448 CHAPTER 16. APPROXIMATION OF FUNCTIONS AND THE INTEGRAL

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) (16.2)

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))|

= |(Pn (b)−Pm (b))− (Pn (a)−Pm (a))|

=

∣∣∣∣∫ b

a(pn− pm)dx

∣∣∣∣≤ ∥pn− pm∥|b−a|

Thus the limit exists because{∫ b

a pndx}

nis a Cauchy sequence and R is complete.

From 16.2, 1. holds for a polynomial p(x). Let ∥pn− f∥→ 0. Then by definition,∫ b

af dx≡ lim

n→∞

∫ b

apndx = pn (xn)(b−a) (16.3)

for some xn in the open interval determined by (a,b) . By compactness, there is a fur-ther subsequence, still denoted with n such that xn → x ∈ [a,b] . Then fixing m such that∥ f − pn∥< ε whenever n≥ m, assume n > m. Then ∥pm− pn∥ ≤ ∥pm− f∥+∥ f − pn∥<2ε and so

| f (x)− pn (xn)| ≤ | f (x)− f (xn)|+ | f (xn)− pm (xn)|+ |pm (xn)− pn (xn)|

≤ | f (x)− f (xn)|+∥ f − pm∥+∥pm− pn∥< | f (x)− f (xn)|+3ε

Now if n is still larger, continuity of f shows that | f (x)− pn (xn)|< 4ε. Since ε is arbitrary,pn (xn)→ f (x) and so, passing to the limit with this subsequence in 16.3 yields 1.

Now consider 2. It holds for polynomials p(x) obviously. So let ∥pn− f∥→ 0. Then∫ c

apndx+

∫ b

cpndx =

∫ b

apndx