2.07.PatProcedures - Accumulation Authors Bill Davis Horacio Porta and Jerry Uhl 1996-2007 Publisher Math Everywhere Inc Version 6.0 2.07

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Unformatted text preview: Accumulation Authors: Bill Davis, Horacio Porta and Jerry Uhl 1996-2007 Publisher: Math Everywhere, Inc. Version 6.0 2.07 Traditional Pat Integration Procedures BASICS Traditionally, calculus courses spend a lot of time and effort on practicing the art of evaluating integrals. By now you will have noticed that such practice isn't a central issue in Calculus& Mathematica . It is important to understand the skills, though. Here's a compendium of techniques. B.1) Partial fractions for quotients of polynomials B.1.a) Look at Mathematica 's calculation of 2.1 5.3 1 x 2- 1 x Clear @ f, x D ; f @ x _ D = 1 x 2- 1 ; 2.1 5.3 f @ x D x 0.327079 Explain where this calculation comes from. Answer: Look at the function you are integrating: Clear @ f, x D ; f @ x _ D = 1 x 2- 1 1- 1 + x 2 Factor the denominator of f @ x D : Factor @ Denominator @ f @ x DDD H- 1 + x L H 1 + x L Put: Clear @ newf, a, b D ; newf @ x _ D = a 1 + x + b- 1 + x b- 1 + x + a 1 + x Now put newf @ x D over a common denominator: ExpandAll @ Together @ newf @ x DDD- a- 1 + x 2 + b- 1 + x 2 + a x- 1 + x 2 + b x- 1 + x 2 You want this to be: f @ x D 1- 1 + x 2 You can arrange this by solving for the a and b that make: eqn1 = H- a + b == 1 L ; eqn2 = H a + b == L ; Solve @8 eqn1, eqn2 <D :: a - 1 2 , b 1 2 >> This tells you that when you go with a = - 1 2 and b = 1 2 , then newf @ x D = f @ x D . a = - 1 2 ; b = 1 2 ; newf @ x D 1 2 H- 1 + x L- 1 2 H 1 + x L The point of this is that 2.1 5.3 newf @ x D x is easy to integrate by hand. It is 1 2 Log @- 1 + x D- 1 2 Log @ 1 + x D 2.1 5.3 : top = 1 2 Log @- 1 + x D- 1 2 Log @ 1 + x D . x-> 5.3; bottom = 1 2 Log @- 1 + x D- 1 2 Log @ 1 + x D . x-> 2.1; integral = top- bottom 0.327079 Check: 2.1 5.3 f @ x D x 0.327079 Got it. B.1.b) Look at Mathematica 's evaluation of t 4 x + 1 H x + 2 L H x + 1 L 2 x: Clear @ x, t D ; t 4 x + 1 H x + 2 L H x + 1 L 2 x If B Re @ t D - 1 Im @ t D ! 0,- 3 t 1 + t + Log @ 128 D + 7 Log @ 1 + t D- 7 Log @ 2 + t D , Integrate B 1 H 1 + x L 2 H 2 + x L + 4 x H 1 + x L 2 H 2 + x L , 8 x, 0, t < , Assumptions ! H Re @ t D - 1 Im @ t D ! LFF Explain where this calculation comes from. Answer: Look at the function you are integrating: Clear @ f, x D ; f @ x _ D = 4 x + 1 H x + 2 L H x + 1 L 2 1 + 4 x H 1 + x L 2 H 2 + x L Factor the denominator of f @ x D : Factor @ Denominator @ f @ x DDD H 1 + x L 2 H 2 + x L Put: Clear @ newf, a, b, c D ; newf @ x _ D = a 1 + x + b H 1 + x L 2 + c 2 + x b H 1 + x L 2 + a 1 + x + c 2 + x Now put newf @ x D over a common denominator: Together @ newf @ x DD 2 a + 2 b + c + 3 a x + b x + 2 c x + a x 2 + c x 2 H 1 + x L 2 H 2 + x L You want this to be: f @ x D 1 + 4 x H 1 + x L 2 H 2 + x L You can arrange this by solving for the a, b and c that make: eqn1 = H 2 a + 2 b + c == 1 L ; eqn2 = H 3 a + b + 2 c == 4 L ; eqn3 = H a + c == L ; Solve @8 eqn1, eqn2, eqn3 <D 88 a 7, c - 7, b - 3 <<...
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This note was uploaded on 10/11/2011 for the course MATH 231 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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2.07.PatProcedures - Accumulation Authors Bill Davis Horacio Porta and Jerry Uhl 1996-2007 Publisher Math Everywhere Inc Version 6.0 2.07

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