Calc08_4day2

# Calc08_4day2 - 8.4 day 2 Tests for Convergence Riverfront...

This preview shows pages 1–7. Sign up to view the full content.

8.4 day 2 Tests for Convergence Riverfront Park, Spokane, WA Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 2006

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

View Full Document
Review: 1 P dx x 0 P 1 P x dx - 1 lim b P b x dx - →∞ 1 1 1 lim 1 b P b x P - + →∞ - + 1 1 1 1 1 P P b b P P - + - + - - + - + If then gets bigger and bigger as , therefore the integral diverges . 1 P 1 P b - + b → ∞ If then b has a negative exponent and , therefore the integral converges . 1 P 1 0 P b - + (P is a constant.)
1 x e dx - 1 lim b x b e dx - →∞ 1 lim b x b e - →∞ - ( 29 1 b b e e - - →∞ - - - 1 b e e →∞ - + 0 1 e = Converges

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

View Full Document
Does converge? 2 1 x e dx - Compare: to for positive values of x . 2 1 x e 1 x e For 2 2 x 1 1 1, e x x x x e e e <
2 1 x e 1 x e For 2 2 x 1 1 1, e x x x x e e e < Since is always below , we say that it is “bounded above” by . 2 1 x e 1 x e 1 x e Since converges to a finite number, must also converge! 1 x e 2 1 x e

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

View Full Document
Direct Comparison Test: Let f and g be continuous on with for all , then: [ 29 , a ( 29 ( 29 0 f x g x x a
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 02/12/2011 for the course MAC 2233 taught by Professor Smith during the Spring '08 term at University of Florida.

### Page1 / 11

Calc08_4day2 - 8.4 day 2 Tests for Convergence Riverfront...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online