# Thread: is this double integral setup right? f(x,y) = kxy , x>=0, y>= 0 , 20 <= x + y <= 30

1. ## is this double integral setup right? f(x,y) = kxy , x>=0, y>= 0 , 20 <= x + y <= 30

So f(x,y) = kxy, it's a joint probability density function and I must find k so that the integral over the valid range is 1.

My multivariate calculus is a little rusty and my first attempt which looked like (attached) yielded a k of -0.0012 and an integral of 1.

However I solved for k by double integrating with the calculator and setting the to 1 so of course it will yield 1 for the given interval, what I want to confirm is that the integral was setup properly.

thanks!

2. You shouldn't have an x variable in the inner bounds of integration since you will end up xs when you do the last integration.

3. OOPS! wrote it wrong in word - the inner integral is dy and outer is dx. Correct then?

4. With dy and dx switched, there is still a problem.

If you sketch the region on the plane. It is a trapezoid.
The bounds for inner integral are different in the regions
x=0 to x=20
x=20 to x=30
because of the y >= 0 condition.

5. could you propose how to properly setup the double integral? Is the correct region shown below?

6. Yes, the region is correct.
What are the bounds for y when x is a certain value?
This is the same thing as asking, what are the bounds when you take a vertical line slice of a region.

What are the bounds for y between x = 0 and x = 20?
What are the bounds for y between x = 20 and x = 30?

Almost there

7. so for x between 0 - 20
y can be [20-x, 30-x]

for x between 20 - 30
y can be [0, 30-x]

how can I show this in the integral?

8. would it need to look like this?

and anyone know why only my first attachment is showing itself in the post, and the others need to be clicked on?

9. Yup, so now split into 2 integrals and add them.

The first one integrates from x=0 to 20
and the second integrates from x=20 to 30.

Yes, exactly what you have in the attachment (I have to click on it for it to open)
Excellent work.