PhysicsAndMathsTutor.com

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1 C Integration: Basic Integration. Use calculus to find the value of ( + )d. (Total 5 marks). Evaluate 8 d, giving your answer in the form a + b, where a and b are integers. (Total marks). f() = Find (a) f (), () f ( ) d. () (Total 7 marks). Use calculus to find the eact value of d (Total 5 marks) 6 5. Given that y =, 0, (a) find d y, d () Edecel Internal Review

2 C Integration: Basic Integration evaluate y d. () (Total 6 marks) 6. Given that y = 6, 0, (a) find d y, d () find y d. () (Total 5 marks) 7. (a) Epand ( + ). () Hence evaluate ( + ) d, giving your answer in the form a + b, where a and b are integers. (5) (Total 7 marks) Edecel Internal Review

3 C Integration: Basic Integration 8. (a) Find + d. () Hence evaluate + d. () (Total 5 marks). d = + + M AA + d = + = (6 + 8) ( + ) M Note = 9 (9 + C scores A0) A 5 st M for attempt to integrate k or k. st A for or a simplified version. nd A for or ( ) ( ) or a simplified version. Ignore + C, if seen, but two correct terms and an etra non-constant term scores MAA0. nd M for correct use of correct limits ( top bottom ). Must be used in a changed function, not just the original. (The changed function may have been found by differentiation). Ignore poor notation (e.g. missing integral signs) if the intention is clear. No working: The answer 9 with no working scores M0A0A0MA0 ( mark). [5] Edecel Internal Review

4 C Integration: Basic Integration. (Or equivalent, such as, or ) MA d = 8 = 8 = + [or, or ( ), or ( + )] MA st M: k, k 0. nd M: Substituting limits 8 and into a changed function (i.e. not or ), and subtracting, either way round. nd A: This final mark is still scored if + is reached via a decimal. N.B. N.B. Integration constant +C may appear, e.g. 8 + C = ( 8 + C) ( + C) = + (Still full marks) But... a final answer such as + + C is A0. It will sometimes be necessary to ignore subsequent working (isw) after a correct form is seen, e.g. d = (M A), followed by incorrect simplification d = = (still M A)... The second M mark is still available for substituting 8 and into and subtracting. []. (a) f () = + 6 B f () = M, Acao Edecel Internal Review

5 C Integration: Basic Integration Acceptable alternatives include + 6 ; + ; dy d y Ignore LHS (e.g. use [whether correct or not] of and ) d d c or constant (i.e. the written word constant) is B0 B M Attempt to differentiate their f (); n n. n n seen in at least one of the terms. Coefficient of. ignored for the method mark. and 0 are acceptable. Acceptable alternatives include ; c or constant is A0 M A cao Eamples f () = + 6 B f () = + B0 M0 A0 f () = + M A0 f () = + 6 B f () = 6 M A0 = + 6 B = M A y = f () = B0 dy d = + B0 f () = M A d y = 6 + d MA0 f () = + 6 B f () = c M A0 f () = c B0 f () = M A ( + + 5)d = M, A = ( + + 5) M = 5 o.e. A Edecel Internal Review 5

6 C Integration: Basic Integration Attempt to integrate f(); n n + Ignore incorrect notation (e.g. inclusion of integral sign) o.e. Acceptable alternatives include + + 5; ; c; N.B. If the candidate has written the integral (either or what they think is the integral) in part (a), it may not be rewritten in, but the marks may be awarded if the integral is used in. Substituting and into any function other than and subtracting either way round. So using their f () or f () or their f ()d or their f () d will gain the M mark (because none of these will give + + 5). Must substitute for all s but could make a slip (for eample) is acceptable for evidence of subtraction ( invisible brackets). M A M 6 o.e. (e.g. 5, 5.75, ) Must be a single number (so 6 is A0). A Answer only is M0A0M0A0 Eamples c MA c ( c) M =, + c c M A = 5 A =, 6 + c M0 A0 (no subtraction) f( )d = ( + + 5) M0A0, M0 = 5 9 = 6 A0 (Substituting and into + + 5, so nd M0) Edecel Internal Review 6

7 C Integration: Basic Integration [ ] ( 6 6)d = M0 A0 ( + 6)d = [ + ] M0 A0 = + ( + 6) M A0 = 8 + ( + ) M A M A M (one negative sign is sufficient for evidence of subtraction) = 6 = 5 A (allow recovery, implying student was using invisible brackets ) (a) f() = f () = BM0A MAM = 5 A The candidate has written the integral in part (a). It is not rewritten in, but the marks may be awarded as the integral is used in. [7] d = (= + 5 ) M A A. ( ) [ 5 ] = (8 + 0 ) ( + 5 ), = M, A 5 Integration: Accept any correct version, simplified or not. All terms correct: M A A, Two terms correct: M A A0, One power correct: M A0 A0. The given function must be integrated to score M, and not e.g Limits: M: Substituting and into a changed function and subtracting, either way round. [5] Edecel Internal Review 7

8 C Integration: Basic Integration 5. (a) dy = + 8 d n n M A n n+ M A ( 6 ) d = + [ ] = M 9 =, or equivalent A 9 [6] 6. (a) dy d = M A M is for n n in at least one term, 6 or is sufficient. A is fully correct answer. Ignore subsequent working. 6 y d = + + C M A A M: Correct power of in at least one term (C sufficient) 6 First A: + C Second A: + [5] 7. (a) + 9, + (Allow ) B, B Allow ( ) or only if later work justifies understanding. ( + + 9) d = (ft dep. on terms) M A ft + [...] = (8 + (8 ) 8) ( ) M = (seen or implied) B = A 5 Special case: Misread as ( + ) (a) B for + Just the two M marks are available. [7] Edecel Internal Review 8

9 C Integration: Basic Integration 8. (a) (+ )d = c M A (, 0) ( + ) d = + + = ( ) ( + + ) M = 7 A [5]. While most candidates gained method marks for their integration attempt, some did not realise that was, or had difficulty in simplifying. More common was the inability to evaluate the definite integral correctly, with causing particular problems. Overall, however, it was common to see full marks scored.. Although many candidates scored full marks on this question, others had difficulty dealing with. This was sometimes misinterpreted as or, leading to incorrect integration but still allowing the possibility of scoring a method mark for use of limits. Sometimes there was no integration attempt at all and the limits were simply substituted into. The candidates who performed the integration correctly were usually able to deal with their surds and proceeded score full marks, although a few left their answer as + 8. Occasionally the answer was given as a decimal.. Part (a) was answered well with many correct solutions. A few candidates integrated f(). Some candidates had difficulty differentiating the constant term. The most common incorrect solution was f () = 6. In part a few candidates used f() or f () as their integral. However, most integrated successfully and substituted accurately. Occasional arithmetic slips were seen.. This standard test of definite integration was handled well by the vast majority of candidates. Mistakes, where made, tended to be in the integration of, although errors in simple arithmetic sometimes spoilt otherwise correct solutions. Candidates who differentiated were only able to pick up one method mark, for the substitution of limits. 5. The general standard of calculus displayed throughout the paper was ecellent and full marks were common on this question. A few candidates took the negative inde in the wrong direction, differentiating to obtain and integrating to obtain. Edecel Internal Review 9

10 C Integration: Basic Integration 6. Pure Mathematics P The vast majority of candidates gained the method marks and many went on to score four or five marks; the four mark score was usually due to the omission of the arbitary constant in part. Differentiation and integration of caused problems for some candidates, and those who 6 wrote y as before differentiating were usually defeated. Providing 8 in part (a) and + in part had been seen, subsequent errors such as writing these as or respectively, were not penalised, but it should be noted that such errors 8 with indices were quite common. Core Mathematics C This was another straightforward question for many candidates and the principles of differentiation and integration were understood by most. The negative power caused problems for some again, they realized that the term needed to be written as but did not always apply the rules successfully. In part some could not simplify to + and a sizeable minority lost a mark for failing to include a constant of integration. 7. Most candidates made good attempts to epand the epression in part (a), although a few did not recognise that is equivalent to and some gave ( ). Integration techniques in part were well known. The most common cause of mark loss in this part was the inability to epress the final answer in the required surd form, with not being recognised as Ö, and 8 sometimes simplified to 6. = 8. No Report available for this question. Edecel Internal Review 0

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com . f() = 4 cosec 4 +, where is in radians. (a) Show that there is a root α of f () = 0 in the interval [.,.3]. Show that the equation f() = 0 can be written in the form = + sin 4 (c) Use the iterative formula