Question
Set up an integral that represents the surface area of the surface generated by revolving the graph of $f(x)=\sqrt{5x-8}$ around the $x$-axis over the interval $[2,4]$.
Do not evaluate the integral.

Step 1 ：Given the function $f(x)=\sqrt{5x-8}$, we are asked to find the surface area of the surface generated by revolving this function around the x-axis over the interval [2,4].

Step 2 ：The formula for the surface area of a solid of revolution is $A=2\pi {\int }_a^{b}f(x)\sqrt{1+[f^{\prime }(x){]}^2}dx$, where $f(x)$ is the function being revolved, $f^{\prime }(x)$ is its derivative, and [a, b] is the interval over which the function is being revolved.

Step 3 ：First, we need to find the derivative of $f(x)$, denoted as $f^{\prime }(x)$. For $f(x)=\sqrt{5x-8}$, the derivative $f^{\prime }(x)=\frac{5}{2\sqrt{5x-8}}$.

Step 4 ：Substitute $f(x)$ and $f^{\prime }(x)$ into the formula, we get the integrand for the surface area formula: $A=2\pi \sqrt{1+\frac{25}{4(5x-8)}}\sqrt{5x-8}$.

Step 5 ：Finally, we set up the integral over the interval [2, 4]. The integral that represents the surface area of the surface generated by revolving the graph of $f(x)=\sqrt{5x-8}$ around the x-axis over the interval [2,4] is $\boxed{{\displaystyle -\frac{11\sqrt{33}\pi }{10}+\frac{73\sqrt{73}\pi }{30}}}$.