The area of the region \ (x, y) : x^2 - 8x ≤ y ≤ -x\ is :

The given region is bounded by the parabola $y = x^2 - 8x$ and the line $y = -x$.

To find the points of intersection, equate the two equations:

$x^2 - 8x = -x$

$x^2 - 7x = 0$

$x(x - 7) = 0$

The points of intersection are $x = 0$ and $x = 7$.

In the interval $[0, 7]$, the line $y = -x$ lies above the parabola $y = x^2 - 8x$.

The required area $A$ is given by:

$A = \int_{0}^{7} (-x - (x^2 - 8x)) dx$

$A = \int_{0}^{7} (7x - x^2) dx$

Evaluating the integral:

$A = \left[ \dfrac{7x^2}{2} - \dfrac{x^3}{3} \right]_{0}^{7}$

$A = \dfrac{7(49)}{2} - \dfrac{343}{3}$

$A = \dfrac{343}{2} - \dfrac{343}{3}$

$A = 343 \left( \dfrac{1}{2} - \dfrac{1}{3} \right)$

$A = \dfrac{343}{6}$

Answer: $\dfrac{343}{6}$