Evaluate the integral $\int_{1 / 2}^{1}\left(x^{-3}-6\right) d x$ using the Fundamental Theorem of Calculus. \[ \int_{1 / 2}^{1}\left(x^{-3}-6\right) d x=\square \] (Type an exact answer.)

Step 1 ：First, we need to find the antiderivative of the function \(f(x) = x^{-3} - 6\).

Step 2 ：The antiderivative of \(x^{-3}\) is \(-1/2 * x^{-2}\) and the antiderivative of \(-6\) is \(-6x\).

Step 3 ：So, the antiderivative \(F(x)\) of \(f(x)\) is \(-1/2 * x^{-2} - 6x\).

Step 4 ：We need to evaluate \(F(x)\) at \(x = 1\) and \(x = 1/2\) and subtract the two results to get the value of the integral.

Step 5 ：Final Answer: The value of the integral \(\int_{1 / 2}^{1}\left(x^{-3}-6\right) d x\) is \(\boxed{-1.5}\).