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March 20th, 2015, 02:18 PM   #1
szz
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Doubt with multivariable function

Hi all,

I have a function declared as such:

$\displaystyle f_{xy}(x,y) = \left\{
\begin{matrix}
k & \text{if} & 0 < y < x < 1\\
0 & & \text{otherwise}
\end{matrix}\right .$


And I have to integrate it.

From the fundamental theorem of calculus I know that if the integration limits are $\displaystyle a$ and $\displaystyle b$ with $\displaystyle a < b$ then:

$\displaystyle \int_a^b f(x) \,\text dx$

So, from this I deduce I have to calculate:

$\displaystyle \int_0^y\int_x^1 f_{xy}(x,y)\,\text dx \text dy$

Am I right ?

Thank you in advance, as always.
szz
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March 20th, 2015, 03:51 PM   #2
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Are you integrating over a region or a path?

Over the unit square, we'd want something like
$$\int_0^1 \int_0^1 f(x,y) \,\mathrm d x \,\mathrm d y = \int_0^1 \int_y^1 k \,\mathrm d x \,\mathrm d y$$

You can't have $x$ as a limit when integrating with respect to $x$. Also, you need $0 \lt y \lt 1$ and $y \lt x \lt 1$ because this is where $f(x,y)=k$.
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March 21st, 2015, 01:41 AM   #3
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Hi v8archie !

I recognize that at half past one o' clock in the morning my brain does not work properly...

The funcion comes from a probability problem so I think we are integrating over an area.

If the function is:

$\displaystyle
f_{xy}(x,y) = \left\{
\begin{matrix}
k & \text{if} & 0 < y < x < 1\\
0 & & \text{otherwise}
\end{matrix}\right .$


I would say that should be better to write:

$\displaystyle \int_0^x \int_y^1 k \,\text dx\text dy$

Rather then what I wrote above in my first post.

Am I right ?

Last edited by szz; March 21st, 2015 at 01:44 AM.
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March 21st, 2015, 02:00 AM   #4
szz
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Quote:
Originally Posted by v8archie View Post
Also, you need $0 \lt y \lt 1$ and $y \lt x \lt 1$ because this is where $f(x,y)=k$.
I am a little bit confused on this, respect what I wrote in post #3..
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March 21st, 2015, 03:12 AM   #5
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If I resolve the equation as post # 3 I have

$\displaystyle \begin{aligned}
& f_{xy}(x,y) = \left\{
\begin{matrix}
k & \text{if} & 0 < y < x < 1\\
0 & & \text{otherwise}
\end{matrix}\right .\\
& \int_0^x \int_y^1 k\,\text dx\text dy = \int_0^x (k - ky)\,\text dy = kx - {kx^2\over 2} + c_1 + c_2
\end{aligned}$


Also:

$\displaystyle \int_0^1 \int_y^1 k\,\text dx\text dy = k - {k\over 2} = {k \over 2} = \left.\left(kx - {kx^2\over 2}\right )\right|^1_0 $

So I think that both are equivalent but expressed in a different way.
Right ?

Last edited by szz; March 21st, 2015 at 03:20 AM.
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March 21st, 2015, 04:36 AM   #6
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They are clearly NOT equivalent since one is a function of x and the other is a constant.

Since you are integrating a function of variables x and y, with respect to x and y, over a specific region in the xy-plane the result must be a constant- a number.

Your function is defined by f(x, y)= k if 0< x< y< 1, 0 otherwise. Now, you should be able to see that this region is a right triangle. One way of thinking about this is that x can be any number between 0 and 1, so draw an xy- coordinate system and mark x= 0, x= 1 on the x- axis. Now, for each x we have "x< y< 1" so draw the boundaries, y= 1 and y= x on your graph. The first is, of course, a horizontal straight line at y= 1 and the second is the line through (0, 0) and (1, 1). y must lie between those two lines so shade the area [b]above[b] y= x and below y= 1. The region here is the triangle with vertices at (0, 0), (1, 1), and (0, 1).

Conversely, you could note that y is between 0 and 1 so mark y= 0 and y= 1 on the y-axis. Now, for each y x< y so shade the area bounded on the left by x= 0, the y-axis and on the right by y= x. You can see that this is exactly the same region as above, the triangle with vertices at (0, 0), (1, 1), and (0, 1).

So you can set up this integral in either of two ways. Corresponding to the analysis above, starting with "mark x= 0, x= 1 on the x- axis", the "outer" integral will be with respect to x and go from x= 0 to x= 1. Then "for each x we have "x< y< 1" so the "inner" integral is from x on the bottom to 1 on the top:
$\displaystyle \int_0^1\int_x^1 f(x) dy dx$

Corresponding to the next paragraph, "mark y= 0 and y= 1 on the y-axis", the 'outer integral will be with respect to y and go from y= 0 to y= 1. Then "for each y we have x< y< 1" so that x has a lower limit of 0 and a upper limit of y:
$\displaystyle \int_0^1\int_0^y f(x, y) dx dy$

For any integrable f(x,y) those will give the same answer.

However, in this particular integral, f(x,y)= k, a constant, on this region so you don't need to integrate at all! The integral is just k times the area of the region. Since the region is a right triangle with vertices at (0, 0), (1, 1), and (0, 1), having sides of length 1, (so half the unit square) its area is 1/2.

You should do both integrals above to see that
$\displaystyle \int_0^1\int_x^1 k dy dx= \int_0^1\int_0^y f(x, y) dx dy= k/2$.
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