improving the FTC proof

[cbm755]

Screenshot from CLP-2 on FTC:

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"But since F'(x) = f(x)": this seems to put the cart before the horse.

They way I've always done this is by referencing the Integral Mean Value Theorem (not the regular MVT), which states that there exists some $c$ such that $\int_{x}^{x+h} f(t) dt = f(c) (x + h - x)$. Then your argument about $c$ being squeezed works.

[cbm755]

To clarify (and since I've just learning that GitHub renders $\LaTeX$!): $F'(x) = \lim_{h \to 0} \frac{F(x + h) - F(x)}{h}$. Then take $F(x) = \inta^x f(t) \textrm{d} t$. So $F'(x) = \lim{h \to 0} \frac{\int{x}^{x+h} f(t) \textrm{d} t}{h}$. Invoking the IMVT as I indicate above, we have: $$F'(x) = \lim{h \to 0} \frac{f'(c)h}{h} = f'(c).$$

[joelfeldman]

The problem with this is that the IMVT has not been covered at this point. Furthermore the proof of the IMVT in CLP-2 uses the FTC. I have eliminated this footnote completely.