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. 
 A: $$x=\frac{k}{n}:\quad \sum_{k=1}^n \Delta x\cdot \big(x^3-3x^2\big) = \sum_{k=1}^n \frac{1}{n}\left(\left(\frac{k}{n}\right)^3-3\left(\frac{k}{n}\right)^2\right)$$
Note that $(k/n)^3=k^3/n^\color{Red}3$ not $k^3/n$, and $(k/n)^2=k^2/n^\color{Red}2$ not $k^2/n$. Now this becomes
$$\sum_{k=1}^n \left(\frac{1}{n^4}k^3-\frac{3}{n^3} k^2\right).$$
I suppose you have formulas for $\sum_{k=1}^n k^2$ and $\sum_{k=1}^n k^3$? With these you just have to distribute using linearity and evaluate the resulting limit as $n\to\infty$.
A: Your algebra glitched a bit when you set up the Riemann sum: with $x_k=\dfrac{k}n$ you’ll have $$f(x_k)=\left(\frac{k}n\right)^3-3\left(\frac{k}n\right)^2=\frac{k^3}{n^3}-\frac{3k^2}{n^2}\;,$$ not $\dfrac{k^3}n-\dfrac{3k^2}n$. That makes your Riemann sum
$$R_n=\sum_{k=1}^n\frac1n\left(\frac{k^3}{n^3}-\frac{3k^2}{n^2}\right)=\frac1{n^4}\sum_{k=1}^nk^3-\frac3{n^3}\sum_{k=1}^nk^2\;.$$ If you combine this with the summation formulas in Scott Carter’s answer, you should be on your way.
A: You need to know two sums: $\sum_{k=1}^n k^2 =\frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k^3 = \left[ \frac{n(n+1)}{2} \right]^2$. The proofs can be found here . Now with $\Delta x = 1/n$, figure out what $x_0$ through $x_n$ are, factor out the denominators and go to town. 
A: 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
$$
\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \\ 
\sum_{k=1}^n k^3 = \left( \frac{n(n+1)}{2} \right)^2
$$
