Notes

B1: Correct value 3.35 only, this may be seen in the table or in the script

Notes

B1: Correct step length, may be seen in the application of Simpson’s rule formula. Could be seen as \(1.75 - 1.5\) or \(\dfrac{2.5 - 1.5}{4}\)

M1: Correct application of Simpson’s rule using their step length. If no step length stated and uses \(\dfrac{1}{3}\left[\text{ends} + 2(\text{evens}) + 4(\text{odd})\right]\) this is M0

A1: Awrt 4.8 must be using 3.35 from (a)
Note calculator answer is 4.67 is no marks
Note Answer only with no working is no marks,
Note: If they write \(\dfrac{1}{3} \times h\left[\text{ends} + 2(\text{evens}) + 4(\text{odd})\right]\) which leads to awrt 4.8 this gets B1M1A1
If their answer is incorrect this would be B0 (if no \(h\) stated) M0A0

Notes

M1: Uses a correct identity to express \(\displaystyle\int \mathrm{e}^{\cot^2 x}\,\mathrm{d}x\) in terms of \(\displaystyle\int \mathrm{e}^{\operatorname{cosec}^2 x}\,\mathrm{d}x\) and then uses correct index work to allow \(\displaystyle\int_{1.5}^{2.5} \mathrm{e}^{\cot^2 x}\,\mathrm{d}x\) to be evaluated using part (b) \(\dfrac{1}{\mathrm{e}} \times \text{“}4.8\text{”}\)

A1ft: Awrt 1.8 or follow through \(\dfrac{1}{\mathrm{e}} \times \text{their (b)}\) which must be evaluated and correct for their (b) rounded to 1 d.p.
If uses Simpson’s rule again this is M0 A0

Notes

B1: Correct value, seen in table or in the work. Accept awrt.

Notes

B1: Correct step length used, stated or clearly implied in working.

M1: Correct structure for Simpson’s rule \(\dfrac{1}{3}\text{“}h\text{”}\left[\text{ends} + 2\text{evens} + 4\text{odds}\right]\) using their value of \(h\). If no value of \(h\) is stated nor is implied and the \(\dfrac{1}{3}h\) is not clearly seen, award bod for any multiple in front of the bracket that does not imply a clearly incorrect method – but the insides must be correct. Condone a missing closing bracket at the end, but otherwise bracketing must be correct or implied correct by the answer. Condone minor miscopies of the ordinates as long as they are in correct positions.

A1: For 8.09, correct answer only, must be to 3.s.f. Must have scored the M.
Note calculator gives 8.086 …

Notes

B1ft: Deduces the value by finding \(0.5 \times\) their answer to (b). Allow for awrt 3.s.f. answers giving tolerance with regards to prior rounding. Accept as a fraction, e.g. \(\dfrac{1213}{300}\) as long as it is half their answer to (b), or rounds correctly to it.

Notes

B1: Correct step length of 0.5 which may be implied e.g. by their 0, 0.5, etc.

M1: Attempts to find \(y\) values for all their \(x\) values – may be in terms of e or numerical values. Must be trying to find at least 3 values (e.g. if step length 1 is used). If substitution is not seen at least 2 of the values other than 1 must be correct to at least 2 s.f. rounded or truncated.

M1: Correct application of Simpson’s rule: \(\dfrac{h}{3}\)(ends + 2evens + 4odds) (must have an odd number of ordinates). Must be \(y\) values not \(x\) values.

A1cao: 3.87. Must be to 3 s.f.

Note: must see evidence of Simpson’s Rule so answer with no working scores no marks.

Notes

B1: Must have awrt 3.87 for part (a), makes a sensible comment regarding the accuracy. Must relate to the accuracy, not just the value.

Accept

• It is correct to 2 significant figures
• It is correct to 1 decimal place
• It is not correct to 3 significant figures
• It is awrt 99.6% accurate (using 3.87) or 99.7% accurate (using 3.866…)
• Percentage error is awrt 0.3%/0.4% so very accurate
• Very accurate as only out by 0.015 (requires reference to accuracy and evidence)

Do not accept e.g. it is out by 0.015 with no reference to what this means about the accuracy. Do not accept e.g. “very close to” as evidence, some quantification (absolute error, number of s.f. it agrees to) must be given.

Notes

B1: Uses the model to identify the correct initial conditions and requirements for \(h\). May be implied by use in the equation.

M1: Applies the approximation formula with their values for \(v_0\), \(\left(\dfrac{\mathrm{d}v}{\mathrm{d}t}\right)_{0}\) and \(h\) to find a value for \(v_1\)

A1: \(v_1 = 5\)

M1: Uses their \(v_1\) to find a value for \(\left(\dfrac{\mathrm{d}v}{\mathrm{d}t}\right)_{1}\) and applies the approximation formula with their values for \(v_1\), \(\left(\dfrac{\mathrm{d}v}{\mathrm{d}t}\right)_{1}\) and \(h\) to find a value for \(v_2\)

A1: \(v_2 = 8.75\) or \(8.75\,\mathrm{m\,s^{-1}}\)

Notes

B1: Reduce the value of 10 or explains this is what needs reducing, but do not accept 0 or negative values in place of the 10. Note: “change the 10” is B0 if it does not explain how to change it.

Notes

B1: Identifies a correct step length for the situation – 6 hours is a quarter of a day, so \(h = 0.25\)

M1: Uses “\(y_0\)” \(= x(0) = 3\) and \(t = 0\) to find “\(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0\)” \(= \left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)_0\). Accept with whichever notation used, as long as it is clear they are attempting the correct things.

M1: Applies the approximation formula with their “\(h\)” and their “\(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0\)”

A1: For awrt 3.04 ppm. Accept \(\dfrac{82}{27}\) ppm

Notes

B1: A correct equation relating \(\dfrac{\mathrm{d}u}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) from the chain rule.

M1: Makes a complete substitution for \(x\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) in equation (I) or a complete substitution for \(u\) and \(\dfrac{\mathrm{d}u}{\mathrm{d}t}\) in equation (II)

A1*: Simplifies correctly to achieve the given result.

Notes

B1: Correct integrating factor found or spotted. Allow for \(\mathrm{e}^{\ln\cosh t}\)

M1: Applies IF to achieve \(u\,\text{“}\cosh t\text{”} = \displaystyle\int \text{“}\cosh t\text{”}\left(1 + \frac{3}{\cosh t}\right)\mathrm{d}t\)

M1: A reasonable attempt to integrate the RHS. Need not include constant of integration. If I.F. correct allow for \(\pm\sinh t + 3t\,(+c)\)

A1: Correct general solution, either implicit or explicit form including the context of integration (award when first seen and isw)

Notes

M1: Uses the initial conditions in an appropriate equation to find the constant of integration. Either \(t = 0\) and \(u = 27\) in the answer to (c), or \(t = 0\) and \(x = 3\) if substitution for \(x\) occurs first.

M1: Reverses the substitution and rearranges to find equation for \(x\), with evaluated constant included.

A1: Correct equation, any equivalent form, but must be \(x = \ldots\)

Notes

M1: Uses their model solution to find the value at \(t = 0.25\)

M1: Applies \(\dfrac{\text{actual value} - \text{estimate}}{\text{actual value}} \times 100\) with their values.

A1: States part (a) is overestimate by 1.05%

Notes

B1: Correct strip width for the method chosen \(\dfrac{1}{3}\) for the interval \([-1, 1]\)

M1: Uses the model to find the appropriate values for the method. May use that the function is even to only work out half of them, so may be implied by use in the formula. At least two correct values to 4 s.f. needed for the method.

M1: Applies the “bracket” of Simpson’s rule, “\(y_0 + 4y_1 + 2y_2 + 4y_3 + 2y_4 + 4y_5 + y_6\)”. Coefficients must be correct.

A1: Correct value for the “bracket”. If not explicitly seen, may be implied by awrt 4.689 as a value for the cross section area following correct values.

M1: Correct full method to find the volume. E.g. multiplies their bracket by their \(\dfrac{h}{3}\) and by 0.085.
Accept an attempt in any consistent units, so e.g. in mm³ ie \(85 \times \dfrac{\frac{1}{3}}{3} \times \text{“}42.203\text{”} \times 1000^2\)

A1: Correct answer in m³.

Alternative interval [0,1]

B1: Correct strip width for the method chosen \(\dfrac{1}{6}\) for the interval \([0, 1]\) and later doubled.

M1: Uses the model to find the appropriate values for the method. May use that the function is even to only work out half of them, so may be implied by use in the formula. At least two correct values to 4 s.f. needed for the method.

M1: Applies the “bracket” of Simpson’s rule, “\(y_0 + 4y_1 + 2y_2 + 4y_3 + 2y_4 + 4y_5 + y_6\)”. Coefficients must be correct.

A1: Correct value for the “bracket”. If not explicitly seen, may be implied by awrt 4.680 as a value for the cross section area following correct values.

M1: Correct full method to find the volume. E.g. multiplies their bracket by their \(\dfrac{h}{3}\) and by 0.085.
Accept an attempt in any consistent units, so e.g. in mm³ ie \(85 \times \dfrac{\frac{1}{6}}{3} \times \text{“}42.1206\text{”} \times 1000^2 \times 2\) (corrected from the printed mark scheme, which has “42.203” here)

A1: Correct answer in m³.

Using 6 ordinates

Max score B0 M1 M0 A0 M0 A0

B0: Incorrect strip width

M1: Uses the model to find the appropriate values for the method. At least two correct values to 4 s.f. needed for the method.

Notes

B1: Correct step length of 0.4 which may be implied e.g. by their 0.4, 0.8, etc.

M1: Attempts to find \(y\) values for their \(x\) values – may be in terms of e or numerical values. Must see an attempt to find at least 3 values.

M1: Correct structure for \(y\) values of Simpson’s rule (ends + 2evens + 4odds) (must have an odd number of ordinates). Must be \(y\) values not \(x\) values.

dM1: \(\dfrac{\text{“}0.4\text{”}}{3} \times \text{their } 123.54\ldots\) or for \(\dfrac{h}{3} \times \text{their } 123.54\ldots\) leading to a value and where \(h\) has clearly been defined earlier.
Dependent on both previous method marks

A1: Awrt 16.5

Note that a minimum we would expect to see for full marks is:

\(h = 0.4\)

(Note that a calculator gives \(16.030\ldots\) for the area)