9.1. THE INTEGRAL OF 1700’S 189

Proof: Suppose F,G are both antiderivatives. Then letting

H (x)≡ F (x)−G(x) ,H ′ (x) = 0

it follows by the mean value theorem, H (b)−H (a) = 0(b−a) = 0 so F (b)−G(b) =F (a)−G(a) which implies F (b)−F (a) = G(b)−G(a).

Proposition 9.1.5 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, x̂ /∈ {a,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 Proposition 9.1.4 and Lemma 9.1.2. Consider 1.Let F ′ (x) = f (x) ,F as in Lemma 9.1.2 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 is an antiderivative of f . Since F (x) = G(x)−G(a) , f = G′ = F ′. It follows thatF ′ (x) = f (x) with an appropriate one sided derivative at the ends of the interval. Now letF ′ = 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