f(x)=2x^3-x^2+1 [1,4]=interval

Set up the limit process (once with right endpts, and the other with left entdpts.) to find the area under the curve and above the x-axis on the given interval

Should the right and left endpt limit set ups have the same area? If so what is it?

PLEASE HELP.....I GOT 107.55 BUT IT IS NOT THE RIGHT ANSWER AND I DON'T SEE HOW THERE CAN BE A DIFFERENCE IN SETTING UP

f(x) is always above the x axis in that interval. Therefore just integrate the function from 1 to 4. I don't know what you mean by "the limit process".

The indefinite integral is x^4/2 -x^3/3 +x. That equals 128 -21.33 +4 = 110.67 at x = 4 and 0.50 -0.33 +1 = 1.17 at x = 1.

The area should be 110.67-1.17 = 109.50

My algebra tends to be sloppy, so my answer could be wrong, too.

thanks so much!!!

To find the area under the curve and above the x-axis on the given interval [1, 4], we can use the limit process with both right and left end points.

First, let's start with the right-endpoint limit setup:

1. Divide the interval [1, 4] into n subintervals of equal width.
2. The width of each subinterval, denoted as Δx, is given by Δx = (b - a) / n, where a and b are the endpoints of the interval.
In this case, a = 1 and b = 4, so Δx = (4 - 1) / n.
3. Choose a right endpoint for each subinterval. The right endpoints will be 1 + kΔx, where k takes values from 1 to n.
4. Evaluate the function at each right endpoint, f(1 + kΔx), and multiply it by the corresponding width, Δx.
5. Sum up all the areas: A = Σ[f(1 + kΔx) · Δx] from k = 1 to n.
6. Take the limit of the sum as n approaches infinity: lim(n→∞) Σ[f(1 + kΔx) · Δx].

Now, let's move on to the left-endpoint limit setup:

1. Follow the same steps as above, but instead of choosing right endpoints for each subinterval, choose left endpoints: 1 + (k-1)Δx.
2. Evaluate the function at each left endpoint, f(1 + (k-1)Δx), and multiply it by the corresponding width, Δx.
3. Sum up all the areas: B = Σ[f(1 + (k-1)Δx) · Δx] from k = 1 to n.
4. Take the limit of the sum as n approaches infinity: lim(n→∞) Σ[f(1 + (k-1)Δx) · Δx].

Now, to answer the question of whether the right and left endpoint limit setups should have the same area: No, they generally won't have the same area. The reason for this is that choosing different endpoints for each subinterval yields different rectangles, and the function may vary within each subinterval.

To find the exact difference between the areas obtained from the two limit setups, you need to calculate A - B. This will give you the difference in areas between the two setups.

I hope this explanation helps you understand the process and reasoning behind the setup of the limit process for finding the area under the curve.