Question

How am I supposed to integrate this?

Answer 1

`6x^2-4x+1=6\left(x-\frac13\right)^2+\frac13\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`