Compute upper and lower integral

  • #1
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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]
 

Answers and Replies

  • #2
PeroK
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If ##f(x) = x^2##, what is ##10sin(x)##?
 
  • #3
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@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
PeroK
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@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
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@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
PeroK
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@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
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@PeroK I understand. Just this problem confuses me. I don't know how to even approach solving.
 
  • #8
PeroK
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@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
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@PeroK after graphing it seems fuzzy to understand this problem and what I must do.
 
  • #10
PeroK
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@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
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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
 
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  • #12
LCKurtz
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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
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I got the wrong answer on my program. Wondering what I have done wrong in these calculations. Does anything stand out?
 
  • #14
SammyS
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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
 

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