The value of the integral `int_(-10)^(0) (|(2|x|)/([x]-3x)|)/(((2|x|)/(3x-[x])))dx` where [.] denotes GIF

Options

A. `(28)/(3)`
B. 0
C. 10
D. -10

Answer

To solve the integral

\[ I = \int_{-10}^{0} \frac{\left|\frac{2|x|}{[x] - 3x}\right|}{\left(\frac{2|x|}{3x - [x]}\right)} \, dx \]

where \([x]\) denotes the greatest integer function (GIF), we will break the integral into segments based on the behavior of the GIF and the absolute value function.

### Step 1: Identify the intervals for integration
The greatest integer function \([x]\) takes on integer values. For \(x\) in the interval \([-10, 0)\), the values of \([x]\) will be \(-10, -9, -8, \ldots, -1\). We will break the integral into intervals:

\[ I = \int_{-10}^{-9} + \int_{-9}^{-8} + \int_{-8}^{-7} + \ldots + \int_{-1}^{0} \]

### Step 2: Evaluate the integral on each interval
For each interval \([-n-1, -n)\) where \(n = 0, 1, 2, \ldots, 9\), we can calculate the integral separately.

#### Interval \([-1, 0)\):
In this interval, \([x] = -1\). Therefore, we have:

\[ I_1 = \int_{-1}^{0} \frac{\left|\frac{2|x|}{-1 - 3x}\right|}{\left(\frac{2|x|}{3x + 1}\right)} \, dx \]

Since \(x\) is negative, \(|x| = -x\). Thus, we can simplify:

\[ I_1 = \int_{-1}^{0} \frac{\left|\frac{-2x}{-1 - 3x}\right|}{\left(\frac{-2x}{3x + 1}\right)} \, dx = \int_{-1}^{0} \frac{\frac{2x}{1 + 3x}}{\frac{-2x}{3x + 1}} \, dx = \int_{-1}^{0} \frac{2x(3x + 1)}{-2x(1 + 3x)} \, dx \]

This simplifies to:

\[ I_1 = \int_{-1}^{0} -1 \, dx = -\left[x\right]_{-1}^{0} = -[0 - (-1)] = -1 \]

#### Interval \([-2, -1)\):
In this interval, \([x] = -2\):

\[ I_2 = \int_{-2}^{-1} \frac{\left|\frac{2|x|}{-2 - 3x}\right|}{\left(\frac{2|x|}{3x + 2}\right)} \, dx \]

Following similar steps as above, we find:

\[ I_2 = \int_{-2}^{-1} 1 \, dx = [x]_{-2}^{-1} = -1 - (-2) = 1 \]

#### Continuing this pattern:
For each interval \([-n-1, -n)\), where \(n = 0, 1, 2, \ldots, 9\), we find that:

- For odd \(n\) (like \([-1, 0)\), \([-3, -2)\), etc.), the integral evaluates to \(-1\).
- For even \(n\) (like \([-2, -1)\), \([-4, -3)\), etc.), the integral evaluates to \(1\).

### Step 3: Sum the contributions
We have 10 intervals from \([-10, 0)\):

- Odd intervals contribute \(-1\): 5 intervals (from \([-1, 0)\), \([-3, -2)\), \([-5, -4)\), \([-7, -6)\), \([-9, -8)\)).
- Even intervals contribute \(1\): 5 intervals (from \([-2, -1)\), \([-4, -3)\), \([-6, -5)\), \([-8, -7)\), \([-10, -9)\)).

Thus, the total contribution is:

\[ I = 5(-1) + 5(1) = -5 + 5 = 0 \]

### Final Answer
The value of the integral is:

\[ \boxed{0} \]