Question

How am I supposed to integrate this?

Answer 1

$6x^2-4x+1=6\left(x-\dfrac13\right)^2+\dfrac13\ge0$

which means the parabola lies above the x-axis over its entire domain. This means the area is given by

$\displaystyle\int_{-1}^2(6x^2-4x+1)\,\mathrm dx=2x^3-x^2+x\bigg|_{x=-1}^{x=2}=10-(-5)=15$