Chapter 07.01
Primer on Integration
After reading this chapter, you should be able to:
1.
2.
3.
4.
define an integral,
use Riemann’s sum to approximately calculate integrals,
use Riemann’s sum and its limit to find the exact expression of integrals, and
find exact integrals of different functions such as polynomials, trigonometric
function and transcendental functions.
What is integration?
The dictionary definition of integration is combining parts so that they work together or form
a whole. Mathematically, integration stands for finding the area under a curve from one
point to another. It is represented by
b
 f ( x)dx
a
where the symbol  is an integral sign, and a and b are the lower and upper limits of
integration, respectively, the function f is the integrand of the integral, and x is the variable
of integration. Figure 1 represents a graphical demonstration of the concept.
Riemann Sum
Let f be defined on the closed interval [a, b] , and let  be an arbitrary partition of
[a, b] such as: a  x0  x1  x2  .....  xn1  xn  b , where xi is the length of the i th
subinterval (Figure 2).
If ci is any point in the i th subinterval, then the sum
n
 f (c )x , x
i 1
07.01.1
i
i
i 1
 ci  xi
07.01.2
Chapter 07.01
is called a Riemann sum of the function f for the partition  on the interval [a, b] . For a
given partition  , the length of the longest subinterval is called the norm of the partition. It is
denoted by  (the norm of  ). The following limit is used to define the definite integral.
y
y = f(x)
a
x
b
b
Figure 1 The definite integral as the area of a region under the curve, Area   f ( x)dx .
a
If ci is any point in the i
n
 f (c )x , x
i
i 1
i
i 1
th
subinterval, then the sum
 ci  xi
∆xi
x0
x1
xi-1
...
xi
…
xn-1
xn
Figure 2 Division of interval into n segments.
is called a Riemann sum of the function f for the partition  on the interval [a, b] . For a
given partition  , the length of the longest subinterval is called the norm of the partition. It is
denoted by  (the norm of  ). The following limit is used to define the definite integral.
n
lim
 0
 f (c )x
i 1
i
i
I
Primer on Integration
07.01.3
This limit exists if and only if for any positive number  , there exists a positive number 
such that for every partition  of [a, b] with    , it follows that
n
I   f (ci )xi  
i 1
for any choice of ci in the i th subinterval of  .
If the limit of a Riemann sum of f exists, then the function f is said to be integrable
over [a, b] and the Riemann sum of f on [a, b] approaches the number I .
n
lim
 0
 f (c )x
i 1
i
i
I
where
b
I   f ( x)dx
a
Example 1
Find the area of the region between the parabola y  x 2 and the x -axis on the interval
[0,4.5] . Use Riemann’s sum with four partitions.
Solution
We evaluate the integral for the area as a limit of Riemann sums. We sketch the region
(Figure 3), and partition [0,4.5] into four subintervals of length
4.5  0
x 
 1.125 .
4
25
20
y 15
10
5
0
0
1.125
2.25
x
Figure 3 Graph of the function y  x 2 .
The points of partition are
x0  0, x1  1.125, x2  2.25, x3  3.375, x4  4.5
3.375
4.5
07.01.4
Chapter 07.01
Let’s choose ci ’s to be right hand endpoint of its subinterval. Thus,
c1  x1  1.125, c2  x2  2.25, c3  x3  3.375, c4  x4  4.5
The rectangles defined by these choices have the following areas:
f (c1 )x  f (1.125)  (1.125)  (1.125) 2 (1.125)  1.4238
f (c2 )x  f (2.25)  (1.125)  (2.25) 2 (1.125)  5.6953
f (c3 )x  f (3.375)  (1.125)  (3.375) 2 (1.125)  12.814
f (c4 )x  f (4.5)  (1.125)  (4.5)2 (1.125)  22.781
The sum of the areas then is
4.5
4
0
i 1
2
 x dx   f (ci )x,
 1.4238  5.6953 12.814  22.781
= 42.715
4.5
How does this compare with the exact value of the integral  x 2 dx ?
0
Example 2
Find the exact area of the region between the parabola y  x 2 and the x  axis on the
interval [0, b] . Use Riemann’s sum.
Solution
Note that in Example 1 for y  x 2 that
f ci x  i 2 x
Thus, the sum of these areas, if the interval is divided into n equal segments is
3
n
S n   f (ci ) x
i 1
n
  i 2 (x)3
i 1
n
 (x)3  i 2
i 1
Since
b
, and
n
n
n(n  1)( 2n  1)
i2 

6
i 1
x 
then
b3 n(n  1)( 2n  1)
n3
6
3
2
b 2n  n  2n  1
=
6
n2
Sn 
Primer on Integration
07.01.5
b3 
3 1
2   2 
6
n n 
The definition of a definite integral can now be used

b

f ( x )dx  lim
x 0
a
n
 f (c )x
i
i 1
To find the area under the parabola from x  0 to x  b , we have
b
2
 x dx  lim
0
||0
n
 f (c )x
i 1
i
 lim S n
n
b3 
3 1
2  2 

n 6
n n 

3
b
 2  0  0
6
b3

3
For the value of b  4.5 as given in Example 1,
4.5
4.53
2
x
dx

0
3
= 30.375
 lim
The Mean Value Theorem for Integrals
The area of a region under a curve is usually greater than the area of an inscribed rectangle
and less than the area of a circumscribed rectangle. The mean value theorem for integrals
states that somewhere between these two rectangles, there exists a rectangle whose area is
exactly equal to the area of the region under the curve, as shown in Figure 4. Another
variation states that if a function f is continuous between a and b , then there is at least one
point in [a, b] where the function equals the average value of the function f over [a, b] .
Theorem: If the function f is continuous on the closed interval [a, b] , then there exists a
number c in [a, b] such that:
b
1
f (c ) 
f ( x)dx
b  a a
Example 3
Graph the function f ( x)  ( x  1) 2 , and find its average value over the interval [0, 3] . At what
point in the given interval does the function assume its average value?
07.01.6
Chapter 07.01
Figure 4 Mean value rectangle.
Solution
b
Average( f ) 
1
f ( x)dx
b  a a
3
1

( x  1) 2 dx

30 0
3

1
( x 2  2 x  1)dx

30
1  1
 
   27  9  3   0
3  3
 
1
The average value of the function f over the interval [0, 3] is 1. Thus, the function assumes
its average value at
f (c )  1
(c  1) 2  1
c  0, 2
The connection between integrals and area can be exploited in two ways. When a formula
for the area of the region between the x -axis and the graph of a continuous function is
known, it can be used to evaluate the integral of the function. However, if the area of region
is not known, the integral of the function can be used to define and calculate the area. Table 1
lists a number of standard indefinite integral forms.
Primer on Integration
07.01.7
y
10
9
8
7
6
5
4
3
2
1
0
x
0
1
2
3
4
5
Figure 5 The function f ( x)  ( x  1) 2 .
Example 4
Find the area of the region between the circle x 2  y 2  1 and the x -axis on the interval
[0,1] (the shaded region) in two different ways.
Solution
y
1
0
1
x
Figure 6 Graph of the function x 2  y 2  1 .
The first and easy way to solve this problem is by recognizing that it is a quarter circle.
Hence the area of the shaded area is
1
A  r 2
4
07.01.8
Chapter 07.01

1
 (1) 2
4

4
The second way is to use the integrals and the trigonometric functions. First, let’s simplify
the function x 2  y 2  1 .
x2  y2  1

y2  1 x2
y  1 x2
The area of the shaded region is the equal to
1
A   1  x 2 dx
0
We set x  sin  , dx  cos  d
1
A   1  x 2 dx
0


 1  sin   cos d
2
2
0
 2

 cos   cos d
2
0
 2

 cos
2
 d
0
By using the following formula
1  cos 2
cos 2  
,
2
we have
 2
1  cos 2
A 
d
2
0
 2

1
  2 
0
cos 2 
 d
2 
 2
sin 2 
1
  
4  0
2


   0   0  0
4



4
Primer on Integration
07.01.9
The following are some more examples of exact integration. You can use the brief table of
integrals given in Table 1.
Table 1 A brief table of integrals
 dx  x  C
 sin xdx   cos x  C
 a f ( x) dx  a  f ( x) dx  C
 cos xdx  sin x  C
 u( x)  v( x)dx   u( x)dx   v( x)dx  C  tan xdx   ln cos x  C  ln sec x  C
x n 1
 x dx  n  1  C
n
 u dv  u v   v du  C
dx
1
 ax  b  a ln ax  b  C
1
 sec( ax)dx  a ln sec( ax)  tan( ax)  C
 cot xdx   ln csc x  C  ln sin x  C
 sec
2
axdx 
1
tan( ax)  C
a
x
 a dx 
ax
C
ln a
 sec( x) tan( x)dx  sec( x)  C
ax
 e dx 
e ax
C
a
 csc( x) cot( x)dx   csc( x)  C
Example 5
Evaluate the following integral
1
 2 xe
0
 x2
dx
07.01.10
Chapter 07.01
Solution
Let u   x 2 , du  2 xdx
At x  0, u  (0) 2  0
At x  1, u  (1)2  1
1
1
x
 2 xe dx   (e )(2 xdx)
 x2
0
2
0
1
  (eu )(du)
0
 
1
  eu 0
 e1  (e0 )
=0.6321
Example 6
Evaluate
 /4

0
1  sin x
dx
cos 2 x
Solution
 /4

0
1  sin x
dx 
cos 2 x
 /4

1
  cos
2
0
x

sin x 
dx
cos 2 x 
 /4

 sec
2

x  sec x  tan x dx
0
 /4

 /4
 sec xdx   sec xtan xdx
2
0
0
 tan x0  sec x0
 4
 1  0  

 4

2 1
 2
Example 7
Evaluate  x sec 2 x dx
Solution
We use the formula
 udv  uv   vdu
Let u  x, du  dx , and dv  sec 2 x dx, v  tan x
So the new integral is
Primer on Integration
 x sec
2
07.01.11
xdx  x tan x   tan xdx
 x tan x  ln cos x  C
Example 8
Evaluate
2
 x ln xdx
1
Solution
1
x2
dx and dv  xdx, v 
x
2
Using the formula  udv  uv   vdu , the new integral is
Let u  ln x, du 
2
2

 x 2  1 
x2 
  dx 



x
ln
x
dx

ln
x



1
2 1 1  2  x 

2
2
2

x2 
x
 ln x     dx
2 1 1 2

2
2

x2   x2 
 ln x     
2 1  4 1


22  
12   2 2   12 
  ln 2     ln 1       
2 
2   4   4 


 1   4   1 
 2 ln 2   ln 1      
 2   4   4 


 1   1 
 2 ln 2    0   1  
 2   4 

 0.6362
Example 9
Evaluate
1
5x
 (4  x
0
2 2
)
dx
Solution
b
g (b)
a
g (a)
We use the formula  f ( g ( x)) g ( x)dx 
then integrating from g (a ) to g (b) .
Let
u  g ( x)  4  x 2 ,
 f (u)du , by substituting
u  g (x) , du  g ( x)dx
07.01.12
Chapter 07.01
so
g (0)  4, g (1)  5 , and
du  2 xdx
The new integral is
1
1
5x
1
5
dx

0 (4  x2 )2 0 (4  x2 )2  2  (2x)dx
5
5 1
  2 du
24u
5
5  1
  
2  u 4
5 1
1 
( )  ( )

2 5
4 
5 11
 5
( )

22 20
20

0
.
125
= 0.125

Example 10
Evaluate
4
 2 x 1dx
0
Solution
First, let’s analyze the expression 2 x  1 .
2 x  1  (2 x  1) , x 
 (2 x  1) , x 
4
1/ 2
0
0
 2 x  1 dx 
1
2
1
2
4
  (2x  1)dx   (2x  1)dx

 x x
2

12
0

1/ 2

4
 x  x 12
2
 1 1   
 1 1 
      0  16  4     
 4 2 
 4 2   
 12.5
Example 11
Evaluate
Primer on Integration
2
x

2
07.01.13
2
dx
1
Solution
2
2
2
2
 x 2  1 dx   x  1 x  1 dx
2

x  1  x  1
 x  1 x  1 dx

2

x 1

2

x 1
 x  1 x  1  x  1 x  1 dx
2
1
1
 x  1 dx   x  1 dx
 lim ln x  1 b  lim ln x  1 b
2
b
2
b
b
 x 1 
 lim ln

b
 x  1  2
b 1 
 3
 lim ln
 ln
b  
b  1 
 1
b 1 

 ln 3  ln  lim

 b b  1 
 ln 3  ln 1
 ln 3
 1.0986
Example 12
1
Graph the function y  ( x 2  2) 3 / 2 , and find the length of the curve from x  0 to x  3 .
3
Solution
We use the equation
b
dy
L   1  ( ) 2 dx
dx
a
We have:
1
y  ( x 2  2) 3 / 2
3
So,
3 / 21
dy  1   3 
       x2  2
 2 x 
dx  3   2 

 x x2  2

07.01.14
Chapter 07.01
3


2
L   1  x x 2  2 dx
y
0
50
45
40
35
30
25
20
15
10
5
0
0
1
2
3
x
1
Figure 7 Graph of the function y  ( x 2  2) 3 / 2
3
3
  1  x 2 ( x 2  2) dx
0
3
  1  x 4  2 x 2 dx
0
3
  ( x 2  1) 2 dx
0
3
  ( x 2  1)dx
0
3
 x3

   x
3
0
 12
Example 13
Find the area of the shaded region given in Figure 8.
4
5
6
Primer on Integration
07.01.15
f ( x)  1
1
y
0.8
g ( x)  cos 2 x
0.6
0.4
0.2
0

2
0
Figure 8 Graph of the function cos 2 x .
Solution
For the sketch given,

a  , b   , and
2
f ( x)  g ( x)  1  cos 2 x  sin 2 x

A
 sin x  dx
2
2


1  cos 2 x
dx
2
2




1
 2 
 
2
cos 2 x 
dx
2 

 x sin 2 x 
 
4 
2
2


   
  sin 2    sin 2 2  

 
  
  




4 
4
4
 2






 
 

   0     0 
 4

 2
x

07.01.16
Chapter 07.01

4

Example 14
Find the volume of the solid generated by revolving the shaded region in Figure 9 about the
y-axis.
y
y=1
x  tan(  / 4) y
x
Figure 9 Volume generated by revolving shaded region.
Solution
b
We use the formula V    ( radius ) 2 dy
a
Let
u

y, du 

dy .
4
4
Therefore, at y  0, u  0
y  1, u 

4
1
V    R y  dy
2
0
 
    tan 
4
0
1
4  
     tan 
 0 4
1
2

y  dy

2
 
y  dy
 4
Primer on Integration
07.01.17
 4
 4  tan u  du (Choosing u 
2
0
 4


4
y)

 4   1  sec 2 u du
0
 4 u  tan u 0
 4
 


 4   tan   0  tan 0
4
 4

 


 4    1  0  0

 4

 0.8584
INTEGRATION
Topic
Primer on integration
Summary These are textbook notes of a primer on integration.
Major
General Engineering
Authors
Autar Kaw, Loubna Guennoun
Date
February 12, 2016
Web Site http://numericalmethods.eng.usf.edu