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cmkluza

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Thanks for any help!

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- Thread starter cmkluza
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- #1

cmkluza

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Thanks for any help!

- #2

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What does the fundamental theorem of calculus tells you?

- #3

cmkluza

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Fundamental theorem of calculus is [tex] \int^B_A f(x) dx = F(B) - F(A)[/tex] right? So, ##\int^B_A f(x) dx = 0 = F(B) - F(A)##? This suggests that F(B) = F(A), but doesn't tell me anything about ##f(x) = 0##, right? What am I missing?What does the fundamental theorem of calculus tells you?

- #4

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Thanks for any help!

Well, it's not true. We have

[tex]\int_0^{2\pi} \sin(x) = 0[/tex]

but the sine is not the zero function.

- #5

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Since you haven't made any assumptions on the domain or initial values I assumed it would be valid to all of them.Fundamental theorem of calculus is [tex] \int^B_A f(x) dx = F(B) - F(A)[/tex] right? So, ##\int^B_A f(x) dx = 0 = F(B) - F(A)##? This suggests that F(B) = F(A), but doesn't tell me anything about ##f(x) = 0##, right? What am I missing?

- #6

cmkluza

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In that case, could you help me explain this in a specific situation? I'm looking at this resource for my research, specifically the end of page 6 and start of page 7. It shows equation 11, which I've been able to follow up to, and then it has a paragraph explaining the logic it uses to produce equation 13, which is just equation 11 without the integral. Do you have any idea about why this is possible? If this isn't correct in general, why would it be correct in this situation?Well, it's not true. We have

[tex]\int_0^{2\pi} \sin(x) = 0[/tex]

but the sine is not the zero function.

Thanks!

- #7

- #8

cmkluza

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I'm sorry, I don't think I'm understanding this at a fundamental level yet. Could you tell me why that statement is true on a more conceptual level? That might allow me to understand the proof using the fundamental theorem.OK, so the statement is: Let ##f## be continuous on ##[a,b]## and such that for all ##A,B\in [a,b]## holds that ##\int_A^B f(x)dx=0##, then ##f=0##. This is true indeed. And this seems to be how @fresh_42 interpreted it. His proof using the fundamental theorem should work.

- #9

pasmith

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Let [itex]f[/itex] be continuous on [itex][a,b][/itex] such that [itex]\int_A^B f(x)\,dx = 0[/itex] for every [itex]a \leq A < B \leq b[/itex].

We show, by contradiction, that under these conditions it cannot be the case that [itex]f(c) \neq 0[/itex] for any [itex]c \in (a,b)[/itex]:

Suppose [itex]c \in (a,b)[/itex] is such that [itex]f(c) > 0[/itex]. Then, by continuity of [itex]f[/itex] at [itex]c[/itex], there exists a [itex]\delta > 0[/itex] such that if [itex]A=\max\{a, c - \delta\} < x < \min\{b, c + \delta\}=B[/itex] then [itex]f(x) > 0[/itex]. But then by a basic property of integrals we have

[tex]\int_A^B f(x)\,dx > \int_A^B 0\,dx = 0,[/tex] which is a contradiction. Thus such a [itex]c[/itex] cannot exist. A similar argument shows that there cannot exist any [itex]c \in (a,b)[/itex] such that [itex]f(c) < 0[/itex].

Hence [itex]f[/itex] is constantly zero on [itex](a,b)[/itex] and by continuity [itex]f(a) = f(b) = 0[/itex] also.

This is a more "conceptual" proof than one using the fundamental theorem, and one can rephrase it thus:

"[itex]\int_a^b \frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x}\,dx = 0[/itex] says only that, on average, cars are neither created nor destroyed on the stretch of highway between [itex]a[/itex] and [itex]b[/itex]. But suppose there exists a [itex]c \in [a,b][/itex] where [itex]\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} \neq 0[/itex]. Then, if [itex]\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x}[/itex] is continuous, there is around [itex]c[/itex] a stretch of highway where cars are being spontaneously created if [itex]\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} > 0[/itex] or destroyed if [itex]\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} < 0[/itex]. Both of these we reject as unphysical."

We show, by contradiction, that under these conditions it cannot be the case that [itex]f(c) \neq 0[/itex] for any [itex]c \in (a,b)[/itex]:

Suppose [itex]c \in (a,b)[/itex] is such that [itex]f(c) > 0[/itex]. Then, by continuity of [itex]f[/itex] at [itex]c[/itex], there exists a [itex]\delta > 0[/itex] such that if [itex]A=\max\{a, c - \delta\} < x < \min\{b, c + \delta\}=B[/itex] then [itex]f(x) > 0[/itex]. But then by a basic property of integrals we have

[tex]\int_A^B f(x)\,dx > \int_A^B 0\,dx = 0,[/tex] which is a contradiction. Thus such a [itex]c[/itex] cannot exist. A similar argument shows that there cannot exist any [itex]c \in (a,b)[/itex] such that [itex]f(c) < 0[/itex].

Hence [itex]f[/itex] is constantly zero on [itex](a,b)[/itex] and by continuity [itex]f(a) = f(b) = 0[/itex] also.

This is a more "conceptual" proof than one using the fundamental theorem, and one can rephrase it thus:

"[itex]\int_a^b \frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x}\,dx = 0[/itex] says only that, on average, cars are neither created nor destroyed on the stretch of highway between [itex]a[/itex] and [itex]b[/itex]. But suppose there exists a [itex]c \in [a,b][/itex] where [itex]\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} \neq 0[/itex]. Then, if [itex]\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x}[/itex] is continuous, there is around [itex]c[/itex] a stretch of highway where cars are being spontaneously created if [itex]\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} > 0[/itex] or destroyed if [itex]\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} < 0[/itex]. Both of these we reject as unphysical."

Last edited:

- #10

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Conceptually, if f is not zero at a point and you integrate near that point the integral cannot be zero. That's the basic idea that needs to be formalised.I'm sorry, I don't think I'm understanding this at a fundamental level yet. Could you tell me why that statement is true on a more conceptual level? That might allow me to understand the proof using the fundamental theorem.

- #11

Svein

Science Advisor

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There is a theorem that states: if [itex] \int_{L}\left\lvert f \right\rvert=0[/itex] then [itex]f=0 [/itex] a. e. on L.I've just encountered this somewhere and I need some sort of formal proof for why a continuous function f(x)f(x)f(x) can equal zero because its integral is zero.

Explanation: a.e. (almost everywhere) means "except on a subset of measure zero".

- #12

Edgardo

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http://math.stackexchange.com/a/543800

And two more links for the proof of the sign-preserving property:

https://math.la.asu.edu/~dajones/class/371/ch4.pdf

http://zimmer.csufresno.edu/~mnogin/math171spring05/test2-sol.pdf

Intuitively it says the following. Suppose you wanted to construct a continuous function f that is positive at some point a, i.e. f(a) > 0. So you draw a dot somewhere above the x-axis at x=a for its graph. Now, if you draw the graph a little to the left of that dot and to the right of it, then that graph segment lies above the x-axis.

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