Since the three vertices of the bounded region are known exactly, the area can be found directly using the coordinate geometry shoelace formula instead of integrating.
The triangle has vertices $(x_1,y_1) = (0,0)$, $(x_2,y_2) = (8,0)$ and $(x_3,y_3) = (0,8)$, the origin and the two axis intercepts of the line $y = 8 - x$.
The shoelace (determinant) formula for the area of a triangle with these vertices is:
\[ A = \frac{1}{2}\left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]
Substituting the coordinates:
\[ A = \frac{1}{2}\left| 0(0-8) + 8(8-0) + 0(0-0) \right| = \frac{1}{2}\left| 0 + 64 + 0 \right| = \frac{64}{2} = 32 \]
This confirms, by a purely coordinate-geometry route with no integration at all, that the enclosed area is exactly 32 square units.
\[\boxed{A = 32 \text{ square units}}\]