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
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