sum5.3.23 - by a factor of 9 5 so we make a correction and...

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Math 22, Summer II 2005 B. Dodson
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2 1. Course Info 2. Week 1 Homework: first day 3.2 Quotient rule (product rule) – in-class examples #3, 14 3.4 Trig derivatives – #15 3.5 Chain rule – #37 5.3 Fundamental theorem – see below 5.5 Integration by substitution – #2, 6 and below
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Problem 5.3.23: Use the 2nd part of the Fundamental theorem to evaluate Z 1 0 x 4 5 dx. Solution: We need an anti-derivative for f ( x ) = x 4 5 . While there is a formula that gives the answer quickly, let’s see how the formula works by trying steps. We start from the differentiation formula ( x m ) ± = mx m - 1 . Since f is the derivative, we get x m - 1 = x 4 5 ; so m - 1 = 4 5 , and we solve to get m = 1 + 4 5 = 9 5 . Now we try ( x 9 5 ) ± = 9 5 x 4 5 . We’re off
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Unformatted text preview: by a factor of 9 5 , so we make a correction and take F ( x ) = 5 9 x 9 5 to get F ± = f. 4 Then the fundamental theorem (in this case) gives Z 1 x 4 5 dx = [ F ( x )] 1 = F (1)-F (0) = 5 9 . Note: The usual formula here is Z x n dx = 1 n + 1 x n +1 + c. Problem 5.5.1: Use the substitution u = 3 x to evaluate Z cos 3 x dx. Solution: With u = 3 x, we have du = 3 dx, or dx = du 3 . Substituting we get Z cos 3 x dx = Z cos u du 3 = 1 3 Z cos u du = 1 3 sin 3 x + c, where we have changed back to the original variables from the solution sin u + c using u = 3 x....
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sum5.3.23 - by a factor of 9 5 so we make a correction and...

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