Thank you! The problem states that we can used the fact that
$$\sum_{k=1}^{n-1}(k^m)<\frac{n^{m+1}}{m+1}<\sum_{k=1}^n(k^m)$$
in the problem, which isn't a closed form exactly but the problem implies that it is enough.
I know this is a really involved problem and so I really appreciate everyone's help!I am going to go ahead and type out the work I have in the pdf here:
We want to find the area of the parabolic segment bounded by the x-axis, the y-axis, the line $$x=b$$, and the curve $$y=ax^m+c$$, where $$m\in\mathbb{Z^+}$$ and $$a,c>0$$. To do this we divide the base into $$n$$ equal parts and at each point $$kb/n$$, $$k=1,2,\ldots,n$$, we construct an outer rectangle of height $$a(kb/n) ^m +c$$. Similarly at each point $$kb/n$$, $$k=1,2,\ldots,n-1$$, we construct an inner rectangle of height $$a(kb/n)^m+c$$. The area of the parabolic segment is between the sum $$s_n$$ of the areas of the inner rectangles and the sum $$S_n$$ of the areas of the outer rectangles. The diagram shows the area of the typical outer rectangle.
$$s_n=\sum_{k=1}^{n-1} (\frac{ab^{m+1}}{n^{m+1}}k^m+\frac{bc}{n}) = \frac{ab^{m+1}}{n^{m+1}}\sum_{k=1}^{n-1}\left(k^m\right)+\frac{(n-1)cb}{n}$$
$$S_n=\sum_{k=1}^n ( \frac{ab^{m+1}}{n^{m+1}}k^m+\frac{bc}{n}) = \frac{ab^{m+1}}{n^{m+1}}\sum_{k=1}^{n}(k^m)+ cb$$
We use a previously proved theorem which states that
$$
\sum_{k=1}^{n-1}(k^m)<\frac{n^{m+1}}{m+1}<\sum_{k=1}^n(k^m)$$
and multiply by and add the appropriate terms in order to obtain
$$s_n<\frac{ab^{m+1}}{m+1}+\frac{(n-1)cb}{n} <\frac{ab^{m+1}}{m+1}+cb$$
from the left side and
$$S_n>\frac{ab^{m+1}}{m+1}+cb $$
from the right side so that, combined, we have
$$s_n<\frac{ab^{m+1}}{m+1}+cb<S_n.$$
This is the first part of the proof. The second part is to show that, if $$A$$ is any number satisfying $$s_n<A<S_n$$ for all $$n\in\mathbb{Z^+}$$, then $$A=\frac{ab^{m+1}}{m+1}+cb$$. So assume $$s_n<A<S_n$$.
Adding $$n^m$$ to the left two sides of the given theorem and subtracting $$n^m$$ from the right two sides, we can get
$$\sum_{k=1}^n(k^m)<\frac{n^{m+1}}{m+1}+n^m$$
$$\sum_{k=1}^{n-1}(k^m)>\frac{n^{m+1}}{m+1}-n^m
$$
and, by multiplying by and adding the appropriate terms,
$$S_n<\frac{ab^{m+1}}{m+1}+cb+\frac{ab^{m+1}}{n} $$
and
$$s_n>\frac{ab^{m+1}}{m+1}+\frac{(n-1)cb}{n}-\frac{ab^{m+1}}{n}.$$
The trouble arises here because at this point I would like to show that
$$s_n>\frac{ab^{m+1}}{m+1}+cb-\frac{ab^{m+1}}{n}$$ so that I could have the combined inequality
$$ \frac{ab^{m+1}}{m+1}+cb-\frac{ab^{m+1}}{n}<s_n<A<S_n<\frac{ab^{m+1}}{m+1}+cb+\frac{ab^{m+1}}{n}$$
and then use this to show that $$A\neq\frac{ab^{m+1}}{m+1}+cb$$ implies a bound for $$n$$. This was the way the sample problem (without a constant term) in the book went about it.
