Calc08_3day2 - 8.3 day 2 Tests for Convergence Greg Kelly...

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

8.3 day 2 Tests for Convergence Greg Kelly, Hanford High School, Richland, Washington
Image of page 1

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

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 lim 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.)
Image of page 2
1 x e dx - 1 lim b x b e dx - →∞ 1 lim b x b e - →∞ - ( 29 1 lim b b e e - - →∞ - - - 1 1 lim b b e e →∞ - + 0 1 e = Converges
Image of page 3

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

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 <
Image of page 4
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
Image of page 5

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

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 2 ( 29 a f x dx
Image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern