P9.4 Trapezium rule

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

Examiner's comment

This was a lovely twist on a usually very standard question using the trapezium rule that was overall very well-answered. It was very rare to see a blank response to this question. Candidates usually scored full marks in part (a). Where marks were lost the most common error was working out “h”, the width of each strip, as (4 3) 1 6 6 - = where 6 was the number of ordinates given, rather than (4 3) 1 5 5 - = , where 5 was the number of strips required. Very few candidates failed to give “h” a value. Most applied the standard trapezium rule correctly, although a minority failed to use the area of 17.59 to form an equation. Only a handful of solutions used the method of adding the areas of individual trapezia, which is much more time consuming and does not demonstrate as effectively an understanding the trapezium rule, which is provided in the formula booklet. The brackets were dealt with correctly in most cases although quite a number of candidates expanded the brackets before collecting terms, creating a more difficult equation to cope with. Others multiplied both sides by 10 which made the manipulation much easier. Several candidates had invisible brackets and recovered appropriately but lost the A mark as the answer was given. Although many values needed to be copied from the given table of x and y values, there were very few copying slips seen.

Common mistakes

• h = 0.2
• brackets must be dealt with correctly before reaching the given answer

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

Examiner's comment

This was a lovely twist on a usually very standard question using the trapezium rule that was overall very well-answered. It was very rare to see a blank response to this question. Part (b) was dealt with very well overall, with a variety of different methods used to solve the equations simultaneously, usually via the use of a calculator. Some arithmetic errors were seen in forming the second equation, 28 a b + = , which resulted in the wrong values for a and b. Again, despite requiring the copying of many values from the given table of x and y values to form the required equation, there were few transcription errors. It is also worth noting that nearly all candidates used the given equation in (a) to correctly solve for values of a and b. Although it was made possible for them to gain the method mark for using “their” equation, this was very infrequently utilised, or required. Furthermore, it was pleasing to see candidates who made no attempt at (a) still proceeded to use the given answer to attempt and generally gain full marks in (b).

Common mistakes

• form a + b = 28 from the sum of the y values and solve with a + 2b = 51

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

Examiner's comment

Part (a) of this question required candidates to use a given table of function values in an application of the Trapezium Rule. This was generally done very well with candidates knowing how to apply the rule. Most achieved h = 1.5 although a common mistake was for students to think that the number of strips was the same as the number of x values instead of being one less. Thus dividing by 5 instead of 4 resulting in h = 1.2. Most knew how to apply the trapezium rule and had the bracketing correct and so achieved the correct answer. A few left out some of the brackets followed by an incorrect calculation. 𝟏𝟏. 𝟓𝟓 𝟐𝟐(𝟏𝟏. 𝟔𝟔𝟔𝟔+ 𝟐𝟐. 𝟔𝟔𝟔𝟔) + 𝟐𝟐(𝟐𝟐+ 𝟐𝟐. 𝟐𝟐𝟐𝟐+ 𝟐𝟐. 𝟒𝟒𝟒𝟒) With or without an outer bracket either at the beginning or end – would lead them to then calculate the expression as it is written, often leading to an answer like: - 𝟏𝟏. 𝟓𝟓 𝟐𝟐(𝟒𝟒. 𝟐𝟐𝟐𝟐) + 𝟏𝟏𝟏𝟏. 𝟒𝟒𝟒𝟒= 𝟑𝟑. 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏. 𝟒𝟒𝟒𝟒= 𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔𝟔𝟔 There were a few who worked out the areas of individual trapezia and then summed the areas achieving a correct answer. A minority didn’t know how to apply the rule Most of those who attempted (b)(ii) didn’t realise that they had to split the expression into two parts applying log laws. Part of the expression then had to be integrated and added to their answer to part (a). Often seen was the calculation 9log3(2x) or 9 ´ 13.275. Some seemed to think that log3(18x) meant that the graph/function had been subjected to a stretch of either 9 or 1/9. Some didn’t show sufficient working to be sure that they were applying the correct method whilst others showed the correct use of logs but then just added 2 to their part (a) answer, not integrating this between the limits. Even when a correct relationship of log3 18x = log3 2x + log3 9 was established, the subsequent argument often did not have sufficient detail to show where their final answer came from. The question clearly stated “making your method clear”. Very few achieved full marks on this part.

Common mistakes

• h = 1.5 (5 ordinates = 4 strips, not h = 1.2)
• bracket correctly: 0.75{1.63 + 2.63 + 2(2 + 2.26 + 2.46)} = 13.3
• calculator integral 13.324 needs working

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

Examiner's comment

Part (b) of the question clearly stated that the result from part (a) was to be used and that the method used was to be made clear. Thus, candidates who proceeded to use the Trapezium Rule again with revised values of y received no credit. In (b)(i) many realised that the power law of logs had to be used and realised that their answer to part (a) had to be multiplied by 10, so achieving a correct answer. Many on the other hand unfortunately raised their answer to part (a) by a power of 10.

Common mistakes

• (i) power law: 10 × 13.3 = 133, not 13.3¹⁰
• (ii) log₃18x = 2 + log₃2x, so 12 + 13.3 = 25.3 with the ∫2 dx = 12 shown
• reusing the trapezium rule scores 0

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. Most candidates successfully answered part (a) applying the Trapezium Rule using all five values of y from the given table. Some students, who found the correct estimate of the area of R, failed to round their answer as requested, or occasionally rounded incorrectly, and lost the accuracy mark. Errors, although rare, were usually as a result of an incorrect strip width.

Common mistakes

• h = 0.5
• ½h{y₀ + y₄ + 2(y₁ + y₂ + y₃)} = 2.41 (3 s.f.)
• wrong strip width

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

Examiner's comment

Part (a) was generally answered well, with the majority of candidates using h = 0.5 and finding an area rounding to 1.50, but a number completed it with a final answer of 1.505… and consequently lost the final A mark for part (a). The most common error seen was the use of h= 2 5 which candidates are arriving at because of a misunderstanding in the ‘n’ seen in the formula in 𝑏𝑏-𝑎𝑎 𝑛𝑛 , they presumably think n is the number of entries in a row of the table ignoring that the first column represents x0 and y0. Those candidates who calculated the width of the trapezia independent of the formula 𝑏𝑏-𝑎𝑎 𝑛𝑛 were usually more successful. The formula was set up correctly on most occasions, with the most common error being the lack of exterior brackets, often resulting in 4.95 as an answer. There were some candidates using a variation of the formula, with the 1 2 absorbed into the brackets, and a few worked out the areas of four separate trapezia, usually successfully.

Common mistakes

• h = 0.5, not 2/5 (n is the number of strips, not the number of y values)
• outer brackets needed (missing brackets gives 5.85 or 4.95)
• final answer 1.50 to 3 s.f., not 1.505

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

Examiner's comment

Success in part (b) was mixed, with around half the candidates multiplying their answer by 3 and the others multiplying by 9. A minority of candidates restarted, losing access to marks in this question.

Common mistakes

• the square root of 9x/(1 + x) is 3 times the original integrand, so multiply (a) by 3 (multiplying by 9 is the common error)
• do not restart with the trapezium rule

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

Examiner's comment

In part (c), the mark was only available to those who arrived at an answer rounding to 4.5 in part (b). Answers were often concise and clear, stating that their answer was correct to 2s.f. or 1d.p. and therefore quite accurate. Others went for a percentage error approach, which was also acceptable and usually completed successfully. The main error in this part was that candidates misunderstood what was being asked of them, and instead arguing that the answer was an underestimate, confusing this with another common trapezium rule question.

Common mistakes

• comment on accuracy (e.g. correct to 2 s.f., or percentage error under 1%)
• just "it is an underestimate" scores nothing
• needs awrt 4.5 in (b)

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

Examiner's comment

Q02 proved to be challenging, with a significant number of candidates struggling to realise that the length of the runway could be estimated by the area under the speed- time curve generated from the table of values given in the question. In Q02(a), only a minority of candidates estimated the length of the runway by applying the trapezium rule, which was considered the most appropriate method, with many of these giving the correct answer of 415m. While the table of values clearly shows an b a h n 5, h interval width the application of the formula with instead of 6 n 25. h n led some candidates to use an incorrect strip width Other candidates 5 6 made calculation errors, bracketing errors or transcription errors. Some candidates used the equations of motion with constant acceleration (i.e. the suvat equations) to estimate the length. Many of the candidates who used this method applied the calculation 2 42 (25) to obtain an estimate of 550m. A few used v u at 2 1 2 s ut at to find the acceleration, followed by to estimate the length. Some 2 u candidates used in their suvat calculations, in contradiction to the table of values 0 A few candidates used a method which was 1 2 ms . which indicated an initial speed of equivalent to the trapezium rule, of applying suvat to each of the 5 time-intervals, to give the correct answer of 415m. Other methods seen included: summing up the area of rectangles which enclosed the area above the curve for each of the 5 time intervals; summing up areas of rectangles which enclosed the area below the curve for each of the 5 time intervals; and applying (average speed) (total time). The most common methods which were deemed incorrect included 25 42 1050; 1 (25)(42) (speed)(time) (0)(2) (5)(5) (10)(10) ... (25)(42) 2005; 525; 2 or a 6 rectangle approach (instead of 5 rectangles). As for Q02(b), while many candidates who used a trapezium rule or suvat method in Q02(a) identified their estimate of the length as an overestimate, some found difficulty in articulating a reason for this. Those who were most successful drew a diagram which showed clearly the extra area generated by the trapezium rule in relation to the curve. Some candidates sketched the speed-time curve and explained that the curve was convex or that the acceleration was continually increasing. Only a few candidates who used a rectangle method were able to give a correct reason. Some candidates answered Q02(b) by finding an alternative estimate and comparing it with their estimate in Q02(a). For example, some found 415 in Q02(a) and 550 in Q02(b), and then concluded that ‘415 must be an underestimate’. Other candidates, who also received no credit, gave physical explanations about friction, air resistance or the shape of the aircraft, while others stated that ‘the jet accelerated smoothly’.

Common mistakes

• distance = area under the speed-time graph
• trapezium rule with h = 5 (not 25/6, 6 strips): 415 m
• suvat with constant acceleration (550) earns the method mark only
• 25 × 42 or a sum of speed × time scores nothing

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

Examiner's comment

Q02 proved to be challenging, with a significant number of candidates struggling to realise that the length of the runway could be estimated by the area under the speed- time curve generated from the table of values given in the question. In Q02(a), only a minority of candidates estimated the length of the runway by applying the trapezium rule, which was considered the most appropriate method, with many of these giving the correct answer of 415m. While the table of values clearly shows an b a h n 5, h interval width the application of the formula with instead of 6 n 25. h n led some candidates to use an incorrect strip width Other candidates 5 6 made calculation errors, bracketing errors or transcription errors. Some candidates used the equations of motion with constant acceleration (i.e. the suvat equations) to estimate the length. Many of the candidates who used this method applied the calculation 2 42 (25) to obtain an estimate of 550m. A few used v u at 2 1 2 s ut at to find the acceleration, followed by to estimate the length. Some 2 u candidates used in their suvat calculations, in contradiction to the table of values 0 A few candidates used a method which was 1 2 ms . which indicated an initial speed of equivalent to the trapezium rule, of applying suvat to each of the 5 time-intervals, to give the correct answer of 415m. Other methods seen included: summing up the area of rectangles which enclosed the area above the curve for each of the 5 time intervals; summing up areas of rectangles which enclosed the area below the curve for each of the 5 time intervals; and applying (average speed) (total time). The most common methods which were deemed incorrect included 25 42 1050; 1 (25)(42) (speed)(time) (0)(2) (5)(5) (10)(10) ... (25)(42) 2005; 525; 2 or a 6 rectangle approach (instead of 5 rectangles). As for Q02(b), while many candidates who used a trapezium rule or suvat method in Q02(a) identified their estimate of the length as an overestimate, some found difficulty in articulating a reason for this. Those who were most successful drew a diagram which showed clearly the extra area generated by the trapezium rule in relation to the curve. Some candidates sketched the speed-time curve and explained that the curve was convex or that the acceleration was continually increasing. Only a few candidates who used a rectangle method were able to give a correct reason. Some candidates answered Q02(b) by finding an alternative estimate and comparing it with their estimate in Q02(a). For example, some found 415 in Q02(a) and 550 in Q02(b), and then concluded that ‘415 must be an underestimate’. Other candidates, who also received no credit, gave physical explanations about friction, air resistance or the shape of the aircraft, while others stated that ‘the jet accelerated smoothly’.

Common mistakes

• overestimate: the curve is convex (acceleration increasing) so the tops of the trapezia lie above it
• a diagram showing the extra area is best
• comparing with another estimate, friction/air resistance or "accelerated smoothly" scores nothing