The average value of \(y\) is given by:
$$ \frac{1}{b-a}\int\limits_a^b f(x) dx $$
$$ = \frac{1}{2-0}\int\limits_0^2 x^2\sqrt{x^3+1} dx $$
We can integrate using the substitution \(u=x^3+1\).
$$ u = x^3+1 \hskip{3em} du = 3x^2 dx \hskip{3em} dx = \frac{du}{3x^2}$$
Substituting these values into our integral:
$$ \frac{1}{2}\int\limits_1^9 x^2\sqrt{u} \frac{du}{3x^2} $$
$$ = \frac{1}{6} \int\limits_1^9 \sqrt{u} du $$
$$ = \frac{1}{6} \Big[ \frac{2}{3}u^{\frac{3}{2}} \Big]_1^9 $$
$$ = \frac{1}{9}\left(9^{\frac{3}{2}}-1^{\frac{3}{2}}\right) $$
$$ = \frac{1}{9}(27-1) $$
$$ = \boxed{\frac{26}{9}} $$