7.1. THE DEFINITION OF THE INTEGRAL FROM ANTIDERIVATIVES 175

where k ≡−h and so for F ′ (x) the left derivative, it exists at each point of (a,b] and equalsf (x). Also, the right derivative exists on [a,b) and equals f (x) and by similar reasoning,the left derivative exists on (a,b] and equals f (x). Thus F is continuous and F ′ (x) = f (x)for x ∈ (a,b).

Proposition 7.1.4 The above integral is well defined for f continuous on [a,b] andsatisfies the following properties.

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) . 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: The integral is well defined by Lemma 7.1.3 and Proposition 7.0.2. Consider 1.Let F ′ (x) = f (x) ,F as in Lemma 7.1.3 so∫ b

af (x)dx ≡ F (b)−F (a) = f (x̂)(b−a)

for some x̂ in the open interval determined by a,b. This is by the mean value theorem.Hence

∣∣∣∫ ba f dx

∣∣∣≤ ∥ f∥|b−a| .Now consider 2. Let F ′ = f on a closed interval which contains all necessary intervals.

Then from the definition,∫ c

af dx+

∫ b

cf dx = F (c)−F (a)+F (b)−F (c) = F (b)−F (a)≡

∫ b

af (x)dx

Next consider 3. For F (x) ≡∫ x

a f (x)dx, the definition says that F (x) = G(x)−G(a)where G′ (x) = f (x) and so, since G′ = F ′, it follows that F ′ (x) = f (x) with an appropriateone sided derivative at the ends of the interval. Now let F ′ = f ,G′ = g. Then α f +βg =(αF +βG)′ and so∫ b

a(α f (x)+βg(x))dx ≡ (αF +βG)(b)− (αF +βG)(a)

= αF (b)+βG(b)− (αF (a)+βG(a))

= α (F (b)−F (a))+β (G(b)−G(a))

≡ α

∫ b

af (x)dx+β

∫a

βg(x)dx

7.1. THE DEFINITION OF THE INTEGRAL FROM ANTIDERIVATIVES 175where k = —h and so for F’ (x) the left derivative, it exists at each point of (a, b] and equalsf (x). Also, the right derivative exists on [a,b) and equals f (x) and by similar reasoning,the left derivative exists on (a,b] and equals f (x). Thus F is continuous and F’ (x) = f (x)forx € (a,b). WlProposition 7.1.4 The above integral is well defined for f continuous on |a,b\ andsatisfies the following properties.1. i? fdx = f (&) (b—a) for some & between a and b. ThusI? fax| <|\fllb—al.2. If f is continuous on an interval which contains all necessary intervals,[tact [ tac= [fav SO [tact [ tac= ['pav=o3. If F (x) = fi fdt, Then F' (x) = f (x). Also,[Care +Be)ar=a [ pojar+B [ BearIfa <b, and f (x) >0, then [? fdx > 0. AlsoI? fax| <[flax4. [? ldx=b—a.Proof: The integral is well defined by Lemma 7.1.3 and Proposition 7.0.2. Consider 1.Let F’ (x) = f (x), F as in Lemma 7.1.3 so[fejar=Fo)-F@) =F (b-0)for some ¢ in the open interval determined by a,b. This is by the mean value theorem.Hence |? fds < |if\||b—al.Now consider 2. Let F’ = f on a closed interval which contains all necessary intervals.Then from the definition,c b b[ face [ fdx = F (c)—F (a) +F (b) ~F(c) =F(b)-F (a) = | f(a)aexNext consider 3. For F (x) = J? f (x) dx, the definition says that F (x) = G(x) — G(a)where G’ (x) = f (x) and so, since G’ = F’, it follows that F’ (x) = f (x) with an appropriateone sided derivative at the ends of the interval. Now let F’ = f,G’ = g. Then af + Bg =(@F + BG)’ and so(AF + BG) (b) — (a@F + BG) (a)= a@F (b)+BG(b)—(aF (a) +BG(a))= @(F(b)—F (a))+B(G(b)—G(a)a | F)dr+B [Bear[ (arte) + Bear