Advanced Math/Riemann sums
Expert: Paul Klarreich - 12/4/2009
QuestionI am confused on this problem:
Consider f(x)=X^2 on [0,2] and a particular P given by P= {0, 2/n, 4/n,..., 2-2/n,2} Find U(P,f) and L(P,f) and use the definition to evaluate the intergral of x^2 from 0 to 2.
I know the partition is 2/n so you will eventually have to divide by that. It also envolves the left and right end points.
AnswerQuestioner: Sally
Country: United States
Category: Advanced Math
Private: No
Subject: U(P,f) and L(P,f)
Question: I am confused on this problem:
Consider f(x)=X^2 on [0,2] and a particular P given by P= {0, 2/n, 4/n,..., 2-2/n,2} Find U(P,f) and L(P,f) and use the definition to evaluate the intergral of x^2 from 0 to 2.
I know the partition is 2/n so you will eventually have to divide by that. (NO, I THINK YOU MULTIPLY BY THAT) It also envolves the left and right end points.
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Here is the Riemann sum you must compute:
SUM(for k = 1 to n) f(chosen-x[k]) dx
The 'chosen-x[k]' depends on
1) The function.
2) The type of sum.
In this case, f(x) = x^2 is increasing on [0,2], so
The UPPER-sum-chosen-x[k] will be at the right side of the interval, so it will be k * dx
The LOWER-sum-chosen-x[k] will be at the left side of the interval, so it will be (k-1) * dx[k]
Now the dx is 2/n -- you knew that.
So we are ready to go:
I'll do UPPER, you'll do LOWER.
SUM(for k = 1 to n) f(k * 2/n) 2/n
SUM(for k = 1 to n) (2k/n)^2 2/n
SUM(for k = 1 to n) 8k^2/n^3
8/n^3 SUM(for k = 1 to n) k^2
Now you look up:
n(n + 1)(2n + 1)
SUM(for k = 1 to n) k^2 = ----------------
6
and put that in:
8 n(n + 1)(2n + 1)
= ---- -----------------
n^3 6
and finally take lim[n -> inf] of that, and get 8/3, the correct integral.
When you do the LOWER, you will have
(n-1)(n)(2n - 1)
SUM(for k = 1 to n) (k-1)^2 = ----------------
6
which will give the same limit. Try it.
IN THE FUTURE, be sure to define all the terms you use. DO NOT just write U(...) and assume I am your teacher and use the same notation.