The area of the region $\left\{\left(\mathrm{x},\mathrm{y}\right):\mathrm{x}\mathrm{y}\le 8,\mathrm{ }1\le \mathrm{y}\le {\mathrm{x}}^2\right\}$ is

To draw the inequality, let us draw the equation
$xy=8$ and $y=1$ and $y=x^2$
For point of intersection
(i) $xy=8$ and $y=1\Rightarrow A\left(\mathrm{8,1}\right)$
(ii) $xy=8$ and $y=x^2 \Rightarrow x^3=8 \Rightarrow x=2 \Rightarrow B(2, 4)$
(iii) $y=x^2$ and $y=1\Rightarrow C\left(1, 1\right)$ and $D\left( 1, 1\right)$
Now $xy\le 8\Rightarrow$ region which contains origin
$y\ge 1\Rightarrow$ region above line $y=1$
$y\le x^2\Rightarrow$ region outside the parabola
Now required area
Method I:
Using x-axis:
$A=\underset{1}{\overset{2}{\int }}\left(x^2-1\right)\cdot dx+\underset{2}{\overset{8}{\int }}\left(\frac{8}{x}-1\right)\cdot dx$
$A={\left[\frac{x^3}{3}-x\right]}_1^2+{\left[8\mathrm{l}\mathrm{n}x-x\right]}_2^8$
$A=-\frac{14}{3}+16\mathrm{l}\mathrm{n}2$
Method II:
Using y-axis:
$A=\underset{1}{\overset{4}{\int }}\left(\frac{8}{y}-\sqrt{y}\right)\cdot dy$
$A={\left[8\mathrm{l}\mathrm{n}y-\frac{2}{3}{y}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_1^4\Rightarrow A=-\frac{14}{3}+16\mathrm{l}\mathrm{n}2$
Note: The question should include bounded area term as in $2^{\mathrm{n}\mathrm{d}}$ quadrant there exist a area which satisfy the inequality and is unbounded.