P8.2 Integrating standard functions

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In general, this question was a good source of marks for many candidates. In part (b), most candidates were able to integrate at least one term correctly, even if they had the wrong expression from part (a). Many students scored full marks here, especially those with full marks in part (a). 2 marks was common if their answer to part (a) was incorrect, with the first A mark often awarded as a follow-through. Common mistakes included: • forgetting to include “+ c” • achieving 5 4 or 4 5 instead of 1 20 when simplifying fractions • obtaining an incorrect power for the second term when attempting add 1 to 1 2 - Again, some candidates achieved the correct answer and then attempted to multiply through by some value to achieve a different expression, but this was not penalised in part (b). Those who had done less well on part (a) often managed to pick up the first two marks (M1A1) despite earlier errors with their fractions or indices. However, even amongst those scoring well, there was some poor notation e.g. spurious integration signs, though these were not penalised.

Common mistakes

• raise each power by 1 and divide: (1/20)x⁵ − 6x^(½) + c
• forgetting + c
• 5/4 or 4/5 instead of 1/20
• wrong power from −½ + 1

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was well attempted by the vast majority of learners. Part (b) of the question was also usually answered successfully. This question was a show that and included the warning they solutions relying on calculator technology were not acceptable, so it was required that learners showed all steps of their working. Most understood the method and showed appropriate integration followed by correct substitution of limits and correct answer. There were sometimes errors with the coefficient, the most common being 1 2 2 x x rather than 2 1 2 x . Other errors when integrating usually occurred with the power (converting 2 x to 1 2 2x then integrating or differentiating instead of integrating) or the use of ln ax . A small number of responses integrated correctly but then proceeded straight to the final answer without any evidence of substitution of the given limits and so were unable to gain any marks. Those who had the incorrect limits such as using 0 were also unable to score either mark, although transcription errors of 2.89 and1.44 were condoned for the method mark. There were many cases of poor notation such as retaining the integral symbol after integrating and include c + within the calculation but these were condoned on this occasion. 9

Common mistakes

• integrate 2x^(−1/2) to 4√x and show the limits substituted: 4 × 1.7 − 4 × 1.2 = 2
• coefficient errors (x^(1/2) × ½), differentiating instead, ln
• jumping straight to 2 scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This unstructured question required learners to evaluate a definite integral and it was approached in a wide variety of ways with varying degrees of success. The question included a calculator warning which made it clear that solutions relying on calculator technology were not acceptable. Integration by substitution and integration by parts were the most common methods by far and seen in roughly equal measure. For the substitution method, most learners chose to use 2 1 u x = + . A few proceeded no further than rewriting the integral fully in terms of u as they did not recognise that the resulting expression could be split up into two distinct terms which could be integrated directly. Successful attempts using integration by parts required u x = and ( ) 3 d 2 1 d v x x - = + . Some learners who rely on acronyms and found that the “LATE” method to decide the choice of u and d d v x did not help them in this instance. Many chose to restart the question when they realised that they were not making any progress. Most responses that executed these methods and mistakes in the early stages tended to be made with their coefficients rather than the powers. Less commonly seen was first splitting the integrand into partial fractions or using a less efficient substitution such as 3 (2 1) u x = + or even a trigonometric substitution. Some responses negated to heed the calculator warning and jumping to the given answer without any evaluation of limits. Overall, whilst most learners could begin the question by a viable method, sustained accuracy and attention to fine detail were required in order to achieve full marks, many had to restart their solutions or amend earlier work which made it very challenging at times to determine what work was credible, or which solution they wanted to be marked. Due to no errors being allowed for the final mark learners risked not being awarded this mark by not clearly identifying their final attempt. 20

Common mistakes

• by parts with u = x, dv/dx = (2x + 1)^(−3), or substitution u = 2x + 1 (limits 1 and 5) splitting into u^(−2) − u^(−3)
• the LATE rule misleads here
• show the limits substituted before 2/25
• coefficient slips
• several restarts made work hard to follow

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This proved to be a challenging question demanding good algebraic skills and attention to detail. For a final question on a paper, most made an attempt. In part (b), it was notable that those who did not reach a correct answer in part (a) often did not go on to attempt this, even though these marks were available to them. Even though they were told that the integral was a multiple of 2 cosec u some learners tried unsuccessfully to integrate their very complex answers to (a). A large number did not appear to know the integral of 2 cosec u or did not use the formula booklet correctly and some attempted very convoluted methods of integration. Those that did not score at all got muddled in the initial stage, choosing to use ln or some other given integration from the formula book that they had misread or misinterpreted. A significant number got confused between the integral of 2 cosec u andcosec u , which meant they ended up with expressions involving ln(tan ) u or ln(cosec cot ) u u + . However, several responses obtained the correct answer of 1 cot 4 u - , although a few omitted the negative sign, and others 23 stated the answer was 1 cot . 4 u There was a proportion who did not include the negative sign and consequently amended the resulting negative in their final answer. 24

Common mistakes

• ∫cosec²u du = −cot u (sign often lost
• confused with ∫cosec u)
• −¼cot(π/3) + ¼cot(π/6) = 1/(2√3) = √3/6
• can be attempted even if (a) was not completed

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This routine question on indefinite integration was completed very well in almost all responses, with the majority demonstrating familiarity and confidence in dealing with this type of question. It was rare that this question was not attempted, and for many it proved to be a confidence boosting start to the paper. Learners were mostly efficient in their methods, with the majority knowing the steps for integration and applying them successfully. Of those achieving full marks, many did not need a working step but proceeding directly to the correct answer. The common errors were mostly small, such as omitting the constant of integration (+c) or making arithmetic mistakes when simplifying fractions, particularly when dividing by 3 2 . These errors were often due to confusion during the simplification process. A small number of responses mistakenly differentiated rather than integrated, which led to significant errors. This was a rare issue but still notable. Several responses that achieved a correct answer then went on to do further, often incorrect work with their integrated function, such as multiplying by 5 or setting it equal to 0 and attempting to solve. While this was not mathematically correct, subsequent work was ignored to ensure they were not penalised for these errors. Some responses seen confused the integral notation, either retaining the integral symbol and dx in their answer or including square brackets as if they were expecting to substitute in values, but again, this was condoned.

Common mistakes

• −6x^½ → −4x^(3/2) (dividing by 3/2 causes slips)
• forgetting + c
• differentiating instead of integrating

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The final question, which involved differentiation, integration by substitution and trigonometric identities, provided challenge in dealing with the algebra but, given its layout, proved accessible. The majority of learners attempted at least part of this question with some success due to the restart opportunities available. Part (c) proved to be more accessible for many learners, with many making an attempt after struggling with the earlier sections. Those who attempted this part often successful, regardless of their performance in parts (a) and (b). Common errors were with the coefficient, usually given as 1 2 - or 2, or when substituting the limits. Some tried to use double angle formulae to change the form of the integrand and therefore did not make progress with this part of the question, not realising that the given form in part (b) was designed to help them complete the integration. Despite the question asking the candidate to show all stages of their working, some responses did not show the substitution and so did not gain the last two marks.

Common mistakes

• ∫cos 2θ dθ = ½ sin 2θ (not −½ or 2)
• substitute the θ limits, not ±1
• area 1
• the substitution must be shown (calculator answer alone scores 0)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was a very accessible question on integration to start the paper and most candidates were able to score well. Most candidates attempted to separate the fraction into two terms and achieved at least one correct index. The first term caused problems for some with errors such as 5 3 1 2 2 2 , , x x x - . A few ended up with three terms as they had added or subtracted 2 x-. A few tried to integrate the numerator and denominator or wrote 2 (2 3) x x- - but did not expand the brackets and integrated both inside and outside separately. Those who had scored the first M1 usually were able to raise the power by one on at least one term with a correct index and score the next method mark. Several made errors when processing the indices and ended up subtracting by one on at least one of the terms. A few made errors with the coefficients. The most common mistake was to forget the + c.

Common mistakes

• split into 2x^(−3/2) − 3x^(−2) before integrating (writing −3/x² is not enough)
• integrating numerator and denominator separately scores nothing
• index slips such as x^(−5/2) or subtracting 1
• forgetting + c is the most common error
• answer on one line, simplified

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Marks awarded in this question were generally quite varied. Often candidates made progress in (c) only, by attempting some integration with little correct work seen in (a) or (b). Each part of this question presented a given answer, and candidates who failed to observe the warning in the question about showing all stages of their working, or relying entirely on calculator technology, often lost marks. In part (c) candidates generally knew that they needed to proceed with integration, and often the first two marks were scored for correctly integrating even if they made no further progress. Successful candidates usually attempted to evaluate 2 3 4 14 23 d x x x - - + and then subtracted 2 4 2 7 d x x -- + . Some used the area of a trapezium to give 1 (3 15) 6 2 + instead of integrating the equation of the line. Alternatively, some attempted to evaluate 2 3 4 12 16 d x x x - - + reaching the required area of 108 in one step. Candidates were required to make their method clear, including showing the use of the limits 4 - and 2. Some failed to do so, clearly relying on calculator technology to evaluate the integrals. Others failed to show the complete correct strategy for finding the area and often incorrectly identified the trapezium as a triangle. Also, correct notation was required and appropriate use of the integral sign and the ‘dx’ on either side.

Common mistakes

• ∫ from −4 to 2 of (x³ − 14x + 23) dx minus the trapezium ½(15 + 3) × 6 = 54, or ∫(x³ − 12x + 16) dx
• correct limits 2 and −4
• integral sign and dx needed
• calculator integration scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (c) was not well done and quite often no attempt was made. Some students substituted one or more values of x, setting the result equal to the frequency (or frequency density). When integration was used, it was often done well. It was, however, surprising to see that even when the correct integration was equated to 112, many were unable to accurately solve the equation. Many students made slips with the negative signs, and some were unable to solve the equation, multiplying one side by 8 to get rid of the fraction but failing to multiply 112 by 8 at the same time.

Common mistakes

• area under the frequency density curve is the total frequency: ∫ from 1 to 8 of k/x² dx = 112
• (7/8)k = 112 gives k = 128
• substituting x values instead of integrating
• sign slips with −k/x

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (b), successful candidates integrated the expression for v with respect to t and were usually able to obtain a correct, unsimplified expression for the displacement of the particle at time t. There were then occasional errors with the simplification often due to incorrect coefficients. There was no penalty for not including a constant of integration. Successful candidates then used the limits of t = 1 and t = 2 to calculate the distance XY. A surprising number of candidates thought that finding XY meant that they then had to multiply these two values. Since there was a printed answer, candidates needed to clearly show all stages of their working including the use of surds to get to the required form of the answer. Of those that tried to use surds, some struggled to reach the given answer whilst others did not show enough working to gain the final A mark. Ignoring the instructions at the top of the question, many candidates used calculators and decimals appeared in their solutions which resulted in the solutions being incomplete and losing the final mark. Very few candidates used differentiation in this part.

Common mistakes

• integrate v: t² − (14/3)t^(3/2) + 6t
• XY = s(2) − s(1) (no change of direction so displacement = distance)
• keep surds: 2^(3/2) = 2√2 to reach (41 − 28√2)/3
• decimals lose the final mark
• multiplying the two values

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

part (a) to do so. In part (a), most candidates recognised the need to integrate the expression and were able to integrate an exponential function; various methods were seen here but the majority followed the main approach in the mark scheme. A significant minority of candidates multiplied by –0.2 instead of dividing, resulting in an incorrect model and marks being lost in (b) and (c) due to having an impossible equation to solve. The vast Candidates who had a correct answer in part (a) generally performed well in part (b), with the majority able to rearrange the equation and use logarithms correctly to find the value of t. If these candidates were not awarded full marks, it was often because they were unable to convert their decimal into hours and minutes correctly, and therefore lost the final accuracy mark. Candidates who did not have a correct answer to part (a) A 0.2 t e generally gained the first method mark for making the subject of the equation, but many were not able to achieve the second method mark due to the context of the model. Candidates with B > 1.2 reached the logarithm of a negative value which should have provided an indication that there must have been an earlier error in need of addressing but few re-traced their workings and tended to use incorrect methods to create a positive answer. Part (c) was generally less well attempted than parts (a) and (b) with some candidates failing to understand what the question was asking them to do. The most common error was for candidates to attempt to solve d H t = . A surprising proportion of candidates 0 d gave answers higher than the maximum height. For candidates with successful approaches, marks were frequently lost by forgetting units and therefore not satisfying the demand of the question and finding the height of the hole. Similarly, many decided to subtract an often correct 0.9 from the maximum height, which indicated a misunderstanding of the model.

Common mistakes

• integrate −0.12e^(−0.2t) to 0.6e^(−0.2t) + c (divide by −0.2, not multiply)
• use t = 0, H = 1.5 to get c = 0.9 (not c = 1.5)
• give the complete equation with H =

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

part (c). The final part of the question proved more challenging and differentiated between the candidates, with many producing concise and fully correct answers. Most candidates realised it required the use of integration and often scored the B mark for integrating the curve. Many candidates failed to note that the question required an exact answer and use of algebraic integration. This meant that using a calculator to provide a decimal approximation was insufficient for the final method mark. The quality of notation in this part of the question was very varied with the dx often missing from integrals and spurious integral signs included throughout working. 1 = The majority of successful candidates used or integration to find the area of A bh 2 the triangle and then subtracted the area under the curve. However, a significant proportion of candidates made the question much more difficult than necessary, using a mixture of Pythagoras and trigonometry to find the area of the triangle-this often led to errors or premature rounding. Other common errors when finding the area of the triangle included assuming the triangle was isosceles, and therefore that the intersection x = point was at , or splitting the triangle into two right-angled triangles and assuming 2 they both had the same area. Candidates attempting to use the alternative method using lines – curve were more likely to make errors with signs, brackets or limits. They were also required to correctly 2 2.4 7 ( ) cancel the terms to find an exact answer and so most often were unable to 2 7 score the final method and accuracy marks. The most common invalid method was attempting to combine the lines and curve before integrating using the limits 0 and 4, and this scored no marks unless the point of intersection had been found.

Common mistakes

• l1 and l2 meet at (2.4, 19.2)
• triangle 38.4 minus ∫₀⁴ (8x − x^(5/2)) dx = 192/7
• exact 384/35 (calculator decimal loses marks)
• triangle is not isosceles
• missing dx

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

failed to prove the result in part (a) then went on to omit part (b) and missed out on this opportunity. However, it was not uncommon to see full marks in part (a) followed by no marks in part (b), and vice versa. Most candidates who attempted part (a) recognised the requirement to find d x and d squared trigonometric function were generally successful in part (b) of this question. - 1 cos2 = Occasionally candidates struggled to deal with 2 correctly, with 2 sin 2 2 seen all too frequently. These candidates scored no marks following incorrect use of the double angle identity. The other main errors in this part were in sign errors in the double cos4 - angle identity and in the integration of the trigonometric term: commonly was 1 sin 4 + integrated to , and despite this still leading to the correct answer it was at the 4 expense of the two accuracy marks. A significant proportion of candidates incorrectly sin . Some 3 attempted to increase the power and divide leading to a term in 1 1 ( ) - by parts were also seen. x a x unsuccessful attempts at restarting and integrating 2 2 There was also a significant proportion of candidates who attempted integration by parts which generally gained no credit. Candidates who were successful in using integration and at this point 2 cos 2 by parts often had a succinct approach, while many reached realised the use of double angle identity would help, where it would have been more concise to use the double angle identities in the first place.

Common mistakes

• sin²2θ = (1 − cos 4θ)/2 (sign or angle errors common)
• ∫cos 4θ = ¼ sin 4θ
• k = 1/8
• by parts is long

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Almost all candidates recognised that this question required integration and were able to correctly identify that integration by parts was needed to access any of the marks in this question. The most common method was for candidates to apply by parts twice using the formula given in the formula book, although this did frequently result in candidates getting signs wrong for their terms. A number of candidates used the “DI method” which tended to be more successful and less complicated for those candidates who tried this approach. The word LATE or ILATE written on scripts which serves to assess the priority of the first function was also seen. This seemed to be helpful for candidates to identify which function should be differentiated. A small number of candidates used a substitution, usually u = -3x, with varying degrees of success. Although this resulted in fewer minus signs to manipulate, it did not simplify the processing. For most candidates the first step of by parts was generally correct, although coefficient and sign errors were common. A few integrated the 2 8x which led to incorrect forms. Of those candidates who got the correct form for the first step most got the second stage into the correct form though many made errors with signs. It was disappointing to see how few candidates simplified double negatives. Candidates who completed the integration to a correct form generally understood the substitution of the limits although again errors in signs occasionally cost candidates the final method mark as their working did not imply use of 0e 1 = . Candidates should be encouraged to fully show their working when applying limits so that it is clear that the lower limit is being subtracted. Most responses culminated with an attempt at an answer in its exact form but candidates that obtained the correct values for A and B were in the minority. Although not penalised if the intention was clear, some notation was poor with missing dx’s and spurious or missing integral signs.

Common mistakes

• integration by parts twice (differentiate 8x²)
• sign errors with e^(−3x) common, DI method helps
• limits 0 and 1 with e⁰ = 1 shown and lower limit subtracted
• exact answer 16/27 − (136/27)e^(−3)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Most candidates recognised the need and were able to solve the equation 2 4 3 23 x + = to determine the upper limit, 5 x = , although some candidates did not use the diagram to discard the 5 x = - and ended up integrating between the incorrect limits. It was also very common to see the limits 3 and 23 being used, with these candidates unclear about the area that is found through an integration strategy. The most successful strategy for finding the area under the curve was through finding ( ) 5 2 0 4 3 d x x + and then subtracting that from the area of the rectangle. However, there were plenty of candidates that failed to subtract the area under the curve 29 5 3 from the rectangle, and these candidates generally scored 3 marks out of 5. Those candidates that elected to find the difference between the curve function and the line y = 23 and then integrated their expression were more likely to have slips in their working: some failed to subtract correctly, resulting in ( ) 5 2 0 23 4 3 d x x - + ; while others subtracted the wrong way round, leading to a negative answer that some failed to recover to a positive answer. However, in both cases, once the integrand was determined, the majority of candidates using this strategy were able to achieve method mark for being able to perform integration of at least a term in the integrand. There were very few candidates that tried to integrate the function bounded by the y-axis and those that did were generally unsuccessful: often their rearrangement to make x the subject was poorly done and in these cases they often lacked the experience of Year 2 integration to be able to integrate function of the form ( ) n ax b + , where n is a fraction.

Common mistakes

• upper limit x = √5 from 4x² + 3 = 23 (reject −√5)
• limits 3 and 23 are y values, not x limits
• area = rectangle 23√5 minus ∫(4x² + 3)dx, or ∫(20 − 4x²)dx
• forgetting to subtract from the rectangle
• k = 40/3

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Generally, the candidates that made a full attempt at this question did very well by achieving full marks or at least 4 out of 6. However, many failed to use the information in the second bullet point fully (that the curve has a stationary point at ( ) 4,3 ). The majority of the candidates achieved the first mark by understanding that the stationary point at ( ) 4,3 meant that the first derivative must be equal to zero at 4 x = , arriving at 16 2 0 a b + + = . Again, the majority went on to perform the integration of the gradient function ( ) 4 d x a x b x + + to achieve ( ) 3 2 2 2 f 2 3 x x ax bx c = + + + with the majority able to determine that 5 c = -. A common error was to achieve 3 2 3 2 ax following integration, although those who had written the unsimplified 3 2 3 2 ax were still able to access the intermediate accuracy mark. Common problems were for candidates to misinterpret the information about the stationary point, either using ( ) f 4 3 = or ( ) f 4 0 = , or to miss out a constant of integration, which often resulted in concluding that b was 5 -. It was fairly common to see candidates substitute a rearranged 16 2 0 a b + + = into their integrated expression (usually for b) and this typically did not hinder further progress, although it did result in some sign errors that meant their final expression ( ) f x was incorrect.

Common mistakes

• stationary at (4, 3) means f′(4) = 0 (16 + 2a + b = 0) and f(4) = 3, not f′(4) = 3
• integrate a√x to (2/3)ax^(3/2)
• + c with c = −5 (b = −5 is wrong)
• a = −15, b = 14
• f(x) = 2x² − 10x^(3/2) + 14x − 5

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Pythagoras and so either sqrt (82 + 82) or even sqrt (82 + 32) was sometimes seen. In part (c), most integrated correctly and scored the first two marks. Very few got the differentiating and integrating the wrong way round this year. But after that, the majority just found the value of the integral at t = 4 (with or without subtracting the value at t = 0). Some realised that they had to split the integral into from 0 to 3 then 3 to 4 but several of these then added the two components instead of subtracting them. Those who drew a v-t graph were generally more successful when seeing the area below the axis seemed to help. There were relatively few correct answers to this part.

Common mistakes

• integrate to 15t − t³/3 − t² (must be seen: calculator-only answers score 0)
• P turns at t = 3, so distance = s(3) + (s(3) − s(4)) = 94/3 m
• most just evaluated at t = 4
• adding the two parts instead of subtracting the negative one
• a v-t sketch helps

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was an accessible question for virtually all candidates. The fractional index had already been given to avoid any misconceptions. However, for the first question on the exam paper, this was frequently done poorly compared to previous years. Candidates may have anticipated a tougher start to the paper and wanted to demonstrate one of the more advanced integration techniques. Candidates who spotted that you needed to expand the brackets usually went on to score full marks. The most common reason for losing the final answer mark was forgetting the constant of integration. Many candidates failed to notice the simpler method in the main mark scheme and instead attempted to use integration by parts, with either part taken as u. If a candidate chose to set 2 5 u x = -, they did not always expand the brackets in their integration formula to achieve a form which was ready to integrate where the indices had been combined correctly. Candidates who set 1 2 u x = were more successful, as there was less working to do to allow access to the marks. Candidates who chose to do integration by parts did not always combine the terms in their answer, losing the final mark. Many candidates would divide both their u and 'v by 3, leading to final answers that were 1 3 as large as they should have been. Integration by substitution was occasionally seen, but rarely successfully completed. Most attempts involved setting 2 5 u x = -, and this resulted in a tougher integral that required further work; candidates rarely progressed to a point which would have been worthy of a mark. Across the various methods, a common error seen was initially to factorise out the 1 3 , work through the solution without it, but then to forget to reintroduce it. This error meant that neither of the two accuracy marks were available.

Common mistakes

• multiply out first, then integrate term by term (by parts or substitution is much harder and rarely successful)
• integrating numerator and denominator separately scores nothing
• forgetting + c
• taking out the 1/3 and not putting it back

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (b), candidates generally understood that they were required to integrate the expression and the straightforward integration meant the first method mark was commonly achieved. Most realised that they then needed to find the value of their constant of integration using x = 3 and y = 10 to determine the point where the curve crossed the y-axis and achieved the second method mark. The majority of students were successful in achieving full marks in this part, with most stating their answer in coordinate form (0, -28) though some candidates left their answer as c = 28, not making the link with the y intercept explicit. Very few candidates made numerical slips in their processing to find the value of their constant of integration. Where mistakes were made, a common reason for not scoring the second method mark was due to substituting in a value of 0 instead of 10 for y in the integrated expression. Where the constant of integration had been omitted candidates went directly to substituting x = 0. Less commonly, some candidates differentiated rather than integrated and tried to find the roots of the second derivative rather than the y intercept.

Common mistakes

• integrate with + c, use (3, −10) to find c = −28, answer (0, −28) as a point
• substituting y = 0 instead of −10
• c = −28 alone not linked to the y-axis
• circular arguments avoiding (a)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (b), many candidates correctly multiplied their expansion by 6x and then integrated using the limits of 0.2 and 0.4 reaching 0.03304, gaining full marks. A common error, however, was not giving the final answer to the required accuracy of four significant figures, thus losing the final accuracy mark. Those candidates who had expanded part (a) to more than three terms also lost this final accuracy mark as it led to a different approximation. A more complicated and unnecessary method involving integration by parts was used by a small minority of candidates. Candidates also often tried inappropriate approaches like dividing 6x by their expansion followed by a logarithmic integral. A minority of candidates, despite the warning at the start of the question, clearly used calculator technology to obtain 0.032865, rather than the answer of 0.03304 required. Others made inappropriate attempts at the trapezium rule, thus gaining no credit as algebraic integration was required.

Common mistakes

• multiply the expansion by 6x, integrate term by term and use limits 0.2 and 0.4: 0.03304 (4 s.f.)
• calculator value 0.032865 or trapezium rule scores 0
• extra terms give a different estimate

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The vast majority scored M1 for either 2x4 or 5x. The most common error was dealing with coefficient of 1 2 x with a substantial number of candidates who were unable to process the fractional power. Many candidates rewrote the term as 1 2 3 2 x while others made mistakes with the coefficient and rewrote it as 1 2 6x or 1 2 6x -. A few attempted to integrate both the top and bottom of the fraction. Of those who had rewritten the term correctly as 1 2 3 2 x -some made mistakes when simplifying the coefficient of the integrated term. 1 2 3 4 x was seen several times. Sometimes the correct unsimplified expression was seen first but several candidates just wrote down their answer without showing their workings. Quite a few omitted + c following 3 correct terms and lost the final A1. Some candidates still lack confidence with notation such as an integral sign or dx appearing on the final line which resulted in the final mark being lost.

Common mistakes

• write 3/(2√x) as (3/2)x^(−½) first (fractional power errors common)
• coefficient −3x^½ (−¾ and 6 seen)
• 5 integrates to 5x
• forgetting + c
• integral sign or dx left on the final line loses the last mark

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was, on the whole, well-attempted and provided some differentiation between candidates. Most candidates made an attempt at part (b), and many were able to undertake some integration correctly. There were sometimes slips on the fractional powers and the coefficients. This question was effective at differentiating between those candidates who were able to clearly identify a strategy to determine the required area and many were successful in doing so. Most who were successful, employed the main mark scheme approach of determining the area under the curve between x = 0 and x = 4 followed by subtracting the area of the triangle however there were a number of different approaches seen, including finding the area under the curve from x = 0 to x = 2 together with the area between the curve and line between x = 2 and x = 4. Errors were perhaps a little more common in the second of these two approaches perhaps due sign errors when subtracting the two equations. Unfortunately, though, a number of candidates were not clear on the strategy required and so it was not uncommon to see attempts which stopped after calculating the area under the curve between x = 0 and x = 4 and thus a value of 76 9 was a fairly common incorrect final answer. Others using one of the alternative approaches simply found the integral of the difference between the two equations between x = 0 and x = 4 or between x = 0 and x = 2 and failed to go on the find the area under the curve separately. Most candidates had few issues identifying the intercept of the tangent with the x-axis but there was sometimes confusion in identifying the base and/or the height of the triangle when this was required to find the area. Some candidates lost the final A mark in this question by not giving the exact answer, giving it as a rounded decimal instead.

Common mistakes

• area under C from 0 to 4 (76/9) minus the triangle from x = 2 to 4 with height 13/3
• stopping at 76/9 is common
• confusing the y-coordinate 13/3 with the gradient 13/6
• exact answer 37/9

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Candidates who had attempted part (a) generally went on to attempt part (b). Most candidates provided a fully correct solution and gained both marks. Where errors were made, this was usually at the last mark. Common errors seen were; • incorrectly integrating the expression to ln x or ln 2x differentiating 2 • x rather than integrating it • leaving the answer as 2ln3 rather than ln9

Common mistakes

• [2 ln x] from 2.1 to 6.3 = 2 ln 3 = ln 9, so k = 9
• integrating to ln x², ln 2x or differentiating instead
• leaving 2 ln 3
• rounded decimal work loses the A mark

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The final question on the paper proved to be a challenge for many candidates and there were several blank responses seen. It is possible that this was due to time management. Some candidates attempted unsuccessfully to change to a Cartesian equation, suggesting that possibly parametric integration had not been covered. It was good to see students who were unsuccessful at part (a) still attempting part (b), and in some cases scoring full marks. Most candidates were able to integrate the first two terms correctly, but a significant proportion were unable to integrate the third term to get an expression in 3 sin .t The majority of correct solutions came from the use of the reverse chain rule for the final term. Those who attempted to use integration by substitution or by parts were usually unsuccessful. Candidates who had successfully completed the integration often went on to score full marks.

Common mistakes

• 8t − 2 sin 4t + 16 sin³t by reverse chain rule on 48 sin²t cos t (by parts/substitution rarely worked)
• limits 0 and π/4: 2π + 4√2 exact

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Most students knew that this question was to be solved by integration although a very small number thought that the question was about partial fractions, differentiation or a mixture of differentiation with integration. There were multiple approaches to solving this question including integrating a function with simplified indices, integrating by parts and integrating by substitution. The most common and the simplest method seen was multiplying out and simplifying indices. For the candidates who recognised that 4 sqrt x in the denominator should be rewritten as ¼ x -½ in the numerator there was a lot of success on this question. However, there were many candidates who brought the 4 to the top and left it as 4 rather than ¼ and earned a maximum of 3/6. Some wrote 4x½ in the numerator multiplying out to give incorrect powers and losing the first two method marks. Finding the correct limits and substituting them in a changed function was done by most leading to quite a few correct answers. Quite a few gave part or all of the final answer as a decimal, this was not surprising as most were using their calculators to do the final evaluation and calculator use did not leave the result of e.g. ( sqrt 2)5 as a surd. Some candidates recognised that the area needed to be positive, however the final answer was permissible even with a negative area. There were often errors in solutions constructed using integration by parts. There were mistakes made using the formula correctly, some bracketing errors, and some attempted parts the other way round getting harder expressions in the integral which mostly resulted for them a maximum of 1 mark (3rd M mark) obtained. There were a few incorrect substitutions selected for integration by substitution approaches, however candidates who chose u = sqrt x or u = 4 sqrt x were generally successful. A few candidates who set e.g. u = x – 2 in a substitution method scored few marks, if any, as such attempts did not progress the candidate towards a function they could integrate. In substitution method a few candidates made arithmetic errors in finding the new limits or did not attempt to change them at all.

Common mistakes

• write as ¼x^(3/2) − (3/2)x^(½) + 2x^(−½) (4√x becomes ¼x^(−½), not 4x^(½))
• limits 2 and 4 from the roots
• exact answer (12/5)√2 − 16/5, not a decimal
• by parts is much harder

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

occasionally seen. Many candidates integrated to achieve the first two marks in part (c), but not all used the displacement at t = 4 to find the constant of integration. Those that found a vector equation containing a constant of integration C, went on to find C and then substituted t = 1 to reach the correct answer although some responses contained numerical errors involving the negative signs. A common error was to use suvat formulae or r = r0 + vt. rather than integrate.

Common mistakes

• integrate v and use r = i − 4j at t = 4 to find the constant vector C
• suvat or r = r₀ + vt is invalid (acceleration not constant)
• sign slips
• answer −62i + 24j

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was very accessible; however, it was surprising that the majority were unable to score full marks on this question. This should have been a very straight forward question for those who had understood integration. However, a significant number thought that you could integrate the numerator and denominator of the fraction. Another significant number made errors when splitting the fraction into two separate terms. Errors often occurred in dealing with the 2 on the denominator and, as such, this restricted the maximum marks to 2 out of 4. Those who were able to separate the fraction correctly, were often able to score 3 out of 4, with typically the omission of the constant of integration being the reason for not scoring full marks.

Common mistakes

• divide each term by 2x³ first: (3/2)x − 2x⁻³
• integrating numerator and denominator separately scores nothing
• errors with the 2 in the denominator
• answer (3/4)x² + x⁻² + c
• forgetting + c

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was typically answered well by most candidates and was possibly the most successful question on the paper. Candidates typically were able to manipulate the integral to achieve 1 2 ...x - , although those who omitted the negative did not score any marks on this question. Some candidates made errors dealing with the coefficient, but they were typically able to score half of the available marks on the paper. Whilst most proceeded to the correct answer, some incorrectly square rooted to find k rather than squaring.

Common mistakes

• 6x^(−½) integrates to 12x^(½) (sign of the index)
• 36 − 12√k = 20 so √k = 4/3
• square, do not square root: k = 16/9

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was one of the most successful on the paper. It tested aspects such as completing the square, simple integration to find an area and finding a turning point which candidates seemed comfortable finding in nearly all cases. Part (c) was typically completed correctly. Candidates usually opted to integrate the curve and find the area between 0 x = and 2 x = and then subtracted this area from 40. Those candidates who tried to form a single integral to solve the problem were the ones were errors typically occurred. Some candidates equated the line 20 y = to the curve and rearranged to one side and integrated this expression. This method can work, although some responses did not appear that the candidates were that sure as to why they had achieved a negative value for the integral if they collected terms on the wrong side. In addition, errors in collecting terms resulted in an incorrect integrated expression and, ultimately, an incorrect final answer.

Common mistakes

• R = rectangle 2 × 20 − ∫₀² (−3x² + 12x + 8) dx = 40 − 32 = 8
• use the limits 0 and 2
• sign errors if integrating 20 − f(x) as a single integral

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(c) Some candidates knew that they needed to integrate v but a few tried to use suvat equations, losing all the marks and some did not substitute a numerical value for k. The last two marks were rarely scored as the change of direction was not appreciated.

Common mistakes

• integrate v (suvat is invalid: acceleration not constant)
• v changes sign at t = 4 so split the integral 0 to 4 and 4 to 6 and add magnitudes: 116/3 m
• substitute k = 24 first

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Most students were able to gain some marks in part (c) but a score of two out of four was common. Those who got this, generally integrated the curve, obtaining a correct area of 455/12, but then failed to go any further. Very few students used the area of a trapezium formula, instead preferring to integrate the (2x – 23) or to split the area into a rectangle and triangle. Of those who attempted the area of the trapezium a number struggled to get the correct dimensions. Fully correct solutions using a trapezium were rare. A few students translated the curve and line upwards correctly by 23 units and often scored full marks using this method. Those students who attempted the curve minus the line, often struggled with the constant 23, or incorrectly collected like terms (e.g. an x term with an 𝑥𝑥2 term).

Common mistakes

• integrate (curve − line) from 0 to 5 = 625/12
• 455/12 is only the curve part, the trapezium must be subtracted
• sign of l − C must end positive

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question discriminated well with weaker students generally able to pick up 2 marks and many able students were able to gain full marks. In part (b) it was good to see many students recognised that “hence” meant that the answer in part (a) must be used and proceeded with the correct method of expanding the cubic and dividing by u, proceeding correctly to the correct answer. If this was done correctly these students would usually go on to successfully apply the limits correctly and gain full marks. Unfortunately, a number of students attempted a method of integration by parts or some kind of product rule and therefore gained no marks here. Unfortunately a few did not take heed of the “hence” and proceeded to integrate the original expression scoring no marks in this part. Some weaker students split the integral into an incorrect sum of 2 fractions eg 2 𝑢𝑢+ (𝑢𝑢-1)3 𝑢𝑢 and then used integration by parts which they tended to give up on half way through. The method most successful was to take the 2 outside the integral, then integrate. Those that multiplied by 2 were more likely to make slips on the value of the coefficients for the 3rd and 4th term. Several students lost the final 2 marks as they used incorrect limits either 16 and 0 or 5 and 0.

Common mistakes

• expand (u − 1)³ and divide by u before integrating
• not by parts
• apply limits 5 and 1
• 104/3 − 2 ln 5

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

direction of travel. Part (c) tended to be well answered with a good number of candidates integrating and gaining 2 or 3 marks here. It was disappointing to see a few using constant acceleration formulae. Part (d) allowed those who had made little progress on the first 3

Common mistakes

• r = ∫v dt, both powers increase by 1
• use r = −j at t = 1 to get C = −2i
• suvat is invalid (acceleration not constant)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

7 The first part of this question was generally answered well with many candidates scoring full marks, but the second part proved to be more challenging for a disappointing number of them. In part (a), some candidates struggled with either rearranging the 5 2 x term to the form 1 2 5 2 x - or with getting the correct coefficient when integrating this to 1 2 5x . However, the majority of candidates were then confident in substituting both limits into their expression, subtracting and setting their result equal to 4, then rearranging correctly to get the required equation.

Common mistakes

• write 5/(2√x) as (5/2)x^(−½) and integrate to 5√x
• substitute both limits and set equal to 4

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was answered well by the majority of candidates. However, the question highlights the need for detail in proof and ‘show that’ questions which candidates are clearly still getting used to; it appeared to be even the more able candidates were neglecting to state facts that were needed. Candidates should be encouraged to make sure they provide sufficient reasoning when demonstrating a result or writing a proof. The final part requiring integration, whilst not penalised, brought out a lot of poor notation. Candidates were able to score most marks, however. Those who understood the nature of the curve and the importance of the factors and roots found earlier were able to choose the correct bounds and, usually, go on to achieve full marks. In part (c), those who attempted the question typically integrated correctly; however, it was very mixed when it came to evaluating. There were a number of poor graph sketches and some did not recognise which area they were actually finding and chose the wrong bounds for the integral. When evaluating the integral there were solutions lacking brackets which resulted in sign errors. Although the integral evaluates to be negative, the area is the magnitude of this. If the bounds were used in ascending order, then it was expected to see the negative answer before the positive solution. It was acceptable, following candidates carrying out algebraic integration visibly, that they used their calculator to substitute in their limits and could still score full marks. 10

Common mistakes

• limits −2 and 5 from the factors (region below the axis)
• brackets when substituting
• integral is −1715/3 so area is 571⅔ (do not just write the positive value)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

had consequences in part (b).The majority, for part (b), realised that the particle would change its direction of motion when v = 0 and solved this equation to give t = 3.5. They were then able to integrate v correctly and use their t value to obtain the answer. A few, however, used the t value that had been found in part(a) and

Common mistakes

• changes direction when v = 0, t = 3.5 (not the t from (a))
• integrate v (suvat scores M0)
• s = 931/24 = 38.8 m

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

As would be hoped, most candidates scored some marks on this question. Many found part the accuracy marks. There were many more errors seen when students attempted to values of 𝑥𝑥 to substitute in, any errors usually cost both accuracy marks. In comparison, multiply (Ax + B) by (x + 2) and compare the numerators, and when candidates selected (-3 - +12 = 9 being a common slip) and so secure the first accuracy mark. those who used long division were more likely to make an error only in the remainder In part (b), a pleasingly high number of candidates who had found values for A, B and C were able to successfully integrate to gain the first two marks, often aided by the follow through mark that was available. Candidates generally substituted the limits correctly and there was occasional sign errors. The most common being “-15ln 8 -(-15 ln 2) = -15ln 8 - also a relatively good use of the laws of logarithms to combine their terms, albeit with 15 ln 2”. It was not uncommon for a candidate to then incorrectly combine this to get " - 15ln 8 2”. Most success was seen where candidates changed the ln (8) into ln (23) and then 3 ln(2) before combining the two ln 2 terms.

Common mistakes

• integral of 15/(x + 2) is 15ln(x + 2)
• −15ln8 + 15ln2 = −30ln2 (write ln8 = 3ln2), not −15ln(8 − 2) or −15ln16
• answer 54 − 30ln2

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was very well answered, with many candidates gaining full marks. Candidates who used the factor theorem (solving f(-4) = 0 to find a = 6, as in the main scheme) were more successful than those who attempted long division, which invariably went wrong and in some cases resulted in a remainder containing ‘x’. Typical mistakes included: solving f(4) = 0 instead of f(-4) = 0 (usually forfeiting 2 marks), mistakes in integrating the ‘ax’ term and forgetting the constant of integration or adding it in at the end. It was noted that some candidates were able to score full marks in this question without explicitly finding a i.e. by considering (x + 4) (2x2 ….....-3) and deducing the x coefficient of the quadratic directly.

Common mistakes

• integrate f′(x) with + c
• c = −12 from the y intercept
• f(−4) = 0, not f(4) = 0, gives a = 6
• long division usually went wrong
• f(x) = 2x³ + 3x² − 23x − 12

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (i) involved variable acceleration but a significant minority of candidates tried to apply suvat equations and so achieved no credit. In part (i)(a), many did complete the required integration to find an expression for v and substitute t = 4; a fair number of correct answers were seen with the main errors either being numerical slips or neglecting to include the constant, 36i. In part (i)(b), candidates needed to realise that moving perpendicular to i implies the i component of v is zero; those who equated the j components to zero achieved no marks. Occasionally the i and j components were equated and some candidates did not know how to proceed at all. Some focussed on the displacement vector. Those who knew what to do generally obtained the correct answer. Part (ii) was generally found more accessible and many of the candidates who had used suvat in (i) now realised that they had to differentiate here to find the velocity. Most knew how to use the given speed to create an equation in t, although a few just tried to equate their vector to 5 and then either gave up or just dropped the i and j. A fair number achieved full marks for this part of the question.

Common mistakes

• variable acceleration so suvat scores nothing
• integrate each component and include the constant 36i
• v = 8i − (28/3)j

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (i) involved variable acceleration but a significant minority of candidates tried to apply suvat equations and so achieved no credit. In part (i)(a), many did complete the required integration to find an expression for v and substitute t = 4; a fair number of correct answers were seen with the main errors either being numerical slips or neglecting to include the constant, 36i. In part (i)(b), candidates needed to realise that moving perpendicular to i implies the i component of v is zero; those who equated the j components to zero achieved no marks. Occasionally the i and j components were equated and some candidates did not know how to proceed at all. Some focussed on the displacement vector. Those who knew what to do generally obtained the correct answer. Part (ii) was generally found more accessible and many of the candidates who had used suvat in (i) now realised that they had to differentiate here to find the velocity. Most knew how to use the given speed to create an equation in t, although a few just tried to equate their vector to 5 and then either gave up or just dropped the i and j. A fair number achieved full marks for this part of the question.

Common mistakes

• perpendicular to i means the i component of v is zero: t − 2t² + 36 = 0 gives t = 4.5
• equating the j component to zero (or i to j) scores nothing
• use velocity, not displacement

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a) the majority of candidates found this part straightforward, although the omission of +c remains a common flaw. Errors tended to lie with the first term; the minus sign was almost always present but the coefficient was sometimes left unsimplified or the index ‘increased’ to 4. In part (b) most candidates were able to substitute the correct limits into their result from part (a) in an attempt to find k from the resulting linear equation. Common errors in part (b) consisted of

Common mistakes

• write 4/x³ as 4x⁻³
• −2x⁻² not 4x⁻⁴/−4 unsimplified or index raised to 4
• + c needed

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

present but the coefficient was sometimes left unsimplified or the index ‘increased’ to 4. In part (b) most candidates were able to substitute the correct limits into their result from part (a) in an attempt to find k from the resulting linear equation. Common errors in part (b) consisted of a lack of bracketing when substituting in 0.5 equating their integrated expression equal to 0 algebraic slips in solving their linear equation in k.

Common mistakes

• substitute both limits 2 and 0.5 with brackets and subtract
• do not set the integral to 0
• solve the linear equation 7.5 + (15/8)k = 8 to k = 4/15

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

f 4 4 2 3 x x x x f 0 x 4 x 0 , 2 This was an open ended question, types of which are expected in the new specification. Almost all candidates were able to achieve some credit in this question. Poor notation (usually the omission of dx from the integral) caused many to forfeit the final mark. Some candidates seemed unaware that it was unacceptable to rely entirely on a calculator and that they needed to show detailed working in this “show that” question. Many candidates attempted to integrate (usually correctly) but with no other work of any value, some using 0 and k as limits and some even showing the substitution of k and 0 and subtracting. But without a value of k found via differentiation, only the second B mark was available to them. Other candidates thought they could find k by finding the roots of the original cubic despite it being clear from the diagram that the origin was the only root. Of those who realised differentiation was required, most arrived at x = 5/3, 4. Some of these candidates spotted from the graph that k = 4, but many wasted time, either by finding the corresponding y- coordinates (and using the x value corresponding to the lower of these) or using the second differential (where d2y/dx2 > 0) to determine which was k. A few candidates having found the second differential, equated this to zero and erroneously used the resulting value as k. Occasional mistakes with limits included using 5/3 instead of 0 as the bottom limit and subtracting the wrong way round (or even adding) after substituting. Some candidates forfeited the final two marks for failing to show intermediate working.

Common mistakes

• find k by dy/dx = 0: 6x² − 34x + 40 = 0 gives x = 5/3 or 4, choose k = 4 (the minimum, not a root of the cubic)
• integrate from 0 to 4 showing substituted values
• dx and correct notation needed for the given answer

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Most candidates knew they had to integrate for part (b) which they did mostly successfully and many went on to substitute t = 3 as required to find the distance travelled. There were a few numerical slips in carrying out the substitution but there were many correct solutions seen to this part of the question.

Common mistakes

• integrate v: s = 12t + 2t² − t³/3
• v > 0 for 0 ≤ t ≤ 3 so distance = s(3) = 45 m

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The first six marks in this question relating the definite integral with the area under the curve were straightforward. Only the best candidates were able to access the last four with many struggling to explain the significance of the root 5.442. Most students gained all 4 marks in part (a). Some unnecessarily complicated their proof by attempting integration by parts but in the main, all calculations were accurately performed.

Common mistakes

• expand to x³ − 2x² − 8x before integrating (by parts is unnecessary)
• limits −2 to 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The first six marks in this question relating the definite integral with the area under the curve were straightforward. Only the best candidates were able to access the last four with many struggling to explain the significance of the root 5.442. As with all proofs, part (b) proved to be very discriminating. Many candidates picked up the first mark for setting 4 3 2 1 2 20 4 4 3 3 b b b but there was much confusion as to the sign of 20 3 on the right hand side. Most students attempted to proceed by the method shown in the scheme. It should be noted that division by 2 4 4 b b was much easier and faster than dividing by ( 2) b twice. Fully correct solutions to (b) were rare with many picking up 2 out of the 4 marks.

Common mistakes

• integral from 0 to b equals −20/3 (sign confusion common)
• 3b⁴ − 8b³ − 48b² + 80 = 0
• dividing by b² + 4b + 4 is quicker than by (b + 2) twice

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Q05 proved challenging for many candidates and some candidates were unfamiliar with the notation that was used. Most candidates scored either all three marks or zero marks in this question. There were a substantial number of blank responses. It is noted that section 8.4 (guidance) of the Pure Specification states ‘Recognise 0 f( )d lim f( ) b b x a x a x x x x ’. It is also noted that a similar question had not appeared on the 9MA0 June 2018 papers, the SAMs, the specimen papers or the mock papers. Those candidates who recognised that 9 0 4 lim x x x x was another way of writing 9 4 d x x generally produced a correct solution to score all 3 marks. Only a few made errors in their integration of x or in the application of the limits 4 and 1. Most candidates, however, made no creditable progress. Some applied the trapezium rule with 1, h but did not link their trapezium rule attempt with a stated 9 4 d . x x Other candidates attempted to sum x using integer values from 4 x to 9, x resulting in 4 5 6 ... 9 ( 15.1597...). Some candidates mistakenly interpreted ‘ x ’ as a trigger to differentiate , x with a few of these candidates attempting to differentiate x from first principles. A few candidates applied the incorrect method 9 4 4 5 6 ... 9 d x x to give an answer of 2.49 to 3 significant figures.

Common mistakes

• recognise the limit of the sum as ∫₄⁹ √x dx = 38/3
• unfamiliar notation – many blank or used the trapezium rule or summed √4 + … + √9
• δx is not a cue to differentiate

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Q14 discriminated well between the higher ability candidates. Lower and medium ability candidates, however, found this question demanding with most of these candidates scoring no more than two marks. There were a few candidates who made no attempt at this question, either because the question was too demanding, or because their time had run out. Most higher ability candidates made good progress in Q14(a) and Q14(c) but they often struggled to make progress in Q14(b). 4 u h In Q14(a), most candidates who attempted this part differentiated to give a 1 d 1 u with a few finding a correct d h u , h 2(4 ) u correct and some 2 d d 2 h 1 d 1 u . h differentiating incorrectly to give At this stage, some candidates failed to 2 d 2 h make further creditable progress. Many candidates struggled to apply the substitution (4 ) d , k u completely to give an integral of the form with some making sign errors u u in the numerator of this integral. Those who progressed this far generally went on to A divide each term in the numerator by u to give an integral of the form d , B u u 8 with many of them obtaining a correct Most integrated correctly and 2 d .u u 4 u h applied to obtain a correct Some 8ln 4 2(4 ) . h h c candidates could not deal with the transition from 8 c to k to achieve the given and a large number of these candidates incorrectly stated 8ln 4 2 , h h k 2 8 d u k u Those few candidates who applied integration by parts on rarely 8. u progressed to give a correct solution to this part. In Q14(b), some candidates correctly set d h t 0 but only a few deduced the equation d h 4 0. h Some candidates used to state a correct allowable answer such as 16 h h 4 0 16, h 16. or A common mistake was to solve to give 0 16 h h 2. Material for Q14(c) was sometimes seen in Q14(b). Most candidates who attempted 0.25 t Q14(c) separated the variables correctly. Some candidates integrated incorrectly 1 5 0.25 0.75 1.25 , t t 4 t to give or while others integrated without reference to (4 ) h to find their constant of 0, 1 t h the result from Q14(a). Many candidates applied integration. Common errors at this stage included not applying a constant of integration to find their constant of integration. Most 0, 0 t h 1, 1 t h and using either or candidates who progressed this far applied a complete process of substituting their h constant of integration and into their integrated equation and solving this 12 equation to find their value for t. In many cases, sign errors, bracketing errors, manipulation errors or an incorrect method of solving a correct 1.25 221.2795202..., t t prevented candidates from obtaining the correct answer years. 75.2

Common mistakes

• du/dh = −½h^(−½) so dh = −2(4 − u)du
• split −2(4 − u)/u into −8/u + 2
• −8ln u + 2u + c, then replace 8 + c by k
• sign errors in the numerator

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Q14 discriminated well between the higher ability candidates. Lower and medium ability candidates, however, found this question demanding with most of these candidates scoring no more than two marks. There were a few candidates who made no attempt at this question, either because the question was too demanding, or because their time had run out. Most higher ability candidates made good progress in Q14(a) and Q14(c) but they often struggled to make progress in Q14(b). 4 u h In Q14(a), most candidates who attempted this part differentiated to give a 1 d 1 u with a few finding a correct d h u , h 2(4 ) u correct and some 2 d d 2 h 1 d 1 u . h differentiating incorrectly to give At this stage, some candidates failed to 2 d 2 h make further creditable progress. Many candidates struggled to apply the substitution (4 ) d , k u completely to give an integral of the form with some making sign errors u u in the numerator of this integral. Those who progressed this far generally went on to A divide each term in the numerator by u to give an integral of the form d , B u u 8 with many of them obtaining a correct Most integrated correctly and 2 d .u u 4 u h applied to obtain a correct Some 8ln 4 2(4 ) . h h c candidates could not deal with the transition from 8 c to k to achieve the given and a large number of these candidates incorrectly stated 8ln 4 2 , h h k 2 8 d u k u Those few candidates who applied integration by parts on rarely 8. u progressed to give a correct solution to this part. In Q14(b), some candidates correctly set d h t 0 but only a few deduced the equation d h 4 0. h Some candidates used to state a correct allowable answer such as 16 h h 4 0 16, h 16. or A common mistake was to solve to give 0 16 h h 2. Material for Q14(c) was sometimes seen in Q14(b). Most candidates who attempted 0.25 t Q14(c) separated the variables correctly. Some candidates integrated incorrectly 1 5 0.25 0.75 1.25 , t t 4 t to give or while others integrated without reference to (4 ) h to find their constant of 0, 1 t h the result from Q14(a). Many candidates applied integration. Common errors at this stage included not applying a constant of integration to find their constant of integration. Most 0, 0 t h 1, 1 t h and using either or candidates who progressed this far applied a complete process of substituting their h constant of integration and into their integrated equation and solving this 12 equation to find their value for t. In many cases, sign errors, bracketing errors, manipulation errors or an incorrect method of solving a correct 1.25 221.2795202..., t t prevented candidates from obtaining the correct answer years. 75.2

Common mistakes

• separate variables
• ∫t^0.25 dt = 0.8t^1.25
• use the result of (a)
• constant from t = 0, h = 1 (not h = 0)
• t^1.25 = 221.28 so t = 75.2 years

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

vectors throughout and some in i and j components; both are equally acceptable. In part (b) integration was required in order to find the position vector. Mostly the indices were dealt with correctly although errors in the coefficients were more common. Some stopped there without either using the initial conditions or substituting t = 4. Many failed to achieve the final mark as a result of numerical/sign errors. A minority attempted to use suvat formulae to solve the problem despite having used calculus

Common mistakes

• r = ∫v dt then add r₀ = −20i + 20j
• coefficient slips (3t² and 2t^(5/2))
• suvat is invalid as acceleration is not constant
• answer 28i − 44j

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

A good starting question. Almost all students attempted this question with some success. Students who scored no marks often differentiated instead of integrating. A significant number of students scored full marks. For students who integrated, the majority scored at least one mark for increasing one of the powers by one. The most common error was to forget the '+c'. Other errors included: (1) Incorrect simplification of the coefficients with the first term having a coefficient of 8/3 or the second term having a coefficient of 9 (2) Incorrect index work especially on term two (3) Incorrect sign on term two.

Common mistakes

• forgetting + c is the most common error
• differentiating instead of integrating
• coefficient slips (8/3 on the first term, 9 on the second)
• index errors on 6√x (x^(3/2) ÷ 3/2 gives −4x^(3/2))
• answer must be simplified

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Some students were clearly running out of time when they reached this question, as there were many unfinished answers and some blanks. However, there were a great many accurate, concise and well written solutions. Few attempts at the line minus the curve approach were seen. This was an unstructured question in which the students were required to find their own route through the problem. There were two discrete elements to the solution, one via differentiation and one via integration that were required to be tied together to show the given answer. A good number completed the differentiation part correctly. There were slips in the differentiation of 2 x , but generally the method of finding the equation of a normal was well known. An equal number completed the integration part finding the area under the curve between 2 and 4. Similar slips were made on the integration of 2 x . Tying up these two techniques into finding the area of R was less well done. Standard errors seen here involved: (1) attempting the area of a triangle between 2 and 16 (2) integrating under the curve from 2 to 16 (3) finding the equation of the tangent to the curve at P.

Common mistakes

• dy/dx = −64/x³ + 3 gives gradient 2 at P, so the normal has gradient −½ and meets the x-axis at x = 16
• area under the curve from 2 to 4 = 10
• triangle ½ × 12 × 6 = 36
• total 46
• slips differentiating and integrating 32/x²
• many ran out of time

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Both parts of this question were answered well, allowing students to demonstrate their confidence in integrating fractional expressions in x and working with unknown constants. The majority of students failed to make any statement regarding the integral in (a) being independent of k, or in (b) inversely proportional to k. Whilst not a requirement for the final part (b) as well as part (a). Additionally, despite the successful implementation of limits, some careless and disappointing algebra was seen in simplifying their final expressions in an attempt to answer the question. A more unusual but not uncommon error was related to the fact that ‘k’ occurred both in the integral and in the limits. Some students integrated with respect to k whilst others substituted the limits for the term in k rather than the term in x. Other unusual attempts were seen by students attempting to use partial fractions within part (b).

Common mistakes

• ∫2/(3x − k) dx = (2/3)ln(3x − k)
• substitute the limits into x, not into k (some integrated w.r.t. k)
• (2/3)ln(8k/2k) = (2/3)ln4 and state independent of k

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Both parts of this question were answered well, allowing students to demonstrate their confidence in integrating fractional expressions in x and working with unknown constants. independent of k, or in (b) inversely proportional to k. Whilst not a requirement for the final A mark in either part, the logical completion of the question would have been to make these conclusions and show an understanding of these mathematical terms. Common method errors in this question included integrating to obtain a natural logarithm in part (b) as well as part (a). Additionally, despite the successful implementation of limits, some careless and disappointing algebra was seen in simplifying their final expressions in an attempt to answer the question. A more unusual but not uncommon error was related to the fact that ‘k’ occurred both in the integral and in the limits. Some students integrated with respect to k whilst others substituted the limits for the term in k rather than the term in x. Other unusual attempts were seen by students attempting to use partial fractions within part (b).

Common mistakes

• integrate to −1/(2x − k), not a logarithm
• limits 2k and k into x
• 2/(3k) is of the form A/k so inversely proportional (do not isw)
• partial fractions not needed

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was another modelling question, this time modelling the height of a roller coaster above the ground by a differential equation. In part (a), the majority of students were able to separate the variables of H and t and thus integrate both sides. Integrating cos0.25t was done well with most dividing by 0.25 or multiplying by 4. Almost all students were then able to ln 0.1sin 0.25 H t proceed with ease to . The last two marks proved harder to score. Quite a sizeable proportion of students failed to include the constant of integration in and so were unable to apply the boundary condition ln 0.1sin 0.25 H t c in 0, 5 t H an attempt to find c. The final mark was the hardest to gain. The answer was given and many students merely copied down the result on the examination paper. There had to be clear evidence that the constant they found was processed correctly and that their log work was sound. The requirement of at least one correct line of working between their working and the given equation, with no incorrect working, was fair to those who showed a clear argument. Part (b) was found by almost all students, even those who could not attempt part (a). In this case the exact answer 5e 0.1 was accepted as well as 5.53 metres. Fewer students scored both t marks in (c), but many were aware that it would occur when sin(0.25 ) The most 1.

Common mistakes

• separate: ∫1/H dH = ∫cos(0.25t)/40 dt
• ln H = 0.1 sin(0.25t) + c – constant often omitted
• t = 0, H = 5 gives c = ln5
• at least one correct log line before the given answer

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question discriminated well between the average and more able students. There waere a significant minority of students, however, that made no creditable progress. In part (a), many students struggled to translate the description of the model into a mathematical statement. Some incorrect statements, such as 2 d 1 , d r t r 2 d d r r t or d 1 , d r t r did not contain a constant of proportionality; while other incorrect statements 2 d , d r kr t 2 d d r r t k or d d r k t r contained a constant of proportionality. Many students used a correct method to separate the variables and most proceeded to integrate their 2 d d r r k t to give an equation linking r and t which contained a constant of integration. At this stage, some students incorrectly integrated 2r to give 2 .r The majority of students applied the boundary conditions 0, 5 t r and 4, 3 t r (or 240, 3 t r ) to find both constants and proceeded to give an equation linking r with t. Some students who worked in minutes obtained correct answers of either 3 250 49 , 2 t r 3 49 125 2 r t or 3 250 2 . 49 r t A few students who worked in seconds obtained correct answers of either 3 15000 49 , 120 t r 3 49 125 120 r t or 3 15000 120 . 49 r t Many students lost the final mark in this part by not including sufficient detail in their definition of their variables. Some students defined r as the radius of the mint, in mm, but it was rare to see the time, t, defined from a correct starting point (i.e. from when the mint was placed in the mouth) and with the correct unit of time used.

Common mistakes

• dr/dt = −k/r² (constant of proportionality needed)
• separate to ∫r² dr = −∫k dt
• two conditions (t = 0, r = 5
• t = 4, r = 3) for k and c
• r³/3 = −49t/6 + 125/3
• define r (mm) and t (minutes from placing in the mouth)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The vast majority of students recognised the requirement to integrate to find the position vector and subsequently to substitute the given values of t. Occasionally, these were substituted straight into the velocity vector but such instances were rare and much correct working was seen. Most proceeded to find the magnitude of the vector joining the two points to give the exact distance as a surd as required; only a very small minority went straight to a rounded decimal value.

Common mistakes

• integrate v to get r = 2t^½ i − 2t² j (+ C)
• positions at t = 1 and t = 4, not substituting into v
• exact distance √904 = 2√226, not a rounded decimal