Consider ∫R∫syDA, where R is the square R=[0,1]x[0,1]. Let us compute some Riemann sums. For any positive integer n we partition R into n^2 little squares Rij each with sides 1/n long. Within each little square we evaluate the integradn at the point (x*ij,y*ij), which is the upper right corner.

The first thing i'm having problems understanding with this problem is why that did it need to tell me that each square has an area of n^2 and that the lengths and width is 1/n long.

I have no problem calculating riemann sums if they let me pick how much I partition it myself or actually giving me a set number to partition them.

so if each side is 1/n long and the area of the squares is n^2 does that mean i can pick how to partition it as long as both ways i partition it (n and m) are equal?

also is there any deeper meaning to say that (x*ij,y*ij) are in the upper right corner or does that not even matter?

i meant ∫R∫xydA

In order to understand the problem and calculate Riemann sums properly, it is important to know the characteristics of the little squares in the partition. Let's break it down:

1. Each square has an area of n^2: This means that the total area of the region R (the square) can be calculated by multiplying the number of little squares in the partition (n^2) by the area of each square. This is a way to represent the size of the partition.

2. The lengths and width of each square are 1/n long: This information tells you about the dimensions of each little square. Since the total length of one side of the square is 1, dividing it into n smaller squares means each side of the little square will be 1/n long.

3. Can you pick the partition size? In this problem, the partition size is given as n^2 little squares. You don't have the freedom to choose a specific value for n or m independently. The author of the problem has chosen n^2 as the partition size to compute the Riemann sums.

4. Does the specific location of (x*ij, y*ij) matter? In Riemann sums, the specific choice of the point within each little square where you evaluate the integrand (here, (x*ij, y*ij)) is important. The author has specifically chosen the upper right corner of each little square as the point for evaluation. This is a choice made to simplify the calculations and make the Riemann sums more manageable. The choice of a different point within each square may lead to slightly different results, but the overall concept remains the same.

Understanding these details helps you set up and compute Riemann sums more accurately in this specific problem.