Question

Evaluate ${\int }_0^9\left\{\sqrt{x}\right\}dx$, where $\left\{x\right\}$ denotes the fractional part of $x$.

Answer 1

$LetI={\int }_0^9\left\{\sqrt{x}\right\}dx={\int }_0^9(\sqrt{x}-[\sqrt{x}\left]\right)dx$
$={\int }_0^9\left(x^{\frac{1}{2}}\right)dx-{\int }_0^9\left[\sqrt{x}\right]dx$
$=\frac{2}{3}{\left(x^{\frac{3}{2}}\right)}_0^9-{\int }_0^9\left[\sqrt{x}\right]dx$
$=18-{\int }_0^9\left[\sqrt{x}\right]dx$
As ${\int }_0^9\left[\sqrt{x}\right]dx⟹0\le x\le 9$
So, $0\le \sqrt{x}\le 3$
Thus, it should be divided into three parts $0\le \sqrt{x}\le 1$, $1\le \sqrt{x}\le 2$, $2\le \sqrt{x}\le 3$
$=18-[{\int }_0^1\left[\sqrt{x}\right]dx+{\int }_1^4\left[\sqrt{x}\right]dx+{\int }_4^9\left[\sqrt{x}\right]dx]$
$=18-({\int }_0^{1}0dx+{\int }_1^{4}1dx+{\int }_4^{9}2dx)$
$=18-(0+(x{)}_1^4+\left(2x{)}_4^9\right)=18-(3+10)$
$I=5$