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:

$${\int }_a^{b}f(t)𝑑t=\left(b-a\right)\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{2^n-1}{\left(-1\right)}^{m+1}{2}^{-n}f\left(a+m(b-a)/2^n\right).$$

Given that

$$F(x)={\int }_a^{x}f(t)𝑑t,$$

and

$$F^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{F(x+\mathrm{\Delta }x)-F(x)}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{1}{\mathrm{\Delta }x}{\int }_x^{x+\mathrm{\Delta }x}f(t)𝑑t,$$

the above formula leads to:

$$F^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{(x+\mathrm{\Delta }x-x)}{\mathrm{\Delta }x}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{2^n-1}{\left(-1\right)}^{m+1}{2}^{-n}f\left(x+m\mathrm{\Delta }x/2^n\right),$$

or

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

Since it can be shown that

$$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{2^n-1}{\left(-1\right)}^{m+1}{2}^{-n}=\sum _{n=1}^{\mathrm{\infty }}{2}^{-n}=1,$$

It follows that

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

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