How to do this interesting integration? 
$$\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+\Delta x}^{k+1-\Delta x}x^m dx$$

How to integrate the above integral?
Edit1: 
$$\lim_{\Delta x\rightarrow0}\int_{2-\Delta x}^{2+\Delta x}x^m dx$$
Does this intergral give $\space\space\space\space$ $2^m\space\space$  as the output?
Edit2:
Are my following steps correct?
$\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+\Delta x}^{k+1-\Delta x}x^m dx$ =
$\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+\Delta x}^{k+1-\Delta x}x^m dx$ $+$ $\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+1-\Delta x}^{k+1+\Delta x}x^m dx$  $-$ $\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+1-\Delta x}^{k+1 +\Delta x}x^m dx$ =
$\lim_{\Delta x\rightarrow0}\int_{1+\Delta x}^{n+\Delta x}x^m dx$ $-$ $\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+1-\Delta x}^{k+1 +\Delta x}x^m dx$ = 
$\lim_{\Delta x\rightarrow0}\int_{1}^{n}x^m dx$ $+$$\lim_{\Delta x\rightarrow0}\int_{n}^{n+\Delta x}x^m dx$ $-$  $\lim_{\Delta x\rightarrow0}\int_{1}^{1+\Delta x}x^m dx$ $-$
$\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+1-\Delta x}^{k+1 +\Delta x}x^m dx$ =
$\lim_{\Delta x\rightarrow0}\int_{1}^{n}x^m dx$ + 0 - 0 - 0 = $\lim_{\Delta x\rightarrow0}\int_{1}^{n}x^m dx$
= $\int_{1}^{n}x^m dx$
 A: HINT:
$$\sum_{k=1}^{n-1}\int_{k+\Delta x}^{k+1-\Delta x}x^m dx=\int_1^nx^mdx-\left(\int_1^{1+\Delta x}x^mdx+\sum_{k=2}^{n-1}\int_{k-\Delta x}^{k+\Delta x}x^mdx+\int_{n-\Delta x}^nx^mdx\right)$$
Let $M$ be any upper bound for $x^m$ on $[1,n]$; then
$$\begin{align*}
\int_1^{1+\Delta x}x^mdx+\sum_{k=2}^{n-1}\int_{k-\Delta x}^{k+\Delta x}x^mdx+\int_{n-\Delta x}^nx^mdx&\le M\left(\Delta x+2(n-2)\Delta x+\Delta x\right)\\
&=2M\Delta x(n-1)\;.
\end{align*}$$
Added: The same reasoning shows that
$$\lim_{\Delta x\rightarrow0}\int_{2-\Delta x}^{2+\Delta x}x^m dx=0\;:$$
if $M$ is any bound on $x^m$ over the interval $[1,3]$, say, the integral is bounded by $2M\Delta x$.
By the way, you really shouldn’t use $x$ both in $\Delta x$ and as the variable of integration. The new limit, for instance, ought to be
$$\lim_{\Delta x\rightarrow0}\int_{2-\Delta x}^{2+\Delta x}t^m dt$$
or the like, with similar changes in the original problem.
A: For the first question:
$$
\begin{align}
\lim_{\Delta x\to0}\,\left|\,\int_k^{k+1}x^m\,\mathrm{d}x-\int_{k+\Delta x}^{k+1-\Delta x}x^m\,\mathrm{d}x\,\right|
&=\lim_{\Delta x\to0}\,\left|\,\int_k^{k+\Delta x}x^m\,\mathrm{d}x+\int_{k+1-\Delta x}^{k+1}x^m\,\mathrm{d}x\,\right|\\
&\le\lim_{\Delta x\to0}2\Delta x(k+1)^m\\
&=0
\end{align}
$$
we get
$$
\begin{align}
\lim_{\Delta x\to0}\sum_{k=1}^{n-1}\int_{k+\Delta x}^{k+1-\Delta x}x^m\,\mathrm{d}x
&=\sum_{k=1}^{n-1}\int_k^{k+1}x^m\,\mathrm{d}x\\
&=\int_1^nx^m\,\mathrm{d}x
\end{align}
$$
For the Edit:
For $\Delta x<1$,
$$
\left|\,\int_{2-\Delta x}^{2+\Delta x}x^m\,\mathrm{d}x\,\right|\le2\cdot3^m\Delta x
$$
Therefore,
$$
\lim_{\Delta x\to0}\,\int_{2-\Delta x}^{2+\Delta x}x^m\,\mathrm{d}x=0
$$
For your steps:
If you would give the justification for each step, it would help us in commenting on what is correct and what might be wrong and help you in seeing what is right.
A: $$\lim_{s\rightarrow0}\frac1s\int_{2-s}^{2+s}x^m\mathrm dx=2\cdot2^m$$
