The area (in sq.

Question

The area (in sq. units) of the region described by ${(x,y):{y}^{2}\le 2xy\ge 4x-1}$ is

Accepted Answer

$=\int _{-\frac{1}{2}}^{1}(\frac{y+1}{4}-\frac{{y}^{2}}{2})dy$

$\begin{gathered} ={[\frac{{y}^{2}}{8}+\frac{y}{4}-\frac{{y}^{3}}{6}]}_{1}^{-\frac{1}{2}} \ =(\frac{1}{8}+\frac{1}{4}-\frac{1}{6})-(\frac{1}{32}-\frac{1}{8}+\frac{1}{48}) \ =\frac{5}{24}+\frac{7}{96} \ =\frac{27}{96} \ =\frac{9}{32} \end{gathered}$

Solutions · 1