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Unformatted text preview: 6 Techniques of Integration 6.1 Integration by Parts In section 5.2 we discussed solving integrals by making usubstitutions. It was explained then that this is the integration analogue of the Chain Rule for differentiation. In this section we discuss the integration analogue of the product rule. Suppose u and v are functions of x . Then the product rule states ( uv ) = u v + uv We can integrate both sides of this Z ( uv ) d x = Z u v d x + Z uv d x Now the left hand side is just an antiderivative of ( uv ) , ie, uv . Thus uv = Z u v d x + Z uv d x This can be rewritten Z uv d x = uv Z u v d x This is called the integration by parts formula. It is useful when we have to find the integral of a product of two functions. In that case, we can let one of the functions be u and the other one be v and then use the formula. Examples: 1. Evaluate R xe 3 x d x . solution: This integral cant be done with any kind of simple substitution, so we try integration by parts. We have to choose a u and a v such that xe 3 x = uv and that R u v d x will be easier to solve. In this case, we let u = x v = e 3 x To use the formula we need u and v , so we solve for them by taking the derivative of u and the antiderivative of v : u = 1 v = 1 3 e 3 x Now we can use the integration by parts formula: Z uv d x = uv Z u v d x Z xe 3 x d x = x 1 3 e 3 x Z (1) 1 3 e 3 x d x = 1 3 xe 3 x 1 9 e 3 x + C 2. Evaluate R x ln( x ) d x . solution: Once again we would like to use integration by parts, and its clear that one of the parts should be x and the other should be ln( x ). We shouldnt make v = ln( x ) because we dont yet know how to take lns antiderivative. So we set u = ln( x ) v = x Which gives us u = 1 x v = x 2 2 So integration by parts gives us Z uv d x = uv Z u v d x...
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This note was uploaded on 01/07/2010 for the course MAT mat1300 taught by Professor Pieterhofstra during the Fall '09 term at University of Ottawa.
 Fall '09
 PIETERHOFSTRA
 Chain Rule, Integrals, Integration By Parts, The Chain Rule

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