Unit3 - Unit 3: Integration Techniques We have learned how...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Unit 3: Integration Techniques We have learned how to integrate functions of the following friendly forms, and have learned how to use the method of substitution to transform some problems into these forms: i u n du = u n +1 n +1 + C, n n = - 1 i 1 u du = ln | u | + C i e u du = e u + C But sometimes this is not enough. For instance, if we wish to evaluate something like i x ln( x ) dx , nothing we have learned so far will work. For something like this we turn to a new method, namely that of integration by parts , which is sometimes called “double substitution”. The derivation of the formula is easily seen to be a consequence of the product rule. When u and v are functions of x , we have d dx [ uv ] = uv p + vu p where, of course, v p = dv dx and u p = du dx . Antidi±erentiating both sides of this equation with respect to x we get uv = I uv p dx + I vu p dx which can be rearranged to I uv p dx = uv - I vu p dx This is the formula we use. Notice that since v p = dv dx , then v p dx = dv dx dx = dv and since u p = du dx , then u p dx = du dx dx = du . We get: 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Defnition 3.1. The integration by parts formula: If u and v are diFerentiable functions then, i udv = uv - i vdu This formula is useful when i vdu is an integral which is easier to evaluate than the one we started with, i udv . So when facing a di±cult integral, we consider whether we can recognize the form i udv for some u and v . The following steps may make it a bit easier to use this method. Step 1: Write the given integral in the form i f ( x ) g ( x ) dx so that you can identify the two functions f ( x ) and g ( x ). Note that if there appears to be only one function in the integrand, then you may need to identify the second one to be the constant function, g ( x ) = 1. Step 2: Introduce the functions u ( x ) and v ( x ) by letting: u = f ( x ) and dv = g ( x ) dx Now, to use the integration by parts formula, we need to ²nd du and v . That is, we diFerentiate u to get: du dx = f p ( x ) du = f p ( x ) dx and we integrate dv to get: v = i dv v = i g ( x ) dx Step 3: Use the formula to re-express the original integral: replace i udv with uv - i vdu Step 4: Evaluate the new expression (by ²nding i vdu ). Note: When we ²nd v = i dv = i v p dx in Step 2, we don’t need to include an arbitrary constant. This is because when we use v in our formula, we will still have an inde²nite integral appearing in the expression. All arbitrary constants will be collected together as one when we perform the last integra- tion. 2
Background image of page 2
Of course, the main diFculty in this method is determining the choice of u and dv . In Step 1, we recognize the integrand as the product of 2 terms. But how, in Step 2, do we determine which term should be u and which should be dv ? There is no general rule to follow. It is truly a matter of experience. However, there are 2 guiding principles you should keep in mind: (1) dv must be something you know how to integrate, and (2) we want i vdu to be less complicated than i
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/03/2009 for the course MATH 0110A taught by Professor Olds,vicky during the Fall '09 term at UWO.

Page1 / 9

Unit3 - Unit 3: Integration Techniques We have learned how...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online