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\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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{\bf Question} Find a reduction formula for $I_n=\ds\int x^n
\exp(-x) \,dx$ and evaluate $I_3$.


\medskip

{\bf Answer} $I_n=\ds\int x^n e^{-x} \,dx \ \ (\star\star)$

Integrate by parts to reduce the power of $n$.

$$\begin{array} {cc} u=x^n &amp; \ds\frac{dv}{dx}=e^{-x}\\
\ds\frac{du}{dx}=nx^{n-1} &amp; v=-e^{-x} \end{array}$$

\medskip
Therefore

$I_n=-x^n e^{-x}-\ds\int nx^{n-1}x(-e^{-x}) \,dx$

$I_n=-x^n e^{-x}+n\underbrace{\ds\int x^{n-1} e^{-x} \,dx}$

\ \ \ \ \ \ \ \ \ \ but this is $I_{n-1}$ by comparison with
($\star\star$).

\medskip
Therefore \un{$I_n=-x^n e^{-x}+nI_{n-1}$} is the reduction
formula.

\medskip
Use this reduction formula with $n=3$:

$\left. \begin{array} {rcl} I_3 &amp; = &amp; -x^3 e^{-x}+3I_2\\ I_2 &amp; = &amp;
-x^2 e^{-x}+2I_1\\ I_1 &amp; = &amp; -xe^{-x}+I_0 \end{array}\right\}$
What's $I_0$?

$I_0=\ds\int e^{-x} \,dx = -e^{-x}+c$ where $c$ is an arbitrary
constant.

\medskip
Thus recursively substituting $I_1 \rightarrow I_2 \rightarrow
I_3$

$I_3 = -x^3e^{-x}+3(-x^2e^{-x}+2[-xe^{-x}+\{-e^{-x}+c\}])$

PICTURE \vspace {1in}

$\begin{array} {rcl} I_3 &amp; = &amp; -e^{-x}
x^3-3x^2e^{-x}+6[-xe^{-x}-e^{-x}+c]\\ &amp; = &amp;
-e^{-x}x^3-3x^2e^{-x}-6e^{-x}+6c\\ &amp; = &amp;
-e^{-x}[x^3+3x^2+6x+6]+6c\\ &amp; = &amp; \un{-e^{-x}[x^3+3x^2+6x+6]+c'}
\end{array}$

Where $c'$ is a new arbitrary constant.
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