- #1
"Don't panic!"
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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): [tex]F(x)=F(a)+\int_{0}^{1}F'(t(x-a)+a)dt[/tex]
Is this result true simply because [tex]\int_{0}^{1}F'(t(x-a)+a)dt=F(t(x-a)+a)\bigg\vert_{t=0}^{t=1}=F(x)-F(a)[/tex]
And so [tex]F(x)=F(a)+\int_{0}^{1}F'(t(x-a)+a)dt=F(a)+\left(F(x)-F(a)\right)[/tex]
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?!
Is this result true simply because [tex]\int_{0}^{1}F'(t(x-a)+a)dt=F(t(x-a)+a)\bigg\vert_{t=0}^{t=1}=F(x)-F(a)[/tex]
And so [tex]F(x)=F(a)+\int_{0}^{1}F'(t(x-a)+a)dt=F(a)+\left(F(x)-F(a)\right)[/tex]
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?!