# Chapter6 - 6 Techniques of Integration 6.1 Integration by...

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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 u-substitutions. 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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Chapter6 - 6 Techniques of Integration 6.1 Integration by...

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