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

The objective is to compute the value of the expression \( 9 \int_{0}^{9} \left\lfloor \sqrt{\frac{10x}{x+1}} \right\rfloor \, dx \), where \( \lfloor t \rfloor \) represents the greatest integer less than or equal to \( t \).

Key Concept:

Evaluating a definite integral involving the greatest integer function (floor function) typically involves partitioning the integration interval. This partitioning occurs at points where the argument of the floor function evaluates to an integer. Within each resulting subinterval, the floor function yields a constant value, simplifying the integration process.

We will examine the function \( f(x) = \sqrt{\frac{10x}{x+1}} \) to identify the points within the interval \( [0, 9] \) where it attains integer values.

Step-by-Step Derivation:

Step 1: Characterize the function within the floor function.

Let \( f(x) = \sqrt{\frac{10x}{x+1}} \). We first determine the range of this function over the integration interval \( [0, 9] \).

The expression under the square root can be rewritten as \( \frac{10x}{x+1} = \frac{10(x+1) - 10}{x+1} = 10 - \frac{10}{x+1} \). This expression is monotonically increasing for \( x \ge 0 \). Consequently, \( f(x) \) is also monotonically increasing.

Evaluating \( f(x) \) at the interval's boundaries:

• At \( x = 0 \): \( f(0) = \sqrt{\frac{10 \times 0}{0 + 1}} = \sqrt{0} = 0 \).
• At \( x = 9 \): \( f(9) = \sqrt{\frac{10 \times 9}{9 + 1}} = \sqrt{\frac{90}{10}} = \sqrt{9} = 3 \).

As \( x \) progresses from 0 to 9, \( f(x) \) increases from 0 to 3. The possible integer values for \( \lfloor f(x) \rfloor \) are 0, 1, and 2.

Step 2: Locate points where \( f(x) \) equals integer values.

We need to find the \( x \) values for which \( f(x) = 1 \) and \( f(x) = 2 \). These will serve as the points for splitting the integral.

• Setting \( f(x) = 1 \): \[ \sqrt{\frac{10x}{x+1}} = 1 \implies \frac{10x}{x+1} = 1 \implies 10x = x + 1 \implies 9x = 1 \implies x = \frac{1}{9} \]
• Setting \( f(x) = 2 \): \[ \sqrt{\frac{10x}{x+1}} = 2 \implies \frac{10x}{x+1} = 4 \implies 10x = 4x + 4 \implies 6x = 4 \implies x = \frac{2}{3} \]

The value of \( f(x) \) reaches 3 at \(x = 9\), which is the upper integration limit.

Step 3: Partition the integral into subintervals.

Based on the points determined in Step 2, we can ascertain the constant value of the integrand within each subinterval:

• For \( 0 \le x<\frac{1}{9} \), \( 0 \le f(x)<1 \), thus \( \left\lfloor \sqrt{\frac{10x}{x+1}} \right\rfloor = 0 \).
• For \( \frac{1}{9} \le x<\frac{2}{3} \), \( 1 \le f(x)<2 \), thus \( \left\lfloor \sqrt{\frac{10x}{x+1}} \right\rfloor = 1 \).
• For \( \frac{2}{3} \le x<9 \), \( 2 \le f(x)<3 \), thus \( \left\lfloor \sqrt{\frac{10x}{x+1}} \right\rfloor = 2 \).

Let \( I = \int_{0}^{9} \left\lfloor \sqrt{\frac{10x}{x+1}} \right\rfloor \, dx \). The integral can be decomposed as:

\[ I = \int_{0}^{1/9} 0 \, dx + \int_{1/9}^{2/3} 1 \, dx + \int_{2/3}^{9} 2 \, dx \]

Step 4: Calculate the value of \(I\).

\[ I = [0]_{0}^{1/9} + [x]_{1/9}^{2/3} + [2x]_{2/3}^{9} \] \[ I = 0 + \left( \frac{2}{3} - \frac{1}{9} \right) + \left( 2(9) - 2\left(\frac{2}{3}\right) \right) \] \[ I = \left( \frac{6}{9} - \frac{1}{9} \right) + \left( 18 - \frac{4}{3} \right) \] \[ I = \frac{5}{9} + \left( \frac{54 - 4}{3} \right) = \frac{5}{9} + \frac{50}{3} \] \[ I = \frac{5}{9} + \frac{150}{9} = \frac{155}{9} \]

Final Calculation & Outcome:

The problem requires the value of \( 9I \).

\[ 9I = 9 \times \frac{155}{9} = 155 \]

The computed value of the given expression is 155.