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Simplify definite integral formula

  1. Jan 2, 2012 #1

    jvo

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    1. The problem statement, all variables and given/known data
    Working my way through Zienkiewicz's "FEM its basis & fundamentals" I am stuck on the following example:
    [itex]
    - \int_{y_1}^{y_2} t N1 \left\{ \begin{array}{c} \sigma_x \\ \tau_y \end{array} \right\} dy =
    -\left\{ \begin{array}{c} k_x t (2 y_1 + y_2) (y_2 - y_1) / 6 \\ 0 \end{array} \right\}
    [/itex]

    with values:
    [itex] N1 = \frac{ y_2 - y }{ y_2 - y_1 } [/itex]
    [itex] \sigma_x = k_x y [/itex]
    and [itex] \tau_y = 0 [/itex]

    2. Relevant equations
    n/a

    3. The attempt at a solution
    My result for the above integral is: [itex] k_x t \frac{ y_2^3 - 3 y_2 y_1^2 + 2 y_1^3 }{6 (y_2 - y_1)} [/itex]

    Tried to simplify the formula by dividing by (y2 - y1) and (y2-y1)^2 but I don't see how to get to the same result as in the book. What am I doing wrong here?
    For more detail, see the attachments with the page from the book and my results.
    Thanks in advance.
    Jo
     

    Attached Files:

  2. jcsd
  3. Jan 2, 2012 #2

    Dick

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    Science Advisor
    Homework Helper

    It is the same thing as in the book. [itex]y_2^3 - 3 y_2 y_1^2 + 2 y_1^3=(2 y_1 + y_2) (y_2 - y_1)^2[/itex]. You just have to figure out how to factor the numerator.
     
  4. Jan 3, 2012 #3

    jvo

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    Thanks for your reply but I still don't see how to get there. Can you give a hint on how to approach this? As said in my post I tried long division by [itex] (y_2 - y_1) [/itex] and [itex] (y_2 - y_1)^2 [/itex] but don't get even near your factored result.
    I know factoring quadratics and checked 'recipes' as given here: http://www.analyzemath.com/polynomials/factor_polynomials.html
    It's selfstudy, not homework. I realise missing something essential here, but what is it?
    Tia,
    Jo

    PS this is a scan of some long division trials: http://utopia.knoware.nl/~jovo/images/Scan_img_026.jpg
     
    Last edited: Jan 3, 2012
  5. Jan 3, 2012 #4

    Dick

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    You started off ok on the first trial. [itex]y_2^2[/itex] is good for the first term. Now keep the rest ordered in decending powers of [itex]y_2[/itex]. So what you bring down after the first subtraction is [itex]y_2^2 y_1 - 3 y_2 y_1^2[/itex]. Your next term in the quotient should be dividing [itex]y_2[/itex] into [itex]y_2^2 y_1[/itex] (the higher [itex]y_2[/itex] power term). Or [itex]y_2 y_1[/itex]. Try it that way. It does divide evenly.
     
    Last edited: Jan 3, 2012
  6. Jan 3, 2012 #5

    jvo

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    I got it! Thank you for your excellent help.
    My new year's resolution is to practice factoring more :-)
    Regards,
    Jo
     
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