# alternative proof of the fundamental theorem of calculus

An alternative proof for the first part involves the use of a formula derived by the method of exhaustion:

 $\int_{a}^{b}f(t)dt=\left(b-a\right)\sum_{n=1}^{\infty}\sum_{m=1}^{2^{n}-1}% \left(-1\right)^{m+1}2^{-n}f\left(a+m(b-a)/2^{n}\right).$

Given that

 $F(x)=\int_{a}^{x}f(t)dt,$

and

 $F^{\prime}(x)=\lim_{\Delta x\rightarrow 0}\frac{F(x+\Delta x)-F(x)}{\Delta x}=% \lim_{\Delta x\rightarrow 0}\frac{1}{\Delta x}\int_{x}^{x+\Delta x}f(t)dt,$

 $F^{\prime}(x)=\lim_{\Delta x\rightarrow 0}\frac{(x+\Delta x-x)}{\Delta x}\sum_% {n=1}^{\infty}\sum_{m=1}^{2^{n}-1}\left(-1\right)^{m+1}2^{-n}f\left(x+m\Delta x% /2^{n}\right),$

or

 $F^{\prime}(x)=\sum_{n=1}^{\infty}\sum_{m=1}^{2^{n}-1}\left(-1\right)^{m+1}2^{-% n}f\left(x\right).$

Since it can be shown that

 $\sum_{n=1}^{\infty}\sum_{m=1}^{2^{n}-1}\left(-1\right)^{m+1}2^{-n}=\sum_{n=1}^% {\infty}2^{-n}=1,$

It follows that

 $F^{\prime}(x)=f(x).$

The second part of the proof is identical to the parent.

Title alternative proof of the fundamental theorem of calculus AlternativeProofOfTheFundamentalTheoremOfCalculus 2013-03-22 15:55:24 2013-03-22 15:55:24 ruffa (7723) ruffa (7723) 4 ruffa (7723) Proof msc 26-00