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### sum5.3.23

School: Lehigh
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22, Math Summer II 2005 B. Dodson 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 Problem 5.3.23: Use the 2nd part of the Fundamental theorem 1 to evaluate 0 x dx. 4 5 Solution: We need an anti-derivative for f (x) = x 5 ....

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22, Math Summer II 2005 B. Dodson 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 Problem 5.3.23: Use the 2nd part of the Fundamental theorem 1 to evaluate 0 x dx. 4 5 Solution: We need an anti-derivative for f (x) = x 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 (xm ) = mxm-1 . Since f is the derivative, we get xm-1 = x 5 ; so m - 1 = 4 , and solve we to get m = 1 + 5 Now we try (x ) = 9 5 4 4 4 5 = 9. 5 9 4 5 5x . We're off by a factor of 9 , so we make a correction and 5 take F (x) = 5 9 5 9x to get F = f. 4 Then the fundamental theorem (in this case) 1 gives 0 x dx = 4 5 [F (x)]1 0 5 = F (1) - F (0) = . 9 Note: The usual formula here is 1 xn dx = xn+1 + c. n+1 Problem 5.5.1: Use the substitution u = 3x ...

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