P8.1 Fundamental theorem of calculus

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

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

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 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