# Is this true about integrals?

Hertz
Sorry, the title doesn't match up 100% with the content of the topic, but that's because I've decided to be a little bit more explicit about my question.

I am trying to walk through the proof of Euler's Equation from Calculus of Variations, and I'm a little bit confused by the final step.

Right now I have this:
$\int^{x_{2}}_{x_{1}}{(\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'})\eta(x) dx}=0$

and then they proceed to say that this implies:
$\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}=0$

Could someone please explain why this is a correct generalization? Isn't the left side of Euler's Equation really just a function of x? And since eta is also a function of x, couldn't their product technically be an odd function and couldn't the interval technically be a symmetric interval? Wouldn't this mean that the left side of Euler's Equation does not necessarily have to equal zero in very rare circumstances?

[Original Post]
If
∫F(x)G(x)dx=0
and F(x) is arbitrary
Does it imply that
G(x)=0
?

Last edited:

Staff Emeritus
$$(x_0-\epsilon,x_0+\epsilon)$$
$$\int_{x_1}^{x_2} F(x) G(x) dx > 0$$
The way you have stated it, it is NOT true. If, however, f(t) is continuous and $\int_{x_1}^{x_2} f(t)dt= 0$ for all $x_1$ and $x_2$ in a given interval, then it must be true that f(t)= 0 in that interval. You can prove that by contradiction: if there exist $x_0$ such that $f(x_0)\ne 0$ then there exist some interval around $x_0$ on which $f(x)$ f(x) is always positive (or always negative) and integrating over that interval will give a non-zero result.