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.