Question

For any real number $x$, let $[x]$ denote the largest integer less than or equal to $x$. If
$I=\int\limits_{0}^{10}\left[\sqrt{\frac{10 x}{x+1}}\right] d x,$
then the value of $9I$ is _____

Answer 1

$ I=\int\limits_{0}^{10}\left[\sqrt{\frac{10 x}{x+1}}\right] d x$
$\left[\sqrt{\frac{10 x}{x+1}}\right]=n $
$\Rightarrow \frac{n^{2}}{10-n^{2}} \leq x < \frac{(n+1)^{2}}{10-(n+1)^{2}}$ where $n \in I$
For $n=0,0 \leq x < 1 / 9$
$n=1 ; 1 / 9 \leq x < 2 / 3 $
$ n=2 ; 2 / 3 \leq x < 9, n=3, x \geq 9$
$ I=\int\limits_{0}^{1 / 9} 0 \cdot d x+\int\limits_{1 / 9}^{2 / 3} 1 \cdot d x+\int\limits_{2 / 3}^{9} 2 \cdot d x+\int\limits_{9}^{10} 3 \cdot d x $
$=\left(\frac{2}{3}-\frac{1}{9}\right)+2\left(9-\frac{2}{3}\right)+3(10-9)$
$=\frac{182}{9}=9 I=182$