Presentation on theme: "We develop the formula by considering how to differentiate products."— Presentation transcript:
1We develop the formula by considering how to differentiate products. where u and v are both functions of x.Substituting for y,e.g. If ,
2So,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?
3The 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.
4Here’s the product . . .if we rearrange, we getSo, we’ve integrated !We need to turn this method ( called integration by parts ) into a formula.
5ExampleGeneralisationIntegrating:Simplifying the l.h.s.:Rearranging:
6SUMMARYIntegration by PartsTo integrate some products we can use the formula
7So,Using this formula means that we differentiate one factor, uto get
8So,Using this formula means that we differentiate one factor, uto getand integrate the other ,to get v
9So,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 )
10So,differentiateintegrateandWe can now substitute into the formula
11So,anddifferentiateintegrateWe can now substitute into the formulaThe 2nd term needs integrating
12e.g. 2 FindSolution:anddifferentiateThis is a compound function, so we must be careful.integrateSo,
14Definite 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.
15Using 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.
16e.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.
17x cancels. e.g. 3 Find So, integrate differentiate The r.h.s. integral still seems to be a product!BUT . . .x cancels.So,
18e.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
20Examplee.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.