To find \(\int (3x - 2)^5 \, dx\), use the substitution method. Let \(u = 3x - 2\), then \(\frac{du}{dx} = 3\) or \(dx = \frac{du}{3}\).

Thus, \(\int (3x - 2)^5 \, dx = \int u^5 \cdot \frac{du}{3} = \frac{1}{3} \int u^5 \, du\).

The integral of \(u^5\) is \(\frac{u^6}{6}\), so:

\(\frac{1}{3} \cdot \frac{u^6}{6} = \frac{(3x - 2)^6}{18} + C\).

Now, evaluate \(\int_0^1 (3x - 2)^5 \, dx\):

\(\left[ \frac{(3x - 2)^6}{18} \right]_0^1 = \frac{(3 \cdot 1 - 2)^6}{18} - \frac{(3 \cdot 0 - 2)^6}{18}\).

Calculate the values:

At \(x = 1\), \((3 \cdot 1 - 2)^6 = 1^6 = 1\).

At \(x = 0\), \((3 \cdot 0 - 2)^6 = (-2)^6 = 64\).

So, \(\frac{1}{18} - \frac{64}{18} = \frac{1 - 64}{18} = \frac{-63}{18} = -\frac{7}{2}\).

Therefore, the value is \(-\frac{3}{2}\).