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

The region $R$ is defined by $xy \le 8$, $1 \le y \le x^2$, and $x \ge 0$.
First, find the intersection points of the curves.
Intersection of $y = 1$ and $y = x^2$: $x^2 = 1 \implies x = 1$ (since $x \ge 0$).
Intersection of $y = x^2$ and $xy = 8$: $x(x^2) = 8 \implies x^3 = 8 \implies x = 2$. At $x=2$, $y=4$.
Intersection of $xy = 8$ and $y = 1$: $x(1) = 8 \implies x = 8$.
The region is bounded below by $y = 1$ from $x = 1$ to $x = 8$.
The upper boundary consists of two parts:
From $x = 1$ to $x = 2$, the upper curve is $y = x^2$.
From $x = 2$ to $x = 8$, the upper curve is $y = \frac{8}{x}$.
The area $A$ is given by:
$A = \int_{1}^{2} (x^2 - 1) dx + \int_{2}^{8} (\frac{8}{x} - 1) dx$
$A = [\frac{x^3}{3} - x]_1^2 + [8 \log_e x - x]_2^8$
$A = (\frac{8}{3} - 2) - (\frac{1}{3} - 1) + (8 \log_e 8 - 8) - (8 \log_e 2 - 2)$
$A = (\frac{2}{3}) - (-\frac{2}{3}) + 8(3 \log_e 2) - 8 - 8 \log_e 2 + 2$
$A = \frac{4}{3} + 24 \log_e 2 - 8 \log_e 2 - 6$
$A = 16 \log_e 2 + \frac{4}{3} - 6$
$A = 16 \log_e 2 - \frac{14}{3}$
$A = \frac{2}{3} (24 \log_e 2 - 7)$