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Math Help - Integrals #6 and #7

  1. #1
    Super Member Random Variable's Avatar
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    Integrals #6 and #7

    Challenge Problems:


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


     \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  \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.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Random Variable View Post
    Challenge Problems:


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

    A simple Taylor/Laurent expansion of  e^{x /a} and  e^{b /x} should prove my assertion.
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  3. #3
    Super Member Random Variable's Avatar
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    Quote Originally Posted by chiph588@ View Post
     \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx =0

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

    The solution I had in mind is even simpler than that.
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  4. #4
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    If my calculations are correct , the integrals are both zero .


    For the first one ,

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


    and we will find that it is equal to

     \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

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

    Sub.  x = 1-t and we have

     I = -I

     I = 0
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by simplependulum View Post
    If my calculations are correct , the integrals are both zero .


    For the first one ,

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


    and we will find that it is equal to

     \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

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

    Sub.  x = 1-t and we have

     I = -I

     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.
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Random Variable View Post
     \int^{1}_{0} \sqrt[3]{2x^{3}-3x^2-x+1} \ dx
    Let  x=t+\tfrac12

    We get  \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 =  \int^{\tfrac12}_{-\tfrac12} \sqrt[3]{2t^3-\tfrac52t} \ dx = 0 since  \sqrt[3]{2t^3-\tfrac52t} is odd.
    Last edited by chiph588@; June 1st 2010 at 07:17 AM.
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