Results 1 to 6 of 6

Thread: Integrals #6 and #7

  1. #1
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3

    Integrals #6 and #7

    Challenge Problems:


    $\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx $


    $\displaystyle \int^{1}_{0} \sqrt[3]{2x^{3}-3x^2-x+1} \ dx $



    EDIT: I don't like the second integral. It doesn't even make any sense unless you state that $\displaystyle \sqrt[3]{2x^{3}-3x^2-x+1} $ is a real-valued function. Sorry. It's a bad problem.



    Moderator edit: Approved Challenge question.
    Last edited by mr fantastic; May 31st 2010 at 01:14 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by Random Variable View Post
    Challenge Problems:


    $\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx $
    $\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx =0 $

    A simple Taylor/Laurent expansion of $\displaystyle e^{x /a} $ and $\displaystyle e^{b /x} $ should prove my assertion.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Quote Originally Posted by chiph588@ View Post
    $\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx =0 $

    A simple Taylor/Laurent expansion of $\displaystyle e^{x /a} $ and $\displaystyle e^{b /x} $ should prove my assertion.
    Could you elaborate?

    The solution I had in mind is even simpler than that.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jan 2009
    Posts
    715
    If my calculations are correct , the integrals are both zero .


    For the first one ,

    Sub. $\displaystyle x = \frac{ab}{t} $ in the second integral .


    and we will find that it is equal to

    $\displaystyle \int_a^b \frac{e^{x/a}}{x}~dx - \int_a^b \frac{e^{t/a}}{t}~dt $

    And the second integral , i find that it is identical to

    $\displaystyle \sqrt[3]{(2x-1)[(x-\frac{1}{2})^2 - \frac{5}{4}]} $

    Sub. $\displaystyle x = 1-t $ and we have

    $\displaystyle I = -I $

    $\displaystyle I = 0 $
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Quote Originally Posted by simplependulum View Post
    If my calculations are correct , the integrals are both zero .


    For the first one ,

    Sub. $\displaystyle x = \frac{ab}{t} $ in the second integral .


    and we will find that it is equal to

    $\displaystyle \int_a^b \frac{e^{x/a}}{x}~dx - \int_a^b \frac{e^{t/a}}{t}~dt $

    And the second integral , i find that it is identical to

    $\displaystyle \sqrt[3]{(2x-1)[(x-\frac{1}{2})^2 - \frac{5}{4}]} $

    Sub. $\displaystyle x = 1-t $ and we have

    $\displaystyle I = -I $

    $\displaystyle I = 0 $

    Nice.

    If for the first integral you made the same substitution but didn't break it up into two integrals, you would get I = -I.

    And for the second integral, Maple claims the answer is a complex number and Mathematica doesn't know what to do with it. The issue is the cube root of negative real numbers.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by Random Variable View Post
    $\displaystyle \int^{1}_{0} \sqrt[3]{2x^{3}-3x^2-x+1} \ dx $
    Let $\displaystyle x=t+\tfrac12 $

    We get $\displaystyle \int^{1}_{0} \sqrt[3]{2x^{3}-3x^2-x+1} \ dx = \int^{\tfrac12}_{-\tfrac12} \sqrt[3]{2(t+\tfrac12)^{3}-3(t+\tfrac12)^2-(t+\tfrac12)+1} \ dx = $ $\displaystyle \int^{\tfrac12}_{-\tfrac12} \sqrt[3]{2t^3-\tfrac52t} \ dx = 0 $ since $\displaystyle \sqrt[3]{2t^3-\tfrac52t} $ is odd.
    Last edited by chiph588@; Jun 1st 2010 at 07:17 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Contour Integrals (to Evaluate Real Integrals)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Jan 17th 2011, 09:23 PM
  2. integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 20th 2010, 01:54 PM
  3. Replies: 1
    Last Post: Dec 6th 2009, 07:43 PM
  4. Integrals and Indefinite Integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Nov 9th 2009, 04:52 PM
  5. Some integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Aug 20th 2008, 01:41 AM

Search Tags


/mathhelpforum @mathhelpforum