To do this, we'll use the substitution method. Let $u = x^3$, so $du = 3x^2 dx$. Then, the integral becomes:
$$\int_{1}^{\infty} 3 x^{2} e^{-x^{3}} d x = \int_{1}^{\infty} e^{-u} du$$
Now, we can evaluate the integral:
$$\int_{1}^{\infty} e^{-u} du =
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