The area of the surface is given exactly by the integral,
[tex]\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx[/tex]
We have
[tex]y(x)=\dfrac15x^5\implies y'(x)=x^4[/tex]
so the area is
[tex]\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx[/tex]
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 [tex]i[/tex]-th subinterval are, respectively,
[tex]\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2[/tex]
[tex]r_i=\dfrac{5-0}{10}i=\dfrac i2[/tex]
with midpoint
[tex]m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4[/tex]
with [tex]1\le i\le10[/tex].
Over each subinterval, we interpolate [tex]f(x)=\sqrt{1+x^8}[/tex] with the quadratic polynomial,
[tex]p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}[/tex]
Then
[tex]\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx[/tex]
It turns out that the latter integral reduces significantly to
[tex]\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)[/tex]
which is about 651.918, so that the area is approximately [tex]651.918\pi\approx\boxed{2048}[/tex].
Compare this to actual value of the integral, which is closer to 1967.
