How to find $\int_a^bx^mdx$ using the limit sum definition of definite integral? 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?
 A: Assuming $a>0$, I consider the special case $m=-1$. 
Here $f(x)=\frac{1}{x}$, and we divide the interval $a\le x\le b$ so that the points of division are in a geometric progression as $$a,ar,ar^2,\cdots,ar^n=b\,;\quad\text{ so } r=\left(\frac{b}{a}\right)^{1/n}$$
The length of the $k$th ($1\le k\le n$) sub-interval is $$\delta_k=ar^k-ar^{k-1}=ar^{k-1}(r-1)$$
Let $\delta$ be the maximum among $\delta_1,\delta_2,\cdots,\delta_n$. Then as $n\to\infty, r\to1$ and so $\delta\to0$. We take $$\delta_k=ar^{k-1}\quad,\,1\le k\le n$$
So by definition,
\begin{align}
\int_a^b \frac{1}{x}\,dx&=\lim_{\delta\to0}\sum_{k=1}^n f(\xi_k)\delta_k
\\&=\lim_{n\to\infty}\sum_{k=1}^n\frac{ar^{k-1}(r-1)}{ar^{k-1}}
\\&=\lim_{n\to\infty} n(r-1)
\\&=\lim_{1/n\to0}\frac{\left(\frac{b}{a}\right)^{1/n}-1}{1/n}
\\&=\ln \frac{b}{a}\qquad\qquad\left[\because\, \lim_{x\to0}\frac{a^x-1}{x}=\ln a\right]
\end{align}

For the general problem, we can use the general definition of definite integral of a continuous (or integrable) function. I roughly state the definition so that the notation is clear.
Let $f(x)$ be a bounded function defined on a closed interval $[a,b]$. We divide the interval $[a,b]$ into $n$ subintervals $[x_0,x_1],\,[x_1,x_2],\cdots,[x_{n-1},x_n]$ by introducing the points $x_0,x_2,\cdots,x_n$ satisfying $a=x_0<x_1<\cdots<x_n=b$. Let $$\delta_r=x_r-x_{r-1}\quad,r=1,2,\cdots,n$$
Choose any point $\xi_r$ of the $r$th subinterval, i.e. $$x_{r-1}\le \xi_r\le x_r$$
Let $\delta=\max(\delta_1,\cdots,\delta_n)$, the norm of the subdivision of the partition.
We then form the sum $\sum_{r=1}^nf(\xi_r)\delta_r$, which depends on the choice of the points $x_r$'s and $\xi_r$'s.
We now let $\delta$ tend to zero, so that $n$, the number of subdivisions, goes to infinity. If now the sum $\sum_{r=1}^nf(\xi_r)\delta_r$ tends to a definite limit which is independent of the choice of the points $x_r$'s and $\xi_r$'s in $[x_{r-1},x_r]$, then this limit is defined to be the definite (Riemann) integral of $f(x)$ over $[a,b]$. 
That is, $$\int_a^b f(x)\,dx=\lim_{\delta\to0}\sum_{r=1}^n f(\xi_r)\delta_r$$
Many expressions for $\int_a^b f(x)\,dx$ can be found by taking $x_r$'s and $\xi_r$'s in suitable manner. We need one such particular choice here.
We choose $x_r$'s in such a way that they form a G.P. with common ratio $\rho$ (say), like $$a=x_0,x_1=a\rho,x_2=a\rho^2,\cdots,x_n=a\rho^n=b$$
So, $\rho=\left(\frac{b}{a}\right)^{1/n}$ and as $n\to\infty$, $\rho\to 1$.
For $\xi_r$'s, we choose the left hand end point of the $r$th subinterval $[x_{r-1},x_r]$. That is, $$\xi_r=x_{r-1}=a\rho^{r-1}\quad,\,r=1,2,\cdots,n$$
Then for each $r$, $$\delta_r=x_r-x_{r-1}=a\rho^{r-1}(\rho-1)\to0\qquad[\because \rho\to1]$$
Hence we arrive at a new expression for $\int_a^b f(x)\,dx$, given by $$\int_a^b f(x)\,dx=\lim_{n\to\infty\\ (\rho\to1)}a(\rho -1)\sum_{r=1}^n \rho^{r-1}f(a\rho^{r-1})$$
This form is particularly helpful for evaluating integrals of the form $\int_a^b x^m\,dx$, for $0<a<b$ and any rational number $m$. 
For $m\ne -1$, we have by taking $f(x)=x^m$,
\begin{align}
\int_a^b x^m\,dx&=\lim_{n\to\infty\\( \rho\to1)}a(\rho -1)\sum_{r=1}^n \rho^{r-1}(a\rho^{r-1})^m
\\&=a^{m+1}\lim_{n\to\infty}\left[\frac{\rho -1}{\rho^{m+1}}\sum_{r=1}^n \rho^{r(1+m)}\right]
\\&=a^{m+1}\lim_{n\to\infty}\left[\frac{(\rho -1)}{\rho^{m+1}}\rho^{m+1}\frac{\rho^{n(m+1)}-1}{\rho^{m+1}-1}\right]\qquad\left[\because\, m\ne -1,\, \rho^{m+1}-1\ne 0\right]
\\&=a^{m+1}\lim_{n\to\infty}\frac{(\rho-1)}{\rho^{m+1}-1}\left[\left(\frac{b}{a}\right)^{m+1}-1\right]\qquad\quad\left[\because \rho^n=b/a\right]
\\&=(b^{m+1}-a^{m+1})\lim_{\rho\to1}\frac{1}{\left(\frac{\rho^{m+1}-1}{\rho-1}\right)}
\\&=\frac{b^{m+1}-a^{m+1}}{m+1}
\end{align}
In fact, the case $m=-1$ in the beginning has been done in an exactly similar manner. So the case distinction need not be made as mentioned in a comment below.
I had learnt this approach as Walli's method from my high school textbook, although I could not find a reference online.
A: 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 $m\ge 1$
$$\begin{align}
\frac{b-a}{n}\sum_{k=1}^n \color{red}{\left(a+\frac{b-a}{n}k\right)^m}&=\frac{b-a}{n}\sum_{k=1}^n \color{red}{\sum_{\ell=0}^m\binom{m}{\ell}a^\ell \left(\frac{b-a}{n}k\right)^{m-\ell}}\\\\
&=\frac{b-a}{n}\sum_{\ell=0}^m\binom{m}{\ell}a^\ell \left(\frac{b-a}{n}\right)^{m-\ell}\color{blue}{\sum_{k=1}^n k^{m-\ell}}\\\\
&=\frac{b-a}{n}\sum_{\ell=0}^m\binom{m}{\ell}a^\ell \left(\frac{b-a}{n}\right)^{m-\ell}\color{blue}{\left(\frac{n^{m-\ell+1}}{m-\ell+1} +O(n^{m-
\ell})\right)}\\\\
&=\frac{(b-a)^{m+1}}{m+1}\sum_{\ell=0}^m\binom{m+1}{\ell} \left(\frac{a}{b-a}\right)^{\ell}+O\left(\frac1n\right)\\\\
&=\frac{(b-a)^{m+1}}{m+1}\left(\left(1+\frac{a}{b-a}\right)^{m+1}-\frac{a^{m+1}}{(b-a)^{m+1}}\right)+O\left(\frac1n\right)\\\\
&=\frac{b^{m+1}-a^{m+1}}{m+1}+O\left(\frac1n\right)
\end{align}$$
A: I will assume $m>0$. Using MVT, for a partition $P:a=x_1<x_2<...<x_{n+1}=b$, we have $\exists \xi_i \in (x_i,x_{i+1})$ s.t. $\xi_i^{m}(x_{i+1}-x_i)=\frac{x_{i+1}^{m+1}}{m+1}-\frac{x_{i}^{m+1}}{m+1}$. As a result
$$S^{*}(P)=\sum\limits_{i=1}^n \xi_i^{m}(x_{i+1}-x_i)=
\sum\limits_{i=1}^n \left(\frac{x_{i+1}^{m+1}}{m+1}-\frac{x_{i}^{m+1}}{m+1}\right)=\\
\frac{x_{n+1}^{m+1}}{m+1}-\frac{x_{1}^{m+1}}{m+1}=\frac{b^{m+1}-a^{m+1}}{m+1}$$
Because $\inf\limits_{x\in[x_i,x_{i+1}]}\left\{x^m\right\} \leq \xi_i^{m} \leq \sup\limits_{x\in[x_i,x_{i+1}]}\{x^m\}$, we also have 
$$L(x^m,P)\leq S^{*}(P)\leq U(x^m,P)$$
But $x^m$ is continuous on $[a,b]$, thus uniform continuous, so $L(x^m,P)$ and $U(x^m,P)$ will squeeze to $S^{*}(P)$, because
$$0< U(x^m, P) - L(x^m, P)=\sum\limits_{i=1}^n \left(\sup\limits_{x\in[x_i,x_{i+1}]}\{x^m\}-\inf\limits_{x\in[x_i,x_{i+1}]}\{x^m\}\right)(x_{i+1}-x_i)\leq \\
\frac{\varepsilon}{b-a} (b-a) = \varepsilon$$
assuming we choose the partition such that $|y^m-x^m|<\frac{\varepsilon}{a-b}, \forall x,y \in[x_i,x_{i+1}], |x_{i+1}-x_i|<\delta$ (see the definition of uniform continuity). This makes $x^m$ integrable (see Riemann's criterion, theorem 3 here, for more details).
