The integral of a function ##f(x)## from its graph

[brotherbobby]

Homework Statement

For the graph of the function $f(x)$ given below, suppose $\large{g(x) = \int f(x) \mathrm{d}x}$. Select which of the following are true :$\\[10pt]$

1. $g(x)$ is always positive.
2. $g(x)$ is negative at $C$.
3. The slope of $g(x)$ at $B$ is zero.
4. $g(x)$ is greater at $E$ than at $B$.
5. $g(x)$ is smaller at $C$ than at $F$.
6. $g(x)$ is increasing at $F$.

Relevant Equations

1. The integral $g(x)$ of a function $f(x)$ is its antiderivative : If $g(x) = \int f(x) dx\Rightarrow \dfrac{d}{dx} g(x) = f(x)$.
2. The slope of the integral at a point is, therefore, equal to the derivative of the integrand at that point.
3. The amount of an integral between two points $a$ and $b$ on its graph is equal to the area enclosed by that section of the graph and the $x$ axis.
4. A function $f(x)$ is said to be increasing at a point $x_0$ if $f'(x_0)>0$

Problem statement : I start by putting the graph of (the integrand) $f(x)$ as was given in the problem. Given the function $g(x) = \int f(x) dx$.

Attempt : I argue for or against each statement by putting it down first in blue and my answer in red.

1. $g(x)$ is always positive : The exact value of $g(x)$ cannot be determined because an indefinite integral is undetermined upto a constant term $c$ : $\int f(x) dx = g(x)+c$. Hence, the statement in blue is false. $\text{ }\large[$However, if $c$ is given to be 0, then we can see that the area enclosed by the graph (curve) $f(x)$ is positive from $O\rightarrow F$ (throughout). This is because a greater (positive) area is enclosed by the graph in the region $OAB$ and $DEF$ than the smaller (negative) area it encloses in the region $BCD$. However, since nothing is mentioned about the undetermined constant $c$, no definite value of $g$ can be set against any of the designated points on the graph$\large ]$.$\\[10pt]$
   
2. $g(x)$ is negative at $C$ : This statement is false for the same reason as (1) above. $\\[10pt]$
   
3. The slope of $g(x)$ at $B$ is zero : The slope of $g(x)$ is the drawn function $f(x)$. Since $f(B) = 0$, it implies $\left.\frac{dg}{dx}\right|_{B} = 0$. The statement is true.$\\[10pt]$
   
4. $g(x)$ is greater at $E$ than at $B$ : Despite the undermined constant $c$ in point 1 above, which leaves the value of $g(x)$ for all points unknown, it is possible to compare the values of $g(x)$ for a pair of given points because this constant term vanishes on taking differences. Calling the origin as $O$, we find that the area under the graph $f(x)$ from $O\rightarrow E$ is greater that the area under the graph from $O\rightarrow B$. The integral (value) being an area, the functional value of $g(E)>g(B)$, implying that the statement above is true. $\\[10pt]$.
   
5. $g(x)$ is smaller at $C$ than at $F$ : Let us assume, going by "looks" of the regions, we have the $\text{Area of DEF = ABC > BCD}$. Both at $C$ and at $F$, the area of region $OAB$ comes in, which can be removed when taking differences. The area of the function at $F$ includes the entire (negative) contribution from $BCD$, but that is more than compensated by the area of $DEF$ so that $g(F)>g(C)$. Hence the statement above is true. $\\[10pt]$
   
6. $g(x)$ is increasing at $F$ : The increasing (or decreasing) behaviour of a function is given by the sign of its derivative, as is well known. Here $g(x)$ is the function and $f(x)$ its derivative. We find from the graph above that $f(F)>0$. Hence $g(x)$ must be an increasing function at that point, making the statement true. $\text{ }$ [I must add for reasons of clarity that $f(x)$ it itself a decreasing function at $F$ as $f'(F)<0$. But the statement in question is about its integral $g(x)$, not the function $f(x)$ itself.

I would like some hints or suggestions if possible, specially if I am mistaken in my thinking above.

 

[fresh_42]

I think you mean $g(x)=\displaystyle{\int_0^x}f(t)\,dt$. This way, the expression $g(x)$ at $C$ makes sense, namely $g(C)$.

If this was not meant as $g(x)$ then I'm afraid I do not understand what $x$ is.

 

[FactChecker]

I think you mean $g(x)=\displaystyle{\int_0^x}f(t)\,dt$. This way, the expression $g(x)$ at $C$ makes sense, namely $g(C)$.

If this was not meant as $g(x)$ then I'm afraid I do not understand what $x$ is.

In addition, if the integral is known to start at 0, all the mention of unknown constants is irrelevant, so many of the answers would change.

 

[brotherbobby]

I think you mean $g(x)=\displaystyle{\int_0^x}f(t)\,dt$. This way, the expression $g(x)$ at $C$ makes sense, namely $g(C)$.

If this was not meant as $g(x)$ then I'm afraid I do not understand what $x$ is.

Yes, or more simply still, the indefinite integral : $g(x) = \int f(x) \mathrm{d} x$. However, since we are considering limits and points in the problem, I'd agree with your understanding more : let $g(x) = \int_0^x f(t)dt$

 

[fresh_42]

$g(x)=F(u)$ doesn't make sense, because $x$ doesn't occur on the RHS.

 

[Orodruin]

$g(x)=F(u)$ doesn't make sense, because $x$ doesn't occur on the RHS.

You never saw the bastard notation $\int g(x) dx$ used for the antiderivative? I believe this is what is intended and in this case the constants do make sense.

Edit: Just as an example, WolframAlpha (and I therefore suspect Mathematica too, but on mobile so cannot check) uses this notation.

I am not saying it is a good notation. Just that its being used.

Homework Statement:: For the graph of the function $f(x)$ given below, suppose $\large{g(x) = \int f(x) \mathrm{d}x}$. Select which of the following are true :$\\[10pt]$

1. $g(x)$ is always positive.
2. $g(x)$ is negative at $C$.
3. The slope of $g(x)$ at $B$ is zero.
4. $g(x)$ is greater at $E$ than at $B$.
5. $g(x)$ is smaller at $C$ than at $F$.
6. $g(x)$ is increasing at $F$.
Relevant Equations:: 1. The integral $g(x)$ of a function $f(x)$ is its antiderivative : If $g(x) = \int f(x) dx\Rightarrow \dfrac{d}{dx} g(x) = f(x)$.
2. The slope of the integral at a point is, therefore, equal to the derivative of the integrand at that point.
3. The amount of an integral between two points $a$ and $b$ on its graph is equal to the area enclosed by that section of the graph and the $x$ axis.
4. A function $f(x)$ is said to be increasing at a point $x_0$ if $f'(x_0)>0$

Calling the origin as O, we find that the area under the graph f(x) from O→E is greater that the area under the graph from O→B.

Is it though? Did you compare the negative and positive contributions? Note that the area when $g(x) < 0$ counts as negative.

 

[Office_Shredder]

I agree that 4 is probably false, but also that it's not *that* obvious to me.

 

[Orodruin]

I agree that 4 is probably false, but also that it's not *that* obvious to me.

Here are the relevant areas laid out next to each other. Even with my shaky one hand tracing of the curve (I am impaired by a baby sleeping on my legs), it is quite clear which is the larger one.

 

[fresh_42]

You never saw the bastard notation $\int g(x)dx$ used for the antiderivative? I believe this is what is intended and in this case the constants do make sense.

Sure. But if you write $g(x)=\int f(x)dx$ then this is equivalent to $g(x)=\int f(t)dt$ and now it is obvious, that $x$ on the left hand side is undefined: $g$ depends on $x$ whereas the right hand side does not.

Even if you say that $g(x)$ is the anti derivative of $f(x)$, then $\int f(x)dx= g(x)+C$ and $g(x)=\int f(x)dx$ becomes $g(x)=g(x)+C$ which again makes no sense.

And finally, in case $g(x)+Const.=\int f(x)dx$ was meant the question remains: What is $g$ at $C$? $g(C)+Const.$ doesn't make sense either.

Hence I came to the conclusion that I do not know what is meant in any case, except for what I suggested, namely $g(x)=\int_0^xf(x)dx$, and asked whether my interpretation is correct.

 

[Delta2]

I disagree with you only at 4. That's because g(B)=(area OAB)=positive, while g(E)=(area OAB)+(area BCD)+(area DE). But the thing here is that (area BCD) counts as negative.. and it seems to me from the graph that absolute value of (area BCD)>(area DE)

 

[Orodruin]

Even if you say that g(x) is the anti derivative of f(x), then ∫f(x)dx=g(x)+C and g(x)=∫f(x)dx becomes g(x)=g(x)+C which again makes no sense.

It can be made to make sense by considering $\int f(x) dx$ an equivalence class of functions that differ only by a constant. With the corresponding equivalence relation $\sim$, indeed $g(x) \sim g(x) + C$. Regardless, the choice of 0 as the lower boundary is arbitrary and not stated.

 

[Office_Shredder]

@brotherbobby do you feel like you understand the explanations for part 4? I feel like the second half of this thread is probably not helpful, maybe we should just ignore it.

 

[brotherbobby]

@Office_Shredder - I find that I was wrong with No.4. I am the creator of this thread, or the OP as you call it. The question(s) came from a website from where I have since verified my answers.

Yes only (4) was incorrect - the integral $g(x)$ is not higher at E than it is at B; in fact it is lower. That is, to make it formal, $\boxed{g(B)>g(E)}$. This is because the area of the graph (of integrand $f(x)$) and the x-axis is negative on the whole if you move from B to E. If one wonders why so, it is because the area under DE "looks" clearly less than the area from B to D, which is negative.

Thank you for your time and patience.

 

[Orodruin]

The question(s) came from a website from where I have since verified my answers.

Just out of interest, would you mind sharing the link?

 

[brotherbobby]

Just out of interest, would you mind sharing the link?

Yes. You might have to register for free. Please browse by clicking on "NEXT" until you arrive at the page titled CALCULUS. (You have to click thrice to go to the third page).
Go down along the page to the question titled "Derivatives and Anti-Detivatives".

Link

 

[jedishrfu]

In the interests of clarity, I have removed some posts related to the finer points of Calculus notation that may detract from the OP's original questions. My apologies to @fresh_42 and @Orodruin if I have gone a post too far.

Some of your discussion reminded me of the chasm between a mathematicians need for exactness in notation and a physicist's need to solve problems.

Feynman once summed this up in a Messenger Seminar about a physicist asking a mathematician for help on multi-dimensional math. The mathematician offers N-dimensional math and the physicist says he only needs 3 dimensions but after a great deal of work sheepishly returns to ask about handling additional dimensions.

https://www.cantorsparadise.com/ric...-between-mathematics-and-physics-c0847e8a3d75

Closing this thread now, thank you all for posting here.

Jedi