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Definite Integration of a concave upward function- Inequality

mathisfun

New member
Aug 6, 2014
11
IMG-20140805-WA001.jpg
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
On the last integral, is the upper bound "9"?
 

mathisfun

New member
Aug 6, 2014
11

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,623
St. Augustine, FL.
Hello and welcome to MHB! :D

Can you tell us what you've tried or what your thoughts are on how to begin, so our helpers have an idea where you are stuck and then can provide better help?
 

mathisfun

New member
Aug 6, 2014
11
Hello and welcome to MHB! :D

Can you tell us what you've tried or what your thoughts are on how to begin, so our helpers have an idea where you are stuck and then can provide better help?
Thank you so much!
I began with the condition that the secant must lie above the curve at all points if the curve is concave upward but that seem to lead nowhere. I also tried to utilise the fact that the double derivative of f(X) will be positive but to no avail.
Any help will be appreciated.
 

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,623
St. Augustine, FL.
I think I would begin with the minimal case:

\(\displaystyle 0<f^{\prime\prime}(x)\)

So, integrating, what can you say about $f$?
 

mathisfun

New member
Aug 6, 2014
11
I think I would begin with the minimal case:

\(\displaystyle 0<f^{\prime\prime}(x)\)

So, integrating, what can you say about $f$?
Wouldn't integrating this inequality conclude the same thing as what we are given, that f is a concave upward function?
 

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,623
St. Augustine, FL.
Wouldn't integrating this inequality conclude the same thing as what we are given, that f is a concave upward function?
What did you get when integrating?
 

mathisfun

New member
Aug 6, 2014
11
What did you get when integrating?
That f(x) is greater than zero, but that seems to be contradictory as a negative function may also have positive derivative. :(
 

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,623
St. Augustine, FL.
Let's begin with:

\(\displaystyle 0<f^{\prime\prime}(x)\)

And so, on \(\displaystyle [1,k]\) where \(\displaystyle 1<k\in\mathbb{R}\) there must be some constant $C$ such that:

\(\displaystyle C<f^{\prime}(x)\)

What happens if you integrate again?
 

mathisfun

New member
Aug 6, 2014
11
Let's begin with:

\(\displaystyle 0<f^{\prime\prime}(x)\)

And so, on \(\displaystyle [1,k]\) where \(\displaystyle 1<k\in\mathbb{R}\) there must be some constant $C$ such that:

\(\displaystyle C<f^{\prime}(x)\)

What happens if you integrate again?
If f1(X)>f2(x)
doesn't implies that the indefinite integration of f1(x) will be greater than f2(x). This is only applicable when both sides are integrated within limits and thus when integrating from 0 to x, constant gets eliminated.
 

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,623
St. Augustine, FL.
Let's try another approach...you mentioned that if the function is concave up, then it must lie beneath its secant line, or:

\(\displaystyle f(x)<\frac{f(b)-f(a)}{b-a}(x-a)+f(a)\)

What do you find upon integrating both sides from $a$ to $b$?