The area of the region $$R=\left\{(x,y):xy\leq 8,1\leq y\leq x^{2},x\geq 0\right\}$$ is

The area of the region $$R = \{(x, y) : xy \le 8, 1 \le y \le x^2, x \ge 0\}$$ is calculated as follows:

$$Area = \int_{1}^{2} (x^2 - 1) dx + \int_{2}^{8} \left(\frac{8}{x} - 1\right) dx$$

$$= \left[ \frac{x^3}{3} - x \right]_{1}^{2} + \left[ 8 \log_e x - x \right]_{2}^{8}$$

$$= \left( \frac{8}{3} - 2 - \frac{1}{3} + 1 \right) + (8 \log_e 8 - 8 - 8 \log_e 2 + 2)$$

$$= \frac{4}{3} + (24 \log_e 2 - 8 \log_e 2 - 6)$$

$$= \frac{4}{3} + 16 \log_e 2 - 6$$

$$= \frac{48 \log_e 2 - 14}{3} = \frac{2}{3}(24 \log_e 2 - 7)$$

Correct Option: C