Fractional part integral

$\begin{aligned} \int_0^n \{x^2\}\,dx & = \; \sum_{j=1}^{n^2}\int_{\sqrt{j-1}}^{\sqrt{j}} \{x^2\}\,dx \; = \; \sum_{j=1}^{n^2}\int_{j-1}^j \frac{\{u\}}{2u^{\frac12}}\,du \; = \; \sum_{j=1}^{n^2} \int_0^1 \frac{y}{2(j-1+y)^{\frac12}}\,dy \\ & = \; \tfrac12\sum_{j=1}^{n^2}\left[ \tfrac23(j-1+y)^{\frac32} - 2(j-1)(j-1+y)^{\frac12}\right]_0^1 \; = \; \tfrac12\sum_{j=1}^{n^2} \Big[\tfrac23\big(j^{\frac32} - (j-1)^{\frac32}\big) - 2(j-1)\big(j^{\frac12} - (j-1)^{\frac12}\big) \Big] \\ & = \; \tfrac12\sum_{j=1}^{n^2} \Big[\tfrac23\big(j^{\frac32} - (j-1)^{\frac32}\big) + 2j^{\frac12} - 2\big(j^{\frac32} - (j-1)^{\frac32}\big)\Big] \; = \; \sum_{j=1}^{n^2}j^{\frac12} - \tfrac23\sum_{j=1}^{n^2}\big(j^{\frac32} - (j-1)^{\frac32}\big) \\ & = \; \sum_{j=1}^{n^2} j^{\frac12} - \tfrac23n^{3} \end{aligned}$

$\begin{aligned} I & = \int_0^n \left \{x^2\right \} dx \\ & = \sum_{k=0}^{n^2-1} \int_{\sqrt k}^{\sqrt{k+1}} \left(x^2-k\right) dx \\ & = \sum_{k=0}^{n^2-1} \left[\frac {x^3}3 - kx \right]_{\sqrt k}^{\sqrt{k+1}} \\ & = \sum_{k=0}^{n^2-1} \frac {\sqrt{k+1}^3- \sqrt k^3}3 - \sum_{k=0}^{n^2-1} \left(k\sqrt{k+1} - k \sqrt k\right) \\ & = \frac {n^3}3 - \sum_{k=0}^{n^2-1} \left((k+1)\sqrt{k+1} - k \sqrt k - \sqrt {k+1} \right) \\ & = \frac {n^3}3 - n^3 + \sum_{k=0}^{n^2-1} \sqrt {k+1} \\ & = \sum_{k=1}^{n^2} \sqrt k - \frac 23 n^3 \end{aligned}$

Therefore, $I+\dfrac 23n^2 = \displaystyle \boxed{\sum_{k=1}^{n^2} \sqrt k}$ .