The area (in sq. units) of the region described by A = x , y |   y ≥ x 2 - 5 x + 4 ,   x + y…

The given equation is

$A=\left\{\left(x,y\right)| y\ge x^2-5x+4,x+y\ge 1,y\le 0\right\}$ is

The equation $x^2-5x+4$ is cutting the $x-$ axis at $x=1,4$ 

Required area $=\frac{1}{2}\times 2\times 2+\left|\underset{3}{\overset{4}{\int }}\left(x^2-5x+4\right)dx\right|$

$=2+{\left[\frac{x^3}{3}-\frac{5x^2}{2}+4x\right]}_3^4$

$=2+\frac{7}{6}=\frac{19}{6} \mathrm{s}\mathrm{q}.\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}$