Determine a region whose area is equal to the given limit.

lim x-> infinity Sigma (n on top, i=1 on bottom) (2/n)*[5+(2i/n)]^10

I started with delta x=b-a/n =5-0/n =5/n

The width of delta is 2/n so b-a=2 I'm not sure how you got 5-0/n
The integrand looks like (5+x)^10, the limits look like they might be 0 to 2 I'll let you verify if this is right.

I might be mistake on the limits, it could be x= -5 to -3. I haven't done Riemann sums in some time so I'm a little rusty on this.

Now I think I was completely off the mark. The function is f(x) x^10 and the limits are 5 to 7. Shows how out of practice I am!

I reviewed Riemann sums briefly to make sure I know how they're set up. If we let delta-x = dx =(b-a)/n then we have
lim n->infty Sum_i=1-to-n[f(a+i*dx)*dx]
Here a=5, and (b-a)/n = 2/n, b-a=2 so b=7.
As I mentioned, f(x)=x^10
This is the typical form for the sum using the left endpoint for the starting point and the right endpoint of each rectangle for the height, thus these would be circumscribed rectangles since the function is increasing on this interval. The way to determine what function the sum corresponds to is to determine what the dx part is: for this problem it's 2/n. Then determine what the value for a is: here it's 5. You then find b, then f(x).
I'm not sure if this problem is asking you to determine the area of the region or simply the region itself. If you had to determine the area it would be the defintite integral of x^10dx from x=5 to x=7. I'm sure you know how to do this part.
Hopefully this clarifies my previous posts. As I said, it's been awhile since I set these up and I couldn't seem to recall the basics. Looking at a couple sites brought it back fairly quickly though.

To determine the region whose area is equal to the given limit:

1. First, review the concept of Riemann sums.

2. Start by understanding the notation and setting up the Riemann sum. The notation Σ (n on top, i=1 on bottom) represents the sum from i=1 to n. The expression (2/n)*[5+(2i/n)]^10 is the function being summed at each i.

3. Determine the width of each subinterval, Δx. In this case, the expression given is (2/n), which represents the width of each subinterval.

4. Determine the limits of integration. To determine the limits, visualize the function being summed and consider the values of i as it ranges from 1 to n. In this case, the function is (5+(2i/n))^10. The limits could be 0 to 2, -5 to -3, or any other values depending on the context of the problem.

5. Review the formula for the Riemann sum:

lim n->∞ Σ (n on top, i=1 on bottom) [f(a+i*Δx)*Δx]

Here, f(x) = (5+(2i/n))^10 is the function being summed, a represents the starting point of the interval, and b represents the ending point of the interval.

6. Determine the values for a and b based on the given limits. In this case, a = 5 and (b-a) = 2, so b = 7.

7. Once you have determined the function (f(x)), the starting point (a), and the ending point (b), you can proceed with evaluating the Riemann sum using the definite integral. If the problem asks for the area of the region, the Riemann sum can be represented as the definite integral of f(x) from x=5 to x=7.

8. Solve the definite integral to find the area of the region (if that is the given goal).

Remember, practicing Riemann sums and understanding the concepts behind them will help with setting up and solving problems like these.

The problem is asking to determine a region whose area is equal to the given limit.

The given limit expression is lim x-> infinity Sigma (n on top, i=1 on bottom) (2/n)*[5+(2i/n)]^10.

To find the region, we can rewrite the limit as a Riemann sum. Let delta x = (b-a)/n, where a is the lower limit and b is the upper limit.

In this case, a = 5 and (b-a)/n = 2/n. Solving for b, we get b = 7.

The integrand is f(x) = x^10, and the limits of integration are from x = 5 to x = 7.

To find the area of the region, we need to calculate the definite integral of x^10 dx from x = 5 to x = 7.

Therefore, the region whose area is equal to the given limit is the area under the curve y = x^10 from x = 5 to x = 7.