The area of the region $$\{(x, y) : x^2 - 8x \le y \le -x\}$$ is :

$$\text{Area} = \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) \, dx$$

$$x^2 - 8x = -x$$

$$x^2 - 7x = 0$$

$$x(x - 7) = 0 \implies x = 0 \quad \text{and} \quad x = 7$$

$$\text{Area} = \int_{0}^{7} \left( -x - (x^2 - 8x) \right) \, dx$$

$$\text{Area} = \int_{0}^{7} (7x - x^2) \, dx$$

$$\text{Area} = \left[ \frac{7x^2}{2} - \frac{x^3}{3} \right]_{0}^{7}$$

$$\text{Area} = 343 \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{343}{6}$$