Chapter 6

6.1 The concept of definite integral

New WordsIntegrand 被积表达式 Integral sum 积分和definite integral 定积分 Curvilinear trapezoid 曲边梯形Variable of integration 积分变量Interval of integraton 积分区间Integrand sign 积分符号 Integrable 可积的Upper limit of integration 积分上限Lower limit of integration 积分下限

Another of central ideas of calculus is the notion of definite integrals. The definite integral originated from a problem in geometry, that is, the problem of finding area. It was soon found that it also provides a way to calculate other quantities. These problems contain the essential features of the definite integral concept and may help to motivate the general definition of definite integral which is given in this section.

1. Introduction1. Introduction

This chapter will start from two practical problems, and introduce the concept of definite integral, then calculate definite integral by the indefinite integral, and introduce application of definite integral in geometry and physics, etc.

Trapezoidr Curvilinea of Area (1)

called is )( curve theand axis the, , lines by the boundedgraph The

xfyxbxax

a b x

y

o

?A

)(xfy

?0 if trapezoidrcurvilinea of area thefind tohow discuss usLet

xf

:figure see d,r trapezoicurvilinea the

a b x

y

oa b x

y

o

See figure

(4 small rectangles ）

（ 9 small rectangles ）

baxfy , above and under area theofion approximatbetter a becomes area total

their smaller,chosen are rectangles theObviously,

bxxxxxaxxxxxxnba

nn

nn

1210

12110

pointsdividingby ],[,],,[],,[ intervals

small into , interval thedivide Firstly,

Step 1. PartitionStep 1. Partition

a b x

y

o ix1x 1ix 1nx

a b x

y

o ix1x 1ix 1nx

niAn

ynixxx

i

iii

,,2,1,by denoted are area their ,trapezoidsr curvilinea narrow into dr trapezoicurvilinea

thedivide andpoint dividingevery through passing axis the toparallel linestraight a Draw

,,2,1 , are lengthsTheir 1

a b x

y

o i ix1x 1ix 1nx

Step 2. ApproximationStep 2. Approximation

have wetrapezoid,r curvilinea small of area therelpace , is

base and isheight whoserectangle small theform and ,,2,1,,point a Choose 1

i

i

iii

xfnixx

nixfA iii ,,2,1,)(

i

n

ii xfA

)(1

Step 3. SumStep 3. Sum

Step 4. LimitStep 4. Limit

formula sum above oflimit thefind wed,r trapezoicurvilinea theof area theof valueaccurate get the order toIn

.0},,max{ where

)(lim

21

10

n

i

n

ii

xxx

xfA

MotionVelocity Variant of Distance (2)

time?of period in thisbody theof distance

thefind ,,on defined is velocity theand linestraight aon movesbody a that Assume

batfv

We can find the distance by the same method of finding the area of curvilinear trapezoid. First, partition the time interval into n smaller intervals; during a small interval of time, the velocity is considered no change; to obtain the estimate of the distance by sum of the distances covered all small intervals; at last, take the limit to obtain the total distance.

i

n

ii tfS

)(lim distance theThus,10

2. Definition of definite integral2. Definition of definite integral

Definition 1

],,[,],,[],,[ into ,

partition , ],[on bounded is that Suppose

12110 nn xxxxxxbabaxf

Although the practical meaning of above two problems is completely different, they are summed up to find the limit of sum formula.

ii

n

iiii

n

i

xfxxf

)( )( sums theIf1

11

.,in chosen is

point thehowmatter no , 0 towardshrinks , ofpartition ofmesh theasnumber certain aapproach

1 iii xxba

by denoted isIt

. to from of integral definite or the ],[over of integral definite thecalled

islimit theand ,over integrable is then

baxfbaxf

baxf

lim 10 i

n

ii

b

axfdxxf

Remarks:Remarks:

i

n

ii

b

axfdxxfA

baxfy

)(lim

is , above and 0under region theof area the,definition above By the 1

10

ba

xfxf i

n

ii

,over integrablenot is

then exist,not does lim If 210

b

aIdxxf )( ii

n

ixf

)(lim

10

integrand

integration

Variable of integration

nintegratio of interval theis],[ ba

Upper limit of integration

Lower limit of integration

Integral sum

b

a

b

a

b

a

b

a

uufttfxxf

xxf

d)(d)(d)(

is, that n,integratio of variableson thet independen isbut n,integratio of intervals theand integrand on the dependentsonly It number. a isit

limit, therepresents d)( integral Definite (4)

3. Existence of the definite integral3. Existence of the definite integral

Theorem 1

Theorem 2

],[over integrable is then ,first type theof points ousdiscontinu finiteonly

has and ,on bounded is that Suppose

baxf

baxf

ba

xfxf

baxf

i

n

ii

,over integrable

is hence exists, limlimit the

then,,on continuous is function theIf

10

1A2A

3A

4A

4321)( AAAAdxxfb

a

4. Geometric illustration of the definite integral4. Geometric illustration of the definite integral

baxfy

Adxxfxfb

a

, above and under region theof

area theis then ,0 If 1

area negative is

whose, then ,0 If 2 Adxxfxfb

a

See figure

Next, to bring the definition down to earth, let us use it to evaluate the definite integral of some simple functions

Example 1 xx d1 Find1

0

2

Solution:

:figure see ,1,0 above and 1under

region theof area theequals d1 integral,

definite theofon illustrati geometric applyingBy

2

1

0

2

xy

xx

x

y21 xy

O

41

41

d1

2

1

0

2

xx

Example 2

xx d

compute tointegral definite theof definition theUse1

0

2Solution:

For ease of computation, we use the typical partitution in which all sections have the same length. As sampling points, we use right-hand endpoints.

numbers by the 1,0npartitutio andinteger positive a be let weSo n

,121161

6)12)(1(1

11)(

is sum integral The

3

1

23

2

11

2

1

nnnnn

n

innn

ixxfn

i

n

i

n

iiiii

n

i

1,1,,,,2,1,0 nn

ni

nn

.1 is subsectioneach

oflength the, ispoint sampling the takeand

n

ni

i

.31

121161limlim

0 Obviously,

1

2

0

1

0

2

nnxdxx

n

n

n

iii

Example 3

dxx

2

1

1compute tointegral definite theof definition theUse

Solution:

2,,,,,1

numbers by the 2.1n partitutio We120 nn qqqqq

1 is subsectioneach of

length the, ispoint sampling the takeand11

1

qqqqx

qiii

i

ii

)1(11)(

is sum integral The

1

11

11

qqq

xxf in

iii

n

i iii

n

i

)1()1(1

qnqn

i

)12(lim1

nnn ,2ln

dxx

2

1

1i

n

i i

x 10

1lim

)12(lim1

nnn .2ln

nnnn

i

qqnqnq11

1

2,2),12()1()1(

,2ln1

12lim)12(lim

1

1

x

xx

xxx

Example 4

0 and ,1,0on continuous is where

.21lim

thatProve1

0)(ln

xfxf

ennf

nf

nf

dxxfn

n

Proof

nn n

nf

nf

nf

nn

e

nnf

nf

nf

21limln

21lim

.21lim1

0)(ln

dxxfn

ne

nnf

nf

nf

nni

f

ni

fnn

nf

nf

nf

n

in

n

in

nn

e

ee1

lnlim

ln1

lim21

lnlim

1

1

1,0over integrable is ln

,0 and ,on continuous is Sincexf

xfbaxf

1

01)(ln1lnlim Thus, dxxf

nnif

n

in

Example 5 Find

22222

12

11

1limnnnn

nn

Solution:

1

0 21

2

22222

111

1

1lim

12

11

1lim

dxxn

ni

nnnnn

n

in

n

Solution:

I

nn

nn

nnnnsin)1(sin2sinsin1lim

n

in n

in 1

sin1limnn

in

in

1

sinlim1

.sin

210

xdxixi

Example 6 Find

n

nnnn

In

)1(sin2sinsin1lim