<s>[INST] Integral with cube, square and zero starting point I don't know is there is a special rule or trick for this but I am trying to find
$$\int_{0}^1(x^3-3x^2) dx$$
I know that $\dfrac{1}{n}$ is the delta $x$ and this is where I do not know what to do next. I think that I want it to look something like
$$\lim \sum \dfrac{1}{n} \dfrac{k^{3}}{n} - n  \dfrac{3k^{2}}{n}$$ But I am not sure what to do with coefficients yet.
I end up with something that looks like $\dfrac{3}{2} + 2n + \dfrac{2n}{6} + \dfrac{1}{2}$ which doesn't make sense since I have no many infinities. 
 [/INST] So you have $\Delta x = \frac{1}{n}$.
And so$$\begin{align} 
\int_0^1 f(x) dx &= \lim_{n\to \infty} \sum_{k=1}^{n} (\Delta x) f\left(0 + k\Delta x \right) \\
&= \lim_{n\to \infty} \sum_{k=1}^{n} \frac{1}{n} f\left(\frac{k}{n}\right) \\
&= \lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^{n} \left(\frac{k}{n}\right)^3 - 3\left(\frac{k}{n}\right)^2 \\ 
&= \lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^{n} \frac{1}{n^3}k^3 - \frac{3}{n^2}k^2 \\
&= \lim_{n\to \infty} \frac{1}{n}\left[\sum_{k=1}^{n} \frac{1}{n^3}k^3\right] - \frac{1}{n}\sum_{k=1}^{n}\frac{3}{n^2}k^2 \\
&= \lim_{n\to \infty} \frac{1}{n^4}\left[\sum_{k=1}^{n} k^3\right] - \frac{3}{n^3}\sum_{k=1}^{n}k^2 \\
&=  ...
\end{align}
$$
And to find this you can use that
$$
\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \\ 
\sum_{k=1}^n k^3 = \left( \frac{n(n+1)}{2} \right)^2
$$
</s>