Using the shell method, the volume is given exactly by the definite integral,

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
-th subinterval


where
. The midpoint of each subinterval is

Then the Riemann sum approximating the integral above is



(compare to the actual value of the integral of about 14.45)