You made errors in evaluating your function for the Upper and Lower sums.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
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.
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.
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.
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.
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.