The 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 each i. Then: limit as n approaches ∞ Σ with n-1 on top and i=0 on bottom for ((8xi)/n) is equal to what?
xi = 5/n, right? so you have
n-1
∑ (8xi)/n = (8*(5/n)i)/n = 40/n^2 i
i=0
This is a left Riemann sum
as an integral, that would be
∫[0..5] 8/5 x dx
because
(8xi)/n = (8/5 xi) * 5/n
To find the limit of the sum as n approaches infinity, we can convert it into an integral.
The given sum is:
lim(n→∞) Σ(i=0 to n-1) ((8xi)/n)
We can rewrite this sum as:
lim(n→∞) (8/n) * Σ(i=0 to n-1) xi
Now, we want to find the integral of xi over the interval [0, 5]. We can express this integral as:
∫(0 to 5) xi dx
Now, since we are partitioning the interval [0, 5] into n equal subintervals, the width of each subinterval is Δx = (5 - 0)/n = 5/n.
The left-hand endpoint xi of each subinterval can be expressed as (i * 5/n).
Replacing xi and dx in the integral with (i * 5/n) and (5/n) respectively, we get:
∫(0 to 5) (i * 5/n) * (5/n) di
Simplifying this integral, we have:
(25/n^2) * ∫(0 to n-1) (i * di)
Integrating this expression, we get:
(25/n^2) * [(1/2) * i^2] evaluated from i = 0 to i = n-1
Substituting the limits of integration, we have:
(25/n^2) * [(1/2) * (n-1)^2 - (1/2) * (0)^2]
Simplifying further, we get:
(25/n^2) * [(1/2) * (n^2 - 2n + 1)]
(25/2n) * (n - 1)
Now, taking the limit of this expression as n approaches infinity, we get:
lim(n→∞) [(25/2n) * (n - 1)]
As n approaches infinity, the term (2n) dominates the denominator, so we get:
lim(n→∞) [(25/2n) * (n - 1)] = lim(n→∞) [(25/2) * (1 - 1/n)] = 25/2
Therefore, the limit of the given sum as n approaches infinity is 25/2.
To find the limit of the sum as n approaches infinity, we need to evaluate the Riemann sum.
Given that the interval [0, 5] is partitioned into n equal subintervals, the length of each subinterval will be ((5-0)/n) = 5/n.
Now, let's express the left-hand endpoint of the i^th subinterval, xi, in terms of i and n. Since the interval starts at 0, the left-hand endpoint of each subinterval will simply be the starting point of that subinterval. We can express xi as:
xi = (i * (5/n))
Now, let's rewrite the sum using the expression for xi:
Σ((8xi)/n) = Σ((8 * i * (5/n))/n)
We can pull out the constant factors outside the sum:
= (8 * 5/n^2) * Σ(i)
The sum Σ(i) is the sum of consecutive integers from 0 to n-1. The sum of consecutive integers can be found using the formula:
Σ(i) = (n * (n-1))/2
Plugging this into our expression:
= (8 * 5/n^2) * ((n * (n-1))/2)
Simplifying further:
= (40 * (n-1))/n
Now, we can take the limit as n approaches infinity:
lim(n→∞) (40 * (n-1))/n
To evaluate this limit, we can divide every term by n and apply the limit:
= 40 * lim(n→∞) (n-1)/n
Since (n-1)/n approaches 1 as n approaches infinity:
= 40 * 1
= 40
Therefore, the limit as n approaches infinity of the given sum, ((8xi)/n), is equal to 40.