Finding volume of the solid of revolution? Can anyone help me with finding the volume of a solid of revolution of f(x) about the x axis for the interval [1,6].  It's supposed to be able to be done without needing calculus but I am having trouble figuring it out.
$f(x) =
\begin{cases}
1  &  1 \leq x< 2\\ 
1/2 &  2 \leq x< 3\\ 
. &          .\\ 
. &          .\\
1/n &  n\leq x< n+1\\ 
\end{cases}$
I know the volume would be found like this $\pi$ $\int_{1}^{6}(f(x))^2dx$ but I am unsure about how to go about it with this function.
Any help is appreciated.
Thanks

$f(x)$ is constant in $n$ intervals. Hence, $\int^b_a (f(x))^2dx=(f(x))^2(b-a)$. So, the volume is simply
$$\pi\left(1^2(2-1)+2^2(3-2)+\cdots+\dfrac{1}{n^2}(n+1-n)\right)=\pi\left(1^2+2^2+\cdots+\dfrac{1}{n^2}\right)$$
Though I believe that the $2$ should actually be $\dfrac{1}{2}$.
