32-substitution-day-1

32-substitution-day-1 - Math 1a Substitution Fall, 2009 ’...

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Unformatted text preview: Math 1a Substitution Fall, 2009 ’ & $ % The Substitution Rule: Suppose u = g ( x ) is a differentiable function with domain an interval I and f ( x ) is continuous on I . Then Z f ( g ( x )) g ( x ) dx = Z f ( u ) du. Thus we can make substitutions and treat du and dx like differentials: du = du dx dx . Some Integration Formulas you might like to have at your fingertips: Z u n du = u n +1 n + 1 + C n 6 =- 1 Z 1 u du = ln | u | + C Z e u du = e u + C Z sin( u ) du =- cos( u ) + C Z cos( u ) du = sin( u ) + C Compute the following indefinite integrals: 1 Z ( x 2 + 2 x + 1) dx 2 Z ( x + 1) 2 dx 3 Z (2 x + 1) 2 dx 4 Z x 2 ( x 3- 4 ) 6 dx 5 Z x √ x 2 + 1 dx 6 Z x √ x 2 + 1 dx 7 Z 1 (2 x + 3) 4 dx 8 Z sin(4 x- 3) dx 9 Z x cos( x 2 + 1) dx 10 Z e 4 x +1 dx 11 Z xe x 2 +1 dx 12 Z e √ x √ x dx 13 Z 1 2 x dx 14 Z ln( x ) x dx 15 Z 1 x ln( x ) dx Some More Integration Formulas that might help with some of the integrals on this page: Z sec 2 ( u ) du = tan( u ) + C Z sec( u ) tan( u ) du...
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This note was uploaded on 03/27/2012 for the course MATH 1a taught by Professor Benedictgross during the Fall '09 term at Harvard.

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32-substitution-day-1 - Math 1a Substitution Fall, 2009 ’...

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