Conversion of Riemann Sum to Integral with Square Root

Question

I'm given the following Riemann Sum to convert to Integral:

$$\lim_{n\to \infty}\sum_{i=1}^n {1\over n} \sqrt {i-1\over n}$$

I first did the following:

$\Delta x$ = $\dfrac{1}{n},\;$ $f(x)=\sqrt{x},\;$ and $x_i= \dfrac{i-1}{n}$

I tried to solve $a$ by doing:

$a+i \Delta x =\dfrac{i-1}{n}$

$a = \dfrac{-1}{n}$

And now I'm stuck at that. I do not know if I could take the limit of that and assume that $a = 0$ and get $b = 1$, or is that now how you solve it?

Answer 1

redefine $x-1=t$ to get $$ \lim_{n \to \infty} \sum_{t=0}^{n}\frac{1}{n} \sqrt{\frac{t}{n}} $$ Can you handle from here?

Answer 2

Hint: For all $n>1,$ we have $$\begin{align}\sum_{i=1}^n\frac1n\sqrt{\frac{i-1}n} &= \sum_{i=2}^n\frac1n\sqrt{\frac{i-1}n}\\ &= \sum_{i=1}^{n-1}\frac1n\sqrt{\frac{i}n}\\ &= -\frac1n+ \sum_{i=1}^n\frac1n\sqrt{\frac{i}n}.\end{align}$$