9.1Concepts of Definite Integrals 9.2Finding Definite Integrals of Functions 9.3Further Techniques of Definite Integration Chapter Summary Case Study Definite.

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,



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.



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.



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.



,

2sin2

21

cos2

cossin

n

n

.sinπ

0 tdt

Thus, .0

nnx

xiazi

ni0

n

i

Solution:

P. 8

Example 9.2T

Solution:



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.



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:


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


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


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 ............(*)


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


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


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


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


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


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


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



P. 23

Example 9.10T



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



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).



P. 26


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


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



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



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



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


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



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



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



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



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.



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



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



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



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.


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:



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:



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:



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:



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:



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:



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:



,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


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


.)()())(()(

)(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


.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

)()(

9.1Concepts of Definite Integrals 9.2Finding Definite Integrals of Functions 9.3Further Techniques of Definite Integration Chapter Summary Case Study Definite.

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