lecture31

# lecture31 - du is a factor in the integrand 2 Express the...

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Math 006 (Lecture 31) Reversing the Chain Rule Recall that d dx [ f ( x )] n = n ( f ( x )) n - 1 f 0 ( x ) , d dx e f ( x ) = e f ( x ) f 0 ( x ) , d dx ln[ f ( x )] = f 0 ( x ) f ( x ) and integration is the reverse process of diﬀerentiation. Therefore, we ﬁnd that (a) Z [ f ( x )] n f 0 ( x ) dx = [ f ( x )] n +1 n + 1 + C, n 6 = - 1 (b) Z e f ( x ) f 0 ( x ) dx = e f ( x ) + C (c) Z f 0 ( x ) f ( x ) dx = ln | f ( x ) | + C Example 1. Find each indeﬁnite integral: (a) Z (2 x 3 - 3) 20 (6 x 2 ) dx (b) Z 5 e 5 x dx (c) Z 2 x 4 + x 2 dx 1

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Method of Substitution Example 2. Find Z ( x 2 + 2 x + 5) 5 (2 x + 2) dx Procedure: 1. Select a substitution that appears to simplify the integrand. In particular, try to select u so that
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Unformatted text preview: du is a factor in the integrand. 2. Express the integrand in terms of u and du ONLY! 3. Evaluate the integral in term of u . 4. Express the integral in terms of the original variable. Example 3. Find each indeﬁnite integral: (a) Z 3 x 2 e x 3 dx (b) Z x 1 + x 2 dx (c) Z 4 x 3 √ 1 + x 4 dx 2 (d) Z x-1 √ x + 1 dx (e) Z 1 x (ln x ) 2 dx 3...
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lecture31 - du is a factor in the integrand 2 Express the...

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