#1**+4 **

(c) (i) Recall the definition of Riemann integration:

\(\displaystyle\int^b_a f(x) dx = \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\dfrac{b-a}{n}f\left(a+\dfrac{k(b-a)}{n}\right)\)

As \(\dfrac{b-a}{n} > 0 \wedge f\left(a+\dfrac{k(b-a)}{n}\right) > 0\), every term of the Riemann sum is greater than 0.

Hence, the integral is greater than 0.

MaxWong May 20, 2019

#2**+1 **

This proof is incorrect. You are claiming that the limit that is the integral, is strictly positive because it is a limit of strictly positive numbers. That reasoning is incorrect (think about the sequence of the reciprocals of the natural numbers).

Guest May 20, 2019

#5**+3 **

Hectictar,

Guest #2 is referring to a situation that occurs in the context of the analytic continuation of the Zeta function (see https://www.youtube.com/watch?v=YuIIjLr6vUA). However, I doubt that that context is relevant for the questioner here!

Alan
May 21, 2019

#8**+1 **

I was referring to the sequence 1, 1/2, 1/3,...

Every single one of its terms is strictly positive but the limit is 0 (which is not strictly positive)

Guest May 21, 2019

#10**+2 **

Hmmmm..I'm still confused....are you talking about

\(\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}\frac1k\)

https://www.wolframalpha.com/input/?i=limit+n-%3Einfinity+(sum+1%2F(k),+k%3D1+to+n)

?

hectictar
May 21, 2019

#11**0 **

I think we're getting off track, my point was that the proof made by max is incorrect, because the limit of the sequence a_{n }representing the integral is not one infinite series, but a sequence where each term of the sequence is a finite sum. Different terms of the sequence a_{n} are sums that are made of different elements, as opposed to a series where a sequence of numbers gets added up one by one.

Max's argument boils down to "the limit of a_{n} is strictly positive because each member of a_{n} is strictly positive" but that argument is incorrect (for example: 1, 1/2, 1/3, 1/4....)

Guest May 24, 2019

edited by
Guest
May 24, 2019

#3**+4 **

(ii) False

As a counterexample,

Let f(x) = x - 4 and a = 0 and b = 6

\(\int_{a}^{b}f(x)\,dx\,=\,\int_{0}^{6}(x-4)\,dx\,=\,(\frac{x^2}{2}-4x)\Big|_{0}^{6}\,=\,-6\,<\,0\)

But.... 5 is an element of the interval [0, 6] and

\(f(5)\,=\,5-4\,=\,1\nless0\)

The integral is negative because the triangle under the x-axis is bigger than the triangle above the x-axis,

but not all the values of f(x) are negative for x in the interval [0, 6] .

hectictar May 21, 2019