Integration - fraction part of square root of x

Guide:

\begin{align}\int_0^{100} \left[\sqrt{x} \right]\, dx &= \sum_{i=1}^{10} \int_{(i-1)^2}^{i^2} \left[\sqrt{x} \right]\,dx \\ &=\sum_{i=1}^{10}(i^2-(i-1)^2)\sqrt{(i-1)^2}\\ &= \sum_{i=1}^{10} ( 2i+1) (i-1)\end{align}

$$\{\sqrt x\}=\sqrt x - [\sqrt x]$$

Now see when $1 \le x < 4$. The integer part $[\sqrt x]=1$ Similarly $$1 \le x < 4 \implies [\sqrt x]=1\\ 4 \le x < 9 \implies [\sqrt x]=2 \\ 9 \le x < 16 \implies [\sqrt x]=3 \\ ... \\ 81 \le x < 100 \implies [\sqrt x]=9\\$$ So you have to do the following integral $$\int_0^{100}\{\sqrt x\}=\int_0^{100}\sqrt xdx-\left(\int_1^41dx+\int_4^92dx+...+\int_{81}^{100}9dx \right)$$