# Thread: Order of integration again

1. ## 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?

2. Originally Posted by Mathstud28
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.

3. Originally Posted by mr fantastic

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.

4. Originally Posted by Mathstud28
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.