How to Evaluate as a Riemann Sum

Question

In the given sum ...

$\lim_{n\to \infty} \frac{1}{n} (\sqrt{\frac{1}{n}} + \sqrt{\frac{2}{n}} + \sqrt{\frac{3}{n}} + ... + \sqrt{\frac{n}{n}})$

It states in the book that it is recognized as a Riemann sum for a fn defined on [0,1]. How do you determine that it is defined on [0, 1] and what would the sigma notation look like for this?

I understand how to calculate a Riemann sum, I am just not understanding how they get [0,1] from the given information :/

Reference - Stewart: Calculus Early Trancendentals 7e - Pg 396 #70

Edit (For more info on the given answer):

$\int_a^bf(x)dx=\lim_{n\to \infty}\sum_{i=1}^nf(x_i)\Delta{x}$

$\Delta{x}=\frac{b-a}{n}$ and $x_i=a+i\Delta{x}$

Answer 1

We have $$\lim_{n\to\infty}\frac1n\sum_{r=1}^n\sqrt{\dfrac rn}$$

Now, $$\lim_{n\to\infty}\frac1n\sum_{r=1}^nf\left(\dfrac rn\right)=\int_0^1 f(x)\ dx$$

Answer 2

The points where the square root is evaluated are of the form $k/n$ for $k=1,2,\ldots,n$. This is the Riemann sum you would get if you broke up the interval $[0,1]$ into $n$ equal subintervals and evaluated $\sqrt{x}$ at the right end point of each interval. (The divide by $n$ comes from the fact that each subinterval has length $1/n$.) For a Riemann sum whose limit exists unconditionally it doesn't matter where you evaluate the function in each subinterval, or how long each subinterval is, as long as the length of all the subintervals go to $0$ as the the number of subintervals goes to $\infty$. So your limit is equal to $\int_0^1 \sqrt{x} dx$.