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*Chumbert: Yeah, he said in class today (Wed.) to assume that, right? | *Chumbert: Yeah, he said in class today (Wed.) to assume that, right? | ||
| + | *Ctuchek - I do remember him saying that we will need to use the Mean Value Theorem. | ||
==Interesting Articles about Calculus== | ==Interesting Articles about Calculus== | ||
Revision as of 19:35, 18 September 2008
Contents
Math 181 Honors Calculus
Getting started editing
Lecture Notes
Homework Help
Hello, this is gary from ma181. let's solve the extra credit problem. Here is the problem in italics:
Extra Credit Problem
Suppose that f(x) is continuously differentiable on the interval [a,b]. Let N be a positive integer and let $ M = Max { |f'(x)| : a \leq x \leq b } $. Let $ h = \frac{(b-a)}{N} $ and let $ R_N $ denote the "right endpoint" Riemann Sum for the integral $ I = \int_a^b f(x) dx . $ In other words, $ R_N = \sum_{n=1}^N f(a + n h) h . $
Explain why the error, $ E = | R_N - I | $, satisfies $ E < \frac{M(b-a)^2}{N}. $
- So what does this equation "E < M(b-a)^2/N" mean. This reads that the error is less than the Maximum value of the derivative of the function of x multiplied by the interval squared from x=a to x=b all divided by the total number of subintervals N.
- I don't understand why this must be true. Maybe I'm wrong, but if f(x) were a horizontal line, wouldn't E=0 and M(b-a)^2/N also be =0. That would mean it is a false statement that E < M(b-a)^2/N. Are we to assume that E <= M(b-a)^2/N?
- Chumbert: Yeah, he said in class today (Wed.) to assume that, right?
- Ctuchek - I do remember him saying that we will need to use the Mean Value Theorem.
Interesting Articles about Calculus
The minimum volume happens at the average_MA181Fall2008bell
