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.)

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Answer

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

Steps

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 $\boxed{-1.5}$.