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I need help with this question please! 

 

 May 20, 2019
 #1
avatar+7650 
+3

(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.

 May 20, 2019
 #2
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+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
 #4
avatar+8215 
+2

Guest, how can a sum of positive numbers ever be not positive??

hectictar  May 21, 2019
 #5
avatar+28029 
+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
 #6
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Alan I am sorry but this is not what I was referring to smiley

Guest May 21, 2019
 #7
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+2

Ok.  To what were you referring?  (I misread your earlier reply as the sum of the natural numbers!).

Alan  May 21, 2019
edited by Alan  May 21, 2019
edited by Alan  May 21, 2019
 #8
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+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
 #9
avatar+28029 
+1

I see.

Alan  May 21, 2019
 #10
avatar+8215 
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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
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I think we're getting off track, my point was that the proof made by max is incorrect, because the limit of the sequence arepresenting 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....)

Guest May 24, 2019
edited by Guest  May 24, 2019
 #3
avatar+8215 
+3

(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] .

 May 21, 2019

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