$1+2+3... =-\frac{1}{12}$ - Question regarding this So, I'm not a big expert in this subject but I know $1+2+3...=-\dfrac{1}{12}$ isn't to do with 'real' maths but it's all to do with the zeta function; however I was watching a maths video and the equation:

$$
\frac{x(x+1)}{2}
$$

... is actually a perfect equation for the series $1+2+3...$ etc. where $x$ represents $n$ in a series and $y$ is the sum of the series up to $n$. So, you can conclude that:

$$
\sum^{n}_{i=1}1+2+3...=\frac{x(x+1)}{2}
$$

However, this is where it gets weird; as you have probably guessed, the roots of the equation is $x=0,-1$  but if I want to find the integral of the roots from $-1$ to $0$ which is under the $x$ axis, I get the following:

$$
\int_{-1}^{0} \frac{x(x+1)}{2}\:dx=-\frac{1}{12}
$$

So, my question is why is this the case; what connection is there between the value of the integral under the $x$ axis that this graph has compared to the summation of the series?

Link to Desmos graph for more clarity
Your result for this integral is just a consequence of a general formula wich gives the value of the zeta function at negative integer. Integrating Faulhabert* (or maybe bernoulli's, i'm not sure) formula between -1 and 0 leads to the formula. 

*The formula is the one that gives the n-th partial sum of k-th power of integers in term of polynomials of degree k+1.
Your result for this integral is just a consequence of a general formula wich gives the value of the zeta function at negative integer. Integrating Faulhabert* (or maybe bernoulli's, i'm not sure) formula between -1 and 0 leads to the formula. 

*The formula is the one that gives the n-th partial sum of k-th power of integers in term of polynomials of degree k+1.