Так как [m]f (x)=F `(x)[/m]
[m]f(x)=(\frac{1}{π}arctg \frac{x}{π}+\frac{1}{2})` [/m]
[m]f(x)=\frac{1}{π}\cdot \frac{1}{1+(\frac{x}{π})^2}\cdot (\frac{x}{π})`+0 [/m]
[m]f(x)=\frac{1}{π^2}\cdot\frac{π^2}{π^2+x^2} [/m]
[m]f(x)=\frac{1}{π^2+x^2} [/m]
2)
P( α< X < β)=F( β)-F( α )
P( π< X <π*sqrt(3))=F(π*sqrt(3))-F( π )=[m](\frac{1}{π}arctg \frac{π\sqrt{3}}{π}+\frac{1}{2})[/m]- [m](\frac{1}{π}arctg \frac{π}{π}+\frac{1}{2})=[/m]
[m]=\frac{1}{π}arctg\sqrt{3}-\frac{1}{π}arctg 1=\frac{1}{π}\cdot \frac{π}{3}-\frac{1}{π}\cdot \frac{π}{4}=\frac{1}{3}-\frac{1}{4}=\frac{1}{12} [/m]