TITLE: How to find $\int_a^bx^mdx$ using the limit sum definition of definite integral?
QUESTION [5 upvotes]: We have to evaluate

$$\int_a^bx^mdx$$

Converting it to a sum as limit tends to infinity 

$$=\lim_{h\to0}h\sum_{r=1}^{n} f(a+rh)=\lim_{h\to0}h\sum_{r=1}^{n} (a+rh)^m$$
  with $h=(b-a)/n$ with $n\to\infty$.

Now, I have having trouble going on with this. What to do next?

REPLY [2 votes]: From the Euler-Maclaurin Summation Formula, $ \color{blue}{\sum_{k=1}^n k^{m-\ell}=\frac{n^{m-\ell+1}}{m-\ell+1}+O\left(n^{m-\ell}\right)}$.  Using this result along with the binomial expansion $ \color{red}{(x+y)^m=\sum_{\ell=0}^m x^\ell y^{m-\ell}}$ and the relationship $ \binom{m}{\ell}\frac{1}{m-\ell+1}=\frac1{m+1}\binom{m+1}{\ell}$, we have for $m\ge 1$
$$\begin{align}
\frac{b-a}{n}\sum_{k=1}^n \color{red}{\left(a+\frac{b-a}{n}k\right)^m}&=\frac{b-a}{n}\sum_{k=1}^n \color{red}{\sum_{\ell=0}^m\binom{m}{\ell}a^\ell \left(\frac{b-a}{n}k\right)^{m-\ell}}\\\\
&=\frac{b-a}{n}\sum_{\ell=0}^m\binom{m}{\ell}a^\ell \left(\frac{b-a}{n}\right)^{m-\ell}\color{blue}{\sum_{k=1}^n k^{m-\ell}}\\\\
&=\frac{b-a}{n}\sum_{\ell=0}^m\binom{m}{\ell}a^\ell \left(\frac{b-a}{n}\right)^{m-\ell}\color{blue}{\left(\frac{n^{m-\ell+1}}{m-\ell+1} +O(n^{m-
\ell})\right)}\\\\
&=\frac{(b-a)^{m+1}}{m+1}\sum_{\ell=0}^m\binom{m+1}{\ell} \left(\frac{a}{b-a}\right)^{\ell}+O\left(\frac1n\right)\\\\
&=\frac{(b-a)^{m+1}}{m+1}\left(\left(1+\frac{a}{b-a}\right)^{m+1}-\frac{a^{m+1}}{(b-a)^{m+1}}\right)+O\left(\frac1n\right)\\\\
&=\frac{b^{m+1}-a^{m+1}}{m+1}+O\left(\frac1n\right)
\end{align}$$