I was wondering whether one can prove brute force without the notion of the Riemann sums the fact that
$$ \underline{\int}_0^1xdx=\frac{1}{2} $$
I am trying to do this without any theorems about integrals just by definition, but I am struggling with this for some reason. A brief search has also yielded no proof for this fact. Has anyone encountered such a proof for this or perhaps provide me with some?
Is this fine? Let $\mathcal{P}_n=\{0,1/n,2/n,...,(n-1)/n,1\}$ be a partition with $n$ subintervals. Then if $f(x)=x$ on $[0,1]$ then we have $$L(\mathcal{P}_n,f)=\sum_{i=1}^{n}\min_{x\in[\frac{i-1}{n},\frac{i}{n}]}f(x)\Delta x_i=\sum_{i=1}^{n}\frac{i-1}{n^2}=\frac{1}{n^2}\sum_{i=1}^{n}(i-1)=\frac{n(n-1)}{2n^2}.$$ Now taking $n\rightarrow\infty$ gives the desired result.