Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 4 down vote accepted

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

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 $n\to\infty$.

share|cite|improve this answer
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.

share|cite|improve this answer

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.

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.