Defining the Definite Integral

Definition¶

\begin{align} S(a,b,x^p,n)&=\sum_{i=1}^n\left(a+\frac{b-a}{n}\cdot i\right)^p\cdot\frac{b-a}{n}\\ &=\frac{b-a}{n}\sum_{i=1}^n\sum_{d=0}^p\binom{p}{d}a^{p-d}\left(\frac{b-a}{n}\right)^di^d\\ &=\frac{b-a}{n}\sum_{d=0}^p\binom{p}{d}a^{p-d}\left(\frac{b-a}{n}\right)^d\sum_{i=1}^ni^d. \end{align}

Now, by Faulhaber's Formula, we can simplify $\sum_{i=1}^ni^d$ :

\begin{align} S(a,b,x^p,n)&=\frac{b-a}{n}\sum_{d=0}^p\binom{p}{d}a^{p-d}\left(\frac{b-a}{n}\right)^d\frac{1}{d+1}\sum_{r=0}^d\binom{d+1}{r}B_rn^{d+1-r}\\ &=\sum_{d=0}^p\binom{p}{d}a^{p-d}\left(b-a\right)^{d+1}\frac{1}{d+1}\sum_{r=0}^d\binom{d+1}{r}B_rn^{-r}\\ &=\sum_{r=0}^pB_r\cdot\left[\sum_{d=r}^p \binom{p}{d}a^{p-d}\left(b-a\right)^{d+1}\frac{1}{d+1}\binom{d+1}{r}\right]n^{-r} \end{align}