Question

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

Answer 1

Answer: 182

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