Chapter 4 The Integral
The area of a circle: Egyptian knew how to calculate before 1650 B. C.
The general method for calculating the area: Archimedes (287 B.
C.~212 B. C.) proposed the method of exhaustion.
4.1 Antiderivatives
Definition 1:
F is an antiderivative of f if F ' x   f x  .
Example 1:
f x   3x 2  F x   x 3  c , where c is a constant.
◆
Theorem 1:
If F and G are two differentiable functions that have the same
derivative in
a, b ,
'
'
i.e., F x   G x  . Then, F x   Gx   c ,
where c is any constant.
Definition 2:
Let F be an antiderivative for f , then the indefinite integral of f is
written
 f x dx  F x   c , where
f is referred to as the integrand,
x is referred to as the variable of integration and c is any constant. If f
has an antiderivative, then f is said to be integrable.
1
Theorem 2:
x r 1
 kdx  kx  c,  x dx  r  1  c,
 sin x dx   cosx   c,  cosx dx  sin x   c
r
Theorem 3:
1.
 kf x dx  k  f x dx .
2.
  f x   g x dx   f x dx   g x dx .
[justifications:]
1. Let
 f x dx  F .
Since
d kF  kdF

 kf  x  ,
dx
dx
kF  k  f  x dx   kf  x dx
by theorem 1.
2. Let
 f x dx  F
and
 g x dx  G . Since
d F  G  dF dG


 f x   g x  ,
dx
dx dx
F  G   f  x dx   g  x dx    f  x   g x dx
by theorem 1.
Theorem 4:
If f is continuous, then f is integrable.
2
4.2 Approximations to Area and The Definite Integral
Motivating Example :
1
f(x)=x^2
f(x)
Smin
Smax
9/16
4/16
1/16
0
0
1/4
2/4
3/4
1
x
2
In the above figure, the area, A, between f x   x over
0,1
and
the x-axis can be approximated by the rectangles in dashed lines or dotted
lines. The area of the rectangles in dotted lines is
2
S min
2
2
1 1 2 1 3 1
         
4 4 4 4 4 4
1 1 4 1 9 1

    
16 4 16 4 16 4
7

32
while the area of the rectangles in dashed lines is
2
2
2
2
1 1 2 1 3 1 4 1
S max                
4 4 4 4 4 4 4 4
1 1 4 1 9 1 16 1

      
16 4 16 4 16 4 16 4
15

32
3
Thus,
0.22 
7
15
 A
 0.47 .
32
32
In the above approximation, the interval
0,1
is divided into
subintervals of equal length,
0, 14, 14 , 2 4, 2 4 , 3 4, 3 4 ,1.
If the interval
0,1
is divided into more subintervals of equal lengths,
for example,
0, 18 , 18 , 2 8 , 2 8 , 38 , 38 , 4 8 , 4 8 , 58 , 58 , 6 8 , 6 8 , 7 8 , 7 8 ,1
then the area A can be approximated by similar rectangles in dashed lines
or dotted lines. The area of the rectangles in dotted lines is
2
S min
2
2
1 1  2 1
7 1
           
8 8 8 8
8 8
35

128
while the area of the rectangles in dashed lines is
2
2
2
2
1 1  2 1
7 1 8 1
S max                  
8 8 8 8
8 8 8 8
51

128
Thus,
0.27 
35
51
 A
 0.40 .
128
128
A more accurate approximation can be obtained. In general, the interval
0,1
is divided into subintervals with equal length
4
1 .
n
f(x)=x^2
0
0
(k-1)/n
k/n
(n-1)/n
1
x
The area of the rectangles in dotted lines is
1 1 2 1
 n 1 1
         
 
n n n n
 n  n
2
S min
2
2
12  2 2    n  1

n3
1 n1 2
 3 k
n k 1
n  1n2n  1

6n 3
2
while the area of the rectangles in dashed lines is
1 1 2 1
 n 1  1  n  1
         
    
n n n n
 n  n n n
2
S min
2
2
12  2 2    n  1  n 2

n3
1 n 2
 3 k
n k 1
nn  12n  1

n3
2
5
2
Thus,
n  1n2n  1  A  nn  12n  1
6n 3
6n 3
As n tends to infinity, by squeezing theorem,
lim
n
n  1n2n  1  1  lim nn  12n  1
6n 3
3
A  lim A 
n
6n 3
n
,
1
3
Note:
In the above approximations, the same result can be obtained as the
 

heights of the rectangles are replaced by f xi , where
xi 
i  1  c, 0  c  1 , i  1, 2,, n.
n
n
i
 (i  1) 
f
f



 , the values of
That is, rather than using
or
n
 n 
the function evaluating at the inner points of the subintervals are
used as the heights of the rectangles. For example, as using the
middle points of these subintervals as the heights,
xi 
i  1 
n
1 2i  1

, i  1, 2,, n,
2n
2n
 1  1  3  1
 2n  1  1
 approximat ed area            
 
 2n  n  2n  n
 2n  n
2
2
2
2 n 1
1  3    2n  1

4n 3
4n 2  1

12n 2

2
2
2
6
n 1
 k   2k 
2
k 1
k 1
3
4n
2
Thus, as n tends to infinity, the approximated area tends to
1 .
3
◆
Definition 3 (Riemann sum and regular partition):
Let
a  x0  x1  x2    xn1  xn  b .
Let
xi  xi1  xi , i  1, 2,, n , The Riemann sum is
 f x x , x  x
n

i
i 1

i
i
i 1
, xi  .
As
xi 
ba
n
the partition is regular.
Definition 4 (the definite integral):
Let f be defined on
a, b . Then,
b
 f xdx 
a
 f x x ,
n
lim
max xi  0
i 1

i
i
whenever the limit exists.
Note:
b
 f xdx  lim
a
x  0
n
 
n
  b n a , x  b n a .
 f x x  lim  f xi 
i 1

i
n 
i 1
Theorem 5:
f is continuous on a, b , then f is integrable on a, b . That is,
7
b
 f x dx
exists.
a
Note:
f is continuous, then
b
 f xdx  lim
x  0
a
n
 
n
  b n a , x  b n a
 f x x  lim  f xi 

i
i 1
n 
i 1
Theorem 6:
Let c be a constant. Then,
b
 cdx  cb  a  .
a
f(x)=c
c
0
a
b
x
Definition 5:
For any real number a,
a
 f x dx  0.
a
8
Definition 6:
b
If a  b and
 f x dx
exists, then
a
a
b
b
a
 f x dx   f x dx .
Example 2:
0
1
2
 x dx   x dx 
2
1
0
1
3 .
◆
Definition 7 (area):
b
The area bounded by the function y  f x , is denoted by Aa and is
defined by the formula
a
A 
b
a
 f x  dx.
a
- f x 
a
b
f x 
9
Properties of Definite Integral:
Theorem 7:
If
f
is continuous on
integrable on
a, c
a, b
a  c  b , then f is
and if
c, b , and
and on
a
c
b
b
a
c
 f xdx   f xdx   f xdx.
Theorem 8:
f is integrable on
integrable on
a, b
and if k is any constant, then kf is
a, b , and
b
b
a
a
 kf xdx  k  f xdx.
Theorem 9:
If the function f and g are both integrable on
is integrable on
a, b , then
f g
a, b , and
b
b
b
a
a
a
  f x  g xdx   f xdx   g xdx.
Theorem 10:
If f is integrable on
a, b
and f  0 there, then
b
 f x dx  0 .
a
Theorem 11:
If the function
f
and g
are both integrable on
10
a, b
and
f x  g x , then
b
b
a
a
 f xdx   g xdx .
Theorem 12:
If f is integrable on
a, b
and m  f  M , then
b
mb  a    f x dx  M b  a  .
a
[justifications of theorem 7:]
a
x0 x1
c
xk xk+1
xn-1
b
xn
Since
 f x x
n

i
i 1
b

a
 
k
  f x xi 
i

i
i 1
 f x x ,
n

i
i  k 1
i
n
k


f x dx  lim  f x xi  lim  f xi xi   f xi xi 
max xi  0
max xi  0
i 1
i  k 1
 i 1

 
n
 lim
max xi  0
 

i
 f x x 
k

i
i 1
c
b
a
c
i
 f x x
n
lim
max xi  0
  f x dx   f x dx
[justifications of theorem 8:]
11
 
i  k 1

i
i
b
k  f x dx  k lim
max xi 0
a
 lim
max xi 0
 f x x
n

i
i 1
k
 lim
i
max xi 0
 
k  f xi xi
i 1
 kf x x
k

i
i 1
i
b
  kf x dx
a
[justifications of theorem 9:]
Since
f and g are both integrable, then
b
 f xdx 
a
 f x x
n
lim
max xi 0

i
i 1
i
and
b
 g xdx 
a
b
b
 f x dx   g x dx 
a
a
lim
max xi 0
max xi 0

i
i 1

i
i 1
 f x x
n
lim
 g x x .
n
i
i
 lim
max xi 0
 g x x
n
i 1
n
n


 lim  f xi xi   g xi xi 
max xi 0
i 1
 i 1

 
 lim
max xi 0
 
  f x   g x x
k

i
i 1

i
i
b
   f x   g x dx
a
[justifications of theorem 10:]
Since
 

i
f x  0, x  0 
 f x x
n
i 1
12

i
i
 0, xi ,

i
i
b
 f xdx 
a
 f x x
n
lim
max xi 0

i
i 1
i
 0.
[justifications of theorem 11:]
Since hx  g x  f x  0 and
hx  g x  f x  g x   1  f x
is integrable (by theorems 8 and 9), then
b
 hxdx  0 by theorem 10.
a
Thus,
b
 g x    1  f x dx
a
a
0   h x dx 
b
by theorem8 b
b
a
a

 g x dx    1  f x dx
b
 g x dx   1   f x dx

a

by theorem9 b
a
b
b
a
a
 g x dx   f x dx
[justifications of theorem 12:]
Let g x   M and thus g x is integrable. Then, f x   g x  . By
theorems 11 and 6,
b
b
b
a
a
a
 f xdx   g xdx   Mdx  M b  a  .
Similarly,
b
b
a
a
 f xdx   mdx  mb  a  .
4.3 The Fundamental Theorem of Calculus
Theorem 13 (Rolle’s theorem):
Let
f be continuous on
a, b
13
and differentiable on
a, b .
If
f a   f b  0 , then there exists at least one number c in a, b at
'
which f c   0 .
f(x)
[Intuitions:]
(1)
0
a
b
x
f(x)
(2)
0
a
b
x
(3)
14
f(x)
0
a
b
x
'
If (3) (figure), f x   0 in
(1) or (2), suppose
a, b . Thus,
f ' c   0, c  a, b  . If
f takes on some positive values in
a, b .
a, b , such that f x   M  0 ,
Intuitive, there is a number x1 in
1
where M is the maximum value of f x  in
a, b . Then,
f ' x1   0 .
◆
Theorem 14 (mean-value theorem):
Let
f be continuous on
a, b
and differentiable on
a, b .
If
f a   f b  0 , then there exists at least one number c in a, b at
which
f b   f a   f ' c b  a  .
15
[justifications of theorem 14:]
 f b   f a  
h x   
  x  a   f a  .

b

a


Then, let g x   f x   hx  . Since
g a  f a  ha  0, g b  f b  hb  f b  f b  0 ,
by Rolle’s theorem, there is a number c such that
 f b   f a 
g ' c   0  g ' c   f ' c   
0

 ba 
f b   f a 
 f ' c  
.
ba
◆
The fundamental theorem of calculus is the core of calculus. The
following example provides the intuition of the theorem.
Motivating Example (continue):
2
The area, A, bounded by f x   x over
0,1
is 1 3 . Note that the
'
x3
antiderivative of f is F  x   3  c and F x   f x  . As the
interval
0,1
is divided into subintervals with equal length
1 , the
n
approximated area is
1

0
 
f x dx  x1 
2
2 1
2 1
1
i 1 i 
 x2     xn  , xi  
, .
n
n
n
n
n

 
 
By mean-value theorem,
i
 i 1
 i i 1
'
F   F
  F ci    

n 
n
 n 
n
1
 ci2 
n
16
 i 1 i 
c

,  . As xi is chosen such that xi  ci , the

i
where
 n n
approximated area is
1
 f x dx
0
 
 
 
 
 
 
2 1
2 1
2 1
 x1   x 2     x n  ,
n
n
n
2 1
2 1
2 1
 c1   c 2     c n  ,
n
n
n
  1   0    2   1 
  n  1   n  2    n   n  1 
  F    F     F    F       F 
  F
   F    F 

  n   n    n   n 
  n   n    n   n 
1 0
 F 1  F 0  
3 3
1

3
Thus, it is nature to ask if in general for a function
f
with
antiderivative F
b
 f xdx F b  F a .
a
◆
Theorem 15 (fundamental theorem of calculus):
Let f be continuous on
a, b . If
F is any antiderivative of f on
a, b , then
b
 f xdx F b  F a .
a
[justifications of theorem 15:]
17
b
Since f be continuous on
a, b , then  f x dx
exists. Let
a
a  x0  x1  x2    xn1  xn  b .
Then, by mean value theorem,
F b   F a   F x n   F x0 
 F x1   F x0   F x 2   F x1     F x n 1   F x n2   F x n   F x n 1 
 
  F x x
 
 
 F ' x1 x1  x0   F ' x 2 x 2  x1     F ' x n x n  x n 1 
n

i
'
i 1
n
i
 
  f xi xi
i 1
where
xi  xi  xi 1 and xi   xi 1 , xi  . Thus,
n

F b  F a   lim F b  F a   lim  f xi xi    f x dx .
max xi 0
max xi 0
 i 1
 a
 
b
Example 3:
3
 x dx .
3
Calculate
1
[solutions:]
Since the antiderivative of
x
3
is
x4
F x  
c,
4
by the fundamental theorem
of calculus,
 34
 14
 81




x
dx

F
3

F
1


c


c



 4 .
1
4
4

 

3
3
◆
Note:
18
For convenience, the notation,
F  x  a  F b   F a  ,
b
is used.
Theorem 16 (second fundamental theorem of calculus):
a, b , then
If f be continuous on
x
Gx    f t dt ,
a
a, b , and for every x in a, b ,
is continuous, differentiable on
G ' x   f x .
[Intuition of theorem 16:]
Suppose f x  is positive. The area bounded by f x  over
x
A   f x dx  Gx  .
x
a
a
Then,
Aaxx  Aax
G x  x   G x 
G x   lim
 lim
x0
x0
x
x
Axx x
 lim
x0 x
'
In the above figure,
f x  x x  Axx x  f x x
Axx x
 f  x  x  
 f x 
x
By squeezing theorem, since
lim f x  x   f  x   lim f x  ,
x 0
x0
19
a, x
is
Axxx
G x   lim
 f x  .
x0 x
'
4.4 Integration by Substitution and Differentials
Theorem 17:

f g  x g '  x dx  F g  x   c,
where F is an antiderivative of f and c is some constant.
[justifications of theorem 17:]
dF g x 
 F ' g  x g ' x   f g x g ' x .
dx
Example 4:
Calculate
 x
2

2
 1 2 xdx .
[solutions:]
Let
f  x   x 2 , g  x   x 2  1  g '  x   2 x, F  x   x
3


, f g x   x 2  1 .
3
2
By theorem 17,
 x
2

 1 2 xdx  
2
x
f g x g x dx F g x   c 
'
2

3
1
c.
3
◆
Note:
For the purpose of computations, the following procedure can be
used to obtain the integral :
20
u  g x , g ' x  
dg x 
 du  dg x   g ' x dx
dx
  f g x g ' x dx   f u du  F u   c  F g x   c
◆
Example 4 (continue):
Calculate
 x
2

2
 1 2 xdx .
[solutions:]
Let
u  g x   x 2  1 
 x
2
dg x 
 g ' x   2 x  du  dg x   g ' x dx .
dx

 1 2 xdx   f g x g ' x dx   f g x dg x    f u du
2


3
u3
x2  1
  u du 
c 
c
3
3
2
◆
Theorem 18:
If the function u  g x has a continuous derivative on
f is continuous on the range of g x ,
g b 
b
 f g x g x dx   f u du.
'
a
g a
[Intuition of theorem 18:]
21
a, b , and
b
 f g xg xdx
'
f g xgx
a
f g xgx
x0=a x1
△x
b
g b 
 f u du
g a
f g x
f g x
g ' x  △xi
=△xi*
g a 
x 1*
x0*
22
g b 
Let
a  x0  x1  x2    xn1  xn  b
and
g a   x0  x1  g x1   x2  g x2     xn1  g xn1   xn  g b  .
Note that
23
xi  xi  xi 1  0  xi  xi  xi1  g  xi   g xi 1   0

g  xi   g  xi 1 
 g '  xi 1   g '  xi 
xi
 g  xi   g xi 1   g '  xi xi
 xi  g '  xi xi
Then,
b
 f g x g x dx
'
a
 f g  x1 g ' x1 x1  f g  x2 g '  x2 x2    f g xn g '  xn xn
 f g  x1 x1  f g  x2 x2    f g xn xn
 
 
 
 f x1 x1  f x2 x2    f xn xn

x n
 f u du
x0
g b 

 f u du
g a
Example 5:
 x

1
Calculate
2
2
 1 2 xdx .
0
[solutions:]
Let
f x   x 2 , u  g x   x 2  1  F x   x
By theorem 18,
24
3
3
.
 x
1
2
 1 2 xdx  
1
2
0
0
g 1
2
2
u3
f g x g x dx   f u du   u du 
3 1
g 0 
1
'
2
23 13


3 3
7

3
Note:
For the purpose of computations, the following procedure can be
used to obtain the definite integral :
1. The indefinite integral was computed first,
u  g x , g ' x  
dg x 
 du  dg x   g ' x dx
dx
  f g x g ' x dx   f u du  F u   c
2. Evaluate F u  g a   F g b   F g a .
g b 
Example 5 (continue):
1. The indefinite integral is
f x   x 2 , u  g x   x 2  1


u3
x  1 2 xdx 
c.
3
2
2
2.
3 g 1
u
3
g 0 
2
3 1 1
u

3
02 1
2
 23 13  7
u3




3 1  3 2  3 .
25