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,


with midpoint

with
.
Over each subinterval, we interpolate
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.