1. **State the problem:** We want to evaluate the definite integral $$\int_0^2 (12x + 4) f'(x) \, dx$$ given that the area bounded by the graph of $$y = f(x)$$, the x-axis, and the y-axis in the first quadrant is $$\frac{8}{3}$$. 2. **Recall integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ 3. **Choose parts:** Let $$u = 12x + 4 \implies du = 12 \, dx$$ $$dv = f'(x) \, dx \implies v = f(x)$$ 4. **Apply integration by parts:** $$\int_0^2 (12x + 4) f'(x) \, dx = \left[(12x + 4) f(x)\right]_0^2 - \int_0^2 12 f(x) \, dx$$ 5. **Evaluate the boundary term:** At $$x=2$$, $$f(2) = 0$$ (given from the graph). At $$x=0$$, $$f(0) = 4$$ (given from the graph). So, $$\left[(12x + 4) f(x)\right]_0^2 = (12 \cdot 2 + 4) \cdot 0 - (12 \cdot 0 + 4) \cdot 4 = 0 - 16 = -16$$ 6. **Evaluate the integral term:** We know the area under $$f(x)$$ from 0 to 2 is $$\int_0^2 f(x) \, dx = \frac{8}{3}$$. Therefore, $$\int_0^2 12 f(x) \, dx = 12 \cdot \frac{8}{3} = 32$$ 7. **Combine results:** $$\int_0^2 (12x + 4) f'(x) \, dx = -16 - 32 = -48$$ **Final answer:** $$\boxed{-48}$$