chapter 6 6.1 the concept of definite integral. new words integrand 被积表达式 integral sum...
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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
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