The area (in sq. units) of the region described by $$A = \{(x, y) | y \geq x^2 - 5x + 4, x + y \geq 1, y \leq 0\}$$ is

The region $$A$$ is bounded above by $$y = 0$$ (x-axis) and below by the line $$y = 1 - x$$ and the parabola $$y = x^2 - 5x + 4$$.

Intersection points: 

Parabola and Line: $$x^2 - 5x + 4 = 1 - x \implies x = 1, 3$$

X-axis Intersections: Line at $$x = 1$$; Parabola at $$x = 1, 4$$

The lower boundary changes from the line to the parabola at $$x = 3$$.

The total area is the sum of two integrals from $$x = 1$$ to $$x = 4$$:

$$\text{Area} = \int_{1}^{3} [0 - (1 - x)] \, dx + \int_{3}^{4} [0 - (x^2 - 5x + 4)] \, dx$$

$$\int_{1}^{3} (x - 1) \, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{3} = 2$$

$$\int_{3}^{4} (-x^2 + 5x - 4) \, dx = \left[ -\frac{x^3}{3} + \frac{5x^2}{2} - 4x \right]_{3}^{4} = \frac{7}{6}$$

$$\text{Total Area} = 2 + \frac{7}{6} = \mathbf{\frac{19}{6} \text{ sq. units}}$$