1.6. The Fundamental Theorem of Calculus 15

and hence that

x0+h

x0

|f(t) − f(x0)| dt

|h|

ε.

Putting this together with the inequality (1.6.2) above we have that

F (x0 + h) − F (x0)

h

− f(x0) ε

whenever |h| δ, which is exactly what we needed to show.

Corollary 1.6.2. (Fundamental theorem of calculus). If f is a

continuous function on [a, b] and F is any anti-derivative of f, then

b

a

f(x) dx = F (b) − F (a).

Proof. Define the function G(x) =

x

a

f(t) dt. By Theorem 1.6.1 the

derivative of G(x) is f(x) which is also the derivative of F . Hence

F and G differ by a constant, say F (x) = G(x) + C (see Corollary

A.8.4).

Then

F (b) − F (a) = (G(b) + C) − (G(a) + C)

= G(b) − G(a)

=

b

a

f(x) dx −

a

a

f(x) dx

=

b

a

f(x) dx.

Exercise 1.6.3.

(1) Prove that if f : [a, b] → R is a regulated function and

F : [a, b] → R is defined by F (x) =

x

a

f(t) dt, then F is

continuous.

(2) Let S denote the set of all functions F : [a, b] → R which can

be expressed as F (x) =

x

a

f(t) dt for some step function f.

Prove that S is a vector space of functions, each of which