Solution
= Solution
The <domain of integration> is $0\leq x\leq2$, $0\leq y\leq x^3$. Reversing the order in the <double integral> and then putting $u=x^4+1$ gives
$$
\int_0^2\int_0^{x^3}\sqrt{x^4+1}\,dy\,dx
=\int_0^2x^3\sqrt{x^4+1}\,dx
=\left[\frac{(x^4+1)^{3/2}}6\right]_0^2.
$$
Therefore
$$
\boxed{I=\frac{17\sqrt{17}-1}{6}.}
$$