lecture31 - Lecture 31 1 Lecture 31: (Sec. 6.2) Integration...

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Unformatted text preview: Lecture 31 1 Lecture 31: (Sec. 6.2) Integration by Substitution, Part I ex. How can we evaluate 6x(3x2 − 2)5 dx? Lecture 31 2 To see why we can use substitution: Let F be an antiderivative of a given function f . Then if g is a differentiable function of x, d [F (g (x))] = dx Lecture 31 We have the following result: du f (u) dx = dx 3 Lecture 31 4 Evaluate each integral: ex. ex. (3x2 + 1)(x3 + x − 2)15 dx 3 − 5 x4 (4 − x 5 )2 dx Lecture 31 These are examples of the General Power Rule of Integration Let u be a differentiable function of x. Then du un dx = dx NOTE: 5 Lecture 31 6 But consider the following example: Evaluate 5x x2 − 1 dx Lecture 31 7 ex. Evaluate: 1) 5 dx (1 + 2x)2 2) x2 (6 − 3x)(1 + 2x − ) dx 2 Lecture 31 8 Now consider the following examples: ex. x3(2x4 − 3)5 dx ex. x(2x4 − 3)5 dx Lecture 31 9 Sometimes we can substitute further to evaluate an integral: √ ex. Evaluate: x x + 1 dx Lecture 31 10 Application ex. The marginal cost of producing x units of a 400x new product is C (x) = √ (where x is in 2 + 64 x hundreds), and fixed costs are $4000. Find the total cost of producing 600 items. ...
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This note was uploaded on 07/18/2011 for the course MAC 2233 taught by Professor Smith during the Spring '08 term at University of Florida.

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lecture31 - Lecture 31 1 Lecture 31: (Sec. 6.2) Integration...

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