# Thread: Definite integral that is equal to Riemann sums

1. ## Definite integral that is equal to Riemann sums

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?

2. ## Re: Definite integral that is equal to Riemann sums

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.
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?
I do not see in the list.
But $\int_1^3 {\sqrt {1 + x} dx}$ will work.

I think there is a typo in the last one.

3. ## Re: Definite integral that is equal to Riemann sums

Thank you, but how did you figure it out.? I'm stuck with removing that square root.

4. ## Re: Definite integral that is equal to Riemann sums

Thank you, but how did you figure it out.? I'm stuck with removing that square root.
Look at this. But sure to click show steps.

5. ## Re: Definite integral that is equal to Riemann sums

I know how to find the integrals but I don't know how to solve the Riemunns sums?

6. ## Re: Definite integral that is equal to Riemann sums

The interval is $[1,3]$ its length is $3-1=2$. Divide $\frac{2}{n}$.
So the $\Delta_x=\frac{2}{n}$, and the partition points $x_k=(1+k\Delta_x)$ where $k=1,2,\cdots,n$
Thus $\sum\limits_{k = 1}^n {\sqrt {(1 + {x_k})} {\Delta _x}}$