You can find it by integrating twice:
[tex]\displaystyle\int v''(x)\,\mathrm dx=\int(-6x+2)\,\mathrm dx\iff v'(x)=-3x^2+2x+C_1[/tex]
[tex]\displaystyle\int v'(x)\,\mathrm dx=\int(-3x^2+2x+C_1)\,\mathrm dx\iff v(x)=-x^3+x^2+C_1x+C_2[/tex]
Given that [tex]v(0)=0[/tex], you have
[tex]0=-0^3+0^2+0C_1+C_2\implies C_2=0[/tex]
and given [tex]v(1)=-1[/tex],
[tex]-1=-1^3+1^2+C_1\implies C_1=-1[/tex]
[tex]\implies v(x)=-x^3+x^2-x[/tex]
