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.

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So you want to find the integral using "the limit definition"? – Thomas Apr 14 '12 at 2:57
Yes I think that is what it is called. – user138246 Apr 14 '12 at 2:58
You didn't plug the $x_n$ into $f(\cdot)$ correctly, to summarize. – anon Apr 14 '12 at 3:07

$$x=\frac{k}{n}:\quad \sum_{k=1}^n \Delta x\cdot \big(x^3-3x^2\big) = \sum_{k=1}^n \frac{1}{n}\left(\left(\frac{k}{n}\right)^3-3\left(\frac{k}{n}\right)^2\right)$$

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 $n\to\infty$.

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To me it looks like you made a mistake, shouldn't it be $-3/n^3$ infront of the $k^2$ summation? – user138246 Apr 14 '12 at 15:10
@Jordan: Good catch! You are right, I made a mistake. – anon Apr 14 '12 at 18:44

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.

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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.

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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 $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \\ \sum_{k=1}^n k^3 = \left( \frac{n(n+1)}{2} \right)^2$$

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Where's the $\delta$ in your calculations? – anon Apr 14 '12 at 3:04
@anon sorry editing.... – Thomas Apr 14 '12 at 3:04
Now it's even more wrong! :P – anon Apr 14 '12 at 3:08
@anon :) I am trying with this LaTeX .... patience – Thomas Apr 14 '12 at 3:10
Alright, now it's fine. :P – anon Apr 14 '12 at 3:23