1. Jul 10, 2007

### 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)

2. Jul 10, 2007

### 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.

3. Jul 10, 2007

### VietDao29

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: $$\int_a ^ b f(x) dx - \int_a ^ b g(x) dx = 0$$

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

4. Jul 10, 2007

### 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 >.<

5. Jul 11, 2007

He also said h'=o a.e.

6. Jul 11, 2007

### VietDao29

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. =.="

Last edited: Jul 11, 2007
7. Jul 11, 2007

### 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 >.<

8. Jul 11, 2007

### Klaus_Hoffmann

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

9. Jul 11, 2007

### VietDao29

Well, he did come back. ^.^

10. Jul 11, 2007

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.

11. Jul 11, 2007

### VietDao29

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?

12. Jul 11, 2007