Помогите решить!
А1. 1)=(4х-(х^3/3))|^2_(1)=(8-(8/3))-(4-(1/3))=5/3 2)=((-x^3/3)-x^2+3x)|^0_(-3)=9 A2 1)=-cosx|^(π/2)_(0)=1 2)=((x^3/3)+x^2+3x)|^3_(2)=17-(8/3) 3)=(-1/x)|^4_(2)=(-1/4)+(1/2)=1/4 A3 F(x)=x^3+x+C -2=1+1+C C==-4 О т в е т. F(x)=x^3+x-4