Need clarification in Riemann sum I am a self-learner and the book that I am engaged in gave me some basics of Riemann sums but I see that they are not enough to solve most of the problems related to Riemann sums. Can someone give me a little hand as to why $1/n$ always means that we are looking for the integral from $0$ to $1$ (Yes I know that $(b-a)$ here gives $1$ but is not it possible to consider the range not from $0$ to $1$ but from $5$ to $6$ as well?)?
Generally, why does this condition below makes sense?
$$\sum_{i=1}^n\frac{1}{n}f\left(\frac{i}{n}\right)\to\int_0^1f(x)dx$$
 A: If you divide the interval $[0, 1]$ into $n$ pieces, each piece will have width $\frac{1}{n}$. Generally, if you wish to integrate on $[a, b]$ with $n$ pieces you'll still have the factor of $\frac{1}{n}$ present, but your mesh points in your function will be $a + \frac{(b-a)i}{n}$. That is,
$$ \int_a ^b f(x) \ dx \approx \sum_{i=1} ^n \frac{b-a}{n} f\left( a+ \frac{(b-a)i}{n} \right) $$
This just amounts to multiplying the width of a rectangle by its width.
A: Riemann sums start from a partition. Let $(a = x_0, x_1, \dots x_n = b)$ be a partition of $[a,b]$. Then we define $\Delta x_i = |x_i - x_{i-1}|$ and $N_p = \max\{\Delta x_i |i = 1,2,\dots, n\}$
Then, we can write:
$$\int_a^b f(x) dx = \lim_{N_P\to0} \sum_{i=1}^nf(x_i')\Delta x_i$$
where $x_i' \in [x_{i-1},x_i]$
In your case, the partition of $[0,1]$ is given by $(0,1/n,2/n, \dots , n/n = 1)$, and $x_i' = x_i = i/n$. Also, the partition is equidistant so $\Delta x_i = 1/n$ for each $i$. 
Hence, $$\int_0^1 f(x) dx  = \lim_{N_P\to0} \sum_{i=1}^n f\left(\frac{i}{n}\right) \frac{1}{n}$$
A: The $\dfrac1n$ indeed only tells you that $a$ and $b$ are one unit apart.  So, just by looking at $\dfrac1n$, you can't be sure if the interval is $[0,1]$ or $[5,6]$ (to use your example).  You also need to look at the $f\left(\dfrac in\right)$.  Looking at this tells you your interval starts at $0$.  If it started at $5$, then you would have $f\left(5 + \dfrac in\right)$.
In general, you'll have $f\left(a + \dfrac{b-a}n i\right)$ for the interval $[a,b]$ with $n$ subintervals of equal length.  And since the specific problem you posted has $f\left(\dfrac in\right)$, then we can write this as $f\left(\dfrac in\right) = f\left(0 + \dfrac in\right)$, giving us $a = 0$.
