16.4. AN APPROACH TO THE INTEGRAL 447

Theorem 16.4.2 Let f : [a,b]→ R be continuous. Then if the maximum value of f occursat a point x∈ (a,b) , it follows that if f ′ (x) = 0. If f achieves a minimum at x∈ (a,b) wheref ′ (x) exists, it also follows that f ′ (x) = 0.

Proof: By Theorem 10.1.39, f achieves a maximum at some point x. If f ′ (x) exists,then

f ′ (x) = limh→0+

f (x+h)− f (x)h

= limh→0−

f (x+h)− f (x)h

However, the first limit is non-positive while the second is non-negative and so f ′ (x) = 0.The situation is similar if the minimum occurs at x ∈ (a,b). ■

The Cauchy mean value theorem follows. The usual one is obtained by letting g(x) = x.

Theorem 16.4.3 Let f ,g be continuous on [a,b] and differentiable on (a,b) . Then thereexists x∈ (a,b) such that f ′ (x)(g(b)−g(a))= g′ (x)( f (b)− f (a)). If g(x)= x, this yieldsf (b)− f (a) = f ′ (x)(b−a) , also f (a)− f (b) = f ′ (x)(a−b).

Proof: Let h(x)≡ f (x)(g(b)−g(a))−g(x)( f (b)− f (a)) . Then

h(a) = h(b) = f (a)g(b)−g(a) f (b) .

If h is constant, then pick any x ∈ (a,b) and h′ (x) = 0. If h is not constant, then it haseither a maximum or a minimum on (a,b) and so if x is the point where either occurs, thenh′ (x) = 0 which proves the theorem. ■

Recall that an antiderivative of a function f is just a function F such that F ′ = f .

You know how to find an antiderivative for a polynomial.(

xn+1

n+1

)′= xn so

∫∑

nk=1 akxk =

∑nk=1 ak

xk+1

k+1 +C. With this information and the Weierstrass theorem, it is easy to defineintegrals of continuous functions with all the properties presented in elementary calculuscourses. It is an approach which does not depend on Riemann sums yet still gives thefundamental theorem of calculus. Note that if F ′ (x) = 0 for x in an interval, then for x,yin that interval, F (y)−F (x) = 0(y− x) so F is a constant. Thus, if F ′ = G′ on an openinterval, F,G continuous on the closed interval, it follows that F −G is a constant and soF (b)−F (a) = G(b)−G(a).

Definition 16.4.4 For p(x) a polynomial on [a,b] , let P′ (x) = p(x) . Thus, by the meanvalue theorem if P′, P̂′ both equal p, it follows that P(b)− P(a) = P̂(b)− P̂(a) . Thendefine

∫ ba p(x)dx≡ P(b)−P(a). If f ∈C ([a,b]) , define

∫ ba f (x)dx≡ limn→∞

∫ ba pn (x)dx

where limn→∞ ∥pn− f∥ ≡ limn→∞ maxx∈[a,b] | f (x)− pn (x)|= 0.

Proposition 16.4.5 The above integral is well defined and satisfies the following proper-ties.

1.∫ b

a f dx = f (x̂)(b−a) for some x̂ between a and b. Thus∣∣∣∫ b

a f dx∣∣∣≤ ∥ f∥|b−a| .

2. If f is continuous on an interval which contains all necessary intervals,∫ c

af dx+

∫ b

cf dx =

∫ b

af dx, so

∫ b

af dx+

∫ a

bf dx =

∫ b

bf dx = 0