Find $$\int_{a}^{b} x^m dx$$ where $0<a<b$ and $m\neq -1$.
The answer goes as follows : but I get lost in the calculations which I need help with pinning down precisely:
We choose the points of devision of the segment $[a,b]$ as $x_i=x_0q^i; x_n=b; x_0=a; \implies q=({b\over a})^{1\over n};x_i=a ({b\over a})^{i\over n}$
Then: $$S_n=\sum_{i=0}^{n-1}f(x_i)(x_{i+1} - {x_i}) = a^{m+1} ({b\over a})^{1\over n}-1) \sum _{i=0}^{n-1} ({b\over a})^{i(m+1)\over n} $$
On
Because $x\to x^m$ is incresing the min value and max value of this function on$[x_{i-1},x_{i}]$ is : $$A_{\min,i}=a^mq^{(i-1)m}\ \ \ \ \ \ A_{\max,i}=a^mq^{im} $$ now we have to compute the lower sum and the upper one: $$A_{\min}=\sum_{i=1}^nA_{\min,i}(x_i-x_{i-1})=a^{m+1}\sum_{i=1}^nq^{(i-1)r}q^{m}(q-1) =a^{m+1}(q-1)\sum_{i=1}^{n}q^{(i-1)(m+1)}=a^{m+1}(q-1)\frac{q^{i(m+1)-1}}{q^{m+1}-1}$$
and similarly you can compute the upper bound too: $$A_{max}=a^{m+1}(q-1)q^m\sum_{i=1}^{n}q^{(i-1)(m+1)}=a^{m+1}(q-1)q^m\frac{q^{n(m+1)-1}}{q^{m+1}-1} $$
an the two sums can be simplified to: $$A_{\min}=(b^{m+1}-a^{m+1})\frac{1}{\frac{q^{m+1}-1}{q-1}}\ \ \ A_{\max}=(b^{m+1}-a^{m+1})\frac{q^m}{\frac{q^{m+1}-1}{q-1}} $$
from here we have to compute the limit $\lim_{q\to 1} A_{\max}$ and $\lim_{q\to 1} A_{\min}$ and we hwill use: $$\lim_{x\to 1}\frac{x^r-1}{x-1}=r+1 $$
As you note, the points are chosen to form a geometric progression $x_i=aq^i$, with $q=\left(\dfrac ba\right)^{1/n}$. Thus we have $x_{i+1}-x_i=qx_i-x_i=(q-1)x_i$.
The Riemann sum is
$$\sum_{i=0}^{n-1}x_i^m(x_{i+1}-x_i)=(q-1)\sum_{i=0}^{n-1}x_i^{m+1}=(q-1)a^{m+1}\sum_{i=0}^{n-1}q^{i(m+1)}=(q-1)a^{m+1}\frac{q^{n(m+1)}-1}{q^{m+1}-1}\\ =\frac{q-1}{q^{m+1}-1}a^{m+1}\left(\left(\frac ba\right)^{m+1}-1\right).$$
The fraction can be simplified as $\dfrac1{q^m+q^{m-1}+\cdots1},$ which will tend to $\dfrac1{m+1}$ as $q\to1$.
The final value is $$\frac{b^{m+1}-a^{m+1}}{m+1}.$$