The value of `int_(0)^(1)({2x}-1)({3x}-1)dx`, (where {} denotes fractional part opf x) is equal to :

Options

A. a.`19/36`
B. b.`19/144`
C. c.`19/72`
D. d.`19/18`

Answer

To solve the integral \( I = \int_{0}^{1} \{2x - 1\} \{3x - 1\} \, dx \), where \(\{x\}\) denotes the fractional part of \(x\), we will break the integral into intervals where the expressions inside the fractional parts behave consistently.

### Step 1: Understanding the fractional part
The fractional part of a number \(y\) is defined as:
\[ \{y\} = y - \lfloor y \rfloor \]
where \(\lfloor y \rfloor\) is the greatest integer less than or equal to \(y\).

### Step 2: Determine the intervals
We need to find the intervals where \(2x - 1\) and \(3x - 1\) change their integer parts:
- \(2x - 1 = 0\) when \(x = \frac{1}{2}\)
- \(3x - 1 = 0\) when \(x = \frac{1}{3}\)

Thus, we will break the integral into three parts:
1. From \(0\) to \(\frac{1}{3}\)
2. From \(\frac{1}{3}\) to \(\frac{1}{2}\)
3. From \(\frac{1}{2}\) to \(1\)

### Step 3: Evaluate the integral on each interval

**Interval 1: \(x \in [0, \frac{1}{3}]\)**
- Here, \(2x < 1\) and \(3x < 1\), so:
\[ \{2x - 1\} = 2x - 1, \quad \{3x - 1\} = 3x - 1 \]
- The integral becomes:
\[ I_1 = \int_{0}^{\frac{1}{3}} (2x - 1)(3x - 1) \, dx \]

**Interval 2: \(x \in [\frac{1}{3}, \frac{1}{2}]\)**
- Here, \(2x < 1\) and \(3x \geq 1\), so:
\[ \{2x - 1\} = 2x - 1, \quad \{3x - 1\} = 3x - 1 - 1 = 3x - 2 \]
- The integral becomes:
\[ I_2 = \int_{\frac{1}{3}}^{\frac{1}{2}} (2x - 1)(3x - 2) \, dx \]

**Interval 3: \(x \in [\frac{1}{2}, 1]\)**
- Here, \(2x \geq 1\) and \(3x \geq 1\), so:
\[ \{2x - 1\} = 2x - 1 - 1 = 2x - 2, \quad \{3x - 1\} = 3x - 1 - 1 = 3x - 2 \]
- The integral becomes:
\[ I_3 = \int_{\frac{1}{2}}^{1} (2x - 2)(3x - 2) \, dx \]

### Step 4: Calculate each integral

**Calculating \(I_1\):**
\[ I_1 = \int_{0}^{\frac{1}{3}} (2x - 1)(3x - 1) \, dx = \int_{0}^{\frac{1}{3}} (6x^2 - 5x + 1) \, dx \]
Calculating:
\[ = \left[ 2x^3 - \frac{5}{2}x^2 + x \right]_{0}^{\frac{1}{3}} = \left( 2\left(\frac{1}{3}\right)^3 - \frac{5}{2}\left(\frac{1}{3}\right)^2 + \frac{1}{3} \right) = \frac{2}{27} - \frac{5}{18} + \frac{1}{3} \]

**Calculating \(I_2\):**
\[ I_2 = \int_{\frac{1}{3}}^{\frac{1}{2}} (2x - 1)(3x - 2) \, dx = \int_{\frac{1}{3}}^{\frac{1}{2}} (6x^2 - 7x + 2) \, dx \]
Calculating:
\[ = \left[ 2x^3 - \frac{7}{2}x^2 + 2x \right]_{\frac{1}{3}}^{\frac{1}{2}} \]

**Calculating \(I_3\):**
\[ I_3 = \int_{\frac{1}{2}}^{1} (2x - 2)(3x - 2) \, dx = \int_{\frac{1}{2}}^{1} (6x^2 - 10x + 4) \, dx \]
Calculating:
\[ = \left[ 2x^3 - 5x^2 + 4x \right]_{\frac{1}{2}}^{1} \]

### Step 5: Combine the results
Finally, sum \(I_1\), \(I_2\), and \(I_3\) to get the total value of the integral.