FToC Part 2 Proof

Proof

1. Suppose $f$ is cont on $[a,b]$
2. Suppose $\forall x\in [a,b],F(x)={\int }_a^{x}f(t)dt$
3. We WTS: $F$ is Continuous on $[a,b]$ and $F$ diff on $[a,b]$ and $\forall x\in [a,b],F^{\prime }(x)=f(x)$
4. Note that it is sufficient to show $F^{\prime }(x)=f(x)$ as all the other claims follow as Differentiable Implies Continuity Proof, Derivative definition
5. Consider $F^{\prime }(x)={\lim }_{h\to 0}\frac{F(x+h)-F(x)}{h}$ by First Principles
6. $={\lim }_{h\to 0}\frac{{\int }_a^{x+h}f(t)dt-{\int }_a^{x}f(t)dt}{h}$
7. $={\lim }_{h\to 0}\frac{{\int }_a^{x+h}f(t)dt-{\int }_a^{x}f(t)dt}{h}$
8. Note that there are now two cases:
   1. Case 1: Suppose $h\ge 0$
   2. Then, this means that ${\int }_a^{x+h}f(t)dt-{\int }_a^{x}f(t)dt={\int }_x^{x+h}f(x)dt$
   3. Case 2: Suppose $h<0$
   4. Then, this means that ${\int }_a^{x+h}f(t)dt-{\int }_a^{x}f(t)dt={\int }_x^{x+h}f(x)dt$
   5. Note that both equate to the same value
9. Thus, $={\lim }_{h\to 0}\frac{{\int }_x^{x+h}f(t)dt}{h}$ by cases
10. $={\lim }_{x\to 0}\frac{1}{n}{\int }_x^{x+h}f(t)dt$
11. $={\lim }_{x\to 0}\frac{1}{x+h-x}{\int }_x^{x+h}f(t)dt$
12. Notice that $f$ is Continuous on Without Loss of Generality $[x,x+h]\subset [a,b]$
13. Then, by MVT for Integrals,
14. $\exists c\in [x,x+h]$ s.t $f(c)=\frac{1}{x+h-x}{\int }_x^{x+h}f(t)dt$ (Denoted as $(\ast )$)
15. $={\lim }_{h\to 0}f(c)$ by $(\ast )$
16. Note that $c$ is not constant with respect to $h$ as $c\in [x,x+h]$. In other words, $c$ is between this interval and changes as $h$ changes.
17. Then, as $h\to 0,c\to x$
18. This is equivalent to ${\lim }_{c\to x}f(c)=f(x)$