Can you explain how to write the proof of Section 1.15 Exercise 4(a) of Apostol's Calculus I am attempting to self-study Calculus from Apostol's book, but I have gotten stuck on the proof for Exercise 4(a).
There are indeed other solutions, such as the below two links, but they don't explain the steps very well.
http://www.stumblingrobot.com/2015/07/18/compute-some-integrals-of-step-functions/
https://iiserbbookstore.files.wordpress.com/2016/01/tom-apostol-v1-solutions.pdf
Here is the required information.

Definition of the Integral of Step Functions $$\int_a^bs(x)dx = \sum_{k=1}^ns_k \cdot (x_k - x_{k-1})$$
  Where $s_k$ is the value of $s(x)$ along the $k$-th open subinterval of a partition $P$ of $[a, b].$

Here is the actual question:

If $n \in \mathbb Z^+$, prove that $\int_0^n[t]dt = \frac{n(n-1)}{2}$ where $[t]$ denotes the greatest integer $\le t$.

I also don't completely understand the relationship between $n$ from the Definition of the Integral of Step Functions, and $b$, the upper limit of integration.
Any aid you could provide would be greatly appreciated. Thank you.
 A: By definition\begin{align}\int_0^n\lfloor t\rfloor\,\mathrm dt&=0\times(1-0)+1\times(2-1)+\cdots+(n-1)\times\bigl(n-(n-1)\bigr)\\&=1+2+\cdots+(n-1)\\&=\frac{n(n-1)}2.\end{align}
A: Well, a step function is one where there exists $a =x_{0} < x_1 < .... < x_n = b$ and if $x_{k-1} < x < x_{k}$ then $f(x) = s_k$ for some constant.  (and $f(x_k) = $ either $s_{k}$ or $s_{k-1}$ but those values don't affect the integral).
$[t]$ is precisely such a step function.  Where $a=0= x_0; b= n=x_n$ and on the interval $t\in (x_{k-1}, x_k) = (k-1, k)$ then $[t]  = k-1 = s_k$.  And each $x_{k} - x_{k-1} = 1$.
So by that definition and those intervals and those indexes:
$\int_{0}^n [t]dt = \sum_{k=1}^n s_k*(x_k - x_{k-1}) =\sum_{k=1}^n (k-1)*1 = \sum_{k=1}^n (k-1)$.
And that's it.
Proving that $\sum_{k=1}^n k = \frac {n(n+1)}2$ is a well known trick and is beside the point of what this exercise is supposed to teach you.  You may take that as a given as it has nothing to do with calculus.  
So $\sum_{k=1}^{n} (k-1) = \sum_{j=0}^{n-1} j = \frac {(n-1)n}2$
Alternatively $\sum_{k=1}^{n} (k - 1)= \sum_{k=1}^n k - \sum_{k=1}^n 1 = \frac {n(n+1)}2 - n = \frac {n(n-1)}2$.
