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I really need to know

  1. Jan 10, 2004 #1
    i really need to know....

    integration of:
    x/(x^4 + 2x^2 + 1)

    i tried solving that using (ln), u=g(x), and alot of ways, but i couldn't get an answer

    Is the following argument valid? Explain
    f(x) =
    1 if (x) is rational and (x)>0
    -1 if (x) is rational and (x)<0
    0 if (x) is irrational

    is defined for all numbers in [-1,1] and has the property that f(-x)=-f(x) for all (x) in [-1,1].
    Thus, definite integration{0 to 1 of f(x).dx}=0
    i couldn't understand the question, and what is meant by "irrational", i think all values (even integers) can be rational

    thank u alot, i wish to reply as soon as possible, my exam is TOMORROW.... and any efforts will be appreciated
  2. jcsd
  3. Jan 10, 2004 #2
    For first part
    [tex]\int \frac{x dx}{x^4+2x^2+1} = \int \frac{x dx}{(x^2+1)^2}[/tex]

    now substitute x^2+1=t u will get 2xdx=dt and so on u get
    [tex]\int \frac{dt}{2t^2}=-\frac{1}{2t}[/tex]
  4. Jan 10, 2004 #3
    Man for Exams and these types of math go and consult:


    I think here you'll find much help :)
  5. Jan 10, 2004 #4
    I believe it should be in Homework Section And nothing else
    Last edited: Jan 10, 2004
  6. Jan 10, 2004 #5
    Re: i really need to know....

    A number that can be expressed as a fraction where p and q are integers and , is called a rational number with numerator p and denominator q. Numbers that are not rational are called irrational numbers.

    For the validness...I am not sure...first you've to define the function!

    But as the function seems odd, of course the integration will be 0
  7. Jan 10, 2004 #6
    Nonperiodic decimal fractions are called irrational numbers.

    It would be zero only when integration is of the form [tex]\int^a_{-a}[/tex]

    For the second part

    Any number on x axis has two possibility either rational OR irrational for rational f(x)=1 whereas fo irational it is = 0,

    A rational point is surrounded by two irrational points and vice versa So clearly the function is discontinuous

    And function consists of a collection of points as defined by function

    integral wont be zero in this range i.e from [0,1]
    Last edited: Jan 10, 2004
  8. Jan 10, 2004 #7
    The second function is definitely not integrable. On the interval [0,1] any upper sum will be 1 and any lower sum will be 0. Thus there is no value for the integral.
  9. Jan 10, 2004 #8

    This is a fascinating tidbit I don't recall learning. Is it taught in calculus? I'd like to learn more about this idea, maybe see a proof if it isn't too complicated. Can you suggest a reference?
  10. Jan 10, 2004 #9
    Thanks! Himanda! I read once and didn't watched the limits :(

    and mmwave as I told you go to sosmath and there you find lots of students who can share math ideas with you :)
  11. Jan 10, 2004 #10
    Suppose that you have a rational point not surrounded by two irrational points. So you have two rational numbers "a" and "b" with no irrational number between them. Thus the interval [a,b] contains only rational numbers, even over the reals.

    But we clearly have a bijection between [a,b] and [0,1] (just apply a translation and scale the interval). Thus both [a,b] and [0,1] have the same cardinality.

    Yet [a,b] only contains rational numbers and so it must be countable. But we can show that [0,1] is uncountable. Thus there cannot exist a bijection between [a,b] and [0,1]. This is a contradiction. Thus every rational is surrounded by two irrationals.
  12. Jan 10, 2004 #11
    The function is not Riemann integrable. But there are other sorts of integration, namely Lebesgue integration, that can handle such functions.
  13. Jan 10, 2004 #12
    I didn't want to confuse the original poster too much. I figured bringing up Lebesgue integration (or other types of integration) would do that.
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