Using the shell method, the volume is given exactly by the definite integral,
[tex]2\pi\displaystyle\int_0^1x(1+9x^3)\,\mathrm dx[/tex]
Splitting up the interval [0, 1] into 5 subintervals gives the partition,
[0, 1/5], [1/5, 2/5], [2/5, 3/5], [3/5, 4/5], [4/5, 5]
with left and right endpoints, respectively, for the [tex]i[/tex]-th subinterval
[tex]\ell_i=\dfrac{i-1}5[/tex]
[tex]r_i=\dfrac i5[/tex]
where [tex]1\le i\le5[/tex]. The midpoint of each subinterval is
[tex]m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}{10}[/tex]
Then the Riemann sum approximating the integral above is
[tex]2\pi\displaystyle\sum_{i=1}^5m_i(1+9{m_i}^3)\frac{1-0}5[/tex]
[tex]\dfrac{2\pi}5\displaystyle\sum_{i=1}^5\left(\frac{2i-1}{10}+9\left(\frac{2i-1}{10}\right)^4\right)[/tex]
[tex]\dfrac{2\pi}{5\cdot10^4}\displaystyle\sum_{i=1}^5\left(16i^4-32i^3+24i^2+1992i-999\right)=\frac{112,021\pi}{25,000}\approx\boxed{14.08}[/tex]
(compare to the actual value of the integral of about 14.45)
