The Fundamental Theorem is often presented as a single proof. In today's blog, I break it down into two theorems.
The content is taken from Edwards and Penney's Calculus and Analytic Geometry.
Definition 1: Average Value of a Function
If a function f is integrable on [a,b], then the average value y of y = f(x) on [a,b] is:
y = 1/(b-a) ∫ (b,a) f(x)dx
This definition becomes more intuitive, when we realize that:
∫ (b,a) f(x) dx = (b-a)y = (b-a)f(x)
This definition is used in the following theorem:
Theorem 1: Average Value Theorem
If a function f is continuous on [a,b], then:
f(x) = 1/(b-a) ∫ f(x)dx for some point x on [a,b]
(1) Let m = f(c) be the minimum value of f on [a,b] (See Theorem, here for proof that it exists)
(2) Let M = f(d) be the maximum value of f on [a,b]. (See Lemma 3, here for proof that it exists)
(3) Using the Comparison Property (see Lemma 4, here),
m(b-a) ≤ ∫ (b,a) f(x)dx ≤ M(b-a)
Then, dividing all sides by (b-a) gives us:
m ≤ 1/(b-a) ∫ (b,a) f(x)dx ≤ M
(4) Because f is continuous and because y=[1/(b-a) ∫ (b,a) f(x)dx] lies between two values of f: m and M, we can use the Intermediate Value Property (see Theorem, here) to conclude that y must be a value of f.
(5) Thus, there must exists a point x where y = f(x) and where x lies between a and b.
Definition 2: antiderivative
A function F is an antiderivative of a function f if and only if the derivative F'(x) = f(x).
The fundamental theorem of calculus (see Theorem 2 below) gives another notation for this same point:
F(x) = ∫ f(x)dx
Here is the proof.
Theorem 2: The integral is the antiderivative of a continuous function.
d/dx (∫ (x,a) f(t)dt) = f(x) where f(x) is a continuous function
(1) Let F(x) be a continuous function defined on [a,b] by:
F(x) = ∫ (x,a) f(t)dt
(2) By the definition of the derivative (see Definition 1, here):
F'(x) = lim (Δx → 0) [F(x + Δx) - F(x)]/Δx
= lim (Δx → 0) (1/Δx)(∫ (x+Δx,a) f(t)dt - ∫ (x,a) f(t)dt)
(3) Using the interval union property (see Lemma 3, here), we know that:
∫ (x+Δx,a) f(t)dt = ∫ (x,a) f(t)dt + ∫(x+Δx, x) f(t)dt
F'(x) = lim (Δx → 0) (1/Δx) ∫ (x+Δx,x) f(t)dt
(5) The average value theorem (see Thereom 1, above) gives us:
1/Δx ∫ (x+Δx,x) f(t)dt = f(t) for some number t in [x, x + Δx].
(6) It is clear that as Δx → 0, t → x.
(7) Thus, because f is continuous, we see that:
F'(x) = lim(Δx → 0) (1/Δx) ∫ (x+Δx,x) f(t)dt = lim (Δx → 0) f(t) = lim (t → x) f(t) = f(x).
(8) Hence the function F defined above is indeed the antiderivative of f.
Theorem 3: Evaluation of Definite Integrals
The definite integral of a continuous function G'(x) is equal to the difference of the antiderivatives:
∫ (b,a) G'(x)dt = G(b) - G(a)
(1) Let F'(t)=f(t) be a continous function on [a,b].
NOTE: For purposes of clarity, I use f(t) and F'(t) when refering to the continuous function itself and F(x),G(x) when refering to the antiderivative. The reason for this is to make clear the definite integral:
F(x) = ∫ (x,a) f(t)dt.
(2) F(x) is then its antiderivative with:
F(x) = ∫ (x,a) f(t)dx
(3) Let G be any antiderivative of F'(t) such that:
G(x) = F(x) + C on [a,b] where C is a constant.
Since F'(t) is the continuous function, F(x) is the antiderivative and the set of all antiderivatives is characterized by F(x) + C since d/dx(F(x) + C) = d/dx(F(x)) + d/dx(C) = F'(x) + 0 = F'(x) [See Lemma 1, here for details on why F(C)=0]
(4) Since F(a) = ∫ (a,a) f(t)td = 0, we know that:
G(a) = F(a) + C = 0 + C = C
(5) Combining this with step #1 gives us:
G(x) = F(x) + G(a) for all x in [a,b].
which is the same as:
F(x) = G(x) - G(a)
(6) If x=b, this gives us:
G(b) - G(a) = F(b) = ∫ (b,a) f(x)dx.
- Edwards & Penney, Calculus and Analytic Geometry