Interval [0,5] is partitioned into n equal subintervals and a number xi is defined to be the left-hand endpoint of the "i" th subinterval for every i. Then (with n-1 being on top of Σ and i=0 being below it) a limit of n going to infinity Σ (2xi-5)/n is equal to what?
To find the limit of the given sum as n approaches infinity, we can use the definition of a definite integral.
First, let's find the width of each subinterval. The interval [0, 5] is divided into n equal subintervals, so the width of each subinterval is (5-0)/n = 5/n.
Now, let's express the given sum using sigma notation using the new width of the subintervals:
Σ (2xi - 5)/n = Σ (2(xi)(5/n) - 5)/n
Next, let's simplify the expression inside the sigma notation:
= (2/n) * Σ (xi) - (5/n) * Σ (1)
Since Σ (xi) represents the sum of all the left-hand endpoints of the subintervals, it is equivalent to the sum of numbers from 0 to 5 with a step size of 5/n:
Σ (xi) = 0 + 5/n + 10/n + ... + (5 - (5/n))
Using the formula for the sum of an arithmetic series, this sum simplifies to (n+1)(5/n) = 5 + 5/n.
The sum Σ (1) represents the sum of all the terms of 1, and since there are n subintervals, this sum can be written as n(1) = n.
Now, let's substitute these simplified sums back into the original expression:
(2/n) * Σ (xi) - (5/n) * Σ (1) = (2/n)(5 + 5/n) - (5/n)(n)
Now, let's simplify further:
= (10/n) + (10/n²) - 5
Now, as n approaches infinity, (10/n) and (10/n²) both approach 0:
lim (10/n) = 0
lim (10/n²) = 0
Therefore, the limit of the given sum as n goes to infinity is:
lim (Σ (2xi - 5)/n) = lim ((10/n) + (10/n²) - 5)
= 0 + 0 - 5
= -5
So, the answer is -5.