{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2.07.PatProcedures

# 2.07.PatProcedures - 0.327079 Accumulation Authors Bill...

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

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 +

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}