f_(2)(x)=-x
f_(1)(x)=x^3
[m]S= ∫^{0} _{-1}(-x-x^3)dx = (-\frac{x^2}{2}-\frac{x^4}{4})|^{0} _{-1} = (-\frac{0^2}{2}-\frac{0^4}{4})-(-\frac{(-1)^2}{2}-\frac{(-1)^4}{4}) =\frac{1}{2}+\frac{1}{4}=\frac{3}{4}[/m]
F(x)=x⁴/4
F(a)=F(1)=1⁴/4=1/4
F(b)=F(2)=2⁴/4=16/4=4
S=F(2)-F(1)=4-1/4=15/4
S=15/4 ед.кв.