P8.6 Integration with partial fractions

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

Examiner's comment

This question tested the candidates’ skill in various areas, i.e. finding partial fractions, integration, rules of logarithms, manipulating equations to eliminate logarithms using exponentials, indices, and commentating on the expected behaviour of a variable as another tends to infinity. This was a challenging question which in a number of instances, candidates struggled with after part (a). Those however, who were well prepared, were able to score full marks often with relatively concise solutions. There were occasional sign slips in the coefficients, and some candidates failed to write down their final answer, or use it in part (b), losing the second mark here. In part (b) a number of correct forms of integration were possible. Many candidates appreciated that an attempt at separation of variables and integration was required, and many reached correct equations such as ( ) 1 1 1 ln ln 25 25 25 10 V V t c - - = + or ( ) 1 1 1 ln 25 ln 25 25 25 25 10 V V t c - - = + . There were however, many slips with coefficients, and the omission of the minus sign in front of the ln(25 – V) or ln25(25 – V) term was common. Additionally, some omitted a constant of integration entirely, which meant that no further progress could be made. A surprising number of candidates did not see any link with part (a) and instead, integrated a quadratic in V, or did not integrate at all. Most candidates who attempted integration on both sides and included a constant of integration were able to score 2 method marks by using t = 0 and V = 20 to find the value of their constant and then substituting V = 24 to find the time taken. Most who got this far converted their final answer into minutes as required by the question although some candidates left the answer as 0.7167 which lost the final accuracy mark in this part. A few candidates used the alternative method of applying limits of 20 and 24 to their integral with respect to V and 0 and T for their integral with respect to t.

Common mistakes

• separate variables and use the partial fractions from (a)
• −ln(25 − V) sign often lost
• include + c and find it from t = 0, V = 20
• t = 0.7167 hours must be converted: 43 minutes

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

Examiner's comment

Very few candidates managed full marks in part (b), although the majority did gain the first method mark for integrating to obtain …ln(x + 4) or …ln(x 2). Those who did not recognise the logarithmic form for the integration were unable to score any marks in this part. Some achieved the correct form after first using a formal substitution. Of those who integrated their partial fractions successfully, the vast majority were far from strict in their use of the modulus symbol and many lost marks due to lack of appreciation of its importance. Use of modulus notation for integrating reciprocal functions should be picked up by centres as a teaching point with future cohorts. Students who had written the incorrect term ( ) ( ) 3 ln 5 k - - - or even the correct term ( ) 3 ln 5 k - - - after applying the limit 3 x = -, went on to indicate that they believed this term was zero or could just be ignored. Another example of poor practice was missing brackets around the coefficients e.g. ( ) ( ) 2 3ln 4 3ln 2 k x k x + + + - - being surprisingly common. The “invisible” brackets were sometimes recovered but in many cases were not. There were also examples where the brackets around the (x + 4) and the (x 2) were also missing. Those with a systematic well-laid out approach for the substitution of limits in both terms and identifying clear subtraction of the lower limit, went on to score 3 or 4 marks in part (b) and commonly full marks for the question. Many candidates however were hampered by their own layout making the correct interpretation of ln ( 5) as ln |5| even more challenging. The perhaps unfamiliar, exact form needed for k also proved a challenge for even the best candidates. There were some elaborate attempts to achieve an expression for k, some using exponentials, but many gave up along the way. There were, however some very elegant and correct solutions to this question and some different forms of the exact equivalent answer were seen, for example 5 21log e 6 - but most correct answers were given as 21 6. ln5 -

Common mistakes

• (2k + 3)ln(modulus of x + 4) + (k − 3)ln(modulus of x − 2)
• use modulus signs (ln of a negative)
• brackets round the coefficients
• (k + 6)ln5 = 21 so k = 21/ln5 − 6
• terms in ln1 vanish but ln(5) from the second term does not

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

Examiner's comment

The most challenging aspect of part (b) for the candidates was to separate the variables correctly. A large proportion of candidates did not make the link to part (a) and often divided both sides by 3V as a starting point, i.e. ∫ 1 3𝑉𝑉. 𝑑𝑑𝑑𝑑= ∫ 1 (2𝑡𝑡-1)(𝑡𝑡+1) . 𝑑𝑑𝑑𝑑 Where candidates did this they often would incorrectly replace the right hand side with the partial fraction they had found in part (a). For the candidates that got as far as this, the integration of the partial fractions they put in place was generally completed well with the key …ln(2t - 1) - …ln(t + 1) being found more often than not although a common error was leaving a multiplier of 2 when integrating the first fraction to give 2ln(2𝑡𝑡-1). Regardless of the integration a candidate had achieved, there were a large number of candidates that did not consider the constant of integration and tried to jump straight to the answer at this stage. Generally, if a candidate remembered their integration constant, they were more often than not successful in using the given limits to obtain a value which they could use to find an equation for V. A few attempts at solving the first order linear differential equation using an Integrating Factor were seen. Those employing this method were usually reasonably successful.

Common mistakes

• separate as ∫1/V dV = ∫3/((2t − 1)(t + 1)) dt and use (a) (not dividing by 3V then misusing (a))
• ∫2/(2t − 1) dt = ln(2t − 1), not 2ln
• include + c and find it from t = 2, V = 3 before combining logs

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

Examiner's comment

This question proved challenging for the majority of candidates, with many not answering it at all. Students who progressed well in (b) often interpreted their formula correctly to obtain marks in (c). A minority of students who didn’t complete (b) still attempted (c) with some success, e.g. recognising that d d h t was zero at the maximum height or using a ‘ flow in = flow out’ argument.

Common mistakes

• separate variables
• t = −240 ln(24 − 5h) + c
• use t = 0, h = 2 to find c
• correct log work to h = 4.8 − 2.8e^(−t/240)
• forgetting + c

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. Of those who did use partial fractions in part (b), most found the values of "A" and "B" correctly. Unfortunately many integrated B/(3 + 2u) to B ln(3 + 2u), failing to divide the coefficient by 2. The final Method mark was still available for these candidates, but a number failed to simplify their numerical logs to an expression of the required form and so did not gain this mark. A pleasing number of well prepared candidates did gain full marks in this question.

Common mistakes

• spot partial fractions 6/(u(3 + 2u)) = 2/u − 4/(3 + 2u) (many missed this)
• ∫4/(3 + 2u) du = 2ln(3 + 2u), not 4ln
• combine logs to ln(49/36)

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

Only the strongest candidates scored all 11 marks here, with the majority losing at least 1 mark, usually the first mark in (a) or the last mark in (b). In part (b) the vast majority of students knew they had to use partial fractions to attempt the question with many achieving the correct expression and correct integration. The most common error was in simplifying the final logarithmic expression with many candidates leaving some of their answers in terms of ln (3/4) or ln (4/3) for example. It was uncommon therefore to see a fully correct solution in the required form.

Common mistakes

• partial fractions 0.9/(x − 2) − 2.4/(x + 3)
• upper limit 5 from 15 − 3x = 0
• ln 6 = ln 2 + ln 3, ln 8 = 3 ln 2
• final 3.3 ln 3 − 4.8 ln 2, not left with ln(3/4)