Definitions for Riemann sums

  • #1
248
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Just want to see if I actually understand what these all mean.

Partition: is like the x-coordinate values, also gives the number of times the graph was chopped up. We need them in order to find the distance or length of each rectangle. The distance is found by taking the further point minus the point prior to it. The partition is usually written like P = {0,1,2,3,4}

Sub-intervals: would be each of the x-coordinates but broken into a series of intervals like: (0,1),(1,2),(2,3),(3,4) not sure exactly why this is needed. Maybe to get distance easier?

Lower value: is when you take the lowest values between each subinterval and then take that value and plug it into the original function. You then multiply that function with distance between the subinterval and consecutively add them all together and the result gives the total Area.

Upper value: same process for lower value except you take the highest value in the subinterval.

Right-hand sum: same process for lower/upper sum but with the stipulation of taking the right-most point.

Left-hand sum: same process but for the left-most point.

Midpoint sum: (b-a)/n, I don't know what that means. I think it has something with taking a subinterval then subtracting the upper value with the lower value and that generates the midpoint? not sure, then the process would still hold true for the others just don't know how to get the midpoint initially.

Please correct any mistakes or inconsistiencies

Thank you!
 

Answers and Replies

  • #2
34,934
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Just want to see if I actually understand what these all mean.

Partition: is like the x-coordinate values, also gives the number of times the graph was chopped up. We need them in order to find the distance or length of each rectangle. The distance is found by taking the further point minus the point prior to it. The partition is usually written like P = {0,1,2,3,4}
Assuming the interval for the integral is [a, b], a partition is a set of points that breaks up the interval into adjacent subintervals. In general, a partition will be ##\{a = x_0, x_1, x_2, \dots, x_{n - 1}, x_n\}##. Each adjacent pair of numbers in the partition defines one subinterval. The points in the partittion don't have to be equally spaced.
Niaboc67 said:
Sub-intervals: would be each of the x-coordinates but broken into a series of intervals like: (0,1),(1,2),(2,3),(3,4) not sure exactly why this is needed. Maybe to get distance easier?
A portion of the interval for the integration problem. For example, one subinterval is ##[x_3, x_4]##.
Niaboc67 said:
Lower value: is when you take the lowest values between each subinterval and then take that value and plug it into the original function. You then multiply that function with distance between the subinterval and consecutively add them all together and the result gives the total Area.
This might be Lower sum instead of Lower value. The Lower Sum is obtained by using the smallest value in each subinterval (not between, as you wrote) of the integrand function. Each term in the lower sum is the product of the minimum function value in that subinterval times the width of the subinterval. In symbols, each term would be ##f(x_{min}) \cdot (x_{i} - x_{i - 1})##, where ##x_{min}## is the x-value in the subinterval ##[x_{i - 1}, x_i]## at which the function value is smallest.
Niaboc67 said:
Upper value: same process for lower value except you take the highest value in the subinterval.
As above for Lower sum, except that you use the largest function value in each subinterval.
Niaboc67 said:
Right-hand sum: same process for lower/upper sum but with the stipulation of taking the right-most point.
The height of each rectangle used in the sum comes from the function value at the right end of the subinterval. If we're talking about the subinterval ##[x_{i - 1}, x_i]##, the corresponding term in the right-hand sum will be ##f(x_i) \cdot (x_i - x_{i - 1})##.
Niaboc67 said:
Left-hand sum: same process but for the left-most point.
As above, but using the function value at the left end of each subinterval.
Niaboc67 said:
Midpoint sum: (b-a)/n, I don't know what that means. I think it has something with taking a subinterval then subtracting the upper value with the lower value and that generates the midpoint? not sure, then the process would still hold true for the others just don't know how to get the midpoint initially.
Each term in this sum uses the function value at the midpoint of the subinterval. The typical term in this sum is ##f(x_{i - 1} + \frac{x_{i - 1} + x_i}{2}) \cdot (x_i - x_{i - 1})##.
Niaboc67 said:
Please correct any mistakes or inconsistiencies

Thank you!
 
  • #3
mathwonk
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Riemann himself only discussed sums in which the partition was obtained by choosing arbitrary subdivision points, and the evaluations were made in those subintervals at arbitrary points in each subinterval. The integral exists if the limit of the Riemann sum is the same and finite regardless of how those choices are made. If this is the case, then certainly one can make the choice of subdiviions and evaluation points as one like, in particular one can make the subdivisions at equal distances and one can make the evaluations at the endpoints of the subintervals. Riemann then proved that the integral does exist if and only if rhe function is continuous except possibly at a subset of points of "measure zero". I.e. the integral does exist if the subset of points of discontinuity can be covered by a sequence of intervals whose total length is as small as desired.

you might consult the following wiki article and in particular note # 1: which refers you to Riemann's paper on representing functions via trig series and in it defines the Riemann integral, and also proves exactly when the definition exists:

https://en.wikipedia.org/wiki/Riemann_integral
 
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