Fundamental Theorem of Calculus, Part II

 Assume f(x) is a continuous function on [a,b] and also assume that F(x) is any anti- derivative for f(x). Hence,

_a∫^b f(x) dx = F(x) _a|^b = F(b) - F(a)

 Proof

First let g(x) = _a∫^x f (t) dt and then we get from Part I of the Fundamental Theorem of Calculus as g′(x) = f(x) and therefore g(x) is an anti-derivative of f(x) on [a,b]. Then assume that F(x) is any anti-derivative of f(x) on [a,b] which we need to select. Therefore, it means that we should have,

 g′ (x) = F′(x)

 So, by Fact 2 in the Mean Value Theorem section we get that g(x) and F(x) can be different by no more than an additive constant on [a, b].  Conversely, for a < x < b

F(x) = g(x) = c

Now since g(x) and F(x) are continuous on [a,b], if we get the limit of it as x → a⁺ and x → b^-  we can notice that it also holds if x = a and x = b .

Hence, for a ≤ x ≤ b we know that F(x) = g(x) + c.  Let's utilize it and the definition of g(x) to do the subsequent.

F(b) - F(a) = (g(b) + c)- (g(a) + c)

= g(b) - g(a)

= _a∫^b f(t) dt  + _a∫^a f(t) dt

= _a∫^b f(t) dt  + 0

= _a∫^b f(x) dx

Notice that in the final step we used the fact as the variable used in the integral doesn't issue and therefore we could change the t's to x's.