Сalculus2
Please explain in detail if possible

Looks like is the number of subintervals you have to use with the trapezoidal rule, and for Simpson's rule. In the attachments, I take both numbers to be 4 to make drawing simpler.

Split up the integration interval [1, 8] into n subintervals. Each subinterval then has length (8 - 1)/n = 7/n. This gives us the partition

[1, 1 + 7/n], [1 + 7/n, 1 + 14/n], [1 + 14/n, 1 + 21/n], ..., [1 + 7(n - 1)/n), 8]

The left endpoint of the th interval is given by the arithmetic sequence,

and the right endpoint is

both with .

For Simpson's rule, we'll also need to find the midpoints of each subinterval; these are

The area under the curve is approximated by the area of 12 trapezoids. The partition is (roughly)

[1, 1.58], [1.58, 2.17], [2.17, 2.75], [2.75, 3.33], ..., [7.42, 8]

The area of the th trapezoid is equal to

Then the area under the curve is approximately

You first need to use the graph to estimate each value of and .

For example, and . So the first subinterval contributes an area of

For all 12 subintervals, you should get an approximate total area of about 15.9542.

Over each subinterval, we interpolate by a quadratic polynomial that passes through the corresponding endpoints and as well as the midpoint . With , we use the (rough) partition

[1, 1.29], [1.29, 1.58], [1.58, 1.88], [1.88, 2.17], ..., [7.71, 8]

On the th subinterval, we approximate by

(This is known as the Lagrange interpolation formula.)

Then the area over the th subinterval is approximately

As an example, on the first subinterval we have and . The midpoint is roughly , and . Then

Do the same thing for each subinterval, then get the total. I don't have the inclination to figure out the 60+ sampling points' values, so I'll leave that to you. (24 subintervals is a bit excessive)

For part 2, the average rate of change of between the points D and F is roughly

where 5.1 and 2.7 are the x-coordinates of the points F and D, respectively. I'm not entirely sure what the rest of the question is asking for, however...

rounding to the nearest thousandth would be 452,000

this is a very good question

step-by-step explanation:

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