+9675 (c) (i) Recall the definition of Riemann integration:
∫baf(x)dx=limn→∞n∑k=0b−anf(a+k(b−a)n)
As b−an>0∧f(a+k(b−a)n)>0, every term of the Riemann sum is greater than 0.
Hence, the integral is greater than 0.
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).
+33654 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!
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)
+9488 Hmmmm..I'm still confused....are you talking about
limn→∞n∑k=11k
https://www.wolframalpha.com/input/?i=limit+n-%3Einfinity+(sum+1%2F(k),+k%3D1+to+n)
?
I think we're getting off track, my point was that the proof made by max is incorrect, because the limit of the sequence an 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 an 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 an is strictly positive because each member of an is strictly positive" but that argument is incorrect (for example: 1, 1/2, 1/3, 1/4....)
+9488 (ii) False
As a counterexample,
Let f(x) = x - 4 and a = 0 and b = 6
∫baf(x)dx=∫60(x−4)dx=(x22−4x)|60=−6<0
But.... 5 is an element of the interval [0, 6] and
f(5)=5−4=1≮
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] .
