The area of the surface is given exactly by the integral,

We have

so the area is

We split up the domain of integration into 10 subintervals,
[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]
where the left and right endpoints for the  -th subinterval are, respectively,
-th subinterval are, respectively,


with midpoint

with  .
.
Over each subinterval, we interpolate  with the quadratic polynomial,
 with the quadratic polynomial,

Then

It turns out that the latter integral reduces significantly to

which is about 651.918, so that the area is approximately  .
.
Compare this to actual value of the integral, which is closer to 1967.