Consider ∫R∫xydA, 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?
The information provided in the problem about the length and width of the little squares helps in understanding the partitioning of R and how the Riemann sums are calculated.
By saying that each little square has sides 1/n long, it means that the length and width of each little square are equal to 1/n. This information is essential because it helps determine the size of the little squares and the number of squares needed to cover the entire region R. Since R is a square with sides of length 1, dividing each side into n parts ensures that there will be n little squares along each side, resulting in a total of n^2 squares to cover the entire region.
The area of each little square is indeed n^2. This means that the area of the entire region R, which is a square with sides of length 1, can be divided into n^2 little squares, each with an area of n^2. Therefore, it is not possible to freely choose the partition size or the number of little squares while still maintaining the same area for each square. The size and number of little squares are determined by the problem statement.
Regarding the points (x*ij, y*ij) being at the upper right corner of each little square, it is the convention chosen for evaluating the integrand within each square. In this approach, the upper right corner of each square is used as the point to evaluate the function. However, it is worth noting that the choice of the evaluation point within each square does not significantly affect the final result of the Riemann sums as long as the partition becomes finer (i.e., as n approaches infinity).
In this problem, the information about the length of each side of the little squares (1/n) and the area of each square (n^2) is provided to help you understand how the partitioning and evaluation process works. Let's break it down:
1. Length of each side of the little squares (1/n):
The length of each side of the little squares is 1/n, which means that when you divide the square R into n equal parts along both sides, each little square will have a side length of 1/n. This is important because it determines the size of each little square in the partition.
2. Area of each square (n^2):
The area of each little square is given as n^2. This means that each little square has an area of n^2. This information helps you calculate the area of each little square and is relevant when computing Riemann sums.
Now, regarding your question about selecting the partition: Typically, when you are given a specific value for n in a problem, the partition is predetermined. In this case, the problem states that R is partitioned into n^2 little squares, so you do not have the freedom to choose the value of n. However, you can still evaluate different Riemann sums for different values of n to see how the approximations change.
Regarding the location of the evaluation point (x*ij, y*ij): The problem states that the integrand is evaluated at the upper right corner of each little square. This choice of evaluation point is just for convenience and simplicity. In this context, the upper right corner of each small square is chosen as the evaluation point. However, for different problems, the choice of evaluation point can vary.
To summarize, the information about the sides' length of each little square and the area of each square helps you understand the size of the partition and how to calculate the Riemann sums. The specific choice of partition (n^2 little squares) and the evaluation point (upper right corner) are given in this problem to provide a specific setup for calculating the Riemann sums.