obviously 0
∫ b ∫ b ◜∫ ◞b◟ ◝
1g′(t) dt = 1dg(t) + gdf = 1g (b)− 1g (a) = g(b) − g(a)
a a a

This proves:

Theorem 9.2.1If g^{′}is continuous on

[a,b]

, then g

(b)

− g

(a)

= ∫_{a}^{b}g^{′}

(t)

dt.

A version of this presented more directly is the following.

Theorem 9.2.2Suppose∫_{a}^{b}f

(t)

dt exists and F^{′}

(t)

= f

(t)

for each t ∈

(a,b)

forsome F continuous on

[a,b]

. Then

∫
b
a f (t)dt = F (b) − F (a)

Proof:There exists δ > 0 such that if

∥P ∥

< δ, then for P = x_{0},

⋅⋅⋅

,x_{n},

| |
||∫ b ∑n ||
|| f (t)dt − f (tk)(xk − xk− 1)|| < ε, for any tk ∈ [xk−1,xk].
a k=1

Use the mean value theorem to pick t_{k}∈

(xk− 1,xk)

such that f

(tk)

(xk − xk−1)

= F

(xk)

−F

(xk−1)

.
Then

|∫ | |∫ |
|| b ∑n || || b ∑n ||
|| a f (t)dt− f (tk) (xk − xk−1)|| = || a f (t)dt− F (xk)− F (xk−1)||
k=1 ||∫ k=1 ||
= || bf (t)dt− (F (b)− F (a))||< ε
| a |

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

Definition 9.2.3When everything makes sense,

∫ b ∫ a
f (t)dg(t) ≡ − f (t)dg(t)
a b

This also shows the well known change of variables formula.

Theorem 9.2.4Suppose F^{′}

(y)

= f

(y)

for y between g

(a)

and g

(b)

and f, g^{′}arecontinuous. Then

∫ ϕ(b) ∫ b ′
ϕ(a) f (y)dy = a f (ϕ (t))ϕ (t)dt

Proof: This is true because both sides reduce to F

(ϕ (b))

− F

(ϕ (a))

.■

The other form of the fundamental theorem of calculus is also obtained. Note that in case
of the Riemann integral if f ≥ 0, and a ≤ b, then ∫_{a}^{b}f

(t)

dt ≥ 0. Therefore, assuming f is
continuous,

∫ b ∫ b
(|f (t)|− f (t))dt, (|f (t)|+ f (t))dt
a a

and so

∫ b ∫ b ∫ b ∫ b
|f (t)|dt ≥ f (t)dt, |f (t)|dt ≥ − f (t)dt
a a a a

by Theorem 9.1.6. It remains to verify the last part. Let t ∈

(a,b)

and let

|h|

be small enough that everything of interest is in

[a,b]

. First suppose h > 0.
Then

| |
||F (t+ h)− F (t) || ||1 ∫ t+h 1 ∫ t+h ||
||------h-------− f (t)|| = ||h f (s)ds− h f (t)ds||
t t

∫ ∫
≤ -1 t+h|f (s)− f (t)|ds ≤ 1 t+hεds = ε
h t h t

provided that ε is small enough due to continuity of f at t. A similar inequality is obtained
if h < 0 except in the argument, you will have t + h < t so you have to switch
the order of integration in going to the second line and replace 1∕h with 1∕

(− h)

.
Thus

lim F-(t+-h)−-F-(t) = f (t) . ■
h→0 h

An examination of the proof yields the following corollary.