Left endpoint approximation & Riemann Sums (Sigma)

[asz304]

1. The problem statement, all variables and givennown data
1)FInd the nth left endpoint approximation Ln for f(x) = 3x^2-2 on [0,2]. What is the limit as n approaches infinity Ln in this case?


2)Evaluate:
\sum⁴⁵_i=5 (2i-5)

Homework Equations

Ln = \sum^N_j=1
f(c_j)(x_j-x_j-1)

The Attempt at a Solution

1)Not sure where to start. Should I start by substituting 0 and 2 in the function and subtract both of them and get the limit as n approaches infinity?

2)2\sum⁴⁵_i=5 - \sum⁴⁵_i=5

= [ 2(2+1) ]/2 - 225

= -222

I'm not sure with my answer because it says \sum_i=1ⁿ
and in my question i = 5.

Thanks

 

[HallsofIvy]

Get the limit of what "as n approaches infinity"? Just "substituting 0 and 2" doesn't have any thing to do with an "n".

Divide the interval from 0 to 2 into n parts. Then each interval will have length 2/n and will have endpoint 0, 2/n, 4/n, 6/n, ..., i(2/n), ..., n(2/n)= 2. The "left endpoint" of each interval would be 0, 2/n, 4/n, 6/n, ... (n-1)(2/n). That is 0 is the "left endpoint" of the first interval- every point after that is the right endpoint of one interval but the left endpoint of the next, except that the last endpoint, n(2/n) is only a right endpoint. The function, evaluated at those left endpoints would be f(0)= -2, f(2/n)=3(2/n)^2-2= 12/n^2- 2, f(4/n)= 3(4/n)^2- 2= 48/n^2- 2, with the general rule being f(i(2/n))= 3(2i/n)^2- 2= 12i^2/n- 2. The area of the slender rectangle on the ith interval up to that height is "height times width"= \left(12i^2/n - 2\right)\left(2/n\right)= 24i^2/n^2- 4/n The Riemann sum would be
\sum_{i=0}^n 24i^2/n^2- 4/n
Find the sum of that (it will depend on n) and take the limit as n goes to infinity.

For example,
\sum_{i=0}^n -4/n= n\left(-4/n\right)
since a constant, added to itself n times, is just n times that constant.
By this time you should have learned formulas for \sum i and \sum i^2.