11.3 Integral Test-solutions

11.3 Integral Test-solutions - im(gi768 11.3 Integral Test...

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

im (gi768) – 11.3 Integral Test – sadun – (53088) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Determine whether the series summationdisplay k =1 1 k ln(2 k ) is convergent or divergent. 1. series diverges correct 2. series converges Explanation: The function f ( x ) = 1 x ln(2 x ) is continous, positive and decreasing on [1 , ). By the Integral Test, therefore, the series summationdisplay k =1 1 k ln(2 k ) converges if and only if the improper integral integraldisplay 1 f ( x ) dx = integraldisplay 1 1 x ln(2 x ) dx converges, i.e. , if and only if lim t → ∞ integraldisplay t 1 1 x ln(2 x ) dx exists. To evaluate this last integral, set u = ln(2 x ). Then du = 1 x dx , in which case, integraldisplay t 1 1 x ln(2 x ) dx = integraldisplay ln(2 t ) ln2 1 u du = ln(ln(2 t )) ln(ln 2)) . But lim t → ∞ ln(ln(2 t )) = . Consequently, lim t → ∞ integraldisplay t 1 1 x ln(2 x ) dx does not exist. The Integral Test thus ensures that the given series diverges . 002 10.0points Determine whether the series summationdisplay n =7 6 n 2 converges or diverges. 1. series is convergent 2. series is divergent correct Explanation: We apply the integral test with f ( x ) = 6 x 2 . Now f is continuous, positive and decreasing on [7 , ). Thus the series summationdisplay n =7 6 n 2 converges if and only if the improper integral integraldisplay 7 6 x 2 dx converges. But integraldisplay 7 6 x 2 dx = lim t → ∞ integraldisplay t 7 6 x 2
Image of page 1

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

Image of page 2
This is the end of the preview. Sign up to access the rest of the document.
  • Fall '07
  • Sadler
  • lim

{[ 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