To evaluate the integral \(\int_{1}^{2} (6x^2 - 2x) \, dx\), we need to find the antiderivative of the integrand and then apply the Fundamental Theorem of Calculus.

Let's integrate each term separately:

1. Integral of \(6x^2\) with respect to \(x\):

\(\int 6x^2 \, dx = 2x^3 + C_1\)

2. Integral of \(-2x\) with respect to \(x\):

\(\int -2x \, dx = -x^2 + C_2\)

Now, we'll apply the Fundamental Theorem of Calculus to evaluate the definite integral:

\(\int_{1}^{2} (6x^2 - 2x) \, dx = \left[2x^3 - x^2\right]_{1}^{2}\)

Now plug in the upper and lower limits of integration:

\(\left[2(2)^3 - (2)^2\right] - \left[2(1)^3 - (1)^2\right]\)

\(= [16 - 4] - [2 - 1]\)

\(= 12 - 1\)

\(= 11\)