How can one calculate $\int_{0}^{1} x \ dx$ without the fundamental theorem of calculus?
Can I do something like this?
$$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
$$\int_0^1x \ dx=\lim_{n\to\infty} \frac1n\sum_{r=1}^n \frac{r}{n}=\lim_{n\to\infty} \frac1{n^2}\sum_{r=1}^nr$$
$$\text{And then }\sum_{r=1}^nr=\frac{n(n+1)}2$$
Then maybe $\displaystyle\lim_{n\to\infty}\frac{n(n+1)}2$
But then? Is that approach even correct?
In this particular case, area to sum is simply a right triangle with base $1$ and height $1$. This gives us $\text{Area}=\dfrac{1}{2}\cdot \text{Base}\cdot \text{Height}=\dfrac{1}{2}$. However, given that this has tags like "summation" and "limits," I expect that you need to calculate the area using a type of Riemann Sum.
With your approach, you are almost there. Once you have $$\int_0^1 x\,\text{d}x = \lim_{n \to \infty}\left(\dfrac{1}{n^2}\cdot\sum_{r=1}^n r\right)= \lim_{n \to \infty}\left(\dfrac{1}{n^2}\cdot\dfrac{n(n+1)}{2}\right),$$you just need to evaluate that final limit. Note that you cannot evaluate the limits of the factors separately, because this would result in an indeterminate form $0\cdot \infty$. However, this limit simplifies nicely: $$\lim_{n \to \infty}\left(\dfrac{1}{n^2}\cdot\dfrac{n(n+1)}{2}\right)=\dfrac{1}{2}\cdot \lim_{n \to \infty}\left(1+\dfrac{1}{n}\right)=\dfrac{1}{2}.$$