I've one example and I'm wondering where these integration points y and 1 come from. According to the first red rectangular box these integration points should be 0 and 1 but they are not. Is this example correct, if yes, where these unbelivable points (y to 1) come from? I think it means integral over y to 1 but not sure if it's the correct way to say in english.

2. ## interchange of arrange of integration

Dear totalnewbie,

My suggestion to similar problem (binary intagration) is that the first step must be the drawing of the Area. (A good sketch is semi-success.)

In situation this Area is a rectangle triangle with points (0, 0), (0,1) and (1, 1). Check out!

f2(y) = by def = integ(-inf, inf) f(x,y) dx
by the integration the y is fix.
If you look at the drawing of Area, you can see: f(x, y) = 0 if y<0 or y>1, that means f2(y)=0 if y<0 or y>1.
Oke, focus attention on fix y in [0, 1], for example y=0.25.
y=0.25 is a horizontal line.
Where does this line cross the Area? Firstly at the side y=x, secondly at the side x=1, therefore between y and 1 f(x,y)=8xy, else f(x,y)=0.

So f2(y)= integ(-int, y) 0 dx + integ(y,1) 8xy dx + integ(1,inf) 0 dx = 0 + integ(y,1) 8xy dx + 0 = integ(y,1) 8xy dx.

By the way if f is probability function than it must be satisfy two conditions:
1) f(x, y) >= 0 anywhere
2) integ(-inf, inf) integ(-inf, inf) f(x, y) dx dy = 1.

The first condition is evidently true. But the second we have to control.
Check out:
integ(0, 1) f1(x) dx = integ(0, 1) f2(0, 1) dy = 1

I hope you understand anything from my poor English.

3. Originally Posted by totalnewbie
I've one example and I'm wondering where these integration points y and 1 come from. According to the first red rectangular box these integration points should be 0 and 1 but they are not. Is this example correct, if yes, where these unbelivable points (y to 1) come from? I think it means integral over y to 1 but not sure if it's the correct way to say in english.
Hi

Maybe to complete Skalkaz's answer !

If I understood well, the idea is to calculate

$\int_{Area} 8xy \,dx \,dy$

The first way is
$\int_{Area} (8xy \,dy) \,dx = \int_0^1 \int_0^x (8xy\, dy) \,dx = \int_0^1 4x^3 dx = 1$
For a given x, you have $0 \leq y \leq x$ (see green line)

The second way is
$\int_{Area} (8xy\,dx) \,dy = \int_0^1 \int_y^1 (8xy\, dx) \,dy = \int_0^1 (4y - 4y^3) dy = 1$
For a given y, you have $y \leq x \leq 1$ (see red line)