First Reduction Formula

For any $n$ a positive integer, it is always true that

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^n{e}^{x}dx=x^n{e}^x-n{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^{n-1}{e}^{x}dx$

We will first attempt to understand the recursion formula.

$(1)$ When $n=1$ this becomes...

$($a$)$ Work: substituting $1$ in for $n$ on both sides, we obtain $($since the function $f(x)=x^0$ is interpreted as the constant function $f(x)=1)$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^1{e}^{x}dx=x^1{e}^x-1{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^{1-1}{e}^{x}dx$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}x{e}^{x}dx=x{e}^x-{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^{x}dx$

$($b$)$ Answer:

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}x{e}^{x}dx=x{e}^x-{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^{x}dx$

$(2)$ Verify the statement for $n=1$ by using integration by parts. $($Stop when you reproduce your answer in part a$.)$

$($a$)$ Work: We let $u=x$ and $dv=e^{x}dx$ so that

$u,=x,dv,=e^{x}dx$

$du,=dx,v,=e^x$

$($b$)$ Work, cont'd: Performing integration by parts, we have

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}x{e}^{x}dx=$ubrace$(xubrace{)}_u$ubrace$(e^x$ubrace${)}_v-{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}$ubrace$(e^x$ubrace${)}_v$ubrace$(dxubrace{)}_{du}$

This is equal to the answer in part $1$ b $,$ and we are done.

$(3)$ When $n=2$ this becomes $...($substitute $2$ in for $n$ on both sides of the formula above$)$