I created a formula that uses rectangles to approximate the area under a curve. For this instance, the answer to this problem should be equal to the area under the parabola $y=x^2$ from $[0, 2]$, which is $8/3$.
$$\lim_{n\to \infty} \frac 2n \sum_{l=1}^{n-1} \frac{4l^2}{n^2}$$
I don't know how to evaluate this limit, though. Any hints would be appreciated.
Using the sum of squares formula:
$$\lim_{n\to\infty} \frac{2}{n} \sum_{l=1}^{n-1} \frac{4l^2}{n^2} =\lim_{n\to\infty} \frac{8}{n^3} \color{blue}{\sum_{l=1}^{n-1} l^2} = \lim_{n\to\infty} \frac{8}{n^3} \color{blue}{\frac{n(n-1)(2n-1)}{6}}$$
As $n\to\infty$, this limit will approach the ratio of the highest powers of $n$: $16/6 = 8/3$:
$$\lim_{n\to\infty} \frac{8}{n^3} \frac{n(n-1)(2n-1)}{6} = \lim_{n\to\infty} \frac{16n^3 - 24n^2 + 8n}{6n^3} = \lim_{n\to\infty} \frac{8n^2 - 12n + 4}{3n^2}$$
Let $S_n:=\sum_{l=1}^nl^2$. We have $S_{n}-S_{n-1}=n^2$, which is a quadratic polynomial, and $S_n$ must be a cubic polynomial.
Assume $S_n\approx an^3$ and lower powers of $n$. Then
$$S_n-S_{n-1}\approx an^3-a(n-1)^3\approx an^3-an^3+3an^2$$ and lower order terms.
Then $a=\frac13$ and your limit is $\frac83$.
