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Math Help - Order of integration again

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Order of integration again

    Is it possible to set up an integral such that the region is impossible to describe?

    \int_0^1\int_1^x{f(x,y)}~dy~dx

    Because we would then have that

    0\leq{x}\leq{1}

    But

    1\leq{y}\leq{x}

    So basically this makes no sense yeah?

    Because the only value that would make sense is x=1

    Because we have that x is always less than or equal than one but always greater than or equal to one?
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  2. #2
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    Quote Originally Posted by Mathstud28 View Post
    Is it possible to set up an integral such that the region is impossible to describe?

    Mr F says: Obviously it is. The example you offer below is a case in point.

    \int_0^1\int_1^x{f(x,y)}~dy~dx

    Because we would then have that

    0\leq{x}\leq{1}

    But

    1\leq{y}\leq{x}

    So basically this makes no sense yeah?

    Because the only value that would make sense is x=1

    Because we have that x is always less than or equal than one but always greater than or equal to one?
    Note: \int_0^1\int_x^1{f(x,y)}~dy~dx makes perfect sense.

    I could say a bit more but in another thread I promised not to draw attention to your unfortunate handicap.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by mr fantastic View Post

    Note: \int_0^1\int_x^1{f(x,y)}~dy~dx makes perfect sense.
    Following up on this, Mathstud28's integral can be rewritten with a minus sign in front, the rest follows by an inequality game.
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  4. #4
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    Quote Originally Posted by Mathstud28 View Post
    Is it possible to set up an integral such that the region is impossible to describe?
    \int_0^1\int_1^x{f(x,y)}~dy~dx
    Let R be a region (we are defining "region" to be a bounded subset of \mathbb{R}^2). Then \iint_R f is the double integral over f on this region. And we define a double integral to be the limit of the Riemann sums again, this time it is a little more involved because we use double Riemann sums. It turns out that if R = [a,b]\times [c,d] i.e. the rectangle a\leq x\leq b and c\leq y\leq d then \iint_{[a,b]\times [c,d]} f = \int_a^b \int_c^d f(x,y) dy ~ dx (if f is sufficiently well-behaved) this is Fubini's theorem. It tells us how to compute the limiting double Riemann sums in terms of ordinary integrals. Fubini's theorem can be altered to allow non-rectangular regions. My point is it is always possible to have integral defined over arbitrary regions, even if they are crazy to describe. But if you want to use Fubini's theorem you need to set it up in this nice form. Sometimes no, sometimes you need to break up the region into two parts.
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