Compute upper and lower integral

In summary: From the 6 intervals as given, the lower bound should be at about -6.5, and the upper bound should be at about 6.5.
  • #1
Niaboc67
249
3

Homework Statement


Let f(x) = x^2 and let P = { -5/2, -2, -3/2, -1, -1/2, 0, 1/2 }

Then the problem asks me to compute Lf (P) and Uf (P).

Lf (P) =

Uf (P) =

The Attempt at a Solution


Please explain how to solve. I thought that L[f] meant to calculate the lower bound with respect to f(x). and U[f] meant to calculate the upper bound with respect to f(x). Therefore, I plugged in 0 for f(0) = 10sin(0) = 0 for the lower bound. And π/2 for the upper bound. Therefore, f(π/2) = 10sin(π/2) = 10. However, these were both wrong.[/B]
 
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  • #2
If ##f(x) = x^2##, what is ##10sin(x)##?
 
  • #3
@PeroK ok nevermind I got two different questions mixed up there. Really unsure how to even go about solving this problem. Could use a push in the right direction.
 
  • #4
Niaboc67 said:
@PeroK ok nevermind I got two different questions mixed up there. Really unsure how to even go about solving this problem. Could use a push in the right direction.

Isn't it just estimating a function with rectangles? One an over-estimate and one an under-estimate.
 
  • #5
@PeroK I wish I knew. This problem seemed to be either worded weirdly or just isn't providing enough information to solve. Don't I have to take the 6 partitions and then take only the first number in the partition subtracted by the last number in the partition and divide by those two numbers by the number of partitions?
 
  • #6
Niaboc67 said:
@PeroK I wish I knew. This problem seemed to be either worded weirdly or just isn't providing enough information to solve. Don't I have to take the 6 partitions and then take only the first number in the partition subtracted by the last number in the partition and divide by those two numbers by the number of partitions?

I think you need to revise what you've been learning about the theory of integration. You're trying to estimate the area under a function.
 
  • #7
@PeroK I understand. Just this problem confuses me. I don't know how to even approach solving.
 
  • #8
Niaboc67 said:
@PeroK I understand. Just this problem confuses me. I don't know how to even approach solving.

Why not draw a graph? ##f(x)## and the partition ##P##.
 
  • #9
@PeroK after graphing it seems fuzzy to understand this problem and what I must do.
 
  • #10
Niaboc67 said:
@PeroK after graphing it seems fuzzy to understand this problem and what I must do.

If you're looking for homework help, you need to have some idea of the subject matter required for the problem. You have to be able make a serious effort at solving it for yourself.
 
  • #11
Mod note: Merged a separate thread with this one
1. Homework Statement

[tex] Let, f(x) = x^2 and let P = { -5/2, -2, -3/2, -1, -1/2, 0, 1/2 } [/tex]

Compute Lf (P) and Uf (P).
[f] (P) =

U[f] (P)
=

The Attempt at a Solution



I followed the pattern of distance times height. Therefore the further number in the sub-interval minus the closest. Then putting in the lowest touching pt into the function for height for the lower bound and putting in the highest touching point into the function for height.

For the lower bound I did this:
[tex] L[f](p) = (-1/2-0)(0^2)+(-1-(-1/2))((-1/2)^2)+(-3/2-(-1))((-1)^2)+(-2-(-3/2))((-3/2)^2)+(-5/2-(-2))((-2)^2)+(0-1/2)((0)^2) [/tex]

Then for upper bound I did:

[tex] U[f](p) = (-1/2-0)((1/2)^2)+(-1/2-0)((-1/2)^2)+(-1-(-1/2))((-1)^2)+(-3/2-(-1))((-3/2)^2)+(-2-(-3/2))((-3/2)^2)+(-5/2-(-2))((-5/2)^2) [/tex]

Thank you
 
Last edited by a moderator:
  • #12
Niaboc67 said:

Homework Statement


[tex] Let f(x) = x^2 and let P = { -5/2, -2, -3/2, -1, -1/2, 0, 1/2 } [/tex]

Compute Lf (P) and Uf (P).

L[f] (P) =
U[f] (P) =

The Attempt at a Solution



I followed the pattern of distance times height. Therefore the further number in the sub-interval minus the closest. Then putting in the lowest touching pt into the function for height for the lower bound and putting in the highest touching point into the function for height.

For the lower bound I did this:
L[f](p) = (-1/2-0)(0^2)+(-1-(-1/2))((-1/2)^2)+(-3/2-(-1))((-1)^2)+(-2-(-3/2))((-3/2)^2)+(-5/2-(-2))((-2)^2)+(0-1/2)((0)^2)

Then for upper bound I did:

U[f](p) = (-1/2-0)((1/2)^2)+(-1/2-0)((-1/2)^2)+(-1-(-1/2))((-1)^2)+(-3/2-(-1))((-3/2)^2)+(-2-(-3/2))((-3/2)^2)+(-5/2-(-2))((-5/2)^2)

Thank you

Do you have a question?
Did you calculate these sums? What did you get for your answers?
Did they come out negative?
If they did, would that bother you?
 
  • #13
I got the wrong answer on my program. Wondering what I have done wrong in these calculations. Does anything stand out?
 
  • #14
Niaboc67 said:

Homework Statement


Let f(x) = x^2 and let P = { -5/2, -2, -3/2, -1, -1/2, 0, 1/2 }

Compute Lf (P) and Uf (P).

L[f] (P) =
U[f] (P) =

The Attempt at a Solution



I followed the pattern of distance times height. Therefore the further number in the sub-interval minus the closest. Then putting in the lowest touching pt into the function for height for the lower bound and putting in the highest touching point into the function for height.

For the lower bound I did this:
L[f](p) = (-1/2-0)(0^2)+(-1-(-1/2))((-1/2)^2)+(-3/2-(-1))((-1)^2)+(-2-(-3/2))((-3/2)^2)+(-5/2-(-2))((-2)^2)+(0-1/2)((0)^2)

Then for upper bound I did:

U[f](p) = (-1/2-0)((1/2)^2)+(-1/2-0)((-1/2)^2)+(-1-(-1/2))((-1)^2)+(-3/2-(-1))((-3/2)^2)+(-2-(-3/2))((-3/2)^2)+(-5/2-(-2))((-5/2)^2)

Thank you
This looks like a continuation of your recently opened thread on the same subject, and probably should have been posted as part of that thread.

What you have looks better than previous attempts, however all of the widths you have here are negative
 

Related to Compute upper and lower integral

What is the concept of upper and lower integrals?

The concept of upper and lower integrals is a mathematical tool used to find the area under a curve. It involves dividing the area into smaller rectangles and finding the sum of their areas. The upper integral is the largest possible sum of rectangle areas that are entirely under the curve, while the lower integral is the smallest possible sum of rectangle areas that cover the entire curve.

How are upper and lower integrals calculated?

To calculate the upper and lower integrals, you need to divide the interval of the function into smaller subintervals. Then, you find the maximum and minimum values of the function within each subinterval and multiply it by the width of the subinterval. Finally, you sum up all the products for the upper integral and the lower integral.

What is the relation between upper and lower integrals and Riemann sums?

Upper and lower integrals are closely related to Riemann sums, which are used to approximate the area under a curve. The upper integral is equivalent to the upper Riemann sum, which uses the maximum value of the function within each subinterval. Similarly, the lower integral is equivalent to the lower Riemann sum, which uses the minimum value of the function within each subinterval.

When are upper and lower integrals useful?

Upper and lower integrals are useful in many areas of mathematics, such as calculus, analysis, and differential equations. They are particularly helpful when the function being integrated is not continuous, as they provide a more accurate approximation of the area under the curve compared to other methods.

How can upper and lower integrals be used in real-life applications?

Upper and lower integrals have many real-life applications, such as in physics, engineering, and economics. For example, in physics, they can be used to calculate the work done by a variable force. In economics, they can be used to find the total revenue or profit of a business. They are also used in optimization problems, where the goal is to find the maximum or minimum value of a function.

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