Integrating an inequality for two functions of the same variable

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Homework Statement
Please see below
Relevant Equations
Please see below
For this true or false problem,
1717372927804.png

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!
 
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  • #2
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)##.
 
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  • #3
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!
 
  • #4
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.
 
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  • #5
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!
 
  • #6
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).##
 
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  • #7
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).
 
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  • #8
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.
 
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  • #9
##\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.
 
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  • #10
The following inequality is true: if [itex]f(x) \leq g(x)[/itex] on [itex][a,b][/itex] then [tex]
\int_a^b f(x)\,dx \leq \int_a^b g(x)\,dx.[/tex] (We can replace the upper limits with an aarbitrary [itex]t \in [a,b][/itex].) Replacing [itex]f[/itex] and [itex]g[/itex] by derivatives and using the fundamental theorem, this implies [tex]
\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).[/tex] What this is saying is that if [itex]f[/itex] increases more slowly than [itex]g[/itex] and they start in the same place, then [itex]f[/itex] will be less than [itex]g[/itex].
 
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