Integration

$Here $\int_{0}^{1}f(x)dx=2\Leftrightarrow F(1)-F(0)=2\Leftrightarrow F(1)=2+F(0)=2+2=4$\\\\ So $\boxed{F(1)=4}$ and $2F(x+2)+F(x+1)y+F(x+4)=0....................(1)$\\\\ and $24F(x+2)-2y+3=0............................(2)$\\\\ Put $x=-1$,We Get \\\\ $2F(1)+F(0)y+F(3)=0\Leftrightarrow 8+2y+F(3)=0...........(1)$\\\\ Similarly $24F(1)-2y+3=0\Leftrightarrow 24\times 4-2y+3=0.......(2)$\\\\ So $\boxed{F(3)=-107}$\\\\ So $\int_{0}^{1}f(3x)dx=\frac{1}{3}.\int_{0}^{3}f(x)dx=\frac{1}{3}.\left ( F(3)-F(0) \right )=\frac{1}{3}.\left ( -107-2 \right )$\\\\ So $\boxed{\boxed{\int_{0}^{1}f(3x)dx=-\frac{109}{3}}}$

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