Question

Let f(x) = x3 and compute the Riemann sum of f over the interval [6, 7], using the following number of subintervals (n). In each case, choose the representative points to be the midpoints of the subintervals. (Round your answers to two decimal places.) (a) Use two subintervals of equal length (n = 2). (b) Use five subintervals of equal length (n = 5). (c) Use ten subintervals of equal length (n = 10). (d) Can you guess at the area of the region under the graph of f on the interval [6, 7]? _____________ square units

Answer 1

To compute the Riemann sum of
`f(x) = x^3` over the interval [6, 7], we divide the interval into subintervals, find the midpoints and evaluate f(x) at those midpoints. For each subinterval, we multiply the function value by the width of the subinterval and sum up the products. The Riemann sum can be approximated by increasing the number of subintervals, and it approaches the exact area under the curve as the number of subintervals approaches infinity.

Step-by-step explanation:

To compute the Riemann sum of
`f(x) = x^3` over the interval [6, 7], using the midpoint rule, we need to divide the interval into subintervals and find the width of each subinterval. Then, we evaluate f(x) at the midpoints of each subinterval and multiply it by the width of the subinterval. Finally, we sum up all these products to get the Riemann sum.

(a) For n = 2, we have 2 subintervals of equal length. The width of each subinterval is (7-6)/2 = 0.5. The midpoints of the subintervals are 6.25 and 6.75. Evaluating f(x) at these midpoints, we get f(6.25) = (6.25)^3 = 244.14 and f(6.75) = (6.75)^3 = 330.8. Multiplying each of these values by the width of the subinterval and summing them up, we get the Riemann sum as 0.5 * (244.14 + 330.8) = 287.47.

(b) For n = 5, we have 5 subintervals of equal length. The width of each subinterval is (7-6)/5 = 0.2. The midpoints of the subintervals are 6.1, 6.3, 6.5, 6.7, and 6.9. Evaluating f(x) at these midpoints and following the same steps as in part (a), we can find the Riemann sum as 0.2 * (f(6.1) + f(6.3) + f(6.5) + f(6.7) + f(6.9)).

(c) For n = 10, we have 10 subintervals of equal length. The width of each subinterval is (7-6)/10 = 0.1. The midpoints of the subintervals can be found similarly as in part (b), and the Riemann sum can be calculated as 0.1 times the sum of the function values at these midpoints.

(d) As the number of subintervals increases, the Riemann sum becomes a better approximation of the actual area under the curve of f(x) = x^3 on the interval [6, 7]. If we take the limit as the number of subintervals approaches infinity, we get the definite integral of f(x) over the interval [6, 7]. Therefore, to find the exact area under the graph of f on the interval [6, 7], we can compute the integral of f(x) from 6 to 7, which is equal to (
`7^4)/4 - (6^4)/4 = 240.25` square units.

Answer 2

The calculated Riemann sums provide approximations for the area under the curve f(x) = x^3 within the interval [6, 7], incrementally improving accuracy with more subintervals. Additionally, this area can be precisely determined by calculating the definite integral of f(x) as the number of subintervals approaches infinity.

Step-by-step explanation:

To answer the question about the Riemann sum of f(x) = x^3 over the interval [6, 7], we need to follow the process for the different numbers of subintervals specified.

(a) Using two subintervals (n=2)

The interval [6, 7] is divided into two subintervals, each with a width Δx = 0.5.

The midpoints for these subintervals are 6.25 and 6.75.

The Riemann sum can be computed as:

Riemann Sum =
`f(6.25)Δx + f(6.75)Δx = (6.25)^3(0.5) + (6.75)^3(0.5)`

(b) Using five subintervals (n=5)

This time, each subinterval has a width of 0.2.

The midpoint of each subinterval is taken, and the Riemann sum is calculated similarly to part (a).

(c) Using ten subintervals (n=10)

With ten subintervals, the width of each is 0.1.

Again, the Riemann sum is calculated using the midpoints of these subintervals.

(d) Estimating the area under f(x) from x=6 to x=7

As n becomes larger, the Riemann sum approximates the definite integral of f(x) over [6, 7].

The limit as n approaches infinity is the area under f(x) which can be evaluated exactly using integration.