Find the value of the improper integral ${\int }_0^{\infty }{x}^3{e}^{-x}dx$ using integration by parts.

Step-by-step solution

1. Recognize that the integral can be computed via the Gamma function: ${\int }_0^{\infty }{x}^n{e}^{-x}dx=\Gamma (n+1)=n!$ for integer $n$. Here $n=3$, so the value is $3!=6$.
2. Alternatively, evaluate using integration by parts. Let $u=x^3$ and $dv=e^{-x}dx$. Then $du=3x^{2}dx$ and $v=-e^{-x}$. Apply $\int udv=uv-\int vdu$: ${\int }_0^{\infty }{x}^3{e}^{-x}dx=[-x^3{e}^{-x}{]}_0^{\infty }+3{\int }_0^{\infty }{x}^2{e}^{-x}dx$. The boundary term vanishes because $x^3{e}^{-x}\to 0$ as $x\to \infty$ and is $0$ at $x=0$. Repeat integration by parts twice more to reduce the power from $x^2$ to $x$ and then to a constant. Each step introduces a factor of $3$: $\int x^3{e}^{-x}dx=3\int x^2{e}^{-x}dx$, $=3[2\int x{e}^{-x}dx]$, $=3[2\cdot 1\int e^{-x}dx]$. Finally, $\int e^{-x}dx=-e^{-x}$, evaluated from $0$ to $\infty$ gives $1$. Multiplying the factors: $3\times 2\times 1\times 1=6$.

Concept tested: Improper integral, Gamma function

Published 5 September 2026 · How this solution was prepared and checked