[tex]6x^2-4x+1=6\left(x-\dfrac13\right)^2+\dfrac13\ge0[/tex]
which means the parabola lies above the x-axis over its entire domain. This means the area is given by
[tex]\displaystyle\int_{-1}^2(6x^2-4x+1)\,\mathrm dx=2x^3-x^2+x\bigg|_{x=-1}^{x=2}=10-(-5)=15[/tex]
