Use upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width). (Round your answers to three decimal places.)
y = 5/x
To approximate the area of the region using upper and lower sums, we need to divide the interval into subintervals of equal width and evaluate the function at the endpoints of each subinterval.
Let's say we have n subintervals. The width of each subinterval is given by ∆x = (b - a) / n, where a and b are the two endpoints of the interval.
To calculate the upper sum, we take the maximum value of the function on each subinterval and multiply it by the width of the subinterval.
To calculate the lower sum, we take the minimum value of the function on each subinterval and multiply it by the width of the subinterval.
Let's proceed with an example.
Suppose we want to approximate the area of the region bounded by the graph of y = 5/x on the interval [1, 4] using n = 4 subintervals.
Step 1: Calculate the width of each subinterval.
∆x = (4 - 1) / 4 = 0.75
Step 2: Evaluate the function at the endpoints of each subinterval.
For the upper sum, we take the maximum value of the function on each subinterval.
For the lower sum, we take the minimum value of the function on each subinterval.
Subinterval 1: [1, 1.75]
Endpoint 1: x = 1, y = 5/1 = 5 (minimum)
Endpoint 2: x = 1.75, y = 5/1.75 ≈ 2.857 (maximum)
Subinterval 2: [1.75, 2.5]
Endpoint 1: x = 1.75, y = 5/1.75 ≈ 2.857 (minimum)
Endpoint 2: x = 2.5, y = 5/2.5 = 2 (maximum)
Subinterval 3: [2.5, 3.25]
Endpoint 1: x = 2.5, y = 5/2.5 = 2 (minimum)
Endpoint 2: x = 3.25, y = 5/3.25 ≈ 1.538 (maximum)
Subinterval 4: [3.25, 4]
Endpoint 1: x = 3.25, y = 5/3.25 ≈ 1.538 (minimum)
Endpoint 2: x = 4, y = 5/4 = 1.25 (maximum)
Step 3: Calculate the upper and lower sums.
For the upper sum, we add the product of the maximum values and the widths of each subinterval.
For the lower sum, we add the product of the minimum values and the widths of each subinterval.
Upper sum = (2.857 * 0.75) + (2 * 0.75) + (1.538 * 0.75) + (1.25 * 0.75) ≈ 5.241
Lower sum = (5 * 0.75) + (2.857 * 0.75) + (2 * 0.75) + (1.538 * 0.75) ≈ 7.848
Therefore, the upper sum approximation of the area is approximately 5.241, and the lower sum approximation is approximately 7.848.
To approximate the area of a region using upper and lower sums, we need to consider the given function and the number of subintervals (of equal width) specified.
In this case, the function is y = 5/x. Let's assume we have n subintervals.
The width of each subinterval is given by Δx = (b - a)/n, where "a" and "b" represent the interval over which we are calculating the area.
To get the upper sum, we take the maximum value of the function within each subinterval, multiply it by the width of the subinterval, and sum up those values over all the subintervals.
To calculate the lower sum, we take the minimum value of the function within each subinterval, multiply it by the width of the subinterval, and sum up those values over all the subintervals.
Let's go through the steps to find the upper and lower sums:
1. Determine the interval over which you want to calculate the area. For example, if the interval is [a, b] = [1, 5], then a = 1 and b = 5.
2. Calculate the width of each subinterval: Δx = (b - a)/n.
3. Choose sample points within each subinterval to evaluate the function. For the upper sum, we take the right endpoint of each subinterval; for the lower sum, we take the left endpoint.
4. Plug these sample points into the function y = 5/x to find the corresponding function values.
5. Multiply each function value by the width of the subinterval.
6. Sum up these values for the upper sum and lower sum separately.
Note: It is important to mention that the accuracy of the approximation improves as the number of subintervals increases.
By following these steps, you should be able to calculate the approximate area of the region using both the upper and lower sums.
you don't give an interval, but let's say it's [1,5]. If you use n rectangles, then the upper approximation will use right sums. (Draw the graph and a few rectangles to see why.)
n
∑5/(1+k*4/n)
k=1