\begin{tabular}{l} Find the volume of the solid generated by revolving the \\ \( \qquad \begin{aligned} y & =e^{-2 x} \\ y & =0 \\ x & =0 \\ x & =4\end{aligned} \) \\ \hline\end{tabular}

We are given the region bounded by   y = e^(–2x), y = 0, x = 0, x = 4. When this region is revolved about the x-axis, the resulting solid has a volume that can be computed using the disk method. Here’s how we proceed: 1. At a given x, the radius of the disk is given by the distance from the x-axis to the curve, which is R(x) = e^(–2x). 2. The area of a typical disk is A(x) = π[R(x)]² = π(e^(–2x))² = π e^(–4x). 3. The volume is found by integrating this area from x = 0 to x = 4. Thus, the volume V is given by   V = ∫ from 0 to 4 π e^(–4x) dx. 4. Factor out the constant π:   V = π ∫ from 0 to 4 e^(–4x) dx. 5. Evaluate the integral ∫ e^(–4x) dx. The antiderivative is   ∫ e^(–4x) dx = –(1/4) e^(–4x) + C. 6. Now, apply the limits from 0 to 4:   ∫ from 0 to 4 e^(–4x) dx = [ –(1/4) e^(–4x) ] from 0 to 4     = –(1/4)e^(–16) + (1/4)e^(0)     = (1/4)(1 – e^(–16)). 7. Multiply by π to get the volume:   V = π × (1/4)(1 – e^(–16))     = (π/4)(1 – e^(–16)). Thus, the volume of the solid generated by revolving the given region about the x-axis is   V = (π/4)(1 – e^(–16)).