f(x)+f(2-x)=[TEX]x^2-2x+2[/TEX]
[tex]\int_{0}^{2}(f(x)+f(2-x))dx=\int_{0}^{2}x^2-2x+2=\frac{8}{3}\Rightarrow F(x)-F(2-x)|_0^2=\frac{8}{3}\Rightarrow 2F(2)-2F(0)=\frac{8}{3}\Rightarrow \int_{0}^{2}f(x)dx = \frac{4}{3}[/tex]
x=0 =>f(0)+f(2)=2
=>f(2)=-1
[tex]\int_{0}^{2}xf'(x)dx=x.f(x)|_0^2-\int_{0}^{2}f(x)=2f(2) - \frac{4}{3}=\frac{-10}{3}[/tex]