138 CHAPTER 5. FUNCTIONS ON NORMED LINEAR SPACES
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 5.9.4 For p(x) a polynomial on [a,b] , let P′ (x) = p(x) . Thus, by themean value 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
wherelimn→∞∥pn− f∥ ≡ lim
n→∞max
x∈[a,b]| f (x)− pn (x)|= 0
Proposition 5.9.5 The above integral is well defined and satisfies the following prop-erties.
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
3. If F (x)≡∫ x
a f dt, Then F ′ (x) = f (x) so any continuous function has an antideriva-tive, and for any a ̸= b,
∫ ba f dx = G(b)−G(a) whenever G′ = f on the open interval
determined by a,b and G continuous on the closed interval determined by a,b. Also,∫ b
a(α f (x)+βg(x))dx = α
∫ b
af (x)dx+β
∫a
βg(x)dx
If a < b, and f (x)≥ 0, then∫ b
a f dx≥ 0. Also∣∣∣∫ b
a f dx∣∣∣≤ ∣∣∣∫ b
a | f |dx∣∣∣.
4.∫ b
a 1dx = b−a.
Proof: First, why is the integral well defined? With notation as in the above definition,the mean value theorem implies∫ b
ap(x)dx≡ P(b)−P(a) = p(x̂)(b−a) (5.3)
where x̂ is between a and b and so∣∣∣∫ b
a p(x)dx∣∣∣ ≤ ∥p∥|b−a| . If ∥pn− f∥ → 0, then
limm,n→∞ ∥pn− pm∥= 0 and so∣∣∣∣∫ b
apn (x)dx−
∫ b
apm (x)dx
∣∣∣∣= |(Pn (b)−Pn (a))− (Pm (b)−Pm (a))|