I've long ago solved an exercise that was as follows:
This problem provides a new way of finding $ \displaystyle \int\limits_a^b {{x^p}dx} $ for $ 0 < a < b $. It consists of using partitions $t_i$ for which the quotient $\displaystyle \frac{t_i}{t_{i-1}}$ is constant, instead of being $ t_i - t_{i-1} $ constant.
The solutions produced $ t_i = a c^{\frac{i}{n}} $ where $ c = \frac{b}{a} $ and upper and lower sums being:
$ U \left( f, P \right) = \left( {{b^{p + 1}} - {a^{p + 1}}} \right)\frac{{{c^{\frac{p}{n}}}}}{{1 + {c^{\frac{1}{n}}} + {c^{\frac{2}{n}}} + \cdots + {c^{\frac{p}{n}}}}} $
$L\left( f, P \right) = \left( {{b^{p + 1}} - {a^{p + 1}}} \right)\frac{1}{{1 + {c^{\frac{1}{n}}} + {c^{\frac{2}{n}}} + \cdots + {c^{\frac{p}{n}}}}}$
Thus you have $U \left( f, P \right) - L \left( f, P \right) = \left( {{b^{p + 1}} - {a^{p + 1}}} \right)\left[ {\frac{{{c^{\frac{p}{n}}} - 1}}{{1 + {c^{\frac{1}{n}}} + {c^{\frac{2}{n}}} + \cdots + {c^{\frac{p}{n}}}}}} \right]$
which implies that for some $\epsilon > 0$ and $N$ sufficiently large.
$ U \left( f, P \right) - L \left( f, P \right) < \epsilon $
Finally you get the expected result
$\int\limits_a^b {{x^p}dx} = \frac{{{b^{p + 1}} - {a^{p + 1}}}}{{p + 1}} $
I' m wondering if this change in the differential $ t_i - t_{i-1} = \Delta{t_i}$ versus $\displaystyle \frac{t_i}{t_{i-1}}= \Delta{t_i}$ can be interpreted as changing the integrator $d\alpha$. I've started reading about the Riemann Stieltjes integration so it rang a bell, but don't expect me to know a lot about it, just the basics.