Question

The area of the region $(x,y)$ : $xy\le 8,1\le y\le x^2$ is

• A. $16{\log }_{e}2-\frac{14}{3}$
• B. $8{\log }_{e}2-\frac{14}{3}$
• C. $16{\log }_{e}2-6$
• D. $8{\log }_{e}2-\frac{7}{3}$

Answer 1

The correct option is A $16{\log }_{e}2-\frac{14}{3}$


Point of intersection of $y=x^2...\left(i\right)$ and $xy=\mathrm{8...}\left(ii\right)$
solving the equation $\left(i\right)$ and $\left(ii\right)$ we get,
$x=2,y=4$.
Therefore required area of region
$=\underset{1}{\overset{4}{\int }}(\frac{8}{y}-\sqrt{y})dy$
$={[8{\log }_{e}y-\frac{2}{3}{y}^{\frac{3}{2}}]}_1^4$

$=[8{\log }_{e}4-\frac{2}{3}(4{)}^{\frac{3}{2}}-8{\log }_{e}1+\frac{2}{3}]$

$=16{\log }_{e}2-\frac{14}{3}$