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 problem asks for the value of the expression \( 9 \int_{0}^{9} \left\lfloor \sqrt{\frac{10x}{x+1}} \right\rfloor \, dx \), where \( \lfloor t \rfloor \) denotes the greatest integer less than or equal to \( t \).

Concept Used:

To evaluate a definite integral containing the greatest integer function (floor function), the standard method is to split the interval of integration into subintervals. The splitting is done based on the points where the value of the expression inside the floor function becomes an integer. On each of these subintervals, the greatest integer function will have a constant value, making the integral easy to compute.

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

Step-by-Step Solution:

Step 1: Analyze the function inside the floor function.

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

The term inside the square root is \( \frac{10x}{x+1} = \frac{10(x+1) - 10}{x+1} = 10 - \frac{10}{x+1} \). This is a monotonically increasing function for \( x \ge 0 \). Consequently, \( f(x) \) is also a monotonically increasing function.

Let's find the values of \( f(x) \) at the endpoints of the interval:

• 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 \( x \) varies from 0 to 9, \( f(x) \) increases from 0 to 3. The possible integer values that \( \lfloor f(x) \rfloor \) can take are 0, 1, and 2.

Step 2: Find the points where \( f(x) \) takes integer values.

We need to find the values of \( x \) for which \( f(x) = 1 \) and \( f(x) = 2 \). These will be the points where we split our integral.

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

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

Step 3: Split the integral into subintervals.

Based on the points found in Step 2, we can determine the constant value of the integrand on each subinterval:

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

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

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

Step 4: Evaluate the integral \(I\).

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

Final Computation & Result:

The problem asks for the value of \( 9I \).

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

The value of the given expression is 155.