Question
Set up an integral that represents the surface area of the surface generated by revolving the graph of $f(x)=\sqrt{5 x-8}$ 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 \(A = 2\pi \int_a^b f(x) \sqrt{1 + [f'(x)]^2} dx\), where \(f(x)\) is the function being revolved, \(f'(x)\) 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 \(f'(x)\). For \(f(x) = \sqrt{5x - 8}\), the derivative \(f'(x) = \frac{5}{2\sqrt{5x - 8}}\).

Step 4 ：Substitute \(f(x)\) and \(f'(x)\) 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 \(f(x)=\sqrt{5 x-8}\) around the x-axis over the interval [2,4] is \(\boxed{-\frac{11\sqrt{33}\pi}{10} + \frac{73\sqrt{73}\pi}{30}}\).