Answer
$$\frac{59048}{30}$$
Work Step by Step
Given $$f(t)=t^2(1+t^3)^4,\ \ \ [0,2] $$
Then
\begin{align*}
f_{\text{ave}}&=\frac{1}{b-a}\int_{a}^{b}f(t)dt \\
&=\frac{1}{2-0}\int_{0}^{2}t^2(1+t^3)^4dt\\
&=\frac{1}{6} \int_{0}^{2}3t^2(1+t^3)^4dt\\
&=\frac{1}{30}(1+t^3)^{5}\bigg|_{0}^{2}\\
&=\frac{1}{30}(59049-1)\\
&=\frac{59048}{30}
\end{align*}