# 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.

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