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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 can’t 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 it’s clear that one of the parts should be x and the other should be ln( x ). We shouldn’t make v = ln( x ) because we don’t yet know how to take ln’s 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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 Fall '09
 PIETERHOFSTRA
 Calculus, Chain Rule, Derivative, Integrals, Integration By Parts, The Chain Rule, lim, dx, Riemann integral

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