логарифм показательные ребят ! ​

Пошаговое объяснение:

F(x)=0 при x< a f(x)=0 при x> b f(x)=c при a< x< b integral [-беск; +беск] f(x)*dx =integral [a; b] c*dx = c*(b-a)=1 c=1/(b-a) f(x)=1/(b-a) при a< x< b mx= =integral [-беск; +беск] f(x)*x*dx = =1/(b-a)* integral [a; b] x*dx = =1/(b-a)*  x^2/2 [a; b] = =1/(b-a)* ( b^2/2 - a^2/2) = =(b+a)/2 = 3 mx^2= =integral [-беск; +беск] f(x)*x^2*dx = =1/(b-a)* integral [a; b] x^2*dx = =1/(b-a)*  x^3/3 [a; b] = =1/(b-a)* ( b^3/3 - a^3/3) = =(b^2+ab+a^2)/3 dx= =mx^2 - (mx)^2= =(b^2+ab+a^2)/3 - (b+a)^2/4 = = (4b^2+4ab+4a^2-3b^2-6ab-3a^2)/12 = = (b^2-2ab+a^2)/12 = = (b-a)^2/12 = = 4/3 *************************** (b+a)/2 = 3 (b-a)^2/12 = 4/3 **************** b+a = 6 (b-a)^2 = 16 **************** b+a = 6 b-a= 4 **************** b=5; a=1 f(x)=0 при x< 1 f(x)=0 при x> 5 f(x)=1/4 при 1< x< 5 f(x)=integral [-беск; x] f(t)*dt f(x)=0 при x< 1 f(x)=(x-1)/4 при 1< x< 5 f(x)=1 при x> 5 p(x є [2; 5]) = f(5) - f(2) = 1 - (2-1)/4 = 3/4