다항함수 정적분 공식의 일반화

편의상 $\int^{\beta}_\alpha (x-\alpha)^n(\beta-x)^m$를 ${}_n A_m$로 두면,

${}_0 A_{n+m}=\frac{1}{n+m+1}(\beta-\alpha)^{n+m+1}$이고,

\[\begin{align}{}_{n}A_{m}&=\int^{\beta}_\alpha (x-\alpha)^n(\beta-x)^mdx\\&=\frac{n}{m+1}\int^{\beta}_\alpha (x-\alpha)^{n-1}(\beta-x)^{m+1}dx\\&=\frac{n}{m+1}{}_{n-1}A_{m+1} \end{align}\]

($\int (\beta-x)^mdx=\frac{-1}{m+1}(\beta-x)^{m+1}+C$를 이용해 부분적분함)

위 관계식을 연쇄적으로 적용해 ${}_n A_m$의 $n$을 줄이고 $m$을 늘려나가면,

\[ {}_{n} A_{m}=\frac{n!m!}{(n+m)!} {}_{0} A_{n+m}=\frac{n!m!}{(n+m+1)!}(\beta-\alpha)^{n+m+1} \]

따라서 $\int^{\beta}_\alpha (x-\alpha)^n(\beta-x)^m=\frac{n!m!}{(n+m+1)!}(\beta-\alpha)^{n+m+1}$