Verified answer

This is a sequence of Riemann sums of the function f(x) =(6x - 3)/5 over [0 , 5]. Since f is continuous it is integrable and the lenght of the intervals of the partition approaches 0, the limit is Int (0 to 5) (6x - 3)/5 dx = [3x^2 - 3x]/5 from 0 to 5 = (75 -15)/5 - 0 = 12.

What Jazz_will did, without mentioning, was exactly computing such integral based from the definition of integral based on upper and lower sums.

Let's calculate what the maximum and minimum the sum can be.

For the maximum, obviously the sum happens when xi takes the maximum allowed value in each sub interval, so

xi = i (5/n)

For the minimum, xi must take the minimum in each sub interval, so

xi = (i-1) (5/n)

Now plug in the xi's and evaluate the sum

sum(1,n) (6 xi -3)n = -3 + 6/n sum(1,n) xi

For the maximum, sum(1,n) i5/n = 5/n . n(n+1)/2

For the minimum, sum(1,n)(i-1)5/n = 5/n . n(n-1)/2

So the maximum for the sum is

15 (n+1)/n - 3

and the minimum for the sum is

15 (n-1)/n - 3

Both converges to 15 -3 = 12 in the limit.