Question on a particular integral property

["Don't panic!"]

I've been reading Wald's book on general relativity and in one of the questions at the end of chapter 2 he gives a hint which says to make use the following integral identity (for a smooth function in): F(x)=F(a)+\int_{0}^{1}F'(t(x-a)+a)dt
Is this result true simply because \int_{0}^{1}F'(t(x-a)+a)dt=F(t(x-a)+a)\bigg\vert_{t=0}^{t=1}=F(x)-F(a)
And so F(x)=F(a)+\int_{0}^{1}F'(t(x-a)+a)dt=F(a)+\left(F(x)-F(a)\right)
Or is there a deeper explanation to it?
I get that one can arrive at this expression by starting from the fundamental theorem of calculus and making a change of variables, but I'm unsure of what he's trying to say with it?!

 

[HallsofIvy]

Nothing deeper- that's basically why it is true. As to "what he's trying to say with it", I suspect that depends on what he is try to do.

 

[Hawkeye18]

I should mention sloppy notation here. The formula is true if $F'(t(x-a) + a)$ is understood as derivative of the function $\phi$, $\phi(t) = F(t(x-a) + a)$ (evaluated at $t$). However, in the standard notation $F'(t(x-a) + a)$ means the derivative of $F$ evaluated at the point $t(x-a) + a$ , and in this case the formula is false (consider say $F(x) =x$, $a=0$).

 

["Don't panic!"]

Yes, I admit it is a bit sloppy, I was just copying over the notation that Wald gives it in. I'm assuming he meant F'(t(x-a) +a)=\frac{dF(t(x-a)+a)}{dt}?!

He gives it in a question in which he asks the reader to prove by induction that for any smooth function F : \mathbb{R} ^{n} \rightarrow \mathbb{R} ^{n} one can express it as the following power series (i. e. a Taylor series) F(x) =F(a) +\sum_{i=1}^{\infty}(x^{i}-a^{i})H_{i}(x), where x=(x^{1},\ldots,x^{n}) and a=(a^{1},\ldots,a^{n}), and in particular, \frac{\partial F} {\partial x^{i}} \bigg\vert_{x=a} =H_{i} (a). He gives a hint that for n=1, you should use the "known" identity I gave in my first post. I can do the first step in the induction process, by starting from the fundamental theorem of calculus and making a change of variables, but I'm not entirely sure what this is showing. I'm also unsure how to proceed after making the induction step?!

 

[Hawkeye18]

You just should apply the formula $$\phi(1) = \phi(0) +\int_0^1 \phi'(t) dt$$ (the fundamental theorem of calculus) to $$\phi(t) = F(a+(x-a)t). $$