Limit problem by Sutirtha Datta

Evaluation of limits using definite integrals:

$\lim_{n\to\infty} \dfrac1n \sum_{i=1}^n \left \lfloor \sqrt{ \dfrac{4i}n} \right \rfloor = \int_0^1\left\lfloor\sqrt{4x}\right\rfloor{dx} = \int_0^{\frac{1}{4}}0.dx + \int_{\frac{1}{4}}^1.1.dx =[x]_\frac{1}{4}^1 =1-\frac{1}{4}= \frac{3}{4}$

[The graph of $\left\lfloor\sqrt{4x}\right\rfloor$ has jump discontinuities at $x=\frac{1}{4},1,\frac{9}{4}...$ ]