Upper and lower bound Riemann sums

In summary, the upper and lower sums were evaluated incorrectly for the given function and partition. The midpoint sum was calculated correctly and the process of using the given formula for midpoint sums seems to be correct. Further clarification is needed on how upper and lower sums are defined in the textbook/course.
  • #1
Niaboc67
249
3

Homework Statement


Find the upper, lower and midpoint sums for $$\displaystyle\int_{-3}^{3} (12-x^{2})dx$$
$$\rho = \Big\{-3,-1,3\Big\}$$

The Attempt at a Solution



For the upper:
(12-(-1)^2)(-1-(-3)) + (12-(-1))(3-(-1))
=74

For the lower:
(12-(-3)^2)(-1-(-3))+(12-3)(3-(-1))
=42

For midpoint sum:
I used the expression $$\frac{b+a}{n}$$

(12-(-2)^2)(2) + (12-(2)^2)(4)
=48

Are these values correct? Does the process seem good?

Please explain
Thank you
 
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  • #2
I think you have misunderstood slightly. "Upper", "lower" and "midpoint" do not refer to the argument ("x") values, but to the function values.

In your case you need to calculate f(-3), f(-1) and f(3) and use those values to determine "upper" and "lower".
 
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  • #3
@Svein I thought finding upper, lower and midpoints was essentially taking the partition {-3,-1,3} then breaking that into sub-intervals (-3,-1),(-1,3) then referencing the graph to see which is biggest or smallest in that interval. And in (-3,-1) if we are looking for upper that would be -1 then we would evaluate -1 by plugging it back into the original function and multiplying that by the distance between those two points and so on.
 
  • #5
Niaboc67 said:

Homework Statement


Find the upper, lower and midpoint sums for $$\displaystyle\int_{-3}^{3} (12-x^{2})dx$$
$$\rho = \Big\{-3,-1,3\Big\}$$

The Attempt at a Solution



For the upper:
(12-(-1)^2)(-1-(-3)) + (12-(-1)^2)(3-(-1)) ##\ \ \ ## You didn't square the indicated -1 .
=74

For the lower:
(12-(-3)^2)(-1-(-3))+(12-(3^2))(3-(-1)) ##\ \ \ ## You didn't square the indicated 3 .
=42

For midpoint sum:
I used the expression $$\frac{b+a}{n}$$
(12-(-2)^2)(2) + (12-(2)^2)(4)
=48

Are these values correct? Does the process seem good?

Please explain
Thank you
You made errors in evaluating your function for the Upper and Lower sums.

How are upper and lower sums defined in your textbook/course?

The midpoint sum looks to be correct.
 

Related to Upper and lower bound Riemann sums

What is an upper and lower bound Riemann sum?

An upper and lower bound Riemann sum is a method of approximating the area under a curve by dividing the region into smaller rectangles and calculating their areas. The upper bound uses the maximum value of the curve within each rectangle, while the lower bound uses the minimum value.

How do you calculate an upper and lower bound Riemann sum?

To calculate an upper and lower bound Riemann sum, you first divide the region under the curve into smaller rectangles of equal width. Then, you find the maximum and minimum values of the curve within each rectangle. Finally, you multiply the width of each rectangle by the maximum or minimum value to get the area, and add all the areas together.

What is the difference between an upper and lower bound Riemann sum?

The difference between an upper and lower bound Riemann sum lies in the values used to calculate the area of each rectangle. The upper bound uses the maximum value, while the lower bound uses the minimum value. This results in the upper bound being an overestimate of the actual area, and the lower bound being an underestimate.

Why is it important to calculate both an upper and lower bound Riemann sum?

Calculating both an upper and lower bound Riemann sum allows you to approximate the area under a curve with a range of possible values. This can be useful in situations where an exact value is not necessary, and an estimate within a certain range is sufficient. Additionally, calculating both bounds can help to visualize the shape of the curve and understand its behavior.

How does the number of rectangles used in a Riemann sum affect the accuracy of the approximation?

Generally, the more rectangles used in a Riemann sum, the more accurate the approximation will be. This is because using smaller rectangles allows for a better representation of the curve. However, using too many rectangles can also result in a larger margin of error due to rounding and computational limitations. Finding a balance between the number of rectangles and the desired level of accuracy is important when using Riemann sums.

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