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Integral Expressions

  1. Feb 26, 2009 #1


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    I'm asked to evaluate:

    [tex]\int_{-1}^{1} x(x-1)(x+1)dx[/tex]

    The problem is that the question doesn't say anything about area, and we have been notified in class that unless otherwise stated, don't express the answer in sq. units.
    However, I also have a problem that, if this were to find the area on a graph, I would notice the symmetry about the origin and y-axis and thus express the integral as:

    [tex]-2 \int_{0}^{1} x(x-1)(x+1)dx[/tex]

    Now, the first integral (if not area) is equal to 0, while the second is not. Which would be the correct answer?
  2. jcsd
  3. Feb 26, 2009 #2


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    They do not give the same answer, because they are not equivalent.
    You said that there is a symmetry around the origin and y-axis, but note that it is actually an anti-symmetry. That is, if f(x) = x(x - 1)(x + 1), then f(-x) = -f(x) and not f(-x) = f(x). You can note this by sketching the graph, or noticing that (x - 1)(x + 1) is symmetric and x is anti-symmetric, so their product is anti-symmetric.
    So you are arguing
    [tex]\int_{-1}^{1} f(x)dx=\int_{-1}^{0} f(x) dx+\int_{0}^{1} f(x) dx[/tex]
    If you substitute x to -x in the first integral they don't add up, they cancel out.
  4. Feb 26, 2009 #3


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    Thats on behalf of my poor attempt at putting my thoughts into words.
    By symmetry, I meant that they are odd functions, as you stated.

    I see you are heading into the direction of making it equal zero. If the question were asked slightly different. Say, rather than 'evaluate' the question asked 'find the area of', I would instead convert the integral so as not to get a value of zero.

    Basically my question is, if the question doesn't mention area, am I safe to assume this integral equals zero? or would I have to find it as if it were finding area/volume etc. or would I have to find both?

    p.s. my teacher is really picky on the details and will try oppose any 100% marks as I've noticed from a previous assessment task.
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