The area of the region given by x , y : x y ≤ 8 , 1 ≤ y ≤ x 2 is :

Given,

$xy\le 8 \& 1\le y\le x^2$

On solving above curve we get intersecting point as $\left(2,4\right)$

Now on plotting the diagram of the given curves we get,

Required area

$A={\int }_1^2\left(x^2-1\right)dx+{\int }_2^8\left(\frac{8}{x}-1\right)dx$

$\Rightarrow A={\int }_1^2{x}^{2}dx+{\int }_2^8\frac{8}{x}dx-{\int }_1^{8}dx$

$\Rightarrow A={\left[\frac{x^3}{3}\right]}_1^2+8{\left[\ln x\right]}_2^8-{\left[x\right]}_1^8$

$\Rightarrow A=8\ln 4-\frac{14}{3}$ 

$\Rightarrow A=16\ln 2-\frac{14}{3}$ sq. units