Evaluate the following integral using the Fundamental Theorem of Calculus. \( \int_{\frac{1}{2}}^{1}\left(\mathrm{t}^{-3}-6\right) \mathrm{dt} \) \( \frac{1}{\frac{1}{2}}\left(\mathrm{t}^{-3}-6\right) \mathrm{dt}=\square \) (Type an exact answer.)

To evaluate the integral \( \int_{\frac{1}{2}}^{1}\left(t^{-3}-6\right) \mathrm{dt} \) using the Fundamental Theorem of Calculus, we first need to find the antiderivative of the integrand. The integral can be separated into two parts: \[ \int \left(t^{-3} - 6\right) dt = \int t^{-3} dt - \int 6 dt \] Finding the antiderivatives: 1. For \( \int t^{-3} dt \): \[ \int t^{-3} dt = \frac{t^{-2}}{-2} = -\frac{1}{2t^2} \] 2. For \( \int 6 dt \): \[ \int 6 dt = 6t \] So, the antiderivative \( F(t) \) is: \[ F(t) = -\frac{1}{2t^2} - 6t + C \] Now we will evaluate this from \( \frac{1}{2} \) to \( 1 \): \[ \int_{\frac{1}{2}}^{1}\left(t^{-3} - 6\right) dt = F(1) - F\left(\frac{1}{2}\right) \] Calculating \( F(1) \): \[ F(1) = -\frac{1}{2(1^2)} - 6(1) = -\frac{1}{2} - 6 = -\frac{1}{2} - \frac{12}{2} = -\frac{13}{2} \] Calculating \( F\left(\frac{1}{2}\right) \): \[ F\left(\frac{1}{2}\right) = -\frac{1}{2\left(\frac{1}{2}\right)^2} - 6\left(\frac{1}{2}\right) = -\frac{1}{2 \cdot \frac{1}{4}} - 3 = -\frac{1}{\frac{1}{2}} - 3 = -2 - 3 = -5 \] Now substituting back into the definite integral: \[ \int_{\frac{1}{2}}^{1}\left(t^{-3} - 6\right) dt = F(1) - F\left(\frac{1}{2}\right) = \left(-\frac{13}{2}\right) - \left(-5\right) = -\frac{13}{2} + 5 \] Converting \( 5 \) into halves: \[ 5 = \frac{10}{2} \] Thus the final calculation: \[ -\frac{13}{2} + \frac{10}{2} = -\frac{3}{2} \] The evaluated integral is: \[ \int_{\frac{1}{2}}^{1}\left(t^{-3}-6\right) \mathrm{dt} = -\frac{3}{2} \]