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Math Help - Integral-Area of the region

  1. #1
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    Integral-Area of the region

    I have a problem that i am completely stuck on, stuck to the point that i don't even know how to approach it. If someone could simply get me started i can do the rest. Any help is appreciated, thanks.


    Problem: The area of the region bounded by the graphs of y=x^3 and y=x cannot be found by the single integral ∫(ranging from -1 to 1) [X^3-X]Dx. Explain why this is so. Use symmetry to write a single integral that does represent the area.
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  2. #2
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    Quote Originally Posted by Juggalomike View Post
    I have a problem that i am completely stuck on, stuck to the point that i don't even know how to approach it. If someone could simply get me started i can do the rest. Any help is appreciated, thanks.


    Problem: The area of the region bounded by the graphs of y=x^3 and y=x cannot be found by the single integral ∫(ranging from -1 to 1) [X^3-X]Dx. Explain why this is so. Use symmetry to write a single integral that does represent the area.
    \int_{-1}^{1}(x^3-x) dx = 0 can easily be seen both from the symmetry of f(x) = x^3 - x and from simply solving the integral, but we can see that the area of the region is definitely not 0!

    Let A be the region bounded by the graphs of g(x) = x^3 and h(x) = x, then, using the same argument as before (symmetry), we can conclude that A = \int_{0}^{1} (x - x^3) dx + \int_{-1}^{0} (x^3 - x) dx = 2*\int_{0}^{1} (x^3 - x) dx
    Last edited by Defunkt; August 25th 2009 at 12:52 PM. Reason: typo
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  3. #3
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     \int^{0}_{-1}(x^{3}-x) dx + \int^{1}_{0} (x-x^{3}) \ dx = 2\int^{1}_{0}(x-x^{3}) dx

    too slow
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  4. #4
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    thanks a lot for the help, was a lot simpler then i was making it out to be
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