Evaluate the integral by reversing the order $$\int_0^2 \int_{\sqrt{y}}^1 \sqrt{1+x^3}\,dx\,dy$$
What happens to the limit $\sqrt{y} \leq t \leq 1$ for $0 \leq y \leq 2$ in the region of integration. Should we split the ranges of $x$ as $0 \leq x \leq 1$ and $1 \leq x \leq \sqrt{2}$ ?
Integration region is
and then you can write your integral as follows
$$\int_0^1\int_0^{x^2}\sqrt{1+x^3}dy\,dx-\int_1^\sqrt{2}\int_{x^2}^2\sqrt{1+x^3}dy\,dx$$