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Fundamental Theorem of Calculus I
Suppose that a,b∈ℝ, a<b, and f:[a,b]→ℝ is Riemann integrable. Define F:[a,b]→ℝ by F(x):=∫axf. Then F is continuous on [a,b]. If f is continuous at x0∈(a,b), then F is differentiable at x0 and F′(x0)=f(x0).
Since f is Riemann integrable, it is bounded. Choose M∈ℝ such that |f(t)|≤M for every t∈[a,b]. If x,y∈[a,b] and x<y, then by the bounds on a Riemann integral, |F(y)−F(x)|=|∫xyf|≤M(y−x). The same inequality with x and y interchanged gives |F(y)−F(x)|≤M|y−x|, so F is continuous.

Now suppose that f is continuous at x0∈(a,b). For h≠0 small enough that x0+h∈(a,b),

F(x0+h)−F(x0)h−f(x0)=1h∫x0x0+h(f(t)−f(x0))dt. Let ϵ∈ℝ+. Since f is continuous at x0, there is δ∈ℝ+ such that |f(t)−f(x0)|<ϵ whenever |t−x0|<δ. Hence, whenever |h|<δ, |F(x0+h)−F(x0)h−f(x0)|≤ϵ. Therefore the derivative exists and is equal to f(x0).
Zero Derivative Implies Constant
Suppose that a,b∈ℝ, a<b, and F:[a,b]→ℝ is continuous on [a,b] and differentiable on (a,b). If F′(x)=0 for every x∈(a,b), then F is constant on [a,b].
Let x,y∈[a,b] with x<y. By the Mean Value Theorem, there is c∈(x,y) such that F(y)−F(x)y−x=F′(c)=0. Since y−x≠0, we have F(y)=F(x). Thus F is constant.
Fundamental Theorem of Calculus II
Suppose that a,b∈ℝ, a<b, and f:[a,b]→ℝ is continuous. If F:[a,b]→ℝ is continuous on [a,b], differentiable on (a,b), and F′(x)=f(x) for every x∈(a,b), then ∫abf=F(b)−F(a).
Define G:[a,b]→ℝ by G(x):=∫axf. By Fundamental Theorem of Calculus I, G′(x)=f(x) for every x∈(a,b). Since F′=f on (a,b), the function H:=F−G has derivative 0 on (a,b). By zero derivative implies constant, H is constant. Thus F(b)−G(b)=F(a)−G(a). Since G(a)=∫aaf=0 and G(b)=∫abf, rearranging gives ∫abf=F(b)−F(a).