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Definite Integral

  1. Jan 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider ##f(x)=4x^4-24x^3+31x^2+6x-8## be a polynomial function and ##\alpha, \beta, \gamma, \delta## are the roots of the equation ##f(x)=0##, where ##\alpha < \beta < \gamma < \delta##. Let sum of two roots of the equation f(x)=0 vanishes. Then the value of
    [tex]\int_{2\alpha}^{2\beta} \frac{x^{\delta+1}-5x^{\gamma+1}+2\beta |x|+1}{x^2+4\beta |x|+1}dx[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I succeeded in finding the roots. ##\alpha=\frac{-1}{2}, \beta=\frac{1}{2}, \gamma=2## and ##\delta=4##.
    Therefore, the integral is
    [tex]\int_{-1}^{1} \frac{x^5-5x^3+|x|+1}{x^2+2|x|+1}dx[/tex]
    I don't know how to proceed further from here. The only thing I can think of is to write the integral in two parts,
    [tex]\int_{-1}^{0} \frac{x^5-5x^3-x+1}{(x-1)^2}dx +\int_{0}^{1} \frac{x^5-5x^3+x+1}{(x+1)^2}dx [/tex]

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Jan 30, 2013 #2

    haruspex

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    That's a very sensible step. Next, can you recombine them, by change of variable? You should get some cancellation.
     
  4. Jan 30, 2013 #3
    Great! Thanks a lot. That thing never came into my mind. :smile:

    I substituted t=-x in the first integral and the terms cancelled. :)

    Is the answer 2ln2?
     
    Last edited: Jan 30, 2013
  5. Jan 30, 2013 #4

    Dick

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    Yes. But what a strange question. Why did they tell you to suppose two of the roots sum to zero? If you can solve the equation as you did you don't need to be told that.
     
  6. Jan 30, 2013 #5
    I am not sure about why did they give that piece of information. This question is a part of paragraph based questions. There are three questions. I was able to solve the other two so I did not post them here. Do you want me to post the other two questions too?

    Can you help me in this thread: https://www.physicsforums.com/showthread.php?t=668114
     
    Last edited: Jan 30, 2013
  7. Jan 30, 2013 #6

    Dick

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    Sure, post them. Maybe that will give us clue about the strange assumption. I did look at the other thread. And I don't know. g'(1/2)=f(1/2). Haven't figured that one out yet either.
     
  8. Jan 30, 2013 #7
    Here goes the questions:

    Q.1) The value of the expression
    [tex]\delta^{\beta}+\frac{1}{\delta^{\alpha}}+\delta^{ \gamma }+\gamma^{\delta}[/tex]
    is:

    Q.2) [tex]\int \left(\frac{x-\delta}{x-\gamma}\right)^{\alpha+\beta+\delta}dx[/tex]
     
  9. Jan 31, 2013 #8

    jbunniii

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    I thought that was pretty strange, too. But I guess it does help you find the roots: if [itex]c[/itex] and [itex]-c[/itex] are both roots, then we have
    $$0 = f(c)=4c^4-24c^3+31c^2+6c-8 = 4c^4 + 24c^3 + 31c^2 - 6c - 8 = f(-c)$$
    so
    $$48c^3 - 12c = 0$$
    As ##c = 0## is not a root of ##f##, we must have
    $$4c^2 - 1 = 0$$
    and so ##c = \pm 1/2##
     
  10. Jan 31, 2013 #9
    Through observation, I was able to find one root. Then figuring out the other roots was easy. But thanks for the alternative way, if all the roots would have been fractional, I would never be able to find the roots through observation.
     
  11. Jan 31, 2013 #10

    jbunniii

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    I bet you could, if you use the rational root theorem:
    http://en.wikipedia.org/wiki/Rational_root_theorem
     
  12. Jan 31, 2013 #11
    Ah, I always forget about that.
     
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