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Published byKellen Lowry Modified over 2 years ago

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**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 ,

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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?

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**The function in the integral on the l.h.s. . . . **

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.

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

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Example Generalisation Integrating: Simplifying the l.h.s.: Rearranging:

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

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So, Using this formula means that we differentiate one factor, u to get

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So, Using this formula means that we differentiate one factor, u to get and integrate the other , to get v

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

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So, differentiate integrate and We can now substitute into the formula

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

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

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Exercises Find 1. 2. 1. Solutions: = ò dx xe x 2.

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**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.

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**Using Integration by Parts**

Integration by parts cannot be used for every product. 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.

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e.g. 3 Find Solution: What’s a possible problem? ANS: We can’t integrate Can 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.

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**x cancels. e.g. 3 Find So, integrate differentiate**

The r.h.s. integral still seems to be a product! BUT . . . x cancels. So,

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

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e.g. 4 Solution: ( 1 ) Let and So, Substitute in ( 1 )

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Example e.g. 5 Find Solution: 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.

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e.g. 5 Find Solution: We write this as:

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e.g. 5 Find So, where and We next use integration by parts for I2

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e.g. 5 Find So, where and We next use integration by parts for I2

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**2 equations, 2 unknowns ( I1 and I2 ) !**

e.g. 5 Find So, ( 1 ) ( 2 ) 2 equations, 2 unknowns ( I1 and I2 ) ! Substituting for I2 in ( 1 )

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**2 equations, 2 unknowns ( I1 and I2 ) !**

e.g. 5 Find So, ( 1 ) ( 2 ) 2 equations, 2 unknowns ( I1 and I2 ) ! Substituting for I2 in ( 1 )

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**2 equations, 2 unknowns ( I1 and I2 ) !**

e.g. 5 Find So, ( 1 ) ( 2 ) 2 equations, 2 unknowns ( I1 and I2 ) ! Substituting for I2 in ( 1 )

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

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Solutions: and Let 1. ( 1 ) and Let For I2: Subs. in ( 1 )

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

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