The area of the region $\{(x,y):xy \leq 8,\,1 \leq y \leq x^{2}\}$ is (1) $8\log_{e}2-\frac{14}{3}$ (2) $16\log_{e}2-\frac{14}{3}$ (3) $8\log_{e}2-\frac{7}{3}$ (4) $16\log_{e}2-6$

The area of the region {(x,y):xy ≤ 8,,1 ≤ y ≤ x²} is

$xy \le 8, \quad 1 \le y \le x^{2}$

$\text{boundaries: } y = 1, \ y = x^{2}, \ xy = 8$

$\text{Find the area of the region}$

$1.\ \text{plot } y=1,\ y=x^2,\ xy=8$

$2.\ \text{find the intersection points (limits)}$

$3.\ \text{shade and integrate}$

$y=x^2 \cap y=1: \ (1,1)$

$y=x^2 \cap xy=8: \ x^3=8 \Rightarrow (2,4)$

$xy=8 \cap y=1: \ (8,1)$

$1 \le x \le 2: \ \text{upper } y=x^2, \ \text{lower } y=1$

$2 \le x \le 8: \ \text{upper } y=\dfrac{8}{x}, \ \text{lower } y=1$

$A_1 = \int_1^2 (x^2 - 1)\,dx = \left[\dfrac{x^3}{3} - x\right]_1^2$

$= \left(\dfrac{8}{3}-2\right) - \left(\dfrac{1}{3}-1\right) = \dfrac{4}{3}$

$A_2 = \int_2^8 \left(\dfrac{8}{x} - 1\right)dx = \big[8\ln x - x\big]_2^8$

$= (8\ln 8 - 8) - (8\ln 2 - 2) = 16\ln 2 - 6$

$A = A_1 + A_2 = \dfrac{4}{3} + 16\ln 2 - 6 = 16\ln 2 - \dfrac{14}{3}$

$= 16\log_e 2 - \dfrac{14}{3} \Rightarrow \text{option (2)}$