Transcript Module 2 Chapter 9 Definite Integrals

9 Definite Integrals
Case Study
9.1
Concepts of Definite Integrals
9.2
Finding Definite Integrals of Functions
9.3
Further Techniques of Definite Integration
9.4
Definite Integrals of Special Functions
Chapter Summary
Case Study
How can we calculate the
area of the shaded region
under the curve from x  1
to x  3?
My teacher said the solution is
related to indefinite integrals ...
but I don’t know why.
Suppose we want to find the area
of the region bounded by the curve
y  x2, the x-axis and the vertical
lines x  1 and x  3.
As the bounded region is irregular,
we cannot calculate the area by
using a simple formula.
Suppose we divide the interval 1  x  3 into
four equal parts at the points 1.5, 2 and 2.5 as
shown in the figure.
3 1
1
Each part has a width of
unit  unit.
4
2
We then use the areas of the four rectangles to
estimate the area under the curve.
P. 2
9.1 Concepts of Definite Integrals
A. Definition of Definite Integrals
In the figure, f (x) is a non-negative function in the interval a  x  b.
We can find the area of the region bounded by the curve y  f (x),
the x-axis, the lines x  a and x  b by the following steps.
Divide the interval a  x  b
into n equal subintervals by
the points x0 ( a), x1, x2, ... ,
xn ( b), such that each
interval has a width of
ba
Dx 
.
n
2. For i  1, 2, ..., n, choose a point zi in the ith subinterval and draw a
rectangle with height f (zi) and width Dx in the ith subinterval, as
shown in the figure.
Hence the sum of the areas of the n rectangles is given by:
1.
n
f ( zi )Dx

n
f ( z1 )Dx  f ( z2 )Dx  ...  f ( zn1 )Dx  f ( zn )Dx  lim
i 1
P. 3
9.1 Concepts of Definite Integrals
A. Definition of Definite Integrals
Although the total area of the rectangles is just an
approximation of the area of the bounded region,
we can observe that when the width of each rectangle is getting
smaller (as n approaches infinity), the sum of the areas of the
rectangles becomes closer to the area under the curve.
Area under the curve  lim
n
 f ( z i ) Dx
n   i 1
Here, we define the definite integral of f (x) as follows:
Definition 9.1
If f (x) is defined in the interval a a  x  b , the definite
b
integral of f (x) from a to b, which is denoted by a f ( x) dx ,
is defined as:
n
f ( z i ) Dx

a f ( x)dx  nlim

b
i 1
P. 4
9.1 Concepts of Definite Integrals
A. Definition of Definite Integrals
Note:
1. From the above definition, we can observe the meaning
behind the notation for definite integrals: ‘dx’ comes
from ‘Dx’ and the integral sign ‘ ’ is an elongated ‘S’
which means ‘summation’.
In addition, the limit sum is closely related to the indefinite
integral. This will be explained in the next section.
2. In the definite integral, a and b are called the lower limit and the
upper limit respectively.
3. Note that the infinite sum is equal to the area under the curve only
if the curve is non-negative in the interval. In next chapter, we will
learn how to find the area under the curve by using definite integral.
P. 5
9.1 Concepts of Definite Integrals
A. Definition of Definite Integrals
Example 9.1T
1 3
n 2 (n  1) 2
Using the identity 1  2  ...  n 
, evaluate 0 x dx.
4
3
Solution:
3
3
First we divide the interval 0  x  1
into n subintervals of width Dx.
1 0 1
Thus, Dx 
 .
n
n
If we choose zi as the right end point
of each subinterval, then we have
1 i
zi  a  iDx  0  i  
n n
3
n
1 3
i
  1
x
dx

lim
  

0
n  i 1  n  n
 13 23
n3 
 lim  4  4    4 
n  n
n
n 

1 3
3
3
(
1

2



n
)
4
n  n
1 n 2 (n  1) 2
 lim 4 
n  n
4
(n  1) 2
 lim
n  4 n 2
2
1  n  1
 lim  

n  4  n 
1
 (1  0) 2
4
1

4
 lim
P. 6
9.1 Concepts of Definite Integrals
A. Definition of Definite Integrals
Example 9.2T
Using the identity sin q + sin 2q + … + sin nq 
evaluate
π
0
sin tdt.
q
1

cos  cos n  
2
2

2 sin
q
,
2
Solution:
First we divide the interval 0  x  1 into n subintervals of width Dx.
0 
 .
Thus, Dx 
n
n
zi  a  iDx

 0i
n
i

n
P. 7
9.1 Concepts of Definite Integrals
A. Definition of Definite Integrals
Example 9.2T
Using the identity sin q + sin 2q + … + sin nq 
evaluate
π
0
sin tdt.
q
1

cos  cos n  
2
2

2 sin
Solution:
q
,
2

 



cos




0
 cos
2n 

2n
2n  2n

lim




n  
 n
i  

sin
sin
 lim   sin  


2n
2n
n i 1
n n


 
 
 
2
n 
cos
cos


 lim  sin  sin
   sin 
2
n
2
n
2
n
2
n

n  n 

n
n
n   lim 


n 

sin
 sin

1




2n
2n 
cos  cos n  

2n
2 n

 1 cos 0  1 cos 0
 lim 

n  n
2 sin
2
2n

sin tdt
P. 8
9.1 Concepts of Definite Integrals
A. Definition of Definite Integrals
From the above examples, we observe that the definite
b
integral a f ( x) dx is a real number which is independent of
the variable x.
We say that x is a dummy variable and it can be replaced by another
letter, say u, without changing the value of the integral.
b
b
In other words, a f ( x)dx  a f (u )du .
As the two integrals describe the same graph, the only difference is
whether the horizontal axis is labelled ‘x-axis’ or ‘u-axis’.
So their corresponding areas
are the same and the definite
integrals also have the same
value.
P. 9
9.1 Concepts of Definite Integrals
B. Properties of Definite Integrals
In this section, we will study the properties of definite integrals.
Before that, let us introduce the following definitions of definite
integral first:
Definition 9.2
Let f (x) be a continuous function in the interval a  x  b.
Then we define
a
a
(a)
f ( x)dx  0,
a
(b)
b
b
f ( x)dx    f ( x)dx.
a
Now let us consider three useful properties for evaluating integrals:
Properties of Definite Integrals
Let f (x) and g(x) be continuous functions in the interval a  x  b.
9.1.
9.2.
9.3.
b
b
kf ( x)dx  k  f ( x)dx, where k is a constant
a
a
b
b
b


f
(
x
)

g
(
x
)
dx

f
(
x
)
dx

a
a
a g ( x)dx
b
c
b
f
(
x
)
dx

f
(
x
)
dx

a
a
c f ( x)dx, where a  c  b
P. 10
9.1 Concepts of Definite Integrals
B. Properties of Definite Integrals
Example 9.3T
4
7
0
Suppose  f ( x)dx  5,  f ( x)dx  2 and  g ( x)dx  10.
0
4
7
7
7
Evaluate (a)  3 f ( x)dx, (b)  6 f ( x)  2 g ( x)dx.
0
0
Solution:
(a)
7
0
7
3 f ( x)dx  3 f ( x)dx
0
4
7
 3 f ( x)dx   f ( x)dx
 0

4
 3(5  2)
 21
(b)
7
7
7
 0 [6 f ( x)  2 g ( x)]dx  6 0 f ( x)dx  2 0 g ( x)dx
4
7
0
 6 f ( x)dx   f ( x)dx  2 g ( x)dx
 0

4
7
 6(5  2)  2(10)
 22
P. 11
9.2 Finding Definite Integrals of
Functions
In the last section, we saw that it is very tedious and time consuming to
evaluate a definite integral from the definition.
In this section, we will introduce the Fundamental Theorem of
Calculus, which enables us to evaluate a definite integral more
efficiently.
Theorem 9.1 Fundamentals Theorem of Calculus
If f (x) be a continuous function in the interval a  x  b
and F (x) is a primitive function of f (x), then
b
a f ( x)dx  F (b)  F (a).
P. 12
9.2 Finding Definite Integrals of
Functions
Proof:
x
For a x  b, let A( x)  a f (t )dt be the area of the region
enclosed by the curve y  f (t), the t-axis, the vertical lines
t  a and t  x as shown in the figure.
A( x  h)  A( x)
A
'
(
x
)

lim
Then
h 0
h
xh
x
f
(
t
)
dt

f (t )dt


a
a
 lim
h0
h
xh
f (t )dt

x
 lim
h0
h
Now, when h  0, area of PQRT  area of PQRS, that is,
xh
x
f (t )dt  f ( x)  h.
xh

Thus, A'  lim x
h0
f (t )dt
f (x)
h
∴ A(x) is a primitive function of f (x).
P. 13
9.2 Finding Definite Integrals of
Functions
Since F(x) is also a primitive function of f (x), it differs from
A(x) by just a constant, for example C.
Then we have A(x)  F(x) + C ............(*)
Substitute x  a into (*),
a
F (a)  C  A(a)   f (t )dt  0
a
∴
C  –F(a)
∴ A(x)  F(x) – F(a) ............(**)
Substitute x  b into (**),
A(b)  F(b) – F(a)

b
a f ( x)dx  F (b)  F (a)
When applying the above theorem, for simplicity, we use the notation
F ( x)ba
to denote F(b) – F(a).
P. 14
9.2 Finding Definite Integrals of
Functions
Example 9.4T
4
Evaluate  ( x 2  4)dx.
1
Solution:
x3
Since  ( x  4)dx   4 x  C ,
3
3
x
 4 x is a primitive function of x2 + 4x. By the Fundamental
3
Theorem of Calculus,
4
 x3

4 2
1 ( x  4)dx   3  4 x

1
2
 43
 13

   4(4)    4(1)
3
 3

 33
P. 15
9.2 Finding Definite Integrals of
Functions
Example 9.5T
5
Evaluate  (e 3t )  2 dt.
4
Solution:
(e 3t )  2 dt   e 2t 6 dt
5
4
5
4
5
1

  e 2t  6 
2
4
1 10  6 1 8 6
 e
 e
2
2
1
 (e 4  e 2 )
2
P. 16
9.2 Finding Definite Integrals of
Functions
Example 9.6T
0 dx
Evaluate 1 x .
5
Solution:
0 x
dx
 1 5 x   1 5 dx
0
  (e ln 5 )  x dx
0
1
  e (  ln 5) x dx
0
1
0
 1 (  ln 5) x 

e

  ln 5
1
1

(1  5)
 ln 5
4

ln 5
P. 17
9.2 Finding Definite Integrals of
Functions
Example
9.7T
π
Evaluate  6 cos 3qdq .
0
Solution:


6 cos 3q
0

6
1
dq   sin 3q 
3
0
1
   1

  sin 3    sin 3(0)
 6   3

3
1
 0
3

1
3
P. 18
9.2 Finding Definite Integrals of
Functions
Example 9.8T
π
Evaluate  sin 4 qdq .
0
Solution:

0

sin 4 q dq
  1  cos 2q
2

 
 dq
0
2

1 
  (1  2 cos 2q  cos 2 2q )dq
4 0

1 
1  cos 4q 
1

2
cos
2
q


dq

0
4 
2


1  3
1


2
cos
2
q

cos
4
q

dq

0
4 2
2

1 3
1

  q  sin 2q  sin 4q 
4 2
8
0
1  3


    sin 2  sin 4 
4  2


1
3

  (0)  sin 2(0)  sin 4(0) 
8
2

3

8
P. 19
9.2 Finding Definite Integrals of
Functions
Example 9.9T
d
(sin 6 x).
π
dx
5
(b) Hence evaluate 03 sin q cosqdq .
(a) Find
Solution:
d
(sin 6 x)  6 sin 5 x cos x
(a)
dx


1
5
(b)  3 sin x cos xdx   3 6 sin 5 x cos xdx
0
6 0

1
 [sin 6 x]03
6
1


  sin 6  sin 6 0 
6
3

9

128
By (a)
P. 20
9.3 Further Techniques of Definite
Integration
A. Integration by Substitution
In the last chapter, we have learnt the method of integration
by substitution for indefinite integrals.
Similar technique can also be applied when evaluating definite integrals.
Theorem 9.2 Integration by Substitution
Let u  g(x) be a differentiable function in the interval a  x  b.
If y  f(u) is a continuous function in the interval g(a)  u  g(b),
then we have
b
a
f ( g ( x)) g ' ( x)du  
g (b )
g (a)
f (u )du.
P. 21
9.3 Further Techniques of Definite
Integration
A. Integration by Substitution
Proof:
d
F (u )  f (u ).
Let  f (u )du  F (u )  C , that is,
du
By the Fundamental Theorem of Calculus,
g (b )
g (a)
f (u )du  F (u )gg ((ba))  F ( g (b))  F ( g (a)).......(* )
d
d
du




By the Chain Rule,
F (u ) 
F (u )   f(u)g'(x)
dx
du
dx
d
F ( g ( x))  f ( g ( x)) g' ( x)
dx
  f ( g ( x)) g' ( x)dx  F ( g ( x))  C
By the Fundamental Theorem of Calculus,
b
b


f
(
g
(
x
))
g'
(
x
)
dx

F
(
g
(
x
))
a  F ( g (b))  F ( g ( a ))

a
b
g (b )
a
g (a)
  f ( g ( x)) g' ( x)dx  
f (u )du
P. 22
9.3 Further Techniques of Definite
Integration
A. Integration by Substitution
Example 9.10T
e
1
dx.
Evaluate e
x ln x
2
Solution:
Let u  ln x. Then
du 1
 .
dx x
When x  e, u  1.
When x  e2, u  2.

e2
e
2 du
1
dx  
1 u
x ln x
 [ln u]12
 ln 2  ln 1
 ln 2
P. 23
9.3 Further Techniques of Definite
Integration
A. Integration by Substitution
Example
9.11T
π
Evaluate  4 (tan x  tan 2 x) sec 2 xdx.
0
Solution:
Let u  tan x. Then du  sec2xdx.
When x  0, u  0.

When x  , u  1.
4

1
  4 (tan x  tan 2 x) sec 2 xdx   (u  u 2 )du
0
0
1
u 2 u 3 
  
3 0
2
 1 1  0 0
     
 2 3  2 3
5

6
P. 24
9.3 Further Techniques of Definite
Integration
A. Integration by Substitution
In the above example, we can see that after substitution,
the upper limit may become smaller than the lower limit.
This is the reason why we need to learn Definition 9.2(b).
P. 25
9.3 Further Techniques of Definite
Integration
A. Integration by Substitution
We can also use trigonometric substitution to evaluate
definite integrals as in Chapter 8.
When using trigonometric substitution x  f (q), the upper and lower
limits of x have to be changed to that of q, and the range of q follows
that of f –1 (inverse function of f ) as shown below:
Let x be a real number.
1. sin –1x is defined as the angle q such that sin q  x
π
π
(where –1  x  1) and   q  .
2
2
–1
2. cos x is defined as the angle q such that cos q  x
(where –1  x  1) and 0  q  .
π
π
–1


q

.
3. tan x is defined as the angle q such that tan q  x and
2
2
P. 26
9.3 Further Techniques of Definite
Integration
A. Integration by Substitution
Example 9.12T
1 dx
.
Evaluate 0
2
3 x
Solution:
Let x  3 tan q. Then dx  3 sec2 q dθ.
When x  0, q  0.

When x  1, q 
6

2
1 dx
3
sec
q dq
6

 
 0 3  3 tan 2 q
0 3  x2
3 6 sec 2 q dq

3  0 sec 2 q
1 6

dq

0
3

[q ]06
1
3
1  

  0
36




 
P. 27
9.3 Further Techniques of Definite
Integration
A. Integration by Substitution
Example 9.13T
π
π
 x sin x
dx.
Given that 0 f ( x)dx  0 f ( π  x)dx, evaluate 0
2
3  cos x
Solution:
(  x) sin(   x)
x sin x
 0 3  cos2 x dx   0 3  cos2 (  x) dx
(   x ) sin x
sin( – x)  sin x, cos( – x)  –cos x

dx
0 3  cos 2 x
  sin x
 x sin x

dx

 0 3  cos2 x dx
0 3  cos 2 x
 d  cos x)
 x sin x
 

dx
0 3  cos 2 x  0 3  cos 2 x
 x sin x
 d  cos x)
 2
dx


 0 3  cos2 x
0 3  cos 2 x
 x sin x
1  d  cos x)
dx


 0 3  cos2 x
2  0 3  cos2 x

P. 28
9.3 Further Techniques of Definite
Integration
A. Integration by Substitution
Example 9.13T
π
π
 x sin x
dx.
Given that 0 f ( x)dx  0 f ( π  x)dx, evaluate 0
2
3  cos x
Solution:
For
d  cos x )
 0 3  cos 2 x
, let cos x  3 tan q . Then d (cos x)  3 sec2 q dq .


When x  0, q  . When x  , q   .
6
6
 d (cos x)

 
0 3  cos2 x

2

3
sec
q dq
   6

2
3  3 tan q
6
 

6


6
 6
dq
3  
6
     
 
3  6  6 
2

3 3
3 sec 2 q
dq
2
3 sec q


0
x sin x
1  2  2

dx    
2
2  3 3  6 3
3  cos x
P. 29
9.3 Further Techniques of Definite
Integration
B. Integration by Parts
In Chapter 8, we have already learnt the method of integration
by parts for indefinite integrals.
For definite integrals, we can also apply a similar method to
evaluate integrals and it is stated as follows:
Theorem 9.3 Integration by Parts
Let u and v be two differentiable functions. Then,
uv'dx  uvba   vu'dx.
b
a
b
a
We have already proved the corresponding theorem in the last chapter.
P. 30
9.3 Further Techniques of Definite
Integration
B. Integration by Parts
Example 9.14T
1 3 x
Evaluate 0 x e dx.
2
Solution:

1 3 x2
x e dx
0
1 1
  x 2e x d ( x 2 )
2 0
1 1
  x 2 d (e x )
2 0
1
1
 [ x 2 e x ]10   e x d ( x 2 )
0
2
1
1
 (e  0)  [e x ]10
2
2
1
1
 e  (e  1)
2
2
1

2
2
2

2
2

2
P. 31
9.3 Further Techniques of Definite
Integration
B. Integration by Parts
Example 9.15T
5
Evaluate  3x 2 log 5 xdx.
1
Solution:
5
1
1 5 2
3 x ln xdx

1
ln 5
1 5

ln xd ( x 3 )

ln 5 1
5
1  3

[ x ln x]15   x 3d (ln x)

1
ln 5 
5 3 1
1 


(
125
ln
5

0
)

x

dx
1 x 
ln 5 
3 5

1 x
 125 
 
ln 5  3 1
124
 125 
3 ln 5
3x 2 log 5 xdx 
P. 32
9.3 Further Techniques of Definite
Integration
B. Integration by Parts
Example 9.16T
π
Evaluate  e x sin 2 xdx.
0
Solution:

 x
e
0

sin 2 xdx   sin 2 xd (e x )
0
x

 [e sin 2 x]   e x  2 cos 2 xdx

0
 0  2 cos 2 xd (e x )
0


 2[e x cos 2 x]0   e x (2 sin 2 x)dx


0
 x

 2(e  1)  4 e sin 2 xdx
0
 5 e x sin 2 xdx  2  2e 
0

 x
e
0
2  2e 
sin 2 xdx 
5
P. 33
9.4 Definite Integrals of Special Functions
A. Definite Integrals of Even Functions and
Odd Functions
In Chapter 7, we learnt that for a function f (x),
(i) if f (–x)  f (x), then it is called an even function, and
(ii) if f (–x)  –f (x), then it is called an odd function.
For these functions, we have the following theorems for their
definite integrals:
Theorem 9.4 Definite Integrals of Odd and Even Functions
Let k be a constant and f (x) be a continuous function in the interval
–k  x  k.
k
k
(a) If f (x) is an even function, then  f ( x)dx  2 f ( x)dx.
k
0
k
(b) If f (x) is an odd function, then k f ( x)dx  0.
P. 34
9.4 Definite Integrals of Special Functions
A. Definite Integrals of Even Functions and
Odd Functions
Proof:
k
f ( x)dx  
k
0
k
k
f ( x)dx   f ( x)dx
0
Let u  – x. Then du  –dx.
When x  –k, u  k.
When x  0, u  0.

0
k
k
k
0
k
k
k
k
0
0
k
0
0
f ( x)dx    f (u )du   f (u )du   f (  x) dx
f ( x)dx   f ( x)dx   f ( x)dx......(*)
Case 1: f (x) is an even function
Equation (*) becomes:
k
k
k
k
k
0
0
0
f ( x)dx   f ( x)dx   f ( x)dx  2  f ( x)dx
Case 2: f (x) is an odd function
Equation (*) becomes:
k
k
k
k
f ( x)dx    f ( x)dx   f ( x)dx  0
0
0
P. 35
9.4 Definite Integrals of Special Functions
A. Definite Integrals of Even Functions and
Odd Functions
The above theorem can be explained by the area under
a curve.
For an even function, the curve is symmetrical about
the y-axis as shown in the figure.
Therefore, the areas under the curve on the L.H.S.
and R.H.S. of the y-axis should be the same.
The total area is twice that on the R.H.S. of the
y-axis, as stated in the theorem.
In the case of an odd function, the curve is symmetrical
about the origin as shown in the figure.
Therefore, the areas under the curve on the L.H.S.
and R.H.S. of the y-axis are the same, but they are
in opposite signs.
So the integrals on both sides cancel each other
and give a sum of zero.
P. 36
9.4 Definite Integrals of Special Functions
A. Definite Integrals of Even Functions and
Odd Functions
Example 9.17T
6
Evaluate  e x sin xdx.
6
2
Solution:
2
Let f ( x)  e x sin x.
Since f ( x)  e
( x)2
sin(  x)
2
 e x sin x
  f ( x),
 f (x) is an odd function.

6
 6 e
x2
sin xdx  0
P. 37
9.4 Definite Integrals of Special Functions
A. Definite Integrals of Even Functions and
Odd Functions
Example
9.18T
π
Evaluate  4π (tan x sec x  sec 2 x)dx.

Solution: 4


4 (tan x sec x  sec 2


4


x)dx  4 tan x sec xdx  4 sec 2


4
4
and g(x) = sec2x.


xdx
Let f (x) = tan x sec x
f ( x)  tan( x) sec( x)   tan x sec x   f ( x)
 f (x) is an odd function.
g ( x)  sec2 ( x)  sec2 x  g ( x)
 g(x) is an even function.


4 (tan x sec x  sec 2


4

4 sec 2 xdx
0

x]04  2(1  0)
x)dx  0  2
 2[tan
2
P. 38
9.4 Definite Integrals of Special Functions
A. Definite Integrals of Even Functions and
Odd Functions
Example 9.19T
2
Evaluate  x 2 e x dx.
2
Solution:
Let f ( x)  x 2 e .
x
x
Since f ( x)  ( x) 2 e
 x 2 e  f ( x),
x
 f (x) is an even function.
2
 2
2
x 2 e dx  2 x 2 e dx
0
 2(4e 2  0)  4 xd (e x )
 2
2 2 x
x e dx
0
 8e 2  4 [ xe x ]02   e x dx
 2
2 2
x d (e x )
0
x

2
x

0
2
0

 8e 2  4(2e 2  0)  4[e x ]02
2

 2 [ x 2 e x ]02   e x  2 xdx
0
 4e 2  4
P. 39
9.4 Definite Integrals of Special Functions
B. Definite Integrals of Periodic Functions
As we have learnt, the graph of y = sin x repeats itself at intervals of 2
as shown in the figure.
This means that sin (x + 2) = sin x for all x.
Thus we say that y = sin x is a periodic function
with a period of 2.
Definition 9.3
A function f (x) is said to be a periodic function if there is a positive
value T, such as f (x+T) = f (x) for real values of x. in this case, the
smallest value of T which satisfies the relationship above is called the
period of f (x).
P. 40
9.4 Definite Integrals of Special Functions
B. Definite Integrals of Periodic Functions
For the definite integrals of periodic functions, we have
the following theorem:
Theorem 9.5 Definite Integrals of Periodic Functions
If f (x) is a periodic function with period T, then
k T
k
T
f ( x)dx   f ( x)dx for all real constants k.
0
Theorem 9.5 tells us that if a function f (x) is periodic with period T,
then the definite integral of f (x) over an interval of length T is a
constant, regardless of the position of the interval.
P. 41
9.4 Definite Integrals of Special Functions
B. Definite Integrals of Periodic Functions
The following is the proof of the theorem:
Proof:
k T
k
k
T
k T
0
0
T
f ( x)dx   f ( x)dx   f ( x)dx  
f ( x)dx
k
T
k T
0
0
T
   f ( x)dx   f ( x)dx  
f ( x)dx
Let u  x – T. Then du  dx.
When x  T, u  0.
When x  k + T, u  k.

k T
k
k
T
k
0
k
0
T
0
k
0
k
0
T
0
k
0
0
0
f ( x)dx    f ( x)dx   f ( x)dx   f (u  T )du
   f ( x)dx   f ( x)dx   f (u )du
   f ( x)dx   f ( x)dx   f ( x)dx
T
  f ( x)dx
0
P. 42
9.4 Definite Integrals of Special Functions
B. Definite Integrals of Periodic Functions
Example 9.20T
Let f (x) be a periodic function with period 5 and
evaluate
15
5
5
0 f ( x)dx  1
f ( x)dx .
Solution:
15
5
f ( x)dx  
10
f ( x)dx  
5
10 5

55
5
15
10
f ( x)dx  
f ( x)dx
15 10
10 10
f ( x)dx
5
  f ( x)dx   f ( x)dx
0
0
 1 1
2
P. 43
9.4 Definite Integrals of Special Functions
B. Definite Integrals of Periodic Functions
Example 9.21T
Let f (x) be a function and g ( x)   f (t ) f (t )  k dt , where k is a
x
constant. If
T
0
show that k 
0
f (t )dt  0 and g(x) is a periodic function with period T,
2


f
(
t
)
dt
0
T
T
0
.
f (t )dt
Solution:
Let x = x0 where x0 is an arbitrary real constant.
g ( x0  T )  g ( x0 )



x0 T
0
x0
f (t )[ f (t )  k ]dt  
0
x0 T
x0
f (t )[ f (t )  k ]dt  
x0
0
x0 T
x0
f (t )[ f (t )  k ]dt
f (t )[ f (t )  k ]dt  
x0
0
f (t )[ f (t )  k ]dt
f (t )[ f (t )  k ]dt
P. 44
9.4 Definite Integrals of Special Functions
B. Definite Integrals of Periodic Functions
Example 9.21T
Let f (x) be a function and g ( x)   f (t ) f (t )  k dt , where k is a
x
constant. If
T
0
show that k 
0
f (t )dt  0 and g(x) is a periodic function with period T,
2


f
(
t
)
dt
0
T
T
0
.
f (t )dt
Solution:
 g(x) is a periodic function with period T.
g ( x0  T )  g ( x0 )  0
x0 T
x
T
0
0
f (t )[ f (t )  k ]dt  0
T
[ f (t )]2 dt  k  f (t )dt  0
0
k
T
0
[ f (t )]2 dt
T
0
f (t )dt
P. 45
9.4 Definite Integrals of Special Functions
B. Definite Integrals of Periodic Functions
Example 9.22T
Let f (x) be an even and periodic function with period 5. If
k
find the value of k such that 0 f (t )dt  36.
Solution:

5
5 f (t )dt  9,

 5 f (t )dt  2 0 f (t )dt

9
  f (t )dt 
0
2
Let k = 5m + n, where m is a positive integer and 0  n  5.
k
 f (t )dt  36
0
m  n
 0 f (t )dt  36
m
m  n
 0 f (t)dt   5m n f (t )dt  36
m f (t )dt   f (t )dt  36
0
0
n
9m
  f (t )dt  36
0
2
P. 46
9.4 Definite Integrals of Special Functions
B. Definite Integrals of Periodic Functions
Example 9.22T
Let f (x) be an even and periodic function with period 5. If
k
find the value of k such that 0 f (t )dt  36.
5
5 f (t )dt  9,
Solution:
For m = 8, n = 0,
0
98


f (t )dt
L.H.S.

0
2
 36
 R.H.S.
 m = 8, n = 0 are solutions of the equation.
 k  5(8)  0
 40
P. 47
Chapter Summary
9.1 Concepts of Definite Integrals
1.
If f (x) be a function defined in the interval a a  x  b. The definite
integral of f (x) from a to b is given by
n
 f ( zi )Dx,
a f ( x)dx  nlim

b
2.
i 1
ba
where Dx 
and zi is a point in the subinterval
n
a + (i – 1)Dx  x  a + iDx.
Let f (x) and g(x) be continuous functions in the interval a a  x  b.
a
Then
(a)  f ( x)dx  0,
(b)
(c)
(d)
(e)
a
a
b
f ( x)dx    f ( x)dx,
b
a
b
b
kf
(
x
)
dx

k
a
a f ( x)dx, where k is a constant,
b
b
b


f
(
x
)

g
(
x
)
dx

f
(
x
)
dx

a
a
a g ( x)dx,
b
c
b
f
(
x
)
dx

f
(
x
)
dx

a
a
c f ( x)dx, where a  c  b.
P. 48
Chapter Summary
9.2 Finding Definite Integrals of Functions
If f (x) is a continuous function in the interval a a  x  b and
b
a f ( x)dx  F ( x)  C , then
b
b


f
(
x
)
dx

F
(
x
)
a  F (b)  F ( a ).
a
P. 49
Chapter Summary
9.3 Further Techniques of Definite Integration
1.
Let u  g(x) be a differentiable function in the interval a  x  b.
If y  f(u) is a continuous function in the interval g(a)  u  g(b),
then
b
g (b )
 f ( g ( x)) g' ( x)du   f (u)du.
a
2.
g (a)
Let u and v be differentiable functions. Then,
a uv'dx  uv  a vu'dx.
b
b
a
b
P. 50
Chapter Summary
9.4 Definite Integrals of Special Functions
1.
Let f (x) be a continuous function and k be a constant.
(a) If f (x) is an even function, then
k
k
k f ( x)dx  20 f ( x)dx.
(b) If f (x) is an odd function, then
k
 f ( x)dx  0.
k
2.
If f (x) is a periodic function with period T, then
(i)
(ii)
k T
k
nT
0
T
f ( x)dx   f ( x)dx for all real constants k,
0
T
f ( x)dx  n  f ( x)dx for all positive integers n.
0
P. 51