Question

For any real number $x$, let $\left[x\right]$ denote the largest integer less than or equal to $x$. If $I={\int }_0^{10}\left[\sqrt{\frac{10x}{x+1}}\right]dx$, then the value of $9I$ is

Answer 1

$I={\int }_0^{10}\left[\sqrt{\frac{10x}{x+1}}\right]dx$
Let $f\left(x\right)=\left[\sqrt{\frac{10x}{x+1}}\right]$
If $\frac{10x}{x+1}=1\Rightarrow x=\frac{1}{9}$
If $\frac{10x}{x+1}=4\Rightarrow x=\frac{2}{3}$
If $\frac{10x}{x+1}=9\Rightarrow x=9$
$\therefore f\left(x\right)=\{\begin{array}{ccc}0,& 0\le x<\frac{1}{9}& \\ 1,& \frac{1}{9}\le x<\frac{2}{3}& \\ 2,& \frac{2}{3}\le x<9& \\ 3,& 9\le x<10& \end{array}$

$\therefore I={\int }_0^{1/9}0dx+{\int }_{1/9}^{2/3}1dx+{\int }_{2/3}^{9}2dx+{\int }_9^{10}3dx$
$=(\frac{2}{3}-\frac{1}{9})+2(9-\frac{2}{3})+3\left(1\right)$
$=21-\frac{7}{9}=\frac{182}{9}$
$\therefore 9I=182$