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

Correct Answer: 155

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:

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