206 CHAPTER 9. INTEGRATION

+

∣∣∣∣∣ n

∑i=1

f (vi)(g(xi)−g(xi−1))−∫ b

af (t)dg(t)

∣∣∣∣∣< ε +0+ ε = 2ε

Since ε is arbitrary, this shows∫ b

a f (t)dg(t) =∫ b

a f (t)g′ (t)dt.What does the integration by parts formula 9.7 say in case g′ exists and is continuous

and f ′ exists and is continuous? By Proposition 9.4.3 above, both∫ b

a f dg and∫ b

a gd f exist.Now the integration by parts formula says f g(b)− f g(a) =

∫ ba f dg+

∫ ba gd f and from

what was just shown in Proposition 9.4.3, this reduces to

f g(b)− f g(a) =∫ b

af (t)g′ (t)dt +

∫ b

ag(t) f ′ (t)dt (9.8)

which is the usual integration by parts formula from calculus.

Proposition 9.4.4 Let f ,g both be continuous on [a,b] with continuous bounded deriv-atives on (a,b) . Then the usual calculus integration by parts formula 9.8 is valid.

9.5 The Fundamental Theorem of CalculusNote how as a special case, you get the usual fundamental theorem of calculus by lettingf (t)≡ 1. Indeed, from Theorem 9.4.1

∫ b

a1g′ (t)dt =

∫ b

a1dg(t)+

obviously 0︷ ︸︸ ︷∫ b

agd f = 1g(b)−1g(a) = g(b)−g(a)

This proves:

Theorem 9.5.1 If g′ is continuous on [a,b] , then g(b)−g(a) =∫ b

a g′ (t)dt.

A version of this presented more directly is the following.

Theorem 9.5.2 Suppose∫ b

a f (t)dt exists and F ′ (t) = f (t) for each t ∈ (a,b) forsome F continuous on [a,b]. Then

∫ ba f (t)dt = F (b)−F (a).

Proof: There exists δ > 0 such that if ∥P∥< δ , then for P = x0, · · · ,xn,∣∣∣∣∣∫ b

af (t)dt−

n

∑k=1

f (tk)(xk− xk−1)

∣∣∣∣∣< ε, for any tk ∈ [xk−1,xk] .

Use the mean value theorem to pick tk ∈ (xk−1,xk) such that f (tk)(xk− xk−1) = F (xk)−F (xk−1) . Then∣∣∣∣∣

∫ b

af (t)dt−

n

∑k=1

f (tk)(xk− xk−1)

∣∣∣∣∣ =

∣∣∣∣∣∫ b

af (t)dt−

n

∑k=1

F (xk)−F (xk−1)

∣∣∣∣∣=

∣∣∣∣∫ b

af (t)dt− (F (b)−F (a))

∣∣∣∣< ε

Since ε > 0 is arbitrary, this proves the theorem.