1. **State the problem:** Find the definite integral $$\int_0^2 (2x - 1)^3 \, dx$$ using integration by substitution. 2. **Formula and substitution rule:** If $$x = g(t)$$, and $$a = g(\alpha)$$, $$b = g(\beta)$$, then $$\int_a^b f(x) \, dx = \int_{\alpha}^{\beta} f(g(t)) g'(t) \, dt$$ 3. **Choose substitution:** Let $$t = 2x - 1$$. Then, $$x = \frac{t + 1}{2}$$ and $$dx = \frac{1}{2} dt$$. 4. **Change limits:** When $$x = 0$$, $$t = 2(0) - 1 = -1$$. When $$x = 2$$, $$t = 2(2) - 1 = 3$$. 5. **Rewrite the integral:** $$\int_0^2 (2x - 1)^3 \, dx = \int_{-1}^3 t^3 \cdot \frac{1}{2} \, dt = \frac{1}{2} \int_{-1}^3 t^3 \, dt$$. 6. **Integrate:** $$\int t^3 \, dt = \frac{t^4}{4}$$. 7. **Evaluate definite integral:** $$\frac{1}{2} \left[ \frac{t^4}{4} \right]_{-1}^3 = \frac{1}{2} \left( \frac{3^4}{4} - \frac{(-1)^4}{4} \right) = \frac{1}{2} \left( \frac{81}{4} - \frac{1}{4} \right) = \frac{1}{2} \cdot \frac{80}{4} = \frac{1}{2} \cdot 20 = 10$$. **Final answer:** $$10$$