Expocubic

The obvious way is to use the gamma function or by integrating by parts. For the sake of variety I will use a fancier method.

$I(a)=\int e^{ax}x^3 dx =\frac{\partial^3}{\partial a^3} \int e^{ax} dx=\frac{\partial^3}{\partial a^3} \frac{e^{ax}}{a}=\frac{e^{ax}(a^3x^3-3a^2x^2+6ax-6)}{a^4}$ Now let $a=-1$ and plugging in the limits we get $\boxed{6}$ .

$\begin{aligned} \int_0^\infty e^{-x}x^3 \space dx & = \int_0^\infty x^{4-1}e^{-x} \space dx \\ & = \Gamma(4) \quad \quad \small \color{#3D99F6}{\Gamma(z) \text{ is Gamma function} } \\ & = 3! = \boxed{6} \end{aligned}$

Integrating by parts: $\Gamma(z) = \int_{0}^\infty e^{-x}\cdot x^{z-1} dx$ $\Gamma (4) = \int_{0}^\infty e^{-x}\cdot x^{4-1} dx = \int_{0}^\infty e^{-x}\cdot x^{3} dx = \left[- x^3 \cdot e^{-x}\right]^\infty_{0} + 3 \int_{0}^\infty e^{-x}\cdot x^{2} dx =$ $= 3 \int_{0}^\infty e^{-x}\cdot x^{2} dx = 3( \left[- x^2 \cdot e^{-x}\right]^\infty_{0} + 2 \int_{0}^\infty e^{-x}\cdot x dx) =$ $= 6 \int_{0}^\infty e^{-x}\cdot x dx = 6( \left[- x \cdot e^{-x}\right]^\infty_{0} + \int_{0}^\infty e^{-x} dx) =$ $6 \int_{0}^\infty e^{-x} dx = 6( \left[- \cdot e^{-x}\right]^\infty_{0}) = 6 = 3!$