Answer
converges
Work Step by Step
Given $$\int_{0}^{\infty} e^{-x^{3}} d x$$
Since for $x\geq 1$,
\begin{align*}
e^x&\geq x\\
e^{x^3}&\geq x^3\\
e^{-x^3}&\leq x^{-3}
\end{align*}
and $\int_{1}^{\infty } x^{-3}dx$ converges.
Since
\begin{align*}
\int_{0}^{\infty} e^{-x^{3}} d x&=\int_{0}^{1} e^{-x^{3}} d x+\int_{1}^{\infty} e^{-x^{3}} d x
\end{align*}
and $\int_{0}^{1} e^{-x^{3}} d x$ is finite, hence $\int_{0}^{\infty} e^{-x^{3}} d x$ converges.
