Where I'm behind is how to get the integral

You're asking about how to perform the integration of the expression $(-12t + 8)$ with respect to $t$.

Step 1: Apply the power rule for integration. The power rule states that $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$. For a constant $k$, $\int k \, dx = kx + C$. We integrate each term separately: $\int (-12t + 8) \, dt = \int -12t \, dt + \int 8 \, dt$

Step 2: Integrate each term. For the first term, $\int -12t \, dt$: Here, $t$ has a power of $1$. So, $n=1$. $\int -12t^1 \, dt = -12 \frac{t^{1+1}}{1+1} = -12 \frac{t^2}{2} = -6t^2$ For the second term, $\int 8 \, dt$: This is a constant. $\int 8 \, dt = 8t$

Step 3: Combine the integrated terms. So, the indefinite integral is $-6t^2 + 8t + C$. Since we are evaluating a definite integral from $0$ to $1$, we don't need the constant $C$. We write the result in the evaluation bracket: $\left[ -6t^2 + 8t \right]_0^1$

Step 4: Apply the Fundamental Theorem of Calculus. This means we evaluate the antiderivative at the upper limit ($t=1$) and subtract its value at the lower limit ($t=0$). $(-6(1)^2 + 8(1)) - (-6(0)^2 + 8(0))$ $(-6 + 8) - (0 + 0)$ $2 - 0$ $2$

The value of the integral is $\boxed{2}$.