Question
Consider the partition $\mathcal{P}_n = \{q^0,q^1,q^2,...,q^n \}$ where $q^n = 2$. If $f(x) = x^j$ for some positive integer $j$, then evaluate the integral
$$\int_{1}^{2} f(x)\,dx$$
by calculating the limit $\displaystyle \lim_{n \rightarrow \infty}\underline{S}_{\mathcal{P}_n}(f)$ of the corresponding lower Riemann sums.
My Attempt
I began by writing out the series as
$$\sum_{k=1}^{n} (q^k - q^{k-1})(q^{k-1})^j = q^{n(j-1)-j} - 1$$
but clearly, this doesn't really provide any progression to solving the problem. Perhaps there's an issue with the sum, although I've spent a while looking at it and I cannot find any issue.
Is my approach correct? If so, where do I go from here. If it isn't correct, where have I gone wrong?
EDIT: Also, I know that $1<q<2$ and $q \rightarrow 1$ as $n \rightarrow \infty$.
I got a different result for the sum. Since the $q's$ depend on n I used the notation $q_n:=2^{1/n}$ and computed
\begin{align*} \sum_{k=1}^n(q_n^k-q_n^{k-1})(q_n^{k-1})^j&=\sum_{k=1}^n q_n^{k+(k-1)j}-\sum_{k=1}^n q_n^{(k-1)(j+1)}\\ &=(q_n-1)\sum_{k=0}^{n-1}q_n^{k(j+1)}\\ &=(q_n-1)\frac{1-q_n^{(j+1)n}}{1-q_n^{j+1}}\\ &=\frac{q_n-1}{1-q_n^{j+1}}(1-2^{j+1}) \end{align*}
By using l'Hopitals rule we get \begin{align*} \lim_{x\to 1} \frac{x-1}{1-x^{j+1}}=\lim_{x\to 1}\frac{1}{-(j+1)x^{j}}=-\frac{1}{(j+1)} \end{align*} and therefore
\begin{align*} \lim_{n\to \infty}\sum_{k=1}^n(q_n^k-q_n^{k-1})(q_n^{k-1})^j=\frac{2^{j+1}-1}{j+1} \end{align*}