lower sum and upper sum of x^2 ... I hope you're able to understand what I'm writing now:
I have to calculate the lim of lower sum and upper sum for the Integral $\int_0^1 x^2 dx, $ by decomposition the interval into n pieces of the same length.
So, I know the following things:
1.) the width of $ x_{k+1} - x_k = 1/n$ since $(1-0)/n = 1/n. $
This means, all "stripes" have the same width.
2.) the "height" for $ x_0 = f(0) = 0$, $x_{0+k} = f(0+k) = k^2 $ ...
so: $ f(x_k) = (k/n)^2 $ ?
3.) the lower sum is: $ \sum_{k=0}^{n-1} (k/n)^2 * 1/n$ - correct?
4.) the upper sum is: $ \sum_{k=1}^{n} (k/n)^2 * 1/n$ - correct?
Well, I will go on when I know if this is correct or not :)
EDIT:
Hello, thank you all very much (I love this website :-) is it also correct, that the upper sum is from "k=1 to n" while the lower sum is from "k=0 to n-1" ? I wrote this simply by taking a look at a specific graph.. Okay, now I have to find the limit:
Now for lower sum.. :
5.) $ lim_{n \to \infty}\ \sum_{k=0}^{n-1} (k/n)^2 * 1/n$ 
this means:
$ lim_{n \to \infty} (\sum_{k=0}^{n-1} (k^2)*(1/n^2) * (1/n)) $ 
$=$
$lim_{n \to \infty} ((1/n^3)\sum_{k=0}^{n-1} (k^2) =$
$ lim_{n \to \infty} ((1/n^3) \frac{n*(n+1)*(2n+1)}{6}) $
$ lim \frac{2n^2 +3n +1}{6n^2} = 1/3 + 0 + 0 + 0 = 1/3 $
??
I'm sorry if  it's not written in the mathematical-correct way!
 A: Yes, everything you've written is correct. (except for that one weird chain of equations $x_{0+k} = f(0+k) = k^2$ which doesn't really make sense)
Although this problem is simple enough that what I suggest below may seems silly, sometimes doing such things help in complex problems.
It may be useful to define variables that more closely relate to the actual objects of the problem. Specifically, rather than picking out the $n+1$ points $x_k = k/n$, you might instead pick out the $n$ intervals
$$ I_k = \left[ \frac{k}{n}, \frac{k+1}{n} \right] $$
for $k = 0, \ldots, n-1$. Having a name for the intervals makes it simpler to reason about them.

Re: your edit: you're mostly okay. You've made one main error: you used the formula for $\sum_{k=0}^n k^2$, but the upper limit on the sum is only $n-1$.
It turns out, however, that error doesn't affect the end result.
The number of steps you need to write between
$$ \lim_{n \to \infty} ((1/n^3) \frac{n*(n+1)*(2n+1)}{6}) $$
(after correcting for your error) and 
$$ 1/3 $$
depends on how much you want to show you know the steps between those two. In particular, if you are in a setting where those are "obvious", you can just skip all of the way. But if you are in a class where your professor is assessing your knowledge of that stuff and it can't just be assumed you know, then you need to show more steps!
Of course, if you are unsure of your steps, it's best to write them out. Not only does seeing your own work make it easier to spot errors or help to be confident it's correct, it would let your professor and others see how you do the work and suggest corrections and improvements.
(P.S. use \lim instead of lim when you write the math, and you'll get the better typesetting you see in my post)
