0.1$ to estimate $y$ at $x=0.2$ .

1.Rearranged, $y_{n+1}=y_n+h\left(x_n+y_n\right)$ .
2. $y_1=1+0.1(0+1)=1.1$ .
3. $y_2=1.1+0.1(0.1+1.1)=1.1+0.12=1.22$ .

Answer: $y(0.2)\approx1.22$ .

Common mistakes

Don't fall into the trap of using $x_{n+1}$ instead of $x_n$ inside the forward-difference step.
Don't fall into the trap of starting a central-difference calculation from a single value, when it needs two.
Don't fall into the trap of rounding the intermediate values to the accuracy demanded in the final answer, which loses a digit at every step.

Exam tip

Write the rearranged recurrence on its own line before any arithmetic, then tabulate $n$ , $x_n$ and $y_n$ ; the method mark is for the recurrence.

Tier 1 · Easy

1.
Given $\dfrac{\mathrm{d}y}{\mathrm{d}x}=x+y$ with $y=1$ at $x=0$ , use the approximation $\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_n\approx\dfrac{y_{n+1}-y_n}{h}$ with $h=0.1$ to estimate $y$ at $x=0.1$ and at $x=0.2$ .
(4)
(Total for Question 1 is 4 marks)
2.
Given $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\sqrt{x+y}$ with $y=2$ at $x=1$ , use the forward-difference approximation with $h=0.1$ to estimate $y(1.2)$ , giving your answer to $4$ decimal places.
(5)
(Total for Question 2 is 5 marks)

Tier 2 · Standard

1.
For $\dfrac{\mathrm{d}y}{\mathrm{d}x}=x^2-y$ it is given that $y(1)=0.5$ and $y(1.1)=0.55$ . Use the approximation $\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_n\approx\dfrac{y_{n+1}-y_{n-1}}{2h}$ with $h=0.1$ to estimate $y(1.2)$ .
(5)
(Total for Question 1 is 5 marks)
2.
For $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=x-y$ it is given that $y(0)=1$ and $y(0.1)=1.05$ . Use the approximation $\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_n\approx\dfrac{y_{n+1}-2y_n+y_{n-1}}{h^2}$ with $h=0.1$ to estimate $y(0.2)$ .
(5)
(Total for Question 2 is 5 marks)
3.
Use the forward-difference approximation with $h=0.1$ to estimate $y(0.2)$ for $\dfrac{\mathrm{d}y}{\mathrm{d}x}=y$ with $y(0)=1$ , and compare your answer with the exact value.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Given $\dfrac{\mathrm{d}y}{\mathrm{d}x}=x+y^2$ with $y=0.5$ at $x=0$ , use the forward-difference approximation with $h=0.05$ to estimate $y(0.1)$ , giving your answer to $4$ decimal places.
(6)
(Total for Question 1 is 6 marks)
2.
For $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+\dfrac{\mathrm{d}y}{\mathrm{d}x}+y=0$ it is given that $y(0)=1$ and $y(0.1)=0.995$ . Using $h=0.1$ with the central-difference approximations for both derivatives, estimate $y(0.2)$ .
(7)
(Total for Question 2 is 7 marks)
3.
Use Taylor's series to derive the approximation $\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_n\approx\dfrac{y_{n+1}-2y_n+y_{n-1}}{h^2}$ and show that its error is of order $h^2$ .
(6)
(Total for Question 3 is 6 marks)
4.
Given $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{1+x^2}$ with $y(0)=0$ , use the forward-difference approximation with $h=0.25$ to estimate $y(1)$ , and compare your answer with the exact value.
(7)
(Total for Question 4 is 7 marks)
5.
Explain why the central-difference approximation $\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_n\approx\dfrac{y_{n+1}-y_{n-1}}{2h}$ requires two starting values, and for $\dfrac{\mathrm{d}y}{\mathrm{d}x}=y$ with $y(0)=1$ and $y(0.1)=1.1051709$ use it with $h=0.1$ to estimate $y(0.2)$ , comparing the error with that of the forward difference.
(7)
(Total for Question 5 is 7 marks)

FP1-6.2 · Simpson's rule.

Explanation

Simpson's rule estimates $\displaystyle\int_a^b y\,\mathrm{d}x$ by fitting a quadratic through each consecutive group of three ordinates.
With $n$ strips of width $h=\dfrac{b-a}{n}$ and ordinates $y_0,y_1,\dots,y_n$ , the rule is $\displaystyle\int_a^b y\,\mathrm{d}x\approx\frac{h}{3}\left[\left(y_0+y_n\right)+4\left(y_1+y_3+\cdots\right)+2\left(y_2+y_4+\cdots\right)\right]$ .
The number of strips must be even, because the ordinates are consumed three at a time with the outer ones shared, so the strips pair up.
The end ordinates take coefficient $1$ , the odd-numbered ones take $4$ and the even-numbered interior ones take $2$ ; a quick check is that the coefficients must total $3n$ .
The rule is exact for every polynomial of degree at most three, and its error is of order $h^4$ , so halving the strip width reduces the error by roughly a factor of sixteen.

Worked example

Use Simpson's rule with four strips to estimate $\displaystyle\int_1^2\ln x\,\mathrm{d}x$ , giving your answer to $5$ decimal places.

1. $h=0.25$ and the ordinates are $0$ , $0.2231436$ , $0.4054651$ , $0.5596158$ and $0.6931472$ .
2.The bracket is $(0+0.6931472)+4(0.2231436+0.5596158)+2(0.4054651)$ .
3. $=0.6931472+3.1310376+0.8109302=4.635115$ .
4.Multiplying by $\dfrac{h}{3}=0.0833333$ gives $0.3862596$ .

Answer: $\displaystyle\int_1^2\ln x\,\mathrm{d}x\approx0.38626$ .

Common mistakes

Don't fall into the trap of using an odd number of strips, for which the rule is not defined.
Don't fall into the trap of attaching the coefficient $4$ to the even-numbered ordinates instead of the odd-numbered ones.
Don't fall into the trap of using $\dfrac{h}{2}$ from the trapezium rule instead of $\dfrac{h}{3}$ .

Exam tip

Tabulate $n$ , $x_n$ , $y_n$ and the coefficient in four columns and check that the coefficients sum to $3n$ before multiplying.

Tier 1 · Easy

1.
Use Simpson's rule with four strips to estimate $\displaystyle\int_0^1\sqrt{1+x^3}\,\mathrm{d}x$ , giving your answer to $4$ decimal places.
(4)
(Total for Question 1 is 4 marks)
2.
Use Simpson's rule with four strips to estimate $\displaystyle\int_0^{\pi/2}\sin x\,\mathrm{d}x$ , giving your answer to $6$ decimal places, and state the error.
(5)
(Total for Question 2 is 5 marks)

Tier 2 · Standard

1.
Use Simpson's rule with four strips to estimate $\displaystyle\int_0^1\mathrm{e}^{-x^2}\,\mathrm{d}x$ , giving your answer to $6$ decimal places.
(5)
(Total for Question 1 is 5 marks)
2.
Use Simpson's rule with four strips to estimate $\displaystyle\int_1^3\frac{\ln x}{x}\,\mathrm{d}x$ , giving your answer to $4$ decimal places, and compare it with the exact value $\tfrac12\left(\ln3\right)^2$ .
(6)
(Total for Question 2 is 6 marks)
3.
Explain why Simpson's rule requires an even number of strips, and state the coefficient pattern used for six strips.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Show that Simpson's rule is exact for $\displaystyle\int_0^{2h}x^3\,\mathrm{d}x$ .
(5)
(Total for Question 1 is 5 marks)
2.
Use Simpson's rule with six strips to estimate $\displaystyle\int_0^{\pi/3}\sec x\,\mathrm{d}x$ , giving your answer to $4$ decimal places, and compare it with the exact value $\ln\left(2+\sqrt3\right)$ .
(7)
(Total for Question 2 is 7 marks)
3.
Use Simpson's rule with four strips and then with eight strips to estimate $\displaystyle\int_1^2\frac{\mathrm{d}x}{x}$ , and use the two errors to confirm that the error behaves like $h^4$ .
(8)
(Total for Question 3 is 8 marks)
4.
The speed $v\,\text{m s}^{-1}$ of a particle at times $t=0$ , $0.5$ , $1$ , $1.5$ and $2$ seconds is $0$ , $2.4$ , $4.1$ , $5.0$ and $5.3$ respectively. Use Simpson's rule to estimate the distance travelled in the first $2$ seconds.
(5)
(Total for Question 4 is 5 marks)
5.
A student applies Simpson's rule to $\displaystyle\int_0^1 f(x)\,\mathrm{d}x$ with five strips. Explain why the calculation is invalid, and state the smallest number of strips greater than five that could be used together with the coefficient pattern that results.
(4)
(Total for Question 5 is 4 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FP1-6.1 · Numerical solution of first order and second order differential equations.

Tier 1 · Easy

Mark scheme for FP1-6.1 Tier 1 · Easy
Question	Scheme	Marks
1	$y_{n+1}=y_n+h\left(x_n+y_n\right)$ $y_1=1+0.1(1)=1.1$ $y_2=1.1+0.1(1.2)=1.22$	4
(4 marks)		4
Notes
Rearrange the forward difference and step twice. Independent check against the exact solution $y=2\mathrm{e}^{x}-x-1$ : $y(0.1)=1.11034$ and $y(0.2)=1.24281$ , so the estimates are low, as expected from a first-order method.
2	$y_{n+1}=y_n+h\sqrt{x_n+y_n}$ $y_1=2+0.1\sqrt{3}=2.1732051$ $y_2=2.1732051+0.1\sqrt{3.2732051}=2.3541251$ $y(1.2)\approx2.3541$	5
(5 marks)		5
Notes
Apply the recurrence twice, keeping seven decimal places in the working. Independent check by halving the step to $h=0.05$ and taking four steps: the estimate becomes $2.3580$ , closer to the fourth-order value $2.3620$ , and both step-wise estimates fall short of it as expected for a first-order method on an increasing solution.

Tier 2 · Standard

Mark scheme for FP1-6.1 Tier 2 · Standard
Question	Scheme	Marks
1	At $x=1.1$ : $\dfrac{y(1.2)-y(1.0)}{0.2}\approx1.1^2-0.55$ $=1.21-0.55=0.66$ $y(1.2)\approx0.5+0.2(0.66)$ $=0.632$	5
(5 marks)		5
Notes
Apply the central difference at the middle point $x=1.1$ , where both neighbouring ordinates are involved, and rearrange for the unknown one. Independent check by the forward difference from $x=1.1$ : $y(1.2)\approx0.55+0.1(0.66)=0.616$ , and the central-difference value lies on the more accurate side of it.
2	$y_2=2y_1-y_0+h^2\left(x_1-y_1\right)$ $=2(1.05)-1+0.01(0.1-1.05)$ $=1.1-0.0095$ $y(0.2)\approx1.0905$	5
(5 marks)		5
Notes
Rearrange the second-derivative formula to make $y_{n+1}$ the subject and substitute at $n=1$ . Independent check: the second derivative at $x=0.1$ is $0.1-1.05=-0.95$ , so the sequence should be slightly concave, and $1,1.05,1.0905$ has second difference $-0.0095=h^2\times(-0.95)$ exactly as the formula demands.
3	$y_{n+1}=y_n(1+h)$ $y_1=1.1$ and $y_2=1.21$ The exact solution is $y=\mathrm{e}^{x}$ , so $y(0.2)=1.2214$ The estimate is low by about $0.0114$	5
(5 marks)		5
Notes
The recurrence multiplies by $1+h$ at each step, giving $(1.1)^2=1.21$ . Independent check by the series: $\mathrm{e}^{0.2}=1+0.2+0.02+0.00133+\cdots=1.22140$ , so the forward difference has lost the quadratic term, an error of order $h$ per step as expected.

Tier 3 · Hard

Mark scheme for FP1-6.1 Tier 3 · Hard
Question	Scheme	Marks
1	$y_{n+1}=y_n+h\left(x_n+y_n^{2}\right)$ $y_1=0.5+0.05(0+0.25)=0.5125$ $y_2=0.5125+0.05\left(0.05+0.5125^{2}\right)$ $=0.5125+0.05(0.3126563)=0.5281328$ $y(0.1)\approx0.5281$	6
(6 marks)		6
Notes
Two forward-difference steps with the squared term evaluated at the current value of $y$ . Independent check by the Taylor series method: $y'(0)=0.25$ , $y''=1+2yy'$ gives $y''(0)=1.25$ , so $y(0.1)\approx0.5+0.025+0.00625=0.53125$ , and the step-wise estimate lies just below it, as a first-order method should.
2	At $x=0.1$ : $\dfrac{y_2-2y_1+y_0}{h^2}+\dfrac{y_2-y_0}{2h}+y_1=0$ $100\left(y_2-1.99+1\right)+5\left(y_2-1\right)+0.995=0$ $100y_2-99+5y_2-5+0.995=0$ $105y_2=103.005$ $y(0.2)\approx0.9810$	7
(7 marks)		7
Notes
Substitute both finite-difference approximations at the interior point $x=0.1$ , which leaves a single linear equation in the unknown $y_2$ . Independent check against the exact solution $\mathrm{e}^{-x/2}\left(\cos\tfrac{\sqrt3x}{2}+\tfrac{1}{\sqrt3}\sin\tfrac{\sqrt3x}{2}\right)$ , which gives $y(0.1)=0.99517$ and $y(0.2)=0.98133$ , within $0.0004$ of the estimate.
3	$y(x+h)=y+hy'+\dfrac{h^2}{2}y''+\dfrac{h^3}{6}y'''+\dfrac{h^4}{24}y^{(4)}+\cdots$ $y(x-h)=y-hy'+\dfrac{h^2}{2}y''-\dfrac{h^3}{6}y'''+\dfrac{h^4}{24}y^{(4)}-\cdots$ Adding, the odd-order terms cancel: $y(x+h)+y(x-h)=2y+h^2y''+\dfrac{h^4}{12}y^{(4)}+\cdots$ $\dfrac{y(x+h)-2y(x)+y(x-h)}{h^2}=y''+\dfrac{h^2}{12}y^{(4)}+\cdots$ The leading error term is $\dfrac{h^2}{12}y^{(4)}$ , so the error is of order $h^2$	6
(6 marks)		6
Notes
Expanding both neighbouring values and adding cancels every odd derivative, leaving $h^2y''$ as the first surviving term. Independent check with $y=x^4$ at $x=1$ , $h=0.1$ : the formula gives $\dfrac{1.4641-2+0.6561}{0.01}=12.02$ , against the exact $y''=12$ , an error of $0.02=\tfrac{h^2}{12}\times24$ .
4	$y_{n+1}=y_n+\dfrac{0.25}{1+x_n^{2}}$ $y_1=0.25$ , $y_2=0.4852941$ , $y_3=0.6852941$ $y_4=0.8452941$ The exact solution is $y=\arctan x$ , so $y(1)=\dfrac{\pi}{4}=0.7853982$ The estimate is high by about $0.06$	7
(7 marks)		7
Notes
Four forward-difference steps using the left-hand ordinate at each stage. Because the integrand is decreasing, using its value at the left of each interval overestimates the area, so the estimate exceeds $\arctan1$ . Independent check: the same calculation is the left-hand rectangle rule for $\displaystyle\int_0^1\frac{\mathrm{d}x}{1+x^2}$ , and the trapezium rule with the same ordinates gives $0.7828$ , much closer to $\tfrac{\pi}{4}$ .
5	The formula links three consecutive ordinates, so two must be known before the third can be found $y_2=y_0+2hy_1=1+0.2(1.1051709)$ $=1.2210342$ The exact value is $\mathrm{e}^{0.2}=1.2214028$ , an error of $0.00037$ The forward difference gave $1.21$ , an error of $0.0114$ , about $30$ times larger	7
(7 marks)		7
Notes
Rearranging the central difference gives $y_{n+1}=y_{n-1}+2hf\left(x_n,y_n\right)$ , which involves three ordinates at once and so cannot start from a single value. Independent check of the order of accuracy: the central difference has error of order $h^2$ and the forward difference error of order $h$ , so at $h=0.1$ the ratio of errors should be roughly $h^{-1}=10$ or more, consistent with the observed factor of about $30$ .

FP1-6.2 · Simpson's rule.

Tier 1 · Easy

Mark scheme for FP1-6.2 Tier 1 · Easy
Question	Scheme	Marks
1	$h=0.25$ and the ordinates are $1$ , $1.0077822$ , $1.0606602$ , $1.1924240$ and $1.4142136$ The bracket is $2.4142136+4(2.2002062)+2(1.0606602)=13.3363588$ The estimate is $\dfrac{0.25}{3}\times13.3363588=1.1114$	4
(4 marks)		4
Notes
Tabulate the five ordinates, apply the $1,4,2,4,1$ pattern and multiply by $\tfrac{h}{3}$ . Independent check with eight strips: the estimate becomes $1.1114$ to four decimal places as well, confirming convergence.
2	$h=\dfrac{\pi}{8}$ and the ordinates are $0$ , $0.3826834$ , $0.7071068$ , $0.9238795$ and $1$ The bracket is $1+4(1.3065629)+2(0.7071068)=7.6404652$ The estimate is $\dfrac{\pi}{24}\times7.6404652=1.000135$ The exact value is $1$ , so the error is $0.000135$	5
(5 marks)		5
Notes
Apply the rule with $h=\tfrac{\pi}{8}$ and compare with the exact integral $\left[-\cos x\right]_0^{\pi/2}=1$ . Independent check of the fourth-order error: with eight strips the estimate is $1.0000083$ , about sixteen times closer, exactly as an $h^4$ error predicts.

Tier 2 · Standard

Mark scheme for FP1-6.2 Tier 2 · Standard
Question	Scheme	Marks
1	$h=0.25$ and the ordinates are $1$ , $0.9394131$ , $0.7788008$ , $0.5697828$ and $0.3678794$ The bracket is $1.3678794+4(1.5091959)+2(0.7788008)=8.9622646$ The estimate is $\dfrac{0.25}{3}\times8.9622646=0.746855$	5
(5 marks)		5
Notes
The integrand has no elementary antiderivative, which is why a numerical method is required; apply the standard coefficient pattern. Independent check: the true value of this integral is $0.7468241$ , so Simpson's rule with only four strips is already accurate to five significant figures.
2	$h=0.5$ and the ordinates are $0$ , $0.2703101$ , $0.3465736$ , $0.3665163$ and $0.3662041$ The bracket is $0.3662041+4(0.6368264)+2(0.3465736)=3.6066569$ The estimate is $\dfrac{0.5}{3}\times3.6066569=0.6011$ The exact value is $\tfrac12\left(\ln3\right)^2=0.6035$ , so the estimate is low by about $0.0024$	6
(6 marks)		6
Notes
Apply the rule with $h=0.5$ ; the relatively large strip width and the turning point of the integrand at $x=\mathrm{e}$ make the error visible at the third decimal place. Independent check by substitution: $u=\ln x$ gives $\displaystyle\int_0^{\ln3}u\,\mathrm{d}u=\tfrac12\left(\ln3\right)^2=0.6034745$ .
3	The rule fits a quadratic through each consecutive group of three ordinates Each quadratic spans two strips, so the strips must pair up exactly An odd number of strips would leave one strip without a partner For six strips the pattern is $1,4,2,4,2,4,1$	4
(4 marks)		4
Notes
Each parabolic arc covers two adjacent strips and shares its end ordinates with the neighbouring arcs, which is why an even number is needed. Independent check on the coefficients: they sum to $1+4+2+4+2+4+1=18=3n$ with $n=6$ , the required total.

Tier 3 · Hard

Mark scheme for FP1-6.2 Tier 3 · Hard
Question	Scheme	Marks
1	The ordinates are $y_0=0$ , $y_1=h^3$ and $y_2=8h^3$ Simpson's rule gives $\dfrac{h}{3}\left[0+4h^3+8h^3\right]=\dfrac{h}{3}\left(12h^3\right)=4h^4$ The exact integral is $\left[\dfrac{x^4}{4}\right]_0^{2h}=4h^4$ The two agree, so the rule is exact for this cubic	5
(5 marks)		5
Notes
Evaluate the three ordinates of $x^3$ at $0$ , $h$ and $2h$ , apply the rule and compare with the exact integral. Independent check of the general claim: the rule is exact for $1$ , $x$ and $x^2$ because it integrates the interpolating quadratic exactly, and the error term for a general function is proportional to the fourth derivative, which vanishes for every cubic.
2	$h=\dfrac{\pi}{18}$ and the ordinates are $1$ , $1.0154266$ , $1.0641778$ , $1.1547005$ , $1.3054073$ , $1.5557238$ and $2$ The bracket is $3+4(3.7258510)+2(2.3695851)=22.6425740$ The estimate is $\dfrac{\pi}{54}\times22.6425740=1.3173$ The exact value is $\ln\left(2+\sqrt3\right)=1.3170$ , so the estimate is high by about $0.0003$	7
(7 marks)		7
Notes
Apply the rule with seven ordinates and the $1,4,2,4,2,4,1$ pattern. Independent check of the exact value: $\displaystyle\int\sec x\,\mathrm{d}x=\ln\left\|\sec x+\tan x\right\|$ , and at $x=\tfrac{\pi}{3}$ this is $\ln\left(2+\sqrt3\right)=1.3169579$ , while at $x=0$ it is $\ln1=0$ .
3	With four strips the estimate is $0.6932540$ With eight strips the estimate is $0.6931545$ The exact value is $\ln2=0.6931472$ The two errors are $1.068\times10^{-4}$ and $7.35\times10^{-6}$ Their ratio is about $14.5$ , close to the factor of $16$ that halving $h$ predicts for an $h^4$ error	8
(8 marks)		8
Notes
Apply the rule twice with $h=0.25$ and $h=0.125$ and compare each estimate with $\ln2$ . Independent check of the theory: an error proportional to $h^4$ should shrink by $2^4=16$ when the strip width is halved, and the small shortfall from $16$ is the usual effect of the higher-order terms at these step sizes.
4	$h=0.5$ with five ordinates The bracket is $(0+5.3)+4(2.4+5.0)+2(4.1)$ $=5.3+29.6+8.2=43.1$ The distance is $\dfrac{0.5}{3}\times43.1=7.18\,\text{m}$	5
(5 marks)		5
Notes
Distance is the integral of speed with respect to time, so apply the rule to the tabulated ordinates with $h=0.5$ . Independent check by the trapezium rule on the same data: $\tfrac{0.5}{2}\left[0+5.3+2(2.4+4.1+5.0)\right]=7.08\,\text{m}$ , slightly lower as expected for a concave speed-time graph.
5	Simpson's rule pairs the strips into parabolic arcs, so an even number is required With five strips one strip is left over and no quadratic can be fitted to it The smallest valid number greater than five is six The pattern is $1,4,2,4,2,4,1$ , whose coefficients sum to $18=3\times6$	4
(4 marks)		4
Notes
The rule integrates a quadratic through each consecutive triple of ordinates, and each such quadratic covers exactly two strips, so an odd count cannot be partitioned. Independent check: with $n$ strips there are $n+1$ ordinates, of which $\tfrac{n}{2}$ carry coefficient $4$ and $\tfrac{n}{2}-1$ carry coefficient $2$ ; both counts are whole numbers only when $n$ is even.