Subject: Mathematics — Definite Integration
Question: $\text{For any real number~}x\text{, let~}\lbrack x\rbrack\text{~denote the largest integer}$
$\text{less than or equal to~}x\text{. }$
$\text{If~}I = \int_{0}^{10}\mspace{2mu}\mspace{2mu}\left\lbrack \sqrt{\frac{10x}{x + 1}} \right\rbrack dx,\text{~then the value}$
of 9I is \_\_\_\_\_.
Answer: 9I=182
Solution: Let $y = \frac{10x}{x + 1},\ 0 \leq x \leq 10$\
$${\Rightarrow xy + y = 10x
}{\Rightarrow x = \frac{y}{10 - y}
}{\Rightarrow \ 0 \leq \frac{y}{10 - y} \leq 10
}$$$\Rightarrow \ \frac{y}{y - 10} \leq 0$ and
$\frac{y}{10 - y} \leq 10$.
$\Rightarrow \ \frac{11y - 100}{y - 10} \geq 0$
[IMAGE] $y \in \lbrack 0,10)\text{~and~}y \in \left( - \infty,\frac{100}{11} \right\rbrack U(10,\infty)$\
$$\Rightarrow y \in \left\lbrack 0,\frac{100}{11} \right\rbrack$$
$\sqrt{y} \in \left\lbrack 0,\frac{10}{\sqrt{11}} \right\rbrack\ \Rightarrow \ \lbrack\sqrt{y}\rbrack = \{ 0,1,2,3\}$.\
Case I : $0 \leq \frac{10x}{x + 1} < 1$\
$\therefore\ x \in \left\lbrack 0,\frac{1}{9} \right)$ then
$\left\lbrack \sqrt{\frac{10x}{x + 1}} \right\rbrack = 0$\
Case II : $1 \leq \frac{10x}{x + 1} < 4$\
$\therefore\ x \in \left\lbrack \frac{1}{9},\frac{2}{3} \right)$ then
$\left\lbrack \sqrt{\frac{10x}{x + 1}} \right\rbrack = 1$\
Case III : $4 < \frac{10x}{x + 1} < 9$\
$\therefore\ x \in \left\lbrack \frac{2}{3},9 \right)$ then
$\left\lbrack \sqrt{\frac{10x}{x + 1}} \right\rbrack = 2$\
Case IV :
$x \in \lbrack 9,10) \Rightarrow \text{~then~}\left\lbrack \sqrt{\frac{10x}{x + 1}} \right\rbrack = 3$\
$${\therefore\ I = \int_{0}^{1/9}\mspace{2mu}\mspace{2mu} 0 \cdot dx + \int_{1/9}^{2/3}\mspace{2mu}\mspace{2mu} 1 \cdot dx + \int_{2/3}^{9}\mspace{2mu}\mspace{2mu} 2 \cdot dx + \int_{9}^{10}\mspace{2mu}\mspace{2mu} 3 \cdot dx
}{= \left( \frac{2}{3} - \frac{1}{9} \right) + 2\left( 9 - \frac{2}{3} \right) + 3(10 - 9)
}{= \frac{5}{9} + \frac{50}{3} + 3 = \frac{182}{9}
}{\therefore\ 9I = 182.}$$