<s>[INST] What is the average value of $y=x^2\sqrt{x^3+1}$ on the interval $(0,2)$ I know the average value formula is $$\frac{1}{b-a}\int_a^b f(x)\;dx$$
I have no problem plugging in 0 and 2 for a and b respectively. I think i'm struggling actually taking the integral.  
 [/INST] here is an alternative method to substitution (actually its the same method, just arranged a bit differently).
$$\frac{1}{2-0}\int^2_0 x^2\sqrt{x^3+1}dx$$
$$=\frac{1}{2}\int^2_0 x^2\sqrt{x^3+1} dx\cdot\frac{d(x^3+1)/dx}{d(x^3+1)/dx}$$
$$=\frac{1}{2}\int^2_0 x^2\sqrt{x^3+1}\cdot\frac{d(x^3+1)}{3x^2}$$
$$=\frac{1}{6}\int^2_0\sqrt{x^3+1} d(x^3+1)$$
now we substitute $u=x^3+1$ and $du=d(x^3+1)$
$$=\frac{1}{6}\int^9_1\sqrt{u} du=\frac{1}{6}\bigg(\frac{2}{3} u^{3/2} \bigg)\bigg|_1^9$$
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