y=1/x y=x y=x/4
[m]\frac{1}{2}S= ∫^{1} _{0}(x-\frac{x}{4})dx+ ∫^{2} _{1}(\frac{1}{x}-\frac{x}{4}dx=(\frac{x^2}{2}-\frac{x^2}{8})|^{1} _{0}+(ln|x|-\frac{x^2}{8})|^{2} _{1}=[/m]
[m]=(\frac{1^2}{2}-\frac{1^2}{8})-(\frac{0^2}{2}-\frac{0^2}{8})+(ln|2|-\frac{2^2}{8})-(ln|1|-\frac{1^2}{8})=ln2[/m]
В силу симметрии:
[m] S=2\cdot ln2[/m]