Answer
Therefore, the surface area of the curve when it is rotated around the x-axis is $\pi(\sqrt{2} + \ln(1 + \sqrt{2}))$.
Work Step by Step
\begin{equation*}
\text{Surface Area} = 2\pi\int_{a}^{b} f(x)\sqrt{1 + (f'(x))^2} , dx
\end{equation*}
where $f(x)$ is the function being rotated and $a$ and $b$ are the limits of integration.
In this case, $f(x) = \frac{1}{e^x}$ and $f'(x) = -\frac{1}{e^x}$, so we can substitute these values into the formula and simplify:
\begin{align*}
\text{Surface Area} &= 2\pi\int_{0}^{\infty} \frac{1}{e^x} \sqrt{1 + \left(-\frac{1}{e^x}\right)^2} , dx \
&= 2\pi\int_{0}^{\infty} \frac{1}{e^x} \sqrt{1 + \frac{1}{e^{2x}}} , dx
\end{align*}
To solve this integral, we can use the substitution $u = 1 + \frac{1}{e^{2x}}$, which gives us $\frac{du}{dx} = -\frac{2}{e^{2x}}\left(1 + \frac{1}{e^{2x}}\right)^{-2}$ and $dx = -\frac{1}{2}(u-1)u^{-\frac{3}{2}} , du$. Substituting these values into the integral, we get:
\begin{align*}
\text{Surface Area} &= -\pi\int_{2}^{\infty} (u-1)u^{-\frac{3}{2}} , du \
&= -\pi\int_{\sqrt{2}}^{\infty} (v^2 - 1) , dv \quad \text{(substituting } v = \sqrt{u}\text{)} \
&= -\pi\left[\frac{v^3}{3} - v\right]_{\sqrt{2}}^{\infty} \
&= \pi(\sqrt{2} + \ln(1 + \sqrt{2}))
\end{align*}
