## Calculus (3rd Edition)

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.