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© Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules 25: Integration by Parts.

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Presentation on theme: "© Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules 25: Integration by Parts."— Presentation transcript:

1 © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules 25: Integration by Parts

2 Proof of FormulaeProof of Formulae Link

3 Integration by Parts SUMMARY To integrate some products we can use the formula Integration by Parts This is in the formula book : you have to know how to use it

4 Integration by Parts to get... Using this formula means that we differentiate one factor, u So,

5 Integration by Parts So, and integrate the other, to get v Using this formula means that we differentiate one factor, u to get...

6 Integration by Parts So, e.g. 1 Find and differentiate integrate and integrate the other, to get v Using this formula means that we differentiate one factor, u to get... Having substituted in the formula, notice that the 1 st term, uv, is completed but the 2 nd term still needs to be integrated. ( +C comes later )

7 Integration by Parts We can now substitute into the formula So, differentiate integrate and

8 Integration by Parts So, differentiate integrate We can now substitute into the formula and The 2 nd term needs integrating

9 Integration by Parts differentiate integrate and e.g. 2 Find Solution: So, This is a compound function, so we must be careful.

10 Integration by Parts Exercises Find Solutions:   dxxe x

11 Integration by Parts Definite Integration by Parts With a definite integral it’s often easier to do the indefinite integral and insert the limits at the end. We’ll use the question in the exercise you have just done to illustrate.

12 Integration by Parts Integration by parts cannot be used for every product. Using Integration by Parts It works if  we can integrate one factor of the product,  the integral on the r.h.s. is easier* than the one we started with. * There is an exception but you need to learn the general rule.

13 Integration by Parts The following exercises and examples are harder so you may want to practice more of the straightforward questions before you tackle them.

14 Integration by Parts Solution: What’s a possible problem? Can you see what to do? If we let and, we will need to differentiate and integrate x. ANS: We can’t integrate. Tip: Whenever appears in an integration by parts we choose to let it equal u. e.g. 3 Find

15 Integration by Parts So, The r.h.s. integral still seems to be a product! BUT... x cancels. e.g. 3 Find differentiate integrate So, e.g. 3 Find

16 Integration by Parts e.g. 4 Solution: Let The integral on the r.h.s. is still a product but using the method again will give us a simple function. We write and

17 Integration by Parts e.g. 4 Solution: So, Substitute in ( 1 )..... ( 1 ) Let and

18 Integration by Parts Solution: It doesn’t look as though integration by parts will help since neither function in the product gets easier when we differentiate it. e.g. 5 Find However, there’s something special about the 2 functions that means the method does work. The next example is interesting but is not essential. Click below if you want to miss it out. Omit Example

19 Integration by Parts e.g. 5 Find Solution: We write this as:

20 Integration by Parts e.g. 5 Find where So, and We next use integration by parts for I 2

21 Integration by Parts e.g. 5 Find So, where and We next use integration by parts for I 2

22 Integration by Parts e.g. 5 Find So, 2 equations, 2 unknowns ( I 1 and I 2 ) ! Substituting for I 2 in ( 1 )..... ( 1 )..... ( 2 )

23 Integration by Parts e.g. 5 Find So, 2 equations, 2 unknowns ( I 1 and I 2 ) ! Substituting for I 2 in ( 1 )..... ( 1 )..... ( 2 )

24 Integration by Parts e.g. 5 Find So, 2 equations, 2 unknowns ( I 1 and I 2 ) ! Substituting for I 2 in ( 1 )..... ( 1 )..... ( 2 )

25 Integration by Parts Exercises 2. ( Hint: Although 2. is not a product it can be turned into one by writing the function as. ) 1.

26 Integration by Parts Solutions: 1. andLetandLet For I 2 :..... ( 1 ) Subs. in ( 1 )

27 Integration by Parts 2. and Let This is an important application of integration by parts So,

28 Integration by Parts There is a formula for integrating some products. I’m going to show you how we get the formula but it is tricky so if you want to go directly to the summary and examples click below. Summary

29 Integration by Parts If where u and v are both functions of x. Substituting for y, e.g. If, We develop the formula by considering how to differentiate products.

30 Integration by Parts So, Integrating this equation, we get The l.h.s. is just the integral of a derivative, so, since integration is the reverse of differentiation, we get Can you see what this has to do with integrating a product?

31 Integration by Parts Here’s the product... if we rearrange, we get The function in the integral on the l.h.s is a product, but the one on the r.h.s.... is a simple function that we can integrate easily.

32 Integration by Parts Here’s the product... if we rearrange, we get We need to turn this method ( called integration by parts ) into a formula. So, we’ve integrated !

33 Integration by Parts Integrating: Rearranging: Simplifying the l.h.s.: Generalisation Example

34 Integration by Parts

35 The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

36 Integration by Parts SUMMARY To integrate some products we can use the formula Integration by Parts

37 differentiate integrate and e.g. Find Solution: So,

38 Integration by Parts Integration by parts can’t be used for every product. Using Integration by Parts It works if  we can integrate one factor of the product,  the integral on the r.h.s. is easier* than the one we started with. * There is an exception but you need to learn the general rule.

39 Integration by Parts differentiate We can’t integrate so, The r.h.s. integral still seems to be a product! BUT... So, x cancels. e.g. 3 Find integrate

40 Integration by Parts 2. and Let This is an important application of integration by parts


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