Area Under The Line

This region will be bounded to the left by the positive $y$ axis, above by the line $x + y = 20$ and below by the line $x + y = 18$ and the positive $x$ axis.

The area of this region will then be the area of a right isosceles triangle with legs length $20$ minus the area of a right isosceles triangle with legs length $18.$

This comes out to $\dfrac{20^{2}}{2} - \dfrac{18^{2}}{2} = \dfrac{(20 - 18)(20 + 18)}{2} = \boxed{38}.$

In the 1st Quadrant, the inequalities $18 \leq x+y$ and $x+y \leq 20$ defined similar right triangles with areas $\frac{20^2}{2}$ and $\frac{18^2}{2}$ . The area of region in question is therefore $\frac{20^2 - 18^2}{2} = 38$ sq. units.