Integrating an inequality for two functions of the same variable

[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!

 

[renormalize]

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]

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!

 

[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.

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]

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!

 

[fresh_42]

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 $(a,x).$

 

[FactChecker]

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).

 

[fresh_42]

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.

 

[WWGD]

$\int (f-g)'>c> 0$
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.

 

[pasmith]

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 $g$.