choose relevent length of interval i have question related to   Riemann sum. Is then  length of interval matters? Suppose we want to calculate  Riemann  net sum of function
$f(x)=3-x/2$  in this interval $[2,14]$. 
I first take $n=2$, so that i would have   $(14-2)/2=6$, so we would have following values  $8,14$, so in total we would have  $6(f(6)+f(14))=6(3-3+3-7)=6(-4)=-24$,
so it means that  we should take  absolute value and  get $24$. Is  this correct ? if this represent  as a definite integral, then we can say that width of each is equal to $12/n$, and starting from $3$, we would have points  $2+12/n$, $2+24/n$ and so on. Now we should put this value into function yes? but how? i am confused  in this point,do we have
$12/n(f(2+12/n)+f(2+24/n)....f(14))$? and taking limit? it seems time confusing
 A: The identity that you are using is as follows:
$$\int_a^b f(x)dx=\lim_{n\to\infty}\sum_{i=1}^nf\left(a+\frac{b-a}{n}i\right)\left(\frac{b-a}{n}\right)$$ In this equality, we can set the right point in any sub-interval just to compute the wight of the boxes. The function $f(x)=3-x/2$ is continuous on $[2,14]$, so we can speak about above summation and so about the definite integral on this interval. You put $n=\color{red}{2}$ so you got the following summation: $$f\left(2+\frac{14-2}{\color{red}{2}}\right)\frac{14-2}{\color{red}{2}}+f\left(8+\frac{14-2}{\color{red}{2}}\right)\frac{14-2}{\color{red}{2}}=(f(8)+f(14))\times 6$$ See below:

But this summation with two terms gives us a very bad approximation for the definite integral, so as you noted we consider $n$ sub-interval. Indeed, we need to have a summation like above. I mean: 
$$A=\lim_{n\to\infty}\sum_{i=1}^nf\left(2+\frac{12}{n}i\right)\left(\frac{12}{n}\right)=\lim_{n\to\infty}\sum_{i=1}^n\left(2-\frac{6}{n}i\right)(12/n)$$ The length of interval does matter, cause we put it inside the $f$ as a part. In fact it plays an important role in doing the summation and taking the limit after that.

