Integrating an inequality for two functions of the same variable

• member 731016
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For this true or false problem,

The answer if false, however, I am confused by this result as my working shows that the is true.

My working is, integrating both sides of the inequality we get ##\int f'(x) dx> \int g'(x) dx## for all ##x \in (a,b)## which is the equivalent to ##f(x) > g(x)## for all ##x \in (a,b)##.

Does someone please know what I have done wrong?

Thanks!

ChiralSuperfields said:
For this true or false problem,
View attachment 346371
The answer if false, however, I am confused by this result as my working shows that the is true.
Counterexample: ##f(x)=-e^{-x}## and ##g(x)=1##. Then for all ##x##: ##f'(x)>g'(x)## but ##f(x)<g(x)##.

member 731016
renormalize said:
Counterexample: ##f(x)=-e^{-x}## and ##g(x)=1##. Then for all ##x##: ##f'(x)>g'(x)## but ##f(x)<g(x)##.
Thank you for your reply @renormalize!

Sorry I am tying to generalize this without considering specific counter examples. I am trying to consider the most general case, this is why I integrate.

Thanks!

ChiralSuperfields said:
Sorry I am tying to generalize this without considering specific counter examples. I am trying to consider the most general case, this is why I integrate.
But this single counterexample is sufficient to show that your "general case" cannot be proven. Hence statement 6 is False, with no further work needed.

member 731016
renormalize said:
But this single counterexample is sufficient to show that your "general case" cannot be proven. Hence statement 6 is False, with no further work needed.
Thank you for your reply @renormalize !

Sorry I am still confused. I integrated the inequality and obeyed the laws of algebra and my integration still worked. I'm not sure what is going on, I think maybe the question is wrong.

Thanks!

ChiralSuperfields said:
Thank you for your reply @renormalize!

Sorry I am tying to generalize this without considering specific counter examples. I am trying to consider the most general case, this is why I integrate.

Thanks!
You can take the counterexample and put it in your chain of arguments to see at which step it fails. You can also use - which is what I do first - the fundamental theorem of calculus, either for the entire interval ##(a,b)## or for ##(a,x).##

member 731016
ChiralSuperfields said:
I'm not sure what is going on, I think maybe the question is wrong.
Don't forget about that sneaky constant that you should add when you evaluate an antiderivative.
If ##f' > g'## on (a,b), then what can you say about comparing ##\int {f'} + c_1## to ##\int {g'} + c_2##?
In other words, does the slope of ##f## being larger mean that the value of ##f## will be larger? Compare f(x)=2x with g(x)=x+1000 on (0,1).

Mark44, member 731016 and fresh_42
FactChecker said:
Don't forget about that sneaky constant that you should add when you evaluate an antiderivative.
If ##f' > g'## on (a,b), then what can you say about comparing ##\int {f'} + c_1## to ##\int {g'} + c_2##?
In other words, does the slope of ##f## being larger mean that the value of ##f## will be larger? Compare f(x)=2x with g(x)=x+1000 on (0,1).
That's why I like FTC as a first test. It forces you to use the constants and can help to see where a general argument might fail.

member 731016 and FactChecker
##\int (f-g)'>c> 0 ##
It would work if we had ##\int f'-g' =0 ##, with an equality at some value ##x,## so that ##f(x)=g(x)##, i.e., if they're equal at some point and grow at the same rate . Like Fact Checker mentioned, said, one just grows faster than the other, but you don't have additional knowledge of initial conditions. This is a sort of PDE without additional conditions and an inequality.

Last edited:
member 731016 and FactChecker
The following inequality is true: if $f(x) \leq g(x)$ on $[a,b]$ then $$\int_a^b f(x)\,dx \leq \int_a^b g(x)\,dx.$$ (We can replace the upper limits with an aarbitrary $t \in [a,b]$.) Replacing $f$ and $g$ by derivatives and using the fundamental theorem, this implies $$\forall x \in [a,b] : f'(x) \leq g'(x) \quad \Rightarrow \quad \forall x \in [a,b] : f(x) - f(a) \leq g(x) - g(a).$$ What this is saying is that if $f$ increases more slowly than $g$ and they start in the same place, then $f$ will be less than $g$.

member 731016 and FactChecker

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