Integration by parts and by substitution Pure

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 (a), many responses did not successfully manage to fully substitute for all elements of the integral. The main issue was the correct substitution for dx. Responses commonly successfully substituted u in everything except the ‘du’, with some omitting it. Additionally, many did not achieve marks because they did not successfully substitute and achieve 2cosu in the denominator with others attempting to substitute d d 2cos u x u = . Many responses were not presented in a logical structure and did not gain marks because they did not bring together the elements of the substitution into a single expression which showed the progress of the substitution, the manipulation and simplification to the required result. This occasionally resulted in just amending to the result because they knew what the required form was but did not know how to get there. A large proportion did not spot the need to use the identity 2 2 sin cos 1 x x + = and often just square rooted all the values in the bracket achieving (2 2sin )x - . Some learners misapplied identities, using for example cos2x . Those who did use the identity correctly tended to go on to also gain the accuracy mark, but a few had 2 1 4sin u and ended with 2 4cosec u . Those who did compete the full substitution often did it in stages leaving the substitution for dx until the final stage of their proof which was condoned but should not be encouraged. A good proportion also changed the limits into angles in degrees and so did not gain the mark for changing the limits. The limits were often given as 30 degrees and 60 degrees, costing the final mark.

Common mistakes

• substitute every part including dx = 2cos u du
• √(4 − 4sin²u) = 2cos u using sin² + cos² = 1 (not √4 − √(4sin²u))
• simplify to ∫ ¼cosec²u du
• limits in radians π/6 and π/3, not 30° and 60°

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

Examiner's comment

This question, which involved setting up and solving a differential equation using separation of variables, proved to discriminate across the cohort, with a wide range in quality of responses seen. The restart opportunities provided were well used, allowing many learners to access later parts of the question even if they had struggled in earlier parts. In part (b), some learners recognised the need to separate the variables and were able to do this successfully. A significant number placed the 20 incorrectly and others struggled to separate correctly into the reciprocal form which limited access to further marks. Those who separated the variables were generally able to score the method mark for the integration at the beginning of the process. Common mistakes were omitting the 1 6 - , not including the minus sign or multiplying by –6. Most learners included a constant of integration, and most learners recognised the need to use the given initial conditions to find it - those who made it this far generally made the correct substitution of V = 0.25 and t = 0 and obtained a value for the constant. The final mark proved challenging for many, with errors in rearranging their equation to the given form being particularly common. Many struggled with the logarithm work, not applying sum/product or difference/quotient rules correctly, often taking e to the power of individual terms rather than the whole expression on each side of the equation.

Common mistakes

• separate correctly: ∫ 1/(9 − 6V) dV = ∫ 1/20 dt
• −⅙ ln(9 − 6V) (missing −⅙ or sign)
• use t = 0, V = 0.25 for the constant
• exponentiate the whole of each side, not term by term
• V = 3/2 − (5/4)e^(−0.3t)

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. In part (b), learners generally made a good start and seemed to be confident with the method of substitution. Most were able to score the first two B marks and the first M mark for an attempt at a complete substitution. For those that did not achieve the first M mark, it was almost always due to not dealing with the dx. However, the simplification to the required form using the trigonometric identities proved more challenging for many. Although there were many very neat and succinct solutions using various effective approaches, some responses expanded brackets unnecessarily, applied multiple trigonometric identities or worked with cumbersome fractions until they achieved something close to the required form. A small but noticeable number used degrees.

Common mistakes

• limits −π/4 and π/4 (radians, not degrees)
• dx = sec²θ dθ (dx must be replaced)
• 1 + tan²θ = sec²θ then cos²θ − sin²θ = cos 2θ
• avoid unnecessary expansion

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

Examiner's comment

were available on both parts regardless of success in part (a) but many candidates who 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 attempted to do so, with varying levels of success. Those who differentiated successfully generally went on to substitute into the expression correctly and gained at least the first 2 marks. The flexibility in the first method mark made this accessible to d = most candidates who had tried at differentiation. appeared several d x 2 sin cos a times and benefited from the flexibility of the first method mark. The most common . In many cases the notation was 2 and cos a incorrect derivatives were 2 cos a particularly poor, although d x a was very generously condoned provided the right-hand d side was correct. Candidates struggled with the rearrangement of the integrand into the form required to successfully use the double angle formula. Many were not able to work with or simplify the square root terms successfully, and even those that did sometimes failed to recognise the need to use the double angle formula for sin 2 . A minority of candidates progressed correctly to the final answer, while some of those that achieved the final expression still lost the final mark for failing to address the limits, or for missing the d or the integral sign. For those who did not achieve the first method mark this was commonly due to careless copying or premature simplification such as - - 2 2 instead of . sin a a sin a a Those who recognised the requirement for using double angle formulae to integrate a squared trigonometric function were generally successful in part (b) of this question.

Common mistakes

• dx/dθ = 2a sin θ cos θ
• √(a − a sin²θ) = √a cos θ
• sin²θ cos²θ = ¼ sin²2θ (rearranging the integrand is where most fail)
• change limits to 0 and π/2

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

In part (c), the commonest methods of integration adopted were substitution, using either u = x + 3 or u = (x + 3)2 , integration by parts, integration after applying partial fractions and integration using the reverse chain rule after writing the integrand in appropriate form. Candidates had varying success with all of these approaches, gaining a variety of marks. Integration by parts seemed to lead to most arithmetical mistakes. Errors resulted from using parts ‘the wrong way round’ and others integrated the ( ) 2 3 x - + term incorrectly. When integrating by substitution some students incorrectly thought that 2 18 u - would integrate to a natural logarithm. There was some confusion with the limits; some candidates did not change the limits to “u” values and other who did evaluate new limits for their substitution switched back to an expression in x before substituting their changed limits. When using partial fractions some students struggled with the initial format for the partial fractions and it was not uncommon to see partial fractions of the form 3 3 A B x x + + + . There was a number of candidates who either did not attempt this part of the question at all or who lacked a correct strategy. There were responses where students attempted to integrate numerator and denominator separately, e.g. 6x to 3x2 and ( ) 2 1 3 x + to a variation of ( ) 2 ln 3 x + , and proceeded to multiply both results. A number of candidates were awarded just the first mark as their overall problem solving strategy was correct despite the fact they were unable to fully implement it. There were also many candidates who could carry out the work in this part completely successfully.

Common mistakes

• substitution u = 3 + x (change limits to 3.2 and 3.4), parts, or partial fractions 6/(3 + x) − 18/(3 + x)²
• result 6ln(17/16) − 45/136
• 18/u² does not integrate to a log
• switching back to x but using u limits

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

Examiner's comment

This question also proved to be quite challenging. (c). Some started the wrong way around (using u = e – x) and soon realised they were not heading for the given answer and there was some muddling over the signs when applying the limits. Part (d) was not answered so well with only a minority of students realising to apply the answer to part (c) though some used n = 40 rather than n = 4. The first part of (e) was often answered well and some realised that the answer to part (c) could be used to answer (e)(ii) but some tried using n = 10 and 30 and others stopped after find an estimate for the number of patients as 53.1 and failed to complete the question to find the required probability. In part (f) only a small number of candidates realised that

Common mistakes

• integration by parts with u = x, dv/dx = e^(−x) (starting with u = e^(−x) fails)
• sign errors when applying the limits

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

Examiner's comment

This was a very straight forward question on integration by parts with 40% of the candidates scoring full marks. On the reverse side there were also a sizeable number, some 30%, who did not know how to begin and failed to score any marks. Those candidates who knew the method nearly always scored 3, 4 or 5 marks. Reasons for these candidates dropping 1 or 2 marks nearly always centred around • a failure to integrate the term 4 1 4 x x x correctly, many times to 4 12 x • failing to correctly apply the limits, usually achieving 8 7 1 e 16 16 - • achieving a correct answer of 8 7 1 e 16 16 + then inexplicably ''simplifying'' this to 8 7e 1 + Low scoring candidates who attempted the question, usually resorted to integrating each term of the integrand or else having made a good start, disappointingly integrating the 4 1 4 x x x term to 5 ln 20 x x x

Common mistakes

• by parts with u = ln x: (x⁴/4) ln x − x⁴/16
• ln e² = 2
• answer (7/16)e⁸ + 1/16
• x⁴/4 × 1/x integrated wrongly
• "simplifying" to 7e⁸ + 1

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

The modal mark here was 3, most likely due to the accessibility of part (a). Marks in part (b) were only really awarded to the highest scoring candidates on this exam. In part (b) many candidates failed to recognise that integration by parts was needed to make progress. Several students, who made progress and correctly applied ‘integration by parts’ twice, unfortunately lost the final accuracy mark due to sign or coefficient errors that had crept in. Most who integrated appropriately applied the correct limits and applied ln(4) = 2ln(2). Errors were many and varied here and included • Candidates who believed that ( ) ( ) 2 3 1 d 3 ln x x ln x = ∫ • Candidates who believed that ( ) 2 2 ln x ln x = Candidates who used the substitution ln u x = seemed to more easily recognise 2e d u u u ∫ as integration by parts twice.

Common mistakes

• integration by parts twice (u = (ln x)², then ∫ln x dx = x ln x − x)
• 14(ln 2)² − 12 ln 2 + 4 using ln 4 = 2 ln 2
• ∫(ln x)² is not (ln x)³/3 and (ln x)² is not 2 ln x
• sign and coefficient errors

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. Part (a) was a relatively straightforward substitution question and as such many students made a very positive attempt at this part. There were a mixture of methods used to answer part (a) with most students correctly identifying du/dx or dx/du. Those that found the second of these found the substitution very straightforward and generally went on to achieve full marks. Students who found du/dx instead sometimes struggled with the indices and how to deal with the negative power when dividing and hence made mistakes here. There were a small number of students who simply failed to make any attempt at this question, and generally these students gave up on the question entirely. A few students lost the accuracy mark for not stating the limits or omitting du.

Common mistakes

• dx/du = 2(u − 1) is easier than du/dx
• change limits to 1 and 5
• du must be seen

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

This question proved to be more discriminating than expected with many failing to notice the partial fractions in part (b), thereby losing 6 marks. Part (a) was well answered by many candidates, particularly those who opted to calculate dx/du rather than du/dx, as the latter proved harder to substitute correctly into the expression in terms of u. Some students lost the final accuracy mark for not showing a correct intermediate line with integral signs and some did not finish their working with an expression showing the given integral. A few candidates made arithmetic slips in finding the limits or did not attempt to change them at all.

Common mistakes

• dx = 2u du
• change the limits to p = 2, q = 3
• show an intermediate line with integral signs before the given answer

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

Examiner's comment

Many candidates left the entire question or parts of this question completely blank and clearly found it difficult to access the content being assessed here, perhaps due to a lack of familiarity with integration using parametric equations. Many tried to establish the Cartesian equation to answer various parts. That said, there were many candidates who provided clear and concise solutions to each part. In part (a)(ii), many knew that the integral was a multiple of cos3 t but had trouble getting to the printed answer of 20 with correct steps. As the answer was given, it was a requirement to see the limits calculated separately as 0 – ( -20 ) = 20. Integration by parts was a common approach for unsuccessful candidates, while others attempted to use a substitution, often not using the necessary u = cosx. Those candidates with an understanding of the integration of f’(x)[f(x)]n were generally very successful here. Some lost the penultimate accuracy mark usually due to a sign error.

Common mistakes

• reverse chain rule (or u = cos t) gives −20cos³t
• limits must be shown as 0 − (−20) = 20
• by parts is the wrong approach

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

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 an open ended question on integration. The three most common methods seen are shown the mark scheme, two methods using substitution and one using integration by parts. Integration by parts was popular and well understood by the majority of students. Most knew that 3 2 2 2d 2 3 x x x and were able to integrate to an expression of the form 3 5 2 2 2 2 Ax x C x Slips in the values of the constants A and C were rare. However, as with earlier questions, there was a lack of clarity in moving from 2 3 5 2 2 0 4 8 2 2 3 15 x x x to the given answer of 32 2 2 15 Substitution was also well understood with 2 u x more common than 2. u x Errors here included the omission of finding dxin terms of d ,u poor expansion of expressions such as 2( 2) u u within the integral, and, as with parts, a lack of acceptable working shown when proceeding to the given answer.

Common mistakes

• choose u = x + 2 (or u = √(x + 2)) or integration by parts
• change dx to du and change the limits
• expansion slips such as 2(u − 2)√u
• at least one correct line before the given answer

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

Examiner's comment

This question, despite being unstructured, proved accessible, with many students scoring at least 7 out of the 10 marks available. Those students who annotated the diagram were generally more successful in formulating a correct strategy to find the area of the region R. Most students differentiated ln y x x by using the product rule to give the correct d 1 ln . d y x x A few students found the correct d ln d y x x x x and simplified this incorrectly to give d ln . d y x x Many students used a complete method to find the x-value where the normal to C at (e, e) cuts through the x-axis, with the majority finding the correct 3e. x Most students integrated ln y x x correctly by using integration by parts to give 2 2 1 1 ln . 2 4 x x x Many students applied the correct limits of e and 1 to obtain a correct 2 1 1 e 4 4 with some making a bracketing error when applying the limit of 1 to obtain the incorrect 2 1 1 e . 4 4 The majority of students attempted to find the area of the triangle by using the formula 1 (base)(height). 2 While some students applied the formula correctly to give 1 (3e e)(e), 2 there was a considerable number that applied a base length of 3e by writing 1 (3e)(e). 2 A few students used the equation of the normal to find the area of the triangle by applying the correct 3e e 1 3 e d , 2 2 x x but not many evaluated this to arrive at the correct answer of 2e . Several incorrect strategies were seen such as trying to evaluate e 0 ln d , x x x 3e 1 ln d x x x or even 3e 1 1 3 ln e d . 2 2 x x x x Only a minority of students applied the calculation 2 2 1 1 e e 4 4 to achieve the correct 2 5 1 e . 4 4

Common mistakes

• product rule dy/dx = 1 + ln x
• normal gradient −½ meets x-axis at 3e
• ∫ x ln x dx by parts = ½x² ln x − ¼x² with limits 1 and e (bracketing error at x = 1)
• triangle base 3e − e not 3e
• area = 5e²/4 + 1/4