Have these "calculus-based" consecutive summations been discovered yet? Consider the sum-of-powers series
$$S_i(n) = 1^i + 2^i + \cdots + n^i = \sum_{k=1}^n y_i(k) \quad\text{with}\; y_i(x)=x^i \tag{1}$$
For $i=1, 2, 3$, it turns out that we can write
$$\begin{align}
S_1(n) &= \frac{n+1}{n}\cdot\int_0^{n}y_1(x)\,dx \tag{2a}\\[4pt]
S_2(n) &= \frac{n+1}{n}\cdot\int_0^{n}y_2(x)\,dx \cdot \frac{y_2^\prime(n)+1}{y_2^\prime(n)} \tag{2b}\\[4pt]
S_3(n) &= \left(\frac{n+1}{n}\right)^2\cdot\int_0^{n}y_3(x)\,dx \tag{2c}\\[4pt]
\end{align}$$
which simplify to the well-known formulas
$$\begin{align}
S_1(n) &= \frac12 n(n+1) \tag{3a}\\[4pt]
S_2(n) &= \frac16 n(n+1)(2n+1) \tag{3b}\\[4pt]
S_3(n) &= \frac14 n^2 (n + 1)^2 \tag{3c}\\[4pt]
\end{align}$$
The only thing is that I'm not quite sure why these relations can include derivatives and integrals and somehow even work at all - they were simply a result of an extremely tiresome trial and error process. And can this "template" be extended to any $S_i(n)$ series or even more unorthodox variants?
Credits for the condensation and formatting of my answer goes to Blue (in the comments)
 A: (Too long for a comment.) The reuse and ambiguity of symbols is a little confusing. I'd like to try to restate the problem.

Consider the sum-of-powers series
$$S_i(n) = 1^i + 2^i + \cdots + n^i = \sum_{k=1}^n y_i(k) \quad\text{with}\; y_i(x)=x^i \tag{1}$$
For $i=1, 2, 3$, it turns out that we can write
$$\begin{align}
S_1(n) &= \frac{n+1}{n}\cdot y_1^\prime(n)\cdot\int_0^{n}y_1(x)\,dx  \tag{2a}\\[4pt]
S_2(n) &= \frac{n+1}{n}\cdot \frac{y_2^\prime(n)+1}{y_2^\prime(n)}\cdot\int_0^{n}y_2(x)\,dx \tag{2b}\\[4pt]
S_3(n) &= \left(\frac{n+1}{n}\right)^2\cdot\int_0^{n}y_3(x)\,dx \tag{2c}\\[4pt]
\end{align}$$
which simplify to the well-known formulas
$$\begin{align}
S_1(n) &= \frac12 n(n+1) \tag{3a}\\[4pt]
S_2(n) &= \frac16 n(n+1)(2n+1) \tag{3b}\\[4pt]
S_3(n) &= \frac14 n^2 (n + 1)^2 \tag{3c}
\end{align}$$
[Can this technique be extended? Is it new? etc, etc, etc]

Does this capture the intent of the question?
(BTW: I think you need to be a bit more explicit about why the relations in $(2)$ hold.)
