$\text{ The value of }{\int }_0^9\left\{\sqrt{x}\right\}dx,\text{ where }\left\{x\right\}\text{ denotes the fractional part of }x\text{ , is }$

$\underset{0}{\overset{9}{\int }}\left\{\sqrt{x}\right\}dx=\underset{0}{\overset{9}{\int }}\left(\sqrt{x}-\left[\sqrt{x}\right]\right)dx=\underset{0}{\overset{9}{\int }}{x}^{\frac{1}{2}}dx-\underset{0}{\overset{9}{\int }}\left[\sqrt{x}\right]dx$

$=\frac{2}{3}{\left(x^{3/2}\right)}^9-\left(\underset{0}{\overset{1}{\int }}\left[\sqrt{x}\right]dx+\underset{1}{\overset{4}{\int }}\left[\sqrt{x}\right]dx+\underset{4}{\overset{9}{\int }}\left[\sqrt{x}\right]dx\right)$

$=\frac{2}{3}\left(27\right)-\left(0+3+10\right)=5$