Question

The value $9 \int_{0}^{9} \left\lfloor \frac{10x}{x+1} \right\rfloor \, dx$, where $\left\lfloor t \right\rfloor$ denotes the greatest integer less than or equal to $t$, is ________.

Answer 1

Solution: To evaluate $\int_{0}^{9} \left\lfloor \frac{10x}{x+1} \right\rfloor \, dx$, we analyze the function $\frac{10x}{x+1}$ and find the intervals where it takes integer values.

Solve for values of $x$ where $\frac{10x}{x+1} = k$ (where $k$ is an integer):

1. $\frac{10x}{x+1} = 1 \Rightarrow x = \frac{1}{9}$
2. $\frac{10x}{x+1} = 4 \Rightarrow x = \frac{2}{3}$
3. $\frac{10x}{x+1} = 9 \Rightarrow x = 9$

Thus, we break the integral into parts:

$I = 9 \left( \int_{0}^{1/9} 0 \, dx + \int_{1/9}^{2/3} 1 \, dx + \int_{2/3}^{9} 2 \, dx \right)$

Calculate each integral:

$= 9 \left( 0 + \int_{1/9}^{2/3} 1 \, dx + \int_{2/3}^{9} 2 \, dx \right) = 155$