Any ideas please?

Question

The interval [0,5] is partitioned into n equal subintervals, and a number, xi is arbitrarily chosen in the i^th subinterval for each i. Then lim n
n--> infinity ∑
i=1 (6xi-3)/n

Answer 1

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.

Answer 2

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.