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Math 181 Honors Calculus
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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.
- Chumbert - Logically, I think I got it, but I'm not entirely sure how to prove it mathematically:
The $ M(b-a) $ gives the height of one section(slope=(y/x), so slope*x=y), where $ \frac{(b-a)}{N} $ gives the width, and when multiplied together, they give you a rectangle which, if you remember from class, is the error--take the R-sum, then stack the extra blocks on to of each other. Does anyone else remember that? Or should I explain it better?
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