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 $\dfrac{1}{2}$.