To solve this problem, we need to find the exact surface area of the solid formed by rotating the parametric curve given by $x(t) = e^t - 4t$ and $y(t) = 8e^{\frac{t}{2}}$ over the interval $0 \leq t \leq 7$ about the $x$-axis.

The surface area of a solid of revolution generated by rotating a parametric curve $y(t)$ about the $x$-axis is given by the formula:

$SA = \int_{a}^{b} 2\pi y(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

Step 1: Compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$

Given: $x(t) = e^t - 4t$ $\frac{dx}{dt} = e^t - 4$

Also: $y(t) = 8e^{\frac{t}{2}}$ $\frac{dy}{dt} = 8 \cdot \frac{1}{2} e^{\frac{t}{2}} = 4e^{\frac{t}{2}}$

Step 2: Substitute in the formula for surface area

Now, the integrand requires: