Fundamental Theorem of Calculus Part 1 Proof

Proof

1. Suppose $f$ is cont on $[a,b]$
2. Consider ${\int }_a^{b}f(x)dx$
3. $={\lim }_{n\to \infty }{\sum }_{i=1}^{n}f(x_i^{\ast })△x$
4. Suppose $F$ is any anti-derivative on $[a,b]$
   1. Then, $\forall x\in [a,b],F^{\prime }(x)=f(x)$ by defn of Antiderivative
   2. Since $F^{\prime }(x)=f(x)$, $F(x)$ is Differentiable on $[a,b]$
   3. Thus. $F(x)$ is continuous on $[a,b]$ as Differentiability Implies Continuous
   4. It follows that $F(x)$ is also diffferentiable on $(a,b)$
   5. By MVT, $\exists c\in (x_{i-1},x_i)$ s.t $f^{\prime }(c_i)=\frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}$
   6. $⟹f^{\prime }(c_i)=f(x_i)-f(x_{i-1})(△x)$ by defn of $△x$
   7. Note that $f^{\prime }(c_i)=F(c)=F(x){|}_b^a$
   8. Thus, $f(x_i)-f(x_{i-1})=F(c_i)(△x)$
   9. By MVT, $c_i\in [a,b]$. Choose $x_i^{\ast }=c_i$
   10. Now, ${\lim }_{n\to \infty }{\sum }_{i=1}^{n}f(c_i)△x$
   11. $={\lim }_{n\to \infty }{\sum }_{i=1}^n(F(x_i)-F(x_{i-1}))$
   12. This is a Telescoping Series, then, you should get after computation: ${\lim }_{n\to \infty }F(x_0)-F(x_n)={\lim }_{n\to \infty }F(a)-F(B)$
   13. Thus, ${\int }_a^{b}f(x)dx=F(a)-F(b)$