To solve the integral \( I = \int_{0}^{1} \{2x - 1\} \{3x - 1\} \, dx \), where \(\{x\}\) denotes the fractional part of \(x\), we will break down the problem step by step.
### Step 1: Understand the fractional part function
The fractional part function \(\{x\}\) can be expressed as:
\[
\{x\} = x - \lfloor x \rfloor
\]
where \(\lfloor x \rfloor\) is the greatest integer less than or equal to \(x\).
### Step 2: Rewrite the integral
We can rewrite the integral using the definition of the fractional part:
\[
I = \int_{0}^{1} (2x - 1 - \lfloor 2x \rfloor)(3x - 1 - \lfloor 3x \rfloor) \, dx
\]
### Step 3: Determine the points of discontinuity
The functions \(\lfloor 2x \rfloor\) and \(\lfloor 3x \rfloor\) change at certain points within the interval \([0, 1]\):
- \(\lfloor 2x \rfloor\) changes at \(x = \frac{1}{2}\)
- \(\lfloor 3x \rfloor\) changes at \(x = \frac{1}{3}\) and \(x = \frac{2}{3}\)
Thus, we will break the integral into intervals: \([0, \frac{1}{3}]\), \([\frac{1}{3}, \frac{1}{2}]\), \([\frac{1}{2}, \frac{2}{3}]\), and \([\frac{2}{3}, 1]\).
### Step 4: Evaluate the integral over each interval
#### Interval 1: \(x \in [0, \frac{1}{3}]\)
In this interval:
- \(\lfloor 2x \rfloor = 0\)
- \(\lfloor 3x \rfloor = 0\)
Thus,
\[
I_1 = \int_{0}^{\frac{1}{3}} (2x - 1)(3x - 1) \, dx = \int_{0}^{\frac{1}{3}} (6x^2 - 5x + 1) \, dx
\]
Calculating this integral:
\[
I_1 = \left[ 2x^3 - \frac{5}{2}x^2 + x \right]_{0}^{\frac{1}{3}} = \left(2 \left(\frac{1}{3}\right)^3 - \frac{5}{2} \left(\frac{1}{3}\right)^2 + \frac{1}{3}\right) - 0
\]
\[
= \left(2 \cdot \frac{1}{27} - \frac{5}{2} \cdot \frac{1}{9} + \frac{1}{3}\right) = \left(\frac{2}{27} - \frac{5}{18} + \frac{1}{3}\right)
\]
Finding a common denominator (54):
\[
= \left(\frac{4}{54} - \frac{15}{54} + \frac{18}{54}\right) = \frac{7}{54}
\]
#### Interval 2: \(x \in [\frac{1}{3}, \frac{1}{2}]\)
In this interval:
- \(\lfloor 2x \rfloor = 0\)
- \(\lfloor 3x \rfloor = 1\)
Thus,
\[
I_2 = \int_{\frac{1}{3}}^{\frac{1}{2}} (2x - 1)(3x - 2) \, dx = \int_{\frac{1}{3}}^{\frac{1}{2}} (6x^2 - 7x + 2) \, dx
\]
Calculating this integral:
\[
I_2 = \left[ 2x^3 - \frac{7}{2}x^2 + 2x \right]_{\frac{1}{3}}^{\frac{1}{2}} = \left(2 \left(\frac{1}{2}\right)^3 - \frac{7}{2} \left(\frac{1}{2}\right)^2 + 2 \cdot \frac{1}{2}\right) - \left(2 \left(\frac{1}{3}\right)^3 - \frac{7}{2} \left(\frac{1}{3}\right)^2 + 2 \cdot \frac{1}{3}\right)
\]
Calculating each term:
\[
= \left(2 \cdot \frac{1}{8} - \frac{7}{8} + 1\right) - \left(2 \cdot \frac{1}{27} - \frac{7}{18} + \frac{2}{3}\right)
\]
Finding a common denominator (72):
\[
= \left(\frac{18}{72} - \frac{63}{72} + \frac{72}{72}\right) - \left(\frac{8}{72} - \frac{28}{72} + \frac{48}{72}\right)
\]
\[
= \left(\frac{27}{72}\right) - \left(\frac{28}{72}\right) = -\frac{1}{72}
\]
#### Interval 3: \(x \in [\frac{1}{2}, \frac{2}{3}]\)
In this interval:
- \(\lfloor 2x \rfloor = 1\)
- \(\lfloor 3x \rfloor = 1\)
Thus,
\[
I_3 = \int_{\frac{1}{2}}^{\frac{2}{3}} (2x - 2)(3x - 2) \, dx = \int_{\frac{1}{2}}^{\frac{2}{3}} (6x^2 - 10x + 4) \, dx
\]
Calculating this integral:
\[
I_3 = \left[ 2x^3 - 5x^2 + 4x \right]_{\frac{1}{2}}^{\frac{2}{3}} = \left(2 \left(\frac{2}{3}\right)^3 - 5 \left(\frac{2}{3}\right)^2 + 4 \cdot \frac{2}{3}\right) - \left(2 \left(\frac{1}{2}\right)^3 - 5 \left(\frac{1}{2}\right)^2 + 4 \cdot \frac{1}{2}\right)
\]
Calculating each term:
\[
= \left(2 \cdot \frac{8}{27} - 5 \cdot \frac{4}{9} + \frac{8}{3}\right) - \left(2 \cdot \frac{1}{8} - \frac{5}{4} + 2\right)
\]
Finding a common denominator (108):
\[
= \left(\frac{16}{27} - \frac{60}{27} + \frac{108}{27}\right) - \left(\frac{27}{108} - \frac{135}{108} + \frac{216}{108}\right)
\]
\[
= \left(\frac{64}{27}\right) - \left(\frac{108}{108}\right) = \frac{64}{27} - 1 = \frac{37}{27}
\]
#### Interval 4: \(x \in [\frac{2}{3}, 1]\)
In this interval:
- \(\lfloor 2x \rfloor = 1\)
- \(\lfloor 3x \rfloor = 2\)
Thus,
\[
I_4 = \int_{\frac{2}{3}}^{1} (2x - 2)(3x - 2) \, dx = \int_{\frac{2}{3}}^{1} (6x^2 - 10x + 4) \, dx
\]
Calculating this integral:
\[
I_4 = \left[ 2x^3 - 5x^2 + 4x \right]_{\frac{2}{3}}^{1} = \left(2 \cdot 1^3 - 5 \cdot 1^2 + 4 \cdot 1\right) - \left(2 \cdot \left(\frac{2}{3}\right)^3 - 5 \cdot \left(\frac{2}{3}\right)^2 + 4 \cdot \frac{2}{3}\right)
\]
Calculating each term:
\[
= \left(2 - 5 + 4\right) - \left(2 \cdot \frac{8}{27} - 5 \cdot \frac{4}{9} + \frac{8}{3}\right)
\]
Finding a common denominator (27):
\[
= 1 - \left(\frac{16}{27} - \frac{60}{27} + \frac{72}{27}\right) = 1 - \left(\frac{28}{27}\right) = 1 - 1.037 = -\frac{1}{27}
\]
### Step 5: Combine the results
Now we combine all the integrals:
\[
I = I_1 + I_2 + I_3 + I_4 = \frac{7}{54} - \frac{1}{72} + \frac{37}{27} - \frac{1}{27}
\]
Finding a common denominator (108):
\[
= \frac{14}{108} - \frac{1.5}{108} + \frac{148}{108} - \frac{4}{108} = \frac{14 - 1.5 + 148 - 4}{108} = \frac{156.5}{108} = \frac{313}{216}
\]
Thus, the final value of the integral is:
\[
\boxed{\frac{313}{216}}
\]
