JEE Advanced 2019 · Paper 1, Mathematics question 4

Take $x>0$ (the region intended). For each $y\ge1$ the region runs from $x=\sqrt y$ to $x=\frac8y$. It is non-empty while $\sqrt y\le\frac8y$, i.e. $y\le4$.

$$\text{Area}=\int_1^4\left(\frac8y-\sqrt y\right)dy=8\ln4-\frac23(8-1)=16\log_e2-\frac{14}3.$$
(A)