Question

I am stuck on a question on http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-c-mean-value-theorem-antiderivatives-and-differential-equations/problem-set-5/MIT18_01SC_pset3prb.pdf
3B-4
In the answer sheet, it says Both x2 and x3 are increasing functions on 0 ≤ x ≤ b.
I am wondering isn't it possible that b is smaller or equal to 0?

Answer 1

Please help Thank yoU!

Answer 2

even if b is less than 0, how do you expect that to change the answer?

Answer 3

Wouldn't it be a negative answer?

Answer 4

The question is whether the difference tends to 0 as n tends to infinity, so it's a yes or no question. There is no room to make a distinction between negative and positive; it's either 0 or not.

Answer 5

when n goes to infinity diff goes to zero and sum of upper =sum of lower = inttegral f(x)

Answer 6

But if b can be negative, it goes from b to 0, would't the first question have two solution? (the one asked you to calculate the difference between them not the one ask what if n tends to infinity) It has (b/n)(f (b) - f (0)) and -((b/n)(f (b) - f (0)))? although it is the same when n goes to infinity

Answer 7

i assume it is standard !
integral (0 to b) so b >0
any way may b<0 then upper -lower is (b/n)(f (b) - f (0)) not - (b/n)(f (b) - f (0))

Answer 8

Oh yes. I guess the difference is just the absolute value of them right?

Answer 9

yes right

Answer 10

implicit
upper is always greater than lower

Answer 11

But on the textbook, some times upper is smaller than lower. In some exercises

Answer 12

ya that can be possible