Volume of region bounded by two curves revolved around x-axis $f(x) = x^2 - 4x + 5$ and $g(x) = 1/2$ and $x = 1$ and $x = 3$
I found the area from 1 to 3 of the region bounded by these functions: 
I ended up with $\int_1^3 (x^2-4x+9/2)dx$. I am trying to figure out how to evaluate the integral that will give the volume of the solid generated by revolving the region around the x-axis. Any help on how to set up this problem would be greatly appreciated!

$\int_1^3 \pi(x^2-4x+9/2)^2dx$. Also called the washer method
