Area of Region Bounded by Parabola and Line

To find the area of the region bounded by the curves $\{(x, y) : x^2 - 8x \le y \le -x\}$, we first determine the points of intersection between the parabola $y = x^2 - 8x$ and the straight line $y = -x$.

Setting the two equations equal to each other: $x^2 - 8x = -x$ $x^2 - 7x = 0$ $x(x - 7) = 0$

Thus, the curves intersect at $x = 0$ and $x = 7$.

For $x \in [0, 7]$, the line $y = -x$ lies above the parabola $y = x^2 - 8x$ (i.e., $-x \ge x^2 - 8x$).

The area $A$ of the enclosed region is given by the definite integral: $A = \int_{0}^{7} \left( -x - (x^2 - 8x) \right) dx$

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

Evaluating the integral: $A = \left[ \frac{7x^2}{2} - \frac{x^3}{3} \right]_0^7$

Substitute the limits: $A = \left( \frac{7(7)^2}{2} - \frac{7^3}{3} \right) - 0$ $A = \frac{343}{2} - \frac{343}{3}$ $A = 343 \left( \frac{1}{2} - \frac{1}{3} \right)$ $A = 343 \left( \frac{1}{6} \right) = \frac{343}{6}$

Thus, the area of the bounded region is $\frac{343}{6}$.