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Math Help - Help understanding an integral rule

  1. #1
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    Help understanding an integral rule

    Hi again

    I wonder why

    \int_{-\pi}^{0} |\varphi|\sqrt{\varphi^2 +4}d\varphi = \int_{0}^{\pi} \varphi\sqrt{\varphi^2 +4}d\varphi


    Since one are suppose to always submit the whole problem:
    r=\varphi^2, -\pi\leq\varphi\leq\pi Calculate curvelength.

    I used: ds=\int_{-\pi}^{\pi} \sqrt{\varphi^4 +(2\varphi)^2}d\varphi
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  2. #2
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    Hello,

    Substitute t=-\varphi

    And you'll get the new integral
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  3. #3
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    I didn't really manage to use your method but I called a friend and he said that curvelenght/area for \int_{-a}^0 always is equal to \int_0^{a} so I guess that I will just remeber that rule. But is that really true? What for \int_{-1,9}^0\frac{dx}{x-2}
    Last edited by kallekall; July 24th 2009 at 05:28 AM. Reason: forgot to write dx
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  4. #4
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    Quote Originally Posted by kallekall View Post
    curvelenght/area for \int_{-a}^0 always is equal to \int_0^{a} so I guess that I will just remeber that rule. But is that really true? ]
    Well of course that is not true.

    But notice that f(x) = \left| \varphi  \right|\sqrt {\varphi ^2  + 4} is an even function.

    Then for any even function we have \int\limits_{ - a}^a f  = 2\int\limits_0^a f
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  5. #5
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    Moral, do not use rules you do not understand!

    If f(x) is an even function, that is f(-x)= f(x), then \int_{-a}^0 f(x) dx= \int_a^0 f(-u)(-du)= \int_0^a f(u)du, using the substitution u= -x.

    So, again for an even function, \int_{-a}^a f(x)dx= \int_{-a}^0 f(x)dx+ \int_0^a f(x)dx= \int_0^a f(u)du+ \int_0^a f(x)dx= 2\int_0^a f(x)dx

    If f(x) is an odd function, that is f(-x)= -f(x), then \int_{-a}^0 f(x) dx= \int_a^0 -f(-u)(-du)= -\int_0^a f(u)du, using the substitution u= -x.

    Then, for an odd function, \int_{-a}^a f(x)dx= -\int_{0}^a f(x)dx+ \int_0^a f(x) dx= 0.

    If f(x) is neither even nor odd, neither of those is true.
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  6. #6
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    Thanks guys! You have helped me a lot!
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