§4.5 Application of Definite Integrals and Area Between Curves.
9.1Concepts of Definite Integrals 9.2Finding Definite Integrals of Functions 9.3Further Techniques...
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Transcript of 9.1Concepts of Definite Integrals 9.2Finding Definite Integrals of Functions 9.3Further Techniques...
9.1 Concepts of Definite Integrals
9.2 Finding Definite Integrals of Functions
9.3 Further Techniques of Definite Integration
Chapter Summary
Case Study
Definite Integrals9
9.4 Definite Integrals of Special Functions
P. 2
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.
Case StudyCase Study
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.
My teacher said the solution is related to indefinite integrals ... but I don’t know why.
How can we calculate the area of the shaded region under the curve from x 1 to x 3?
We then use the areas of the four rectangles to estimate the area under the curve.
Each part has a width of unit unit. 4
13 2
1
P. 3
In the figure, f (x) is a non-negative function in the interval a x b.
99 .1 .1 Concepts of Definite IntegralsConcepts of Definite Integrals
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.
2. For i 1, 2, ..., n, choose a point zi in the ith subinterval and draw a rectangle with height f (zi) and width x in the ith subinterval, as sh
own in the figure.
A. A. Definition of Definite IntegralsDefinition of Definite Integrals
1. Divide the interval a x b into n equal subintervals by the points x0 ( a), x1, x2, ... , x
n ( b), such that each interval has a width of
.n
abx
Hence the sum of the areas of the n rectangles is given by:
n
ii
nnn xzfxzfxzfxzfxzf
1121 )(lim)()(...)()(
P. 4
Although the total area of the rectangles is just an approximation of the area of the bounded region,
99 .1 .1 Concepts of Definite IntegralsConcepts of Definite Integrals
A. A. Definition of Definite IntegralsDefinition of Definite Integrals
Area under the curve
n
ii
nxzf
1
)(lim
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.
Here, we define the definite integral of f (x) as follows:
Definition 9.1If f (x) is defined in the interval a a x b , the definite
integral of f (x) from a to b, which is denoted by ,
is defined as:
b
adxxf )(
n
ii
n
b
axzfdxxf
1
)(lim)(
P. 5
Note: 1. From the above definition, we can observe the meaning
behind the notation for definite integrals: ‘dx’ comes fro
m ‘x’ and the integral sign ‘’ is an elongated ‘S’ which means ‘summation’.
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.
99 .1 .1 Concepts of Definite IntegralsConcepts of Definite Integrals
A. A. Definition of Definite IntegralsDefinition of Definite Integrals
In addition, the limit sum is closely related to the indefinite integral. This will be explained in the next section.
P. 6
Example 9.1T
Solution:First we divide the interval 0 x 1 into n subintervals of widthx.
99 .1 .1 Concepts of Definite IntegralsConcepts of Definite Integrals
A. A. Definition of Definite IntegralsDefinition of Definite Integrals
Using the identity evaluate,4
)1(...21
22333 nn
n .1
0
3 dxx
Thus, .101
nnx
If we choose zi as the right end point of each subinterval, then we have
n
i
nixiazi 1
0
nn
idxx
n
in
1lim
1
31
0
3
4
3
4
3
4
3 21lim
n
n
nnn
)21(1
lim 3334
nnn
4
)1(1lim
22
4
nn
nn
2
2
4
)1(lim
n
nn
21
4
1lim
n
nn
2)01(4
1
4
1
P. 7
Example 9.2TUsing the identity sin + sin 2 + … +
evaluate
First we divide the interval 0 x 1 into n subintervals of width x.
99 .1 .1 Concepts of Definite IntegralsConcepts of Definite Integrals
A. A. Definition of Definite IntegralsDefinition of Definite Integrals
,
2sin2
21
cos2
cossin
n
n
.sinπ
0 tdt
Thus, .0
nnx
xiazi
ni0
n
i
Solution:
P. 8
Example 9.2T
Solution:
99 .1 .1 Concepts of Definite IntegralsConcepts of Definite Integrals
A. A. Definition of Definite IntegralsDefinition of Definite Integrals
nn
i
tdt
n
in 1
0
sinlim
sin
n
n
nnnnsin
2sinsinlim
n
nn
nnn
2sin2
2
1cos
2cos
lim
n
nn
n
nnn
2sin
2cos
2
2sin
2cos
2lim
n
nn
n
nnn
2sin
2cos
2
2sin
2cos
2lim
0cos10cos1
2
Using the identity sin + sin 2 + … +
evaluate
,
2sin2
21
cos2
cossin
n
n
.sinπ
0 tdt
P. 9
From the above examples, we observe that the definite
integral 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.
99 .1 .1 Concepts of Definite IntegralsConcepts of Definite Integrals
A. A. Definition of Definite IntegralsDefinition of Definite Integrals
b
adxxf )(
In other words, .
As the two integrals describe the same graph, the only difference is whether the horizontal axis is labelled ‘x-axis’ or ‘u-axis’.
b
a
b
aduufdxxf )()(
So their corresponding areas are the same and the definite integrals also have the same value.
P. 10
In this section, we will study the properties of definite integrals. Before that, let us introduce the following definitions of definite integral first:
99 .1 .1 Concepts of Definite IntegralsConcepts of Definite Integrals
Now let us consider three useful properties for evaluating integrals:
BB. . Properties of Definite IntegralsProperties of Definite Integrals
Definition 9.2Let f (x) be a continuous function in the interval a x b. Then we define
(a) (b)
.)()( b
a
a
bdxxfdxxf,0)(
a
adxxf
Properties of Definite IntegralsLet f (x) and g(x) be continuous functions in the interval a x b.
9.1.
9.2.
9.3.
b
a
b
a
b
adxxgdxxfdxxgxf )()()()(
bcadxxfdxxfdxxfb
c
c
a
b
a where,)()()(
constant a is where,)()( kdxxfkdxxkfb
a
b
a
P. 11
Example 9.3T
Solution:Evaluate
99 .1 .1 Concepts of Definite IntegralsConcepts of Definite Integrals
BB. . Properties of Definite IntegralsProperties of Definite Integrals
.10)( and 2)( ,5)( Suppose0
7
7
4
4
0 dxxgdxxfdxxf
(a)
,)(3 (a)7
0 dxxf .)(2)(6 (b)7
0 dxxgxf
7
0)(3 dxxf
7
0)(3 dxxf
7
4
4
0)()(3 dxxfdxxf
)25(3
21
(b)
7
0)](2)(6[ dxxgxf
7
0
7
0)(2)(6 dxxgdxxf
0
7
7
4
4
0)(2)()(6 dxxgdxxfdxxf
)10(2)25(6 22
P. 12
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.
99 ..22 Finding Definite Integrals ofFinding Definite Integrals of FunctionsFunctions
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
aaFbFdxxf ).()()(
P. 13
Proof: For a x b, let 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.
Then
99 ..22 Finding Definite Integrals ofFinding Definite Integrals of FunctionsFunctions
h
xAhxAxA
h
)()(lim)('
0
h
dttfdttfx
a
hx
a
h
)()(lim
0
h
dttfhx
x
h
)(lim
0
.)()( hxfdttfhx
x
h
dttfA
hx
x
h
)(lim'
0
Now, when h 0, area of PQRT area of PQRS, that is,
Thus, f (x)
∴ A(x) is a primitive function of f (x).
x
adttfxA )()(
P. 14
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 ............(*)
99 ..22 Finding Definite Integrals ofFinding Definite Integrals of FunctionsFunctions
Substitute x a into (*),
∴ A(x) F(x) – F(a) ............(**)
∴ C –F(a)
Substitute x b into (**),
0)()()( a
adttfaACaF
A(b) F(b) – F(a)
When applying the above theorem, for simplicity, we use the notation
to denote F(b) – F(a).
)()()( aFbFdxxfb
a
baxF )(
P. 15
Example 9.4T
Solution:Since
99 ..22 Finding Definite Integrals ofFinding Definite Integrals of FunctionsFunctions
Evaluate .)4(4
1
2 dxx
,43
)4(3
2 Cxx
dxx x
x4
3
3
is a primitive function of x2 + 4x. By the Fundamental
Theorem of Calculus, 4
1
34
1
2 43
)4(
x
xdxx
)1(4
3
1)4(4
3
4 33
33
P. 16
Example 9.5T
99 ..22 Finding Definite Integrals ofFinding Definite Integrals of FunctionsFunctions
Solution:
Evaluate .)(5
4
23 dte t
dtedte tt 5
4
625
4
23 )(5
4
62
2
1
te
68610
2
1
2
1 ee
)(2
1 24 ee
P. 17
Example 9.6T
99 ..22 Finding Definite Integrals ofFinding Definite Integrals of FunctionsFunctions
Solution:
Evaluate .5
0
1 x
dx
dxdx x
x
0
1
0
15
5
0
1
5ln )( dxe x
0
1
)5ln( dxe x
0
1
)5ln(
5ln
1
xe
)51(5ln
1
5ln
4
P. 18
Example 9.7T
99 ..22 Finding Definite Integrals ofFinding Definite Integrals of FunctionsFunctions
Solution:
Evaluate .3cos6
π
0 d
6
0
60
3sin3
13cos
d
)0(3sin
3
1
63sin
3
1
03
1
3
1
P. 19
Example 9.8T
99 ..22 Finding Definite Integrals ofFinding Definite Integrals of FunctionsFunctions
Solution:
Evaluate .sinπ
0
4 d
0
20
4
2
2cos1
sin
d
d
0
2 )2cos2cos21(4
1 d
0 2
4cos12cos21
4
1 d
04cos
2
12cos2
2
3
4
1 d
0
4sin8
12sin
2
3
4
1
)0(4sin8
1)0(2sin)0(
2
3
4sin2sin2
3
4
1
8
3
P. 20
Example 9.9T
99 ..22 Finding Definite Integrals ofFinding Definite Integrals of FunctionsFunctions
Solution:
(a)
(b) Hence evaluate .cossin3
π
0
5 d
(a) Find ).(sin6 xdx
d
xxxdx
dcossin6)(sin 56
(b) 3
0
5 cossin xdxx
30
5 cossin66
1xdxx
30
6 ][sin6
1
x
0sin
3sin
6
1 66
128
9
By (a)
P. 21
Similar technique can also be applied when evaluating definite integrals.
In the last chapter, we have learnt the method of integration by substitution for indefinite integrals.
99 ..33 Further Techniques of DefiniteFurther Techniques of Definite IntegraIntegrationtion
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
.)()('))(()(
)(duufduxgxgf
bg
ag
b
a
AA. . Integration by SubstitutionIntegration by Substitution
P. 22
Proof:
By the Fundamental Theorem of Calculus,
f(u)g'(x)
).()( is,that ,)()(Let ufuFdu
dCuFduuf
))).......((())(()()( )()(
)(
)(*agFbgFuFduuf bg
ag
bg
ag
dx
duuF
du
duF
dx
d )()( Rule,Chain By the
))(())(())(()())(( agFbgFxgFdxxg'xgf ba
b
a
CxgFdxxg'xgf ))(()())((
By the Fundamental Theorem of Calculus,
)(
)()()())((
bg
ag
b
aduufdxxg'xgf
)())(())(( xg'xgfxgFdx
d
99 ..33 Further Techniques of DefiniteFurther Techniques of Definite IntegraIntegrationtion
AA. . Integration by SubstitutionIntegration by Substitution
P. 23
Example 9.10T
99 ..33 Further Techniques of DefiniteFurther Techniques of Definite IntegraIntegrationtion
AA. . Integration by SubstitutionIntegration by Substitution
Solution:Let u ln x. Then
Evaluate .ln
12
e
edx
xx
.1
xdx
du
When x e, u 1.When x e2, u 2.
2
1
2
ln
1
u
dudx
xx
e
e
21][lnu
1ln2ln 2ln
P. 24
Example 9.11T
99 ..33 Further Techniques of DefiniteFurther Techniques of Definite IntegraIntegrationtion
AA. . Integration by SubstitutionIntegration by Substitution
Solution:
Evaluate .sec)tan(tan4
π
0
22 xdxxx
Let u tan x. Then du sec2xdx.When x 0, u 0.
When u 1.
40
22 sec)tan(tan xdxxx 1
0
2 )( duuu1
0
32
32
uu
3
0
2
0
3
1
2
1
6
5
,4
x
P. 25
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).
99 ..33 Further Techniques of DefiniteFurther Techniques of Definite IntegraIntegrationtion
AA. . Integration by SubstitutionIntegration by Substitution
P. 26
99 ..33 Further Techniques of DefiniteFurther Techniques of Definite IntegraIntegrationtion
Let x be a real number. 1. sin –1x is defined as the angle such that sin x
(where –1 x 1) and
2. cos –1x is defined as the angle such that cos x (where –1 x 1) and 0 .
3. tan –1x is defined as the angle such that tan x and .2
π
2
π
AA. . Integration by SubstitutionIntegration by Substitution
When using trigonometric substitution x f (), the upper and lower limits of x have to be changed to that of , and the range of follows that of f –1 (inverse function of f ) as shown below:
We can also use trigonometric substitution to evaluate definite integrals as in Chapter 8.
.2
π
2
π
P. 27
Example 9.12T
99 ..33 Further Techniques of DefiniteFurther Techniques of Definite IntegraIntegrationtion
AA. . Integration by SubstitutionIntegration by Substitution
Solution:
Evaluate .3
1
0 2 x
dx
Let x tan . ThenWhen x 0, 0.
3 .sec3 2 dθdx
When x 1, 6
1
0 23
x
dx
6
0 2
2
tan33
sec3
d
60 2
2
sec
sec
3
3
d
603
1 d
60][
3
1
0
63
1
P. 28
Example 9.13T
99 ..33 Further Techniques of DefiniteFurther Techniques of Definite IntegraIntegrationtion
AA. . Integration by SubstitutionIntegration by Substitution
Solution:
Given that evaluate ,)π()(π
0
π
0 dxxfdxxf .cos3
sin0 2
dxx
xx
0 2cos3
sindx
x
xx
0 2 )(cos3
)sin()(dx
x
xx
0 2cos3
sin)(dx
x
xx
0 20 2 cos3
sin
cos3
sindx
x
xxdx
x
x
0 20 2 cos3
sin
cos3
)cosdx
x
xx
x
xd
0 20 2 cos3
)cos
2
1
cos3
sin
x
xddx
x
xx
sin( – x) sin x, cos( – x) –cos x
0 20 2 cos3
)cos
cos3
sin2
x
xddx
x
xx
P. 29
Example 9.13T
99 ..33 Further Techniques of DefiniteFurther Techniques of Definite IntegraIntegrationtion
AA. . Integration by SubstitutionIntegration by Substitution
Solution:
6
62
2
0 2
tan33
sec3cos3
)(cos
d
x
xd
6
62
2
sec3
sec3
d
6
63
d
663
33
2
36332
1
cos3
sin 2
0
2
2
dxx
xx
Given that evaluate ,)π()(π
0
π
0 dxxfdxxf .cos3
sin0 2
dxx
xx
For , let
0 2cos3
)cos
x
xd.tan3cos x .sec3)(cosThen 2 dxd
When x 0, . When x , .6
6
P. 30
99 ..33 Further Techniques of DefiniteFurther Techniques of Definite IntegraIntegrationtion
BB. . Integration by PartsIntegration by Parts
Theorem 9.3 Integration by Parts Let u and v be two differentiable functions. Then,
. b
a
ba
b
avu'dxuvuv'dx
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:
We have already proved the corresponding theorem in the last chapter.
P. 31
Example 9.14T
99 ..33 Further Techniques of DefiniteFurther Techniques of Definite IntegraIntegrationtion
BB. . Integration by PartsIntegration by Parts
Solution:Evaluate .
1
0
3 2
dxex x
1
0
3 2
dxex x 1
0
22 )(2
1 2
xdex x
1
0
2 )(2
1 2xedx
)(][ 2
1 1
0
210
2 22
xdeex xx
10][
2
1)0(
2
1 2xee
)1(2
1
2
1 ee
2
1
P. 32
Example 9.15T
99 ..33 Further Techniques of DefiniteFurther Techniques of Definite IntegraIntegrationtion
BB. . Integration by PartsIntegration by Parts
Solution:
Evaluate .log35
1 52 xdxx
5
1 52 log3 xdxx
5
1
2 ln35ln
1xdxx
5
1
3 )(ln5ln
1xxd
5
1
351
3 )(ln]ln[5ln
1xdxxx
5
1
3 1)05ln125(
5ln
1dx
xx
5
1
3
35ln
1125
x
5ln3
124125
P. 33
Example 9.16T
99 ..33 Further Techniques of DefiniteFurther Techniques of Definite IntegraIntegrationtion
BB. . Integration by PartsIntegration by Parts
Solution:
Evaluate .2sinπ
0 xdxex
02sin xdxe x
0)(2sin xexd
0
2cos2]2sin[ xdxexe xx
0
)(2cos20 xexd
00 )2sin2(]2cos[2 dxxexe xx
0
2sin4)1(2 xdxee x
exdxex 222sin5
0
5
222sin
0
exdxe x
P. 34
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.
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
A. A. Definite Integrals of Even Functions and Definite Integrals of Even Functions and Odd FunctionsOdd Functions
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.(a) If f (x) is an even function, then .
(b) If f (x) is an odd function, then .0)( k
kdxxf
kk
kdxxfdxxf
0)(2)(
P. 35
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
A. A. Definite Integrals of Even Functions and Definite Integrals of Even Functions and Odd FunctionsOdd Functions
Proof:
k
k
k
kdxxfdxxfdxxf
0
0)()()(
Case 1: f (x) is an even function
Equation (*) becomes:
When x –k, u k. When x 0, u 0.
Let u – x. Then du –dx.
00)()(
kkduufdxxf
kduuf
0)(
kdxxf
0)(
......(*))()()(00
kkk
kdxxfdxxfdxxf
kkk
kdxxfdxxfdxxf
00)()()(
kkk
kdxxfdxxfdxxf
00)()()(
k
dxxf0
)(2
Case 2: f (x) is an odd function
Equation (*) becomes:
0
P. 36
The above theorem can be explained by the area under a curve.
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
A. A. Definite Integrals of Even Functions and Definite Integrals of Even Functions and Odd FunctionsOdd Functions
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.
So the integrals on both sides cancel each other and give a sum of zero.
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.
P. 37
Example 9.17T
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
A. A. Definite Integrals of Even Functions and Definite Integrals of Even Functions and Odd FunctionsOdd Functions
Solution:
Evaluate .sin6
6
2
xdxe x
.sin)(Let 2
xexf x
)sin()( Since2)( xexf x
),(
sin2
xf
xex
f (x) is an odd function.
0sin6
6
2
xdxex
P. 38
Example 9.18T
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
A. A. Definite Integrals of Even Functions and Definite Integrals of Even Functions and Odd FunctionsOdd Functions
Solution:
Evaluate .)secsec(tan4
π
4
π2
dxxxx
4
4
2 )secsec(tan dxxxx
4
4
24
4
secsectan xdxxdxx
Let f (x) = tan x sec x and g(x) = sec2x. )sec()tan()( xxxf )(sectan xfxx
f (x) is an odd function. )(sec)( 2 xxg )(sec2 xgx
g(x) is an even function.
4
4
2 )secsec(tan dxxxx
40
2sec20 xdx
2)01(2][tan2 40
x
P. 39
Example 9.19T
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
A. A. Definite Integrals of Even Functions and Definite Integrals of Even Functions and Odd FunctionsOdd Functions
Solution:
Evaluate .2
2
2 dxex x
.)(Let 2 xexxf xexxf 2)()( Since ),(2 xfex x
f (x) is an even function.
2
0
22
2
2 2 dxexdxex xx
2
0
22 dxex x
2
0
2 )(2 xedx
2
0
20
2 2][ 2 xdxeex xx
2
0
2 )(4)04(2 xexde
2
0
20
2 ][48 dxexee xx
20
22 ][4)02(48 xeee
44 2 e
P. 40
As we have learnt, the graph of y = sin x repeats itself at intervals of 2 as shown in the figure.
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
BB. . Definite Integrals of Periodic FunctionsDefinite Integrals of Periodic Functions
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. 41
For the definite integrals of periodic functions, we have the following theorem:
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
BB. . Definite Integrals of Periodic FunctionsDefinite Integrals of Periodic Functions
Theorem 9.5 Definite Integrals of Periodic FunctionsIf f (x) is a periodic function with period T, then
for all real constants k. TTk
kdxxfdxxf
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. 42
The following is the proof of the theorem:
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
BB. . Definite Integrals of Periodic FunctionsDefinite Integrals of Periodic Functions
Proof:
Tk
T
TkTk
kdxxfdxxfdxxfdxxf )()()()(
00
Tk
T
Tkdxxfdxxfdxxf )()()(
00
Let u x – T. Then du dx.
When x T, u 0.When x k + T, u k.
kTkTk
kduTufdxxfdxxfdxxf
000)()()()(
T
dxxf0
)(
kTk
dxxfdxxfdxxf000
)()()(
kTk
duufdxxfdxxf000
)()()(
P. 43
Example 9.20T
Solution:
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
BB. . Definite Integrals of Periodic FunctionsDefinite Integrals of Periodic Functions
Let f (x) be a periodic function with period 5 and
evaluate .
1)(5
0 dxxf
15
5)( dxxf
15
10
10
5
15
5)()()( dxxfdxxfdxxf
1015
1010
510
55)()( dxxfdxxf
5
0
5
0)()( dxxfdxxf
112
P. 44
Example 9.21T
Solution:
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
BB. . Definite Integrals of Periodic FunctionsDefinite Integrals of Periodic Functions
Let f (x) be a function and , where k is a
constant. If and g(x) is a periodic function with period T,
show that
dtktftfxgx
0
)()()(
0)(0
T
dttf
Let x = x0 where x0 is an arbitrary real constant. )()( 00 xgTxg
00
00])()[(])()[(
xTxdtktftfdtktftf
00
0
0
00])()[(])()[(])()[(
xTx
x
xdtktftfdtktftfdtktftf
Tx
xdtktftf0
0
])()[(
.
)(
)(
0
0
2
T
T
dttf
dttfk
P. 45
Solution:
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
BB. . Definite Integrals of Periodic FunctionsDefinite Integrals of Periodic Functions
g(x) is a periodic function with period T. 0)()( 00 xgTxg
0])()[(0
0
Tx
xdtktftf
0)()]([00
2 TT
dttfkdttf
T
T
dttf
dttfk
0
0
2
)(
)]([
Example 9.21T dtktftfxg
x
0
)()()(
0)(0
T
dttf
.
)(
)(
0
0
2
T
T
dttf
dttfk
Let f (x) be a function and , where k is a
constant. If and g(x) is a periodic function with period T,
show that
P. 46
Example 9.22T
Solution:
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
BB. . Definite Integrals of Periodic FunctionsDefinite Integrals of Periodic Functions
Let f (x) be an even and periodic function with period 5. If
find the value of k such that
,9)(5
5 dttf
.36)(0
k
dttf
05)(2)( dttfdttf
2
9)(
0
dttf
Let k = 5m + n, where m is a positive integer and 0 n 5.36)(
0
kdttf
36)(0
nm
dttf
36)()(50
nm
m
mdttfdttf
36)()(00
n
dttfdttfm
36)(2
90
n
dttfm
P. 47
Example 9.22T
Solution:
99 ..44 Definite Integrals of Special FunctionsDefinite Integrals of Special Functions
BB. . Definite Integrals of Periodic FunctionsDefinite Integrals of Periodic Functions
,9)(5
5 dttf
For m = 8, n = 0,
L.H.S.
0
0)(
2
89dttf
R.H.S.36
m = 8, n = 0 are solutions of the equation.
0)8(5 k
40
.36)(0
k
dttf
Let f (x) be an even and periodic function with period 5. If
find the value of k such that
P. 48
9.1 Concepts of Definite Integrals
Chapter Chapter SummarySummary
,)(lim)(1
n
ii
n
b
axzfdxxf
n
abx
,)()( (b) b
a
a
bdxxfdxxf
,0)( (a) a
adxxf
,)()()()( (d) b
a
b
a
b
adxxgdxxfdxxgxf
. where,)()()( (e) bcadxxfdxxfdxxfb
c
c
a
b
a
constant, a is where,)()( (c) kdxxfkdxxkfb
a
b
a
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
where and zi is a point in the subinterval
a + (i – 1)x x a + ix.2. Let f (x) and g(x) be continuous functions in the interval a a x
b. Then
P. 49
9.2 Finding Definite Integrals of Functions
If f (x) is a continuous function in the interval a a x b and
, then
Chapter Chapter SummarySummary
CxFdxxfb
a )()(
).()()()( aFbFxFdxxf ba
b
a
P. 50
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
2. Let u and v be differentiable functions. Then,
9.3 Further Techniques of Definite Integration
Chapter Chapter SummarySummary
.)()())(()(
)(duufduxg'xgf
bg
ag
b
a
. b
a
ba
b
avu'dxuvdxuv'
P. 51
1. Let f (x) be a continuous function and k be a constant.
2. If f (x) is a periodic function with period T, then
9.4 Definite Integrals of Special Functions
Chapter Chapter SummarySummary
.0)( k
kdxxf
.)(2)(0
kk
kdxxfdxxf
TTk
kdxxfdxxf
0)()(
(a) If f (x) is an even function, then
(b) If f (x) is an odd function, then
(i) for all real constants k,
(ii) for all positive integers n. TnT
dxxfndxxf00
)()(