5.8. AN APPROACH TO THE INTEGRAL 125

Proposition 5.8.5 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) (5.2)

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 5.2, 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) (5.3)

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

5.8. AN APPROACH TO THE INTEGRAL 125Proposition 5.8.5 The above integral is well defined and satisfies the following prop-erties.1. fe fdx = f (£) (b—a) for some & between a and b. ThusI? fax| < | f\l|b—al.2. If f is continuous on an interval which contains all necessary intervals,[tact [tac= [pac so [tact [pace ['pae=o3. If F (x) = J fdt, then F' (x) = f (x) so any continuous function has an antideriva-tive, and for any a # b, i? fdx = G(b) —G(a) whenever G' = f on the open intervaldetermined by a,b and G continuous on the closed interval determined by a,b. Also,[care +Bear=a ['sejar+B [ BearIfa <b, and f (x) >0, then [? fdx > 0. AlsoI fax| <[flax4. [? ldx=b—a.Proof: First, why is the integral well defined? With notation as in the above definition,the mean value theorem implies[re)axv=Po)-Pl@ =r ®)0-a) (5.2)where x is between a and b and so[2 p(x) ds] < |[pll\b—al- If [lpn — fll 0, thenLiMn 200 || Pn — Pm|| = 0 and sob b[ max | Pm (x) dx) = |(Pn (b) — Pn (a) — (Pm (B) = Pn (@))|= (Px (0) ~ Po (0)) ~ (Pua) ~ Po (a))|=| "(Pn Pm) ds] < Pn Pm [PalThus the limit exists because { f ° pra} is a Cauchy sequence and R is complete.nFrom 5.2, 1. holds for a polynomial p (x). Let ||p, — f|| > 0. Then by definition,b b| fax = tim | Pndx = Pn (Xn) (b—a) (5.3)for some x, in the open interval determined by (a,b). By compactness, there is a fur-ther subsequence, still denoted with n such that x, — x € [a,b]. Then fixing m such that\|f — pall < € whenever n > m, assume n > m. Then ||p~m — Pall < ||Pm — FI] + || f — Pall <2¢€ and soIF (2) = Pn (%n)1 SF) — f n)| + LF Gn) — Pm On)| + [Pm (%n) = Pn (%n)|SFO) —f On)| + \LF— Pll + lpm — Pall < Lf) — fn) +3