Let [x] denote the greatest integer less than or equal to x. Then the value of `int_(1)^(2)|2x-[3x]|dx` is ______

Answer

To solve the integral \( \int_{1}^{2} |2x - [3x]| \, dx \), we will break it down step by step.

### Step 1: Understand the greatest integer function
The greatest integer function, denoted as \([x]\), returns the largest integer less than or equal to \(x\). For \(x\) in the interval \([1, 2]\), we can evaluate \([3x]\):
- When \(x = 1\), \(3x = 3\) and \([3x] = 3\).
- When \(x = 2\), \(3x = 6\) and \([3x] = 6\).

Since \(3x\) varies from \(3\) to \(6\) as \(x\) goes from \(1\) to \(2\), we can break this integral into sub-intervals based on the value of \([3x]\).

### Step 2: Identify the intervals
The function \([3x]\) changes at \(x = \frac{4}{3}\) and \(x = \frac{5}{3}\):
- For \(1 \leq x < \frac{4}{3}\), \([3x] = 3\).
- For \(\frac{4}{3} \leq x < \frac{5}{3}\), \([3x] = 4\).
- For \(\frac{5}{3} \leq x < 2\), \([3x] = 5\).

### Step 3: Break the integral into parts
We can now express the integral as:
\[ \int_{1}^{2} |2x - [3x]| \, dx = \int_{1}^{\frac{4}{3}} |2x - 3| \, dx + \int_{\frac{4}{3}}^{\frac{5}{3}} |2x - 4| \, dx + \int_{\frac{5}{3}}^{2} |2x - 5| \, dx \]

### Step 4: Evaluate each integral
1. **For \(1 \leq x < \frac{4}{3}\)**:
- Here, \(2x - 3 < 0\), so \(|2x - 3| = 3 - 2x\).
- The integral becomes:
\[ \int_{1}^{\frac{4}{3}} (3 - 2x) \, dx \]

2. **For \(\frac{4}{3} \leq x < \frac{5}{3}\)**:
- Here, \(2x - 4 < 0\), so \(|2x - 4| = 4 - 2x\).
- The integral becomes:
\[ \int_{\frac{4}{3}}^{\frac{5}{3}} (4 - 2x) \, dx \]

3. **For \(\frac{5}{3} \leq x < 2\)**:
- Here, \(2x - 5 \geq 0\), so \(|2x - 5| = 2x - 5\).
- The integral becomes:
\[ \int_{\frac{5}{3}}^{2} (2x - 5) \, dx \]

### Step 5: Calculate the integrals
1. **First integral**:
\[ \int_{1}^{\frac{4}{3}} (3 - 2x) \, dx = \left[ 3x - x^2 \right]_{1}^{\frac{4}{3}} = \left(3 \cdot \frac{4}{3} - \left(\frac{4}{3}\right)^2\right) - \left(3 \cdot 1 - 1^2\right) \]
\[ = \left(4 - \frac{16}{9}\right) - (3 - 1) = \left(4 - \frac{16}{9}\right) - 2 = \frac{36}{9} - \frac{16}{9} - \frac{18}{9} = \frac{2}{9} \]

2. **Second integral**:
\[ \int_{\frac{4}{3}}^{\frac{5}{3}} (4 - 2x) \, dx = \left[ 4x - x^2 \right]_{\frac{4}{3}}^{\frac{5}{3}} = \left(4 \cdot \frac{5}{3} - \left(\frac{5}{3}\right)^2\right) - \left(4 \cdot \frac{4}{3} - \left(\frac{4}{3}\right)^2\right) \]
\[ = \left(\frac{20}{3} - \frac{25}{9}\right) - \left(\frac{16}{3} - \frac{16}{9}\right) = \left(\frac{60}{9} - \frac{25}{9}\right) - \left(\frac{48}{9} - \frac{16}{9}\right) = \frac{35}{9} - \frac{32}{9} = \frac{3}{9} = \frac{1}{3} \]

3. **Third integral**:
\[ \int_{\frac{5}{3}}^{2} (2x - 5) \, dx = \left[ x^2 - 5x \right]_{\frac{5}{3}}^{2} = \left(2^2 - 5 \cdot 2\right) - \left(\left(\frac{5}{3}\right)^2 - 5 \cdot \frac{5}{3}\right) \]
\[ = (4 - 10) - \left(\frac{25}{9} - \frac{25}{3}\right) = -6 - \left(\frac{25}{9} - \frac{75}{9}\right) = -6 + \frac{50}{9} = -\frac{54}{9} + \frac{50}{9} = -\frac{4}{9} \]

### Step 6: Combine the results
Now, we combine the results of all three integrals:
\[ \int_{1}^{2} |2x - [3x]| \, dx = \frac{2}{9} + \frac{1}{3} - \frac{4}{9} \]
Convert \(\frac{1}{3}\) to ninths:
\[ \frac{1}{3} = \frac{3}{9} \]
Thus,
\[ \int_{1}^{2} |2x - [3x]| \, dx = \frac{2}{9} + \frac{3}{9} - \frac{4}{9} = \frac{1}{9} \]

### Final Answer
The value of the integral \( \int_{1}^{2} |2x - [3x]| \, dx \) is \( \frac{1}{9} \).