Max area

Rubén

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Someone can help me solve a doubt, in this exercise supposing that the function that limits more area is the line y=6, and the minimum area limits is y=3, the results coincide, however I do not understand why these two functions are valid, since they do not fulfill the condition of max=6 and min=3 simultaneously.
(I have done it integrating y=6 between x=2 and x=4 for the maximum area y
integrating y=3 between x=2 and x=4 for the minimum area.)
Thank you.
 
View attachment 10624
Someone can help me solve a doubt, in this exercise supposing that the function that limits more area is the line y=6, and the minimum area limits is y=3, the results coincide, however I do not understand why these two functions are valid, since they do not fulfill the condition of max=6 and min=3 simultaneously.
(I have done it integrating y=6 between x=2 and x=4 for the maximum area y
integrating y=3 between x=2 and x=4 for the minimum area.)
Thank you.

\(\displaystyle \text{over the interval }[2,4],~f(x) \leq 6\Rightarrow \int_2^4 ~f(x)~dx \leq \int_2^4 ~6~dx = 12\)

\(\displaystyle \text{over the interval }[2,4],~f(x) \geq 3 \Rightarrow \int_2^4 ~f(x)~dx \geq \int_2^4 ~3~dx =6 \)
 
View attachment 10624
Someone can help me solve a doubt, in this exercise supposing that the function that limits more area is the line y=6, and the minimum area limits is y=3, the results coincide, however I do not understand why these two functions are valid, since they do not fulfill the condition of max=6 and min=3 simultaneously.
(I have done it integrating y=6 between x=2 and x=4 for the maximum area y
integrating y=3 between x=2 and x=4 for the minimum area.)
Thank you.
Please read

https://www.freemathhelp.com/forum/threads/112086-Guidelines-Summary?p=433156&viewfull=1#post433156

I may misunderstand what you have done. You have no function defined so what in the world did you integrate?

Furthermore, I may misunderstand the question, but what I see so far is that

\(\displaystyle 2 \le x \le 4 \implies 3 \le f(x) \le 6.\)

\(\displaystyle a = \text { the area bounded by the x-axis, } f(x), \text { and the lines } x = 2 \text { and } x = 4. \)

In that case, you are asked to find \(\displaystyle p < a < q.\)
 
View attachment 10624
Someone can help me solve a doubt, in this exercise supposing that the function that limits more area is the line y=6, and the minimum area limits is y=3, the results coincide, however I do not understand why these two functions are valid, since they do not fulfill the condition of max=6 and min=3 simultaneously.
(I have done it integrating y=6 between x=2 and x=4 for the maximum area y
integrating y=3 between x=2 and x=4 for the minimum area.)
Thank you.

I think you are saying that if we take the special case of f(x) = 6, you would get that maximum area of 12, but since that function does not have a minimum of 3, it does not fit the conditions stated for f; so you think there might be some lower upper bound on the area. (And similarly for the minimum.)

Imagine a function that is equal to 6 everywhere except at one point, say x=3, where f(3) = 3. This function will have the required maximum and minimum, but the area will still be 12. Or, if you want to work only with continuous functions, imagine a function that rapidly falls to 3 and then rises back to 6; it might yield an area that is as close as you like to 12.

So we can't set any upper bound on the area less than 12.

And this is what Romsek's inequalities show.
 
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