## Wednesday, September 20, 2006

### Fundamental Theorem of Calculus

In today's blog, I present the Fundamental Theorem of Calculus. This is the insight that the integral, the area under the curve, is the same as the antiderivative (that is, the inverse of the derivative) of the curve itself. This enables evaluation of the integral without relying on a Riemann sum.

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]

Proof:

(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.

QED

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

Proof:

(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

(4) Thus:

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.

QED

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)

Proof:

(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.

QED

References

#### 1 comment :

Pranjal Vijay said...

Thank you for giving such a good explanation. I really appreciate your work . This blog is useful to understand the fundamental theorem of calculus in a far better way.