2. Given $(n+1)^{\alpha+1}-n^{\alpha+1}<n^{\alpha}(\alpha+1)<n^{\alpha+1}-(n-1)^{\alpha+1},-1<\alpha<0$. Let $x=\sum_{k=4}^{10^{6}} \frac{1}{\sqrt[3]{k}}$, then the integer part of $x$ is $\qquad$ .
By setting $\alpha=-\frac{1}{3}$ in the given inequality, we get $(k+1)^{\frac{2}{3}}-k^{\frac{2}{3}}15000-\frac{3 \sqrt[3]{16}}{2}>14996 \Leftrightarrow \frac{3 \sqrt[3]{16}}{2}<4 \Leftrightarrow \sqrt[3]{16}<\frac{8}{3}$ $\Leftrightarrow 16<\frac{512}{27}$, which is obviously true, so the integer part of $x$ is 14996.