The problem is limit as n goes to infinity, riemann sum from i=1 to n of (2/n)(1+(2i/n))^1/2
The problem I have i getting rid of the ^1/2? I was thinking that I could evaluate the sums under the square root like Riemann sum of 1 + Riemann sum of 2i/n.all under the square root. I don't think that's how I'm suppose to do it.
Any help is greatly appreciated.
The 4 integrals to choose from are
integral from 1 to 2 f(x) = (1+x)^1/2dx
integral from 1 to n f(x) = (x)^1/2dx
integral from n to 2 f(x) = (1+x)^1/2dx
integral from 1 to 3 f(x) = (x)^1/2dx
What does the does the integral from 1 to the n mean?
Now you seem confused about Riemann Sums. In the future, please make your questions clear.
The interval is its length is . Divide .
So the , and the partition points where