2 Module C3 Module C4 AQA Edexcel MEI/OCR OCR "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"
3 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
4 We develop the formula by considering how to differentiate products. where u and v are both functions of x.Substituting for y,e.g. If ,
5 So,Integrating this equation, we getThe l.h.s. is just the integral of a derivative, so, since integration is the reverse of differentiation, we getCan you see what this has to do with integrating a product?
6 The function in the integral on the l.h.s. . . . Here’s the product . . .if we rearrange, we getThe function in the integral on the l.h.s. . . is a product, but the one on the r.h.sis a simple function that we can integrate easily.
7 Here’s the product . . .if we rearrange, we getSo, we’ve integrated !We need to turn this method ( called integration by parts ) into a formula.
8 ExampleGeneralisationIntegrating:Simplifying the l.h.s.:Rearranging:
9 SUMMARYIntegration by PartsTo integrate some products we can use the formula
10 Using this formula means that we differentiate one factor, u to get
11 So,Using this formula means that we differentiate one factor, uto getand integrate the other ,to get v
12 So,Using this formula means that we differentiate one factor, uto getand integrate the other ,to get ve.g. 1 FindHaving substituted in the formula, notice that the 1st term, uv, is completed but the 2nd term still needs to be integrated.anddifferentiateintegrate( +C comes later )
13 So,differentiateintegrateandWe can now substitute into the formula
14 So,anddifferentiateintegrateWe can now substitute into the formulaThe 2nd term needs integrating
15 e.g. 2 FindSolution:anddifferentiateThis is a compound function, so we must be careful.integrateSo,
17 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.
18 Using Integration by Parts Integration by parts cannot be used for every product.It works ifwe 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.
19 The following exercises and examples are harder so you may want to practice more of the straightforward questions before you tackle them.
20 e.g. 3 FindSolution:What’s a possible problem?ANS: We can’t integrateCan you see what to do?If we let and , we will need to differentiate and integrate x.Tip: Whenever appears in an integration by parts we choose to let it equal u.
21 x cancels. e.g. 3 Find So, integrate differentiate The r.h.s. integral still seems to be a product!BUT . . .x cancels.So,
22 e.g. 4Solution:LetandThe integral on the r.h.s. is still a product but using the method again will give us a simple function.We write
23 e.g. 4Solution:( 1 )LetandSo,Substitute in( 1 )
24 The next example is interesting but is not essential The next example is interesting but is not essential. Click below if you want to miss it out.Omit Examplee.g. 5 FindSolution:It doesn’t look as though integration by parts will help since neither function in the product gets easier when we differentiate it.However, there’s something special about the 2 functions that means the method does work.
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 SUMMARYTo integrate some products we can use the formulaIntegration by Parts
38 Integration by parts can’t be used for every product. Using Integration by PartsIt works ifwe 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 x cancels. e.g. 3 Find We can’t integrate so, integrate differentiate The r.h.s. integral still seems to be a product!BUT . . .So,x cancels.e.g. 3 Findintegrate
40 2.andLetThis is an important application of integration by parts