[tex]\begin{gathered} f^{\prime}(x)=\int 4x+4\sin xdx=2x^2+-4\cos x+c \\ f^{\prime}(0)=-4+c=2\rightarrow c=6 \\ f(x)=\int 2x^2-4cos(x)+6dx=\frac{2}{3}x^3-4\sin (x)+6x+c \\ f(0)=c=4 \\ f(x)=\frac{2}{3}x^3-4\sin (x)+6x+4\rightarrow \\ f(3)=18-4\sin (3)+18+4=40-4\sin (3) \end{gathered}[/tex]
