Is Integral Equality Paradox Possible in Mathematics?

[Klaus_Hoffmann]

if we have that:

$$\int_{a}^{b}dxf(x) - \int_{a}^{b}dxg(x)= \int_{a}^{b}dx(f(x)-g(x))$$

where the integral over (a,b) of f(x) and g(x) exist separately then my question is if

$$\int_{a}^{b}dx(f(x)-g(x)) =0$$ then

does this imply necessarily that $$f(x)=g(x)+h'(x)$$

where h(a)=h(b)=0 and its derivative is 0 almost everywhere on the interval (a,b)

 

[Gib Z]

We have that $$\int^b_a f(x) dx = \int^b_a g(x) dx$$ and from that condition you ask if that implies $f(x) = g(x) \in [b,a]$(Neglecting small value of h'(x) you proposed, i know don't why you did, though it doesn't end up making a difference). As a simple counter example, f(x) and g(x) may be very different Odd functions, and the integral of these two between b and -b will be 0.

 

[VietDao29]

if we have that:

$$\int_{a}^{b}dxf(x) - \int_{a}^{b}dxg(x)= \int_{a}^{b}dx(f(x)-g(x))$$

where the integral over (a,b) of f(x) and g(x) exist separately then my question is if

$$\int_{a}^{b}dx(f(x)-g(x)) =0$$ then

does this imply necessarily that $$f(x)=g(x)+h'(x)$$

where h(a)=h(b)=0

No, this is not correct, you just need that h(a) = h(b), since:

$$\int_a ^ b f(x) dx - \int_a ^ b g(x) dx = \int_a ^ b \left( f(x) - g(x) \right) dx = \int_a ^ b \left( g(x) + h'(x) - g(x) \right) dx$$

$$= \int_a ^ b h'(x) dx = \left. h(x) \right|_a ^ b = h(b) - h(a)$$

Now, if h(a) = h(b), then we'll have: $$\int_a ^ b f(x) dx - \int_a ^ b g(x) dx = 0$$

and its derivative is 0 almost everywhere on the interval (a,b)

No, why should it's derivative is 0? =.="

 

[Gib Z]

Are you sure that worked out right VietDao29 >.<? He asks if $$\int^b_a f(x) dx = \int^b_a g(x) dx$$ then does that imply f(x) = g(x) + h'(x), where h(a)=h(b)=0. It seems from your post you assumed f(x)=g(x) + h'(x), to get that all we need is h(a)=h(b), but that is somewhat the converse of what he is asking I think >.<

 

[DeadWolfe]

He also said h'=o a.e.

 

[VietDao29]

Are you sure that worked out right VietDao29 >.<? He asks if $$\int^b_a f(x) dx = \int^b_a g(x) dx$$ then does that imply f(x) = g(x) + h'(x), where h(a)=h(b)=0. It seems from your post you assumed f(x)=g(x) + h'(x), to get that all we need is h(a)=h(b), but that is somewhat the converse of what he is asking I think >.<

I was pointing out that his claim that h(a) = h(b) = 0 is wrong, and it's even wronger, when he said that h' = 0.

You can find h'(x) simply by subtracting f(x) from g(x).

Say, we have:
$$\int_0 ^ \frac{\pi}{2} \sin x = \int_0 ^ \frac{\pi}{2} \cos x = 1$$

We also have:
$$\sin x = \cos x + (\sin x - \cos x)$$

So h'(x) = sin(x) - cos(x)
Integrate it with respect to x, we have:
h(x) = -cos(x) - sin(x) + C

$$h(0) = -1 + C$$

$$h \left( \frac{\pi}{2} \right) = -1 + C$$

$$\Rightarrow h(0) = h \left( \frac{\pi}{2} \right)$$ (but not equal to 0, unless C = 1)

However, h(0) = h(pi/ 2) does not imply that h'(x) = 0 on the interval (0; pi/2).

So, in conclusion, both of his assertions (?, can I use plural here) are wrong. =.="

 

[Gib Z]

Ok we'll good we agree on that :) Pretty pointless though, "Klaus" never comes back to his threads when he's proven wrong, and matt grime reckons its some other guy named Jose whose not good >.<

 

[Klaus_Hoffmann]

ok,.. thank you i fooled myself once again.

 

[VietDao29]

Ok we'll good we agree on that :) Pretty pointless though, "Klaus" never comes back to his threads when he's proven wrong, and matt grime reckons its some other guy named Jose whose not good >.<

Well, he did come back. ^.^

 

[DeadWolfe]

If he is Jose he's calmed down a lot. Which is annoying actually, because jose was crazy enough to be funny, where as this person is simply obnoxious.

 

[VietDao29]

If he is Jose he's calmed down a lot. Which is annoying actually, because jose was crazy enough to be funny, where as this person is simply obnoxious.

Well, it must be some of my missing days here, at PF, but who's the well-known Jose that you guys are talking about? I don't think I know anyone named Jose. And, btw, what has he done to make him this famous?

 

[DeadWolfe]

Well his username was eljose. You can search for posts by him. Let's just say he was a genius on the level of gauss, and solved multiple millennium prize problems but the mathematical community kept holding him down because they didn't appreciate proper mathematics.

 

[matt grime]

The snobbish mathematical community, I think you'll find, who only let famous people publish in journals.