I'm trying to find the Riemann integral of $x^m$ between $a$ and $b$ with $b>a$.
So far I have managed to get $$\int_a^b x^m~dx=\lim_{n\rightarrow \infty}\left(a^m \times \frac{b-a}{n} \times \sum_{r=1}^n q^{(r-1)m}\right)$$
$$=\lim_{n\rightarrow \infty}\left(a^m \times \frac{b-a}{n} \times \frac{q^{nm}-1}{q^m-1}\right)$$ but I don't know how to evaluate this limit?
Any help?
I forgot to mention the points of division are $$a=aq^0,aq,aq^2,...aq^{r-1},aq^r,...aq^{n-1},aq^n=b$$ and I'm using the left end point to calculate the sum apologies.
Assume $0 < a < b$. Then for every positive integer $n$ let $q = (b/a)^{1/n} > 1$, and consider the partition $$\bigcup_{r=0}^{n-1} \, [aq^r, aq^{r+1}) = [a,b).$$ Then with $f(x) = x^m$, consider $$\begin{align*} S_n &= \sum_{r=0}^{n-1} (aq^{r+1} - aq^r)f(aq^r) \\ &= a^{m+1}(q-1) \sum_{r=0}^{n-1} (q^{m+1})^r \\ &= a^{m+1}(q-1) \frac{(q^{m+1})^n-1}{q^{m+1} - 1} \\ &= \left((aq^n)^{m+1} - a^{m+1}\right) \frac{q-1}{q^{m+1}-1} \\ &= \frac{b^{m+1} - a^{m+1}}{\sum_{k=0}^m q^k}. \end{align*}$$ This is because $aq^n = b$. Now note that it is only the denominator of $S_n$ that is a function of $n$ through $q$. So, as $n \to \infty$, $q \to 1$ since $b > a$. Hence $$\lim_{n \to \infty} S_n = \frac{b^{m+1} - a^{m+1}}{m+1}.$$
Of course, the evaluation of the limit at the end is problematic if $m$ is not itself a nonnegative integer. I leave it to the reader to consider the general case where $m \ne 0$ of the limit $$\lim_{q \to 1} \frac{q^m - 1}{q-1}.$$