To find the volume of the solid formed by rotating the region under the curve \(y = \frac{8}{x} + 2x\) from \(x = 2\) to \(x = 5\) about the \(x\)-axis, we use the formula for the volume of revolution:
\(V = \pi \int_{a}^{b} y^2 \, dx\)
Substitute \(y = \frac{8}{x} + 2x\) into the formula:
\(V = \pi \int_{2}^{5} \left( \frac{8}{x} + 2x \right)^2 \, dx\)
Expand \(\left( \frac{8}{x} + 2x \right)^2\):
\(\left( \frac{8}{x} + 2x \right)^2 = \frac{64}{x^2} + 4x^2 + 32\)
Integrate each term separately:
\(\int \frac{64}{x^2} \, dx = -\frac{64}{x}\)
\(\int 4x^2 \, dx = \frac{4x^3}{3}\)
\(\int 32 \, dx = 32x\)
Combine the integrals:
\(V = \pi \left[ -\frac{64}{x} + \frac{4x^3}{3} + 32x \right]_{2}^{5}\)
Evaluate from \(x = 2\) to \(x = 5\):
\(V = \pi \left( \left( -\frac{64}{5} + \frac{4(5)^3}{3} + 32(5) \right) - \left( -\frac{64}{2} + \frac{4(2)^3}{3} + 32(2) \right) \right)\)
Calculate the values:
\(V = \pi \left( -12.8 + \frac{500}{3} + 160 - (-32 + \frac{32}{3} + 64) \right)\)
\(V = \pi \left( 271.2 \right)\)
Thus, the volume is \(271.2\pi\) or approximately 852.
