Using the definition of the integral, find $\int_0^a x^2 dx$ The definition of the integral I was given (which after searching around seems like the common definition) is the value of the inf{upper sums across all dissections} (integral exists when this coincides with the sup{lower sums across all dissections}). 
Now, when I searched online of how to do the integral in question, all solutions said: partition $[0,a]$ into strips of equal width $(1/N)$ and then let $N$ tend to infinity to get a limit $L$ etc.
But surely this doesn't cover all the possible dissections and also can't be a refinement of some dissections (e.g. if a is rational and I have a dissection with an irrational point $x$ in it then any dissection given in the above way can never have $x$ as a point in it). So why should $L$ be the value of the integral? Yet, I don't know how else to approach this. 
Help greatly appreciated!
 A: As you have that $x^2$ is continuous all partitions will give the same integral. 
Hence
$$\int_0^a x^2 =\lim_{n\to \infty} \sum_{i=0}^n \frac{a}{n} \cdot \frac{(ai)^2}{n^2}=\lim_{n\to \infty} \frac{a}{n^3} \sum_{i=0}^n (ai)^2=\lim_{n\to \infty} \frac{a^3}{n^3} \cdot \frac{n \cdot (n+1)\cdot (2n+1)}{6} $$
The limit of this is 
$$\frac{a^3}{3}$$ which we expected.
We used that 
$$\sum_{i=1}^n  i^2 = \frac{n \cdot (n+1) \cdot (2n+1)}{6}$$
is.
To see that this partiation is allowed we can use that $x^2$ is strict monoton increasing,
So 
$$\lim_{n\to \infty}\sum_{i=1}^n \frac{a}{n} \left(\frac{i}{n}\right)^2 \leq \int_0^a  x^2 \, \mathrm{d}x \leq \lim_{n \to \infty} \sum_{i=0}^n  \left(\frac{i}{n}\right)^2$$
A: $$\int_0^a f(x)=Lt_{n\rightarrow\infty}^{h\rightarrow0} [1/n\left( f(0)+f(h)+f(2h)+........ +f(nh) \right)]$$
$$where, (nh)\rightarrow a \ as\  n\rightarrow \infty \ and \ h\rightarrow0$$
Now expand and use formula for sum using G.P,A.P. , sum of squares , sum of cubes etc.to get a single term (or more depending on terms in function itself) in only $n$ and $h$ . (sum of square on n terms in the example given)  .
Now, put the limits ,you get the ans.
A: An interesting sidelight which may give some insight here is the following.  Consider an interval $[0,1]$ and $n$ points $x_k$ chosen completely at random, except that $0<x_1<x_2<\ldots<x_{n-1}<x_n<1$.  Then  for any continuous function $f$ defined on this interval, the expected value of the Riemann sum is
$$\mathbb{E}\left[\sum_{k=1}^n f(x_k) (x_k-x_{k-1}) \right] = \int_0^1 dt \: [1-(1-t)^n] f(t)$$
A proof of this may be found on p. 113 of this document.  The point is that, as $n \rightarrow \infty$, any Riemann sum over a function, no matter how the intervals are subdivided, converges to the integral of that function.
A: There is a theorem that proves that continuous functions over closed intervals really are Riemann-integrable. The definition of that term, is that the infimum and supremum you speak of really will equal each other.
Once you have that theorem, then since $x^2$ is continuous on $[0,a]$, it suffices to look at one particular refinement of $[0,a]$. This would explain why it would be OK to use a partition of equal step sizes $1/n$ and let $n\to\infty$.
A: You are describing the Riemann integral. And you are absolutely right, the definition talks about all divisions, not just division into $N$ equal parts. But you can prove that if the length on the shortest slice is larger than $1 /N$, uniform division into $N$ pieces gives a smaller minimal sum, and a larger maximal one. This proves that uniform division is enough.
Then you will need the values of sums like:
$$
\frac{1}{N} \sum_{0 \le k \le N - 1} \frac{k^2}{N^2} = \frac{1}{N^3} \sum_{0 \le k \le N - 1} k^2
$$
$$
\frac{1}{N} \sum_{1 \le k \le N} \frac{k^2}{N^2} = \frac{1}{N^3} \sum_{1 \le k \le N} k^2
$$
It is easy to see (even without computing the sums) that the difference is $\dfrac{N^2 - 0}{N^3} \rightarrow 0$, so the integral exists.
