ху = 4; х+ у = 5
[m]S= ∫ ^{4}_]1|(5-x-\frac{4}{x})dx=(5x-\frac{x^2}{2}-4ln|x|)|^{4}_{1}=(5\cdot 4-\frac{4^2}{2}-4ln|4|)-(5\cdot 1-\frac{1^2}{2}-4ln|1|)=...[/m]
