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

$\frac{19}{6}$

The region described by $A = \{(x, y) : y ≥ x^2-5x+4, x + y ≥1, y ≤ 0\}$ is the shaded region shown in Fig.

Its area S is given by

$⇒S=\int\limits_{-2}^0|x_2-x_1|dy=\int\limits_{-2}^0(x_2-x_1)dy=\int\limits_{-2}^0\left\{\frac{5+\sqrt{9+4y}}{2}-(1-y)\right\}dy$

$⇒S=\frac{1}{2}\left[5y+\frac{1}{6}(9+4y)^{3/2}-2y+y^2\right]_{-2}^0$

$⇒S=\frac{1}{2}\left[\frac{9}{2}-\left(-10+\frac{1}{6}+4+4\right)\right]=\frac{19}{6}$