Would it be a legitimate (valid) proof to use an [itex]\epsilon[/itex]-[itex]\delta[/itex] limit approach to prove the fundamental theorem of calculus?(adsbygoogle = window.adsbygoogle || []).push({});

i.e. as the FTC states that if [itex]f[/itex] is a continuous function on [itex][a,b][/itex], then we can define a function [itex]F: [a,b]\rightarrow\mathbb{R}[/itex] such that [tex]F(x)=\int_{a}^{x}f(t)dt[/tex]

Then [itex]F[/itex] is differentiable on [itex](a,b)[/itex], and [itex]\forall\;\; x\in (a,b)[/itex],

[tex]F'(x)=f(x)[/tex] Is it possible to prove that this is true by showing that for any given [itex]\varepsilon >0[/itex] we can find a [itex]\delta >0[/itex] such that [tex]\Bigg\vert\frac{F(x+h)-F(x)}{h}-f(x)\Bigg\vert <\varepsilon[/tex] whenever [itex]0<\delta < h[/itex]?

(I've found these set of notes https://math.berkeley.edu/~ogus/Math_1A/lectures/fundamental.pdf that uses this approach and it seems legitimate and easy to follow, but I'm unsure whether it's valid or not)?!

(Also, is it valid to say that [itex]F'(x)=f(x)\Rightarrow F'=f[/itex], i.e. the derivative of the function [itex]F[/itex] is equal to the original function [itex]f[/itex], such that [itex]F(x)=F(a)+\int_{a}^{x}\frac{dF}{dt}dt[/itex]?)

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# Proving the fundamental theorem of calculus using limits

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