# How do you set up an integral integrating two functions within a domain?

How do you set up an integral integrating two functions within a domain??

## Homework Statement

I have to integrate (2 + x + y) within the domain that is the area between 0 and 1 and (x+y<=1)
I know how to integrate well, I think it's a double integral but I'm not really sure what the range is so not sure what to integrate from and to? Any help will be much appreciated. Thanks. I haven't done this kind of work since 1996 so it's been a while!

Related Calculus and Beyond Homework Help News on Phys.org
HallsofIvy
Homework Helper

## Homework Statement

I have to integrate (2 + x + y) within the domain that is the area between 0 and 1 and (x+y<=1)
I know how to integrate well, I think it's a double integral but I'm not really sure what the range is so not sure what to integrate from and to? Any help will be much appreciated. Thanks. I haven't done this kind of work since 1996 so it's been a while!
If I read this correctly, you are to integrate f(x,y)= 2+ x+ y over the region bounded by x= 0, x= 1, y+ x= 1 and y= 0. (I added the last: without it or something similar the region is unbounded. Yes, that will be a double integral for two reasons: you are integrating a function of two variables and the region over which you are integrating is two dimensional.

For any problem like this, you should draw a picture. Since the line x+ y= 1 goes throught both (1,0) and (0,1), that, together with x= 0 and y= 0, will give you a triangular region. You now need to decide in which order you want to integrate.

If you decide to integrate with respect to y first, then with respect to x, you know that the limits of the "outer integral" (dx) must be numbers. Clearly x must range from 0 to 1 so the integral is from x= 0 to x= 1. Now, for each x, how must y range? draw a vertical line anywhere inside your triangle and look at it. The lower end is at the x-axis (y= 0) and the upper end is at x+ y= 1 or y= 1- x. The integral is
$$\int_{x=0}^1\int_{y= 0}^{1-x} (2+ x+ y)dydx$$
Although most texts don't do it, I think it is a very good idea to write the "x= " and "y= " on the limits of integration like that.

If you decide to integrate with respect to x first, then with respect to y, you know that the limits of the "outer integral" (dy) must show the total range of y: y must range from 0 to 1. For each y x ranges from x= 0 on the left to x= 1- y on the right:
$$\int_{y=0}^1\int_{x= 0}^{1- y}(2+ x+ y)dxdy$$

Obviously those are exactly the same because of the symmetry of this problem.