1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Large terminal integration

  1. Sep 5, 2015 #1
    1. The problem statement, all variables and given/known data

    ∫1/((x^1/3)-2)dx between b=666 and a=8

    Aim is to prove divergence or convergence

    2. Relevant equations


    3. The attempt at a solution

    Okay so I know I can prove for 4x1/3 < x when 8 < x but I don't know where to go from here. There is obviously very large terminal values, and i know you could say evalute each terminal values using fundamental theorem of calculus..

    Any easier way to do it?
     
  2. jcsd
  3. Sep 5, 2015 #2
    The integrand is not continuous over [a, b], so you can't use the fundamental theorem of calculus (directly).

    Try the substitution u^3 = x and write the integrand in terms of u. Make sure you change the limits of integration, too. Then try v = u - 2. Then things will look a little more obvious.
     
  4. Sep 5, 2015 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The original problem appears to be to integrate from 2 to 666. There is, of course, a problem at x= 2 but since the problem does NOT require integration to infinity, I do not understand Ray Vickson's "(ii) whether the function grows small quickly enough to yield a finite integration as the upper limit goes to infinity."
     
  5. Sep 5, 2015 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I had mid-read the original problem; there is no issue at the lower end ##x = 2##. And, somehow, I thought that the OP was trying to figure out whether the integral ##\int_ a^{\infty} f(x) \, dx## exists, by looking at ##\int_a^b f(x) \, dx## for some large, finite ##b = 666## and trying to get a conclusion or insight from that. However, when I go back and re-read the original, I see that no such statement seems to have been made, so the question is actually a more-or-less pointless exercise except, of course, for actually doing the integral, which is elementary, but tedious. Anyway, I deleted my previous post.
     
  6. Sep 5, 2015 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    But there is an issue at x= 2 where the integrand is not defined.
     
  7. Sep 5, 2015 #6

    Mark44

    Staff: Mentor

    I agree that there is no problem at x = 2, but the original integral is over the interval [8, 666], and there is a problem at x = 8.
    No, the original integral is ##\int_8^{666}\frac {dx}{\sqrt[3]{x} - 2}##
     
  8. Sep 5, 2015 #7
    Okay essentially all I want to prove is convergence or divergence of the integral over the [8,666]. I know no problem exists for x=2 but rather the problem exists at x=8

    Using substitution as u= x1/3 and x= x3 i get ∫3*u2/(u-2)

    I than reapply v = u-2 and and therefore u = v + 2 and integrate respectively. But my challenge is, think it is asking you to use some sort of comparison rather than evaluating the integral to calculate the convergence/divergence? Would the comparison test work?
     
    Last edited: Sep 6, 2015
  9. Oct 2, 2015 #8
  10. Oct 3, 2015 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Once you have the integral [itex]3\int \frac{u^2}{u- 2} du[/itex] do the division to get [itex]3\int (v+ 2+ \frac{4}{v-2})dx[/itex]. The first two terms are easily integrable and you can "compare" the last to integrating [itex]\frac{1}{x}[/itex] in the vicinity of x= 0.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Large terminal integration
  1. Large number (Replies: 4)

Loading...