if $f$ is int on $[a,b]$ and $F'(x) = f(x)$ except for a finite number of points in $[a,b]$ To show that if $f$ is integrable on $[a,b]$ and if $F$ is continuous on $[a,b]$ and $F'(x) = f(x)$ except for a finite number of points in $[a,b]$, then $\int_a^bf=F(b) - F(a)$.
Let the finite number of points in $[a,b]$ where $F'(x) \neq f(x)$ be $x_1,x_2,...,x_n$. Thus By fundamental theorem of Calculus, 
$$\int_a^bf= \int_a^{x_1}f +\int_{x_1}^{x_2}f+...+  \int_{x_{n-1}}^{x_n}f+\int_{x_n}^{b}f.$$
Thus $\int_a^bf= F(x_1) -F(a) +F(x_2)-F(x_1)+...+F(b)-F(x_n) = F(b) - F(a)$.
Is the proof correct?
 A: I assume that $F'(x)$ exists for all $x$ and that $F'(x) = f(x)$ for all $x \in [a, b]$ except possibly a finite number of points. Let these points be called $x_{1}, x_{2}, \dots, x_{n}$ and define the function $g(x) = f(x)$ for all $x \in [a, b]$ except $x = x_{i}$. At these exceptional points we set $g(x_{i}) = F'(x_{i})$. It thus follows that $F'(x) = g(x)$ for all $x \in [a, b]$ and $F(x)$ is continuous on $[a, b]$.
Since $f(x), g(x)$ differ only at a finite number of points and $f$ is integrable it follows that $g$ is also integrable over interval $[a, b]$ and has the same integral as $f$ on $[a, b]$. Now by second fundamental theorem of calculus (see note below) it now follows that $$F(b) - F(a) = \int_{a}^{b}g(x)\,dx = \int_{a}^{b}f(x)\,dx$$ 
Note: There are two fundamental theorems of calculus as far Riemann integration is concerned:
1) If $f$ is Riemann integrable on $[a, b]$ and $$F(x) = \int_{a}^{x}f(t)\,dt$$ then $F$ is continuous on $[a, b]$ and $F$ is differentiable at those points $x \in [a, b]$ where $f$ is continuous and $F'(x) = f(x)$ at these points.
2) If $f$ is integrable on $[a, b]$ and there is a function $F$ such that $F'(x) = f(x)$ for all $x \in [a, b]$ then $$F(b) - F(a) = \int_{a}^{b}f(x)\,dx$$
