what I found from messing around was $$ \int_{-1}^{0}\sum_{n=1}^{x}n^sdx=\zeta (-s) $$ $$ s\in \mathbb{N} $$ when the partial sum is changed to an equivalent polynomial using Faulhaber's formula.
For example $$ \sum_{n=1}^{x}n^3 = \frac{x^2(x+1)^2}{4} $$ and $$ \int_{-1}^{0}\frac{x^2(x+1)^2}{4}dx = \frac{1}{120} = \zeta(-3) $$ This is pretty much my first time in the realm of the zeta function, so i wasn't sure if this was obvious, or a given thing. But I couldn't find any reference to it online. What I'm asking is why this is and what I should search for if I want to learn more about this.
This can be proven using a couple of bits from the internet; use Faulhaber's Formula to define the polynomial: $$\sum _{k=1}^{x}{k}^{p}={\frac {\sum _{j=0}^{p} \left( -1 \right) ^{j}{ p+1\choose j}B_j {x}^{p+1-j}}{p+1}}$$ then integrate to get: $$ \begin{aligned} \int _ {-1}^{0}\!{\frac {\sum _{j=0}^{p} \left( -1 \right) ^{j}{p+1\choose j} B_j {x}^{p+1-j}}{p+1}}{dx}&=-\left( -1 \right) ^{p}\frac{1}{p+1}\sum _{j=0}^{p}{\frac {{p+1\choose j}B_j }{p+2-j}},\\ &=-\left( -1 \right) ^{p}{\frac {B_{p+1} }{p+1}},\\ &=\zeta(-p) \end{aligned} $$ where we have used the recursion relation for Bernoulli numbers to pass from the first to the second line and the known value of the zeta function at negative integers for the last line. The $(-1)^p$ is trivial for most $p$ as even order Bernoulli numbers are zero but the zeroth order is not so we must keep it for that case.