true-false-determine-whether-the-statement-is-true-or-false-explain-your-answer-in-each-exercise-a-4

Question

True-False Determine whether the statement is true or false. Explain your answer. [In each exercise, assume that $f$
and $g$ are distinct continuous functions on $[a, b]$ and that $A$ de-
notes the area of the region bounded by the graphs of $y=f(x),$
$y=g(x), x=a, 	ext { and } x=b .]$

$$
	ext { If } \quad A=\left|\int_{a}^{b}[f(x)-g(x)] d xight|
$$
$$
\begin{array}{l}{	ext { then the graphs of } y=f(x) 	ext { and } y=g(x) 	ext { don't cross on }} \ {[a, b] .}\end{array}
$$

Gregory Higby · Accepted Answer

um, I&#x27;m gonna say true in this problem and the reason why music true is the location of the absolute value. Um, because they tell you the area and since they put the absolute value on the outside of the integral from A to B, if I forget what it is, you have X minus f of x dx. Um, then basically, what this boils down to is if this were not the area and they did so I&#x27;m gonna prove this kind of a contra positive. If they did intersex somewhere on that interval, then some of the area would cancel out with some of the area over here. Um, so if if that happens, if this happens than the absolute value, would Onley equal this portion this white space that I haven&#x27;t really covered, um, which the problem tells you that that&#x27;s not true. The area. So basically, what it boils down to is the only way for this statement to be true is if they don&#x27;t intersect in there. There&#x27;s no canceling out, more or less. Um, now, I&#x27;m kind of has intend to say, commit completely to true, because are they counting the endpoints? Um, if I can go back and look at it. I feel like it is. It is unclear because they use closed brackets, but I think it&#x27;s acceptable. And that&#x27;s where your teacher your book might have. A different answer is if they include A and B, they can intersect it at those end points. And, um, this could still be a true statement. So I&#x27;m committing the true, but I&#x27;m willing to listen to anybody that says it&#x27;s false.

Gregory Higby · Answer

So the best way I could explain this one is Ah, the only way as honest ever, all dissatisfied, everything is satisfied. Um, I mean, if I may have switched everything around, But the only way for these this toe happen is ah for the functions to cross each other at least once. Um, so I know I just said the answer, but if you have a graph of f of X, let&#x27;s say something like this and it never crosses G of X will say something like this. Then there&#x27;s definitely an area between them, So this would be, uh, kind of approved by contra positive. So what I&#x27;m getting at is wolf. They&#x27;re the same. Obviously there&#x27;s no area, but they intersect everywhere. But if you have, let&#x27;s say something like this and then something like this, the area above this one would have to cancel out the area, this one. But then you could see a okay, Well, they intersected. So that&#x27;s why you have areas that cancel hum. So hopefully you like my explanation as to why it&#x27;s true.

Gregory Higby · Answer

I do like this problem, cause, uh, you can actually work this out. Like, if you understand that they differ by a positive constant. Um, that means you could just, like, add a value to it. So if you were to do the integral off ah, from a to B of f of X minus, I don&#x27;t think so. It will do f of X minus g of X from the X um and you&#x27;re too. Then replace FX with that. So let&#x27;s see. In a role from A to B G of X plus C minus G of X, let me I just replaced affects with G of X plus e d. X. So then the G of X could cancel. So what this means is this the integral of C X because that CNN derivative of sea from A to B and as you plug in NBC B minus C A. And you can factor out a C in this, um, so very lovely problem that should hopefully you like my work to explain why it is true. I like it

Gregory Higby · Answer

depending on the author&#x27;s intentions here, I would say that this statement is false. Um, and the reason why is let&#x27;s say you had some function that, um maybe, you know, look, something like this on, like, this is your ex values a And this is your xlu be. And you accidentally did this about this function minus this function, then you&#x27;ll definitely get a negative answer. And so you could then say OK, well, if the area I got was negative, something went wrong. So because area needs to be positive, you take the absolute value. You get the right answer. Now, this example I just showed you would make it a true statement because the example they give us negative three. If you have value that, absolutely, you get positive. Three. The issue, though, and why I&#x27;m committing my answer to false is what if the function looks something more like this trying to make it kind of obnoxious. Here where, um, you did this function minus this? Um, the issue with that statement go is some of this, like this area would cancel out with some of this area. Um, so, like, right here, like the white space I have leftover might be an area of negative three. Um, so just absolute valuing. That is not going to be good enough. Um, is ah, incorrect. And hopefully have explained this in a good enough way. Um, I don&#x27;t confuse you any more if I have, um, I would direct you to a teacher. Another student if, uh, if you still have questions.

Gregory Higby · Answer

you have to be careful with this type of problem, because area in the context of this problem must be positive. Um, so in this case, I would say that this is sometimes true. And the reason why this is not always the case is because Ah, well, what if you had a graph on his drawing graph that looks like this and, Ah, this is f of X. This one right here and this one is g of X. Well, if you do FX minus G of X, you&#x27;re gonna get a negative answer. Um, so that&#x27;s where if you have the answer key, you can see Okay, Well, it is true if FX is the larger function, um, so that bounds aren&#x27;t What&#x27;s wrong? The answer would be either this or if G of X is the upper function like it is in this picture. The answer be the in a row from a to B of G of X minus affects the X go

Tyler Gaona · Answer

true, sometimes true are never true area of the region. Between the grafts of these two continuous functions and the lines X equals A and X equals B is always given by this animal. Well, seeing as how in many of our videos, this is what we had set up to find such an area, it seems plausible that it&#x27;s always true, so we can at least draw, you know, one example where it is true. So let&#x27;s see what would be a situation in which this does give you the correct area. Well, let&#x27;s say here is my function f of X and here is my function G of X. Since F is always above G on this interval, this does give us the correct area. But then maybe this leads us to think of an example where it&#x27;s not necessarily true. Slummy draw another one. So basically, I mean, imagine what happens if F was not always above G along this interval. So if we had maybe some point in between where they intersect and F goes below. So we saw examples like this in the previous problem, and now, in this case, it&#x27;s no longer true that This is the correct expression for the area between the two curves right to the left he would have f is above G into the right of this point. You would have G is above F. So you have to split this up into two in a girls, depending on which functions on top and which is on bottom. So the answer to the question is that the statement is sometimes true. Here&#x27;s an example where it is true. Here&#x27;s an example where it isn&#x27;t.

Problem

Estimate the value of $k(0<k<1)$ so that the regi…

Problem 30 Medium Difficulty

Answer

$$
\text { True }
$$

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Howard Anton, Irl Bivens, Stephen Davis

Calculus Early Transcendentals

Grace H.

Kristen K.

Samuel H.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

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$$\text { True }$$

Calculus Early Transcendentals

Grace H.

Kristen K.

Samuel H.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Howard Anton, Irl Bivens, Stephen DavisCalculus Early Transcendentals

Grace H.

Kristen K.

Samuel H.

Joseph L.

$$
\text { True }
$$

Howard Anton, Irl Bivens, Stephen Davis

Calculus Early Transcendentals