{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2.07.PatProcedures - 0.327079 Accumulation Authors Bill...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 Ÿ 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 + ax - 1 + x 2 + bx - 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 == 0 L ; Solve @8 eqn1, eqn2 <D :: a Ø - 1 2 , b Ø 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 Ÿ 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 ê .x -> 5.3; bottom = 1 2 Log @ - 1 + x D - 1 2 Log @ 1 + x D ê .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 Ÿ 0 t 4 x + 1 H x + 2 L H x + 1 L 2 x: Clear @ x, t D ; 0 t 4x + 1 H x + 2 L H x + 1 L 2 x If B Re @ t D ¥ - 1 »» Im @ t D ! 0, - 3t 1 + t + Log @ 128 D + 7Log @ 1 + t D - 7Log @ 2 + t D , Integrate B 1 H 1 + x L 2 H 2 + x L + 4x H 1 + x L 2 H 2 + x L , 8 x, 0, t < , Assumptions Ø ! H Re @ t D ¥ - 1 »» Im @ t D ! 0 LFF Explain where this calculation comes from. · Answer: Look at the function you are integrating: Clear @ f, x D ; f @ x _ D = 4x + 1 H x + 2 L H x + 1 L 2 1 + 4x 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 2a + 2b + c + 3ax + bx + 2cx + ax 2 + cx 2 H 1 + x L 2 H 2 + x L You want this to be: f @ x D 1 + 4x 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 2a + 2b + c == 1 L ; eqn2 = H 3a + b + 2c == 4 L ; eqn3 = H a + c == 0 L ; Solve @8 eqn1, eqn2, eqn3 <D 88 a Ø 7, c Ø - 7, b Ø - 3 << This tells you that when you go with a = 7, b = - 3 and c = - 7 then newf @ x D = f @ x D . a = 7; b = - 3; c = - 7; newf @ x D - 3 H 1 + x L 2 + 7 1 + x - 7 2 + x Confirm: Apart @ f @ x DD - 3 H 1 + x L 2 + 7 1 + x - 7 2 + x The point of this is that Ÿ 0 t newf @ x D x is easy to integrate by hand. It is 3 1 + x + 7 Log @ 1 + x D - 7 Log @ 2 + x D » 0 t : top = 3 1 + x + 7Log @ 1 + x D - 7Log @ 2 + x D ê .x Ø t; bottom = 3 1 + x + 7Log @ 1 + x D - 7Log @ 2 + x D ê .x Ø 0; integral = top - bottom - 3 + 3 1 + t + 7Log @ 2 D + 7Log @ 1 + t D - 7Log @ 2 + t D This is equivalent to the Mathematica calculation: Clear @ x, t D ; 0 t 4x + 1 H x + 2 L H x + 1 L 2 x If B Re @ t D ¥ - 1 »» Im @ t D ! 0, - 3t 1 + t + Log @ 128 D +
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}