The average value is
[tex]\displaystyle\frac1{2-0}\int_0^27x^2(x^3+1)^6\,\mathrm dx[/tex]
Let [tex]u=x^3+1[/tex], so that [tex]\mathrm du=3x^2\,\mathrm dx[/tex]:
[tex]\displaystyle\frac12\int_1^9\frac73u^6\,\mathrm du=\frac16u^7\bigg|_1^9=\frac{9^7-1}6=\frac{2,391,484}3[/tex]
