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# devexpress asp.net pdf viewer : Cut pages from pdf file SDK Library API wpf asp.net .net sharepoint Calculus16-part746

7.3 SomeProperties s ofIntegrals
161
righttoleft,sothatt
0
=bandt
n
=a. Thent=t
i+1
t
i
isnegativeandin
Z
5
6
v(t)dt=
nX 1
i=0
v(t
i
)t;
thevaluesv(t
i
)arenegativebutalsotisnegative,soalltermsarepositiveagain. On
theotherhand,in
Z
0
5
v(t)dt=
nX 1
i=0
v(t
i
)t;
thevaluesv(t
i
)arepositivebuttisnegative,andwegetanegativeresult:
Z
0
5
v(t)dt=
t
3
3
+
5
2
t
2
0
5
=0
5
3
3
5
2
5
2
125
6
:
Finallywenoteonesimplepropertyofintegrals:
Z
b
a
f(x)+g(x)dx=
Z
b
a
f(x)dx+
Z
b
a
g(x)dx:
Thisiseasytounderstandonceyourecallthat(F(x)+G(x))
0
=F
0
(x)+G
0
(x).Hence,if
F
0
(x)=f(x)andG
0
(x)=g(x),then
Z
b
a
f(x)+g(x)dx=(F(x)+G(x))j
b
a
=F(b)+G(b) F(a) G(a)
=F(b) F(a)+G(b) G(a)
=F(x)j
b
a
+G(x)j
b
a
=
Z
b
a
f(x)dx+
Z
b
a
g(x)dx:
Insummary,wewillfrequentlyusethesepropertiesofintegrals:
Z
b
a
f(x)dx=
Z
c
a
f(x)dx+
Z
b
c
f(x)dx
Z
b
a
f(x)+g(x)dx=
Z
b
a
f(x)dx+
Z
b
a
g(x)dx
Z
b
a
f(x)dx=
Z
a
b
f(x)dx
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162
Chapter 7 7 Integration
andifa<bandf(x)0on[a;b]then
Z
b
a
f(x)dx0
andinfact
Z
b
a
f(x)dx=
Z
b
a
jf(x)jdx:
Exercises 7.3.
1. Anobjectmovessothatitsvelocityattimetisv(t)= 9:8t+20m/s. . Describethemotion
oftheobjectbetweent=0andt=5,ndthetotaldistancetraveledbytheobjectduring
thattime,andndthenetdistancetraveled. )
2. Anobjectmovessothatitsvelocityattimetis s v(t)=sint. . Setupandevaluateasingle
deniteintegraltocomputethenetdistancetraveledbetweent=0andt=2. )
3. Anobjectmovessothatitsvelocityattimetisv(t)=1+2sintm/s. . Findthenetdistance
traveledbytheobjectbetweent=0andt=2,andndthetotaldistancetraveledduring
thesameperiod. )
4. Considerthe e functionf(x)= (x+2)(x+1)(x 1)(x 2)on[ 2;2]. . Findthe e totalarea
betweenthecurveandthex-axis(measuringallareaaspositive). )
5. Considerthefunctionf(x) ) =x
2
3x+2on[0;4]. Findthetotalareabetweenthecurve
andthex-axis(measuringallareaaspositive). )
6. Evaluatethethreeintegrals:
A=
Z
3
0
( x
2
+9)dx
B=
Z
4
0
( x
2
+9)dx
C=
Z
3
4
( x
2
+9)dx;
andverifythatA=B+C. )
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8
Techniques of Integration
Overthenextfewsectionsweexaminesometechniquesthatarefrequentlysuccessfulwhen
seekingantiderivativesoffunctions. Sometimesthisisasimpleproblem,sinceitwillbe
way.Forexample,facedwith
Z
x
10
dx
werealizeimmediatelythatthederivativeofx
11
willsupplyanx
10
: (x
11
)
0
=11x
10
. We
don’twantthe\11",butconstantsareeasytoalter,becausedierentiation\ignores"them
incertaincircumstances,so
d
dx
1
11
x
11
=
1
11
11x
10
=x
10
:
Fromourknowledgeofderivatives,wecanimmediatelywritedownanumberofan-
tiderivatives.Hereisalistofthosemostoftenused:
Z
x
n
dx=
xn+1
n+1
+C; ifn6= 1
Z
x
1
dx=lnjxj+C
Z
e
x
dx=e
x
+C
Z
sinxdx= cosx+C
163
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164
Chapter 8 8 Techniques s ofIntegration
Z
cosxdx=sinx+C
Z
sec
2
xdx=tanx+C
Z
secxtanxdx=secx+C
Z
1
1+x2
dx=arctanx+C
Z
1
p
1 x2
dx=arcsinx+C
Needlesstosay,mostproblemsweencounterwillnotbesosimple.Here’saslightlymore
complicatedexample: nd
Z
2xcos(x
2
)dx:
Thisisnota\simple"derivative,butalittlethoughtrevealsthatitmusthavecomefrom
anapplicationofthechainrule. Multipliedonthe\outside"is2x,whichisthederivative
ofthe\inside"functionx
2
. Checking:
d
dx
sin(x
2
)=cos(x
2
)
d
dx
x
2
=2xcos(x
2
);
so
Z
2xcos(x
2
)dx=sin(x
2
)+C:
Evenwhenthechainrulehas\produced"acertainderivative,itisnotalwayseasyto
see. Considerthisproblem:
Z
x
3
p
1 x2dx:
Therearetwofactorsinthisexpression,x
3
and
p
1 x2,butitisnotapparentthatthe
chainruleisinvolved.Somecleverrearrangementrevealsthatitis:
Z
x
3
p
1 x2dx=
Z
( 2x)
1
2
(1 (1 x
2
))
p
1 x2dx:
Thislooksmessy,butwedonowhavesomething thatlookslikethe resultof thechain
rule: thefunction1 x
2
hasbeensubstitutedinto (1=2)(1 x)
p
x,andthederivative
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8.1 Substitution
165
of1 x
2
, 2x,multipliedontheoutside. . IfwecanndafunctionF(x)whosederivative
is (1=2)(1 x)
p
xwe’llbedone,sincethen
d
dx
F(1 x
2
)= 2xF
0
(1 x
2
)=( 2x)
1
2
(1 (1 x
2
))
p
1 x2
=x
3
p
1 x2
Butthisisn’thard:
Z
1
2
(1 x)
p
xdx=
Z
1
2
(x
1=2
x
3=2
)dx
(8:1:1)
1
2
2
3
x
3=2
2
5
x
5=2
+C
=
1
5
1
3
x
3=2
+C:
Sonallywehave
Z
x
3
p
1 x2dx=
1
5
(1 x
2
1
3
(1 x
2
)
3=2
+C:
Sowesucceeded,butitrequiredacleverrststep,rewritingtheoriginalfunctionso
thatitlookedliketheresultofusingthechainrule.Fortunately,thereisatechniquethat
makessuchproblemssimpler,withoutrequiringclevernesstorewriteafunctioninjustthe
right way. It t doessometimesnot work,ormayrequiremorethanoneattempt,butthe
ideaissimple:guessatthemostlikelycandidateforthe\insidefunction",thendosome
algebratoseewhatthisrequirestherestofthefunctiontolooklike.
Onefrequentlygoodguessisanycomplicatedexpressioninsideasquareroot,sowe
startbytryingu=1 x
2
,usinganewvariable,u,forconvenienceinthemanipulations
thatfollow. Nowweknowthatthechainrulewillmultiplybythederivativeofthisinner
function:
du
dx
= 2x;
soweneedtorewritetheoriginalfunctiontoincludethis:
Z
x
3
p
1 x=
Z
x
3
p
u
2x
2x
dx=
Z
x2
2
p
u
du
dx
dx:
RecallthatonebenetoftheLeibniznotationisthatitoftenturnsoutthatwhatlooks
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166
Chapter 8 8 Techniques s ofIntegration
goingon. Forexample,inLeibniznotationthechainruleis
dy
dx
=
dy
dt
dt
dx
:
Thesameistrueofourcurrentexpression:
Z
x
2
2
p
u
du
dx
dx=
Z
x
2
2
p
udu:
Nowwe’realmostthere: sinceu=1 x
2
,x
2
=1 uandtheintegralis
Z
1
2
(1 u)
p
udu:
It’snocoincidencethatthisisexactlytheintegralwecomputedin(8.1.1),wehavesimply
renamedthevariableutomakethecalculationslessconfusing. Justasbefore:
Z
1
2
(1 u)
p
udu=
1
5
1
3
u
3=2
+C:
Thensinceu=1 x
2
:
Z
x
3
p
1 x2dx=
1
5
(1 x
2
1
3
(1 x
2
)
3=2
+C:
To summarize: if f we suspect that a given function isthe derivative of anothervia the
chainrule,we letu denote a likelycandidate fortheinnerfunction,then translate the
given function so that t it t iswritten entirelyin termsof u, , with no x x remaining in the
expression. If f we e can n integrate e this new function n of f u, , then n the antiderivative e of f the
originalfunctionisobtainedbyreplacingubytheequivalentexpressioninx.
Eveninsimplecasesyoumayprefertousethismechanicalprocedure,sinceitoften
helpstoavoidsillymistakes. Forexample,consideragainthissimpleproblem:
Z
2xcos(x
2
)dx:
Letu=x
2
,thendu=dx=2xordu=2xdx. Sincewehaveexactly2xdxintheoriginal
integral,wecanreplaceitbydu:
Z
2xcos(x
2
)dx=
Z
cosudu=sinu+C=sin(x
2
)+C:
Thisisnot the onlywayto do the algebra, , andtypicallythereare e many y pathsto o the
8.1 Substitution
167
thentheintegralbecomes
Z
2xcos(x
2
)dx=
Z
2xcosu
du
2x
=
Z
cosudu:
Theimportantthingtorememberisthatyoumusteliminateallinstancesoftheoriginal
variablex.
EXAMPLE8.1.1 Evaluate
Z
(ax+b)
n
dx,assumingthataandbareconstants,a6=0,
Z
(ax+b)
n
dx=
Z
1
a
u
n
du=
1
a(n+1)
u
n+1
+C=
1
a(n+1)
(ax+b)
n+1
+C:
EXAMPLE8.1.2 Evaluate
Z
sin(ax+b)dx,assumingthataandbareconstantsand
Z
sin(ax+b)dx=
Z
1
a
sinudu=
1
a
( cosu)+C=
1
a
cos(ax+b)+C:
EXAMPLE8.1.3 Evaluate
Z
4
2
xsin(x
2
)dx.Firstwecomputetheantiderivative,then
evaluatethedeniteintegral.Letu=x
2
sodu=2xdxorxdx=du=2.Then
Z
xsin(x
2
)dx=
Z
1
2
sinudu=
1
2
( cosu)+C=
1
2
cos(x
2
)+C:
Now
Z
4
2
xsin(x
2
)dx=
1
2
cos(x
2
)
4
2
1
2
cos(16)+
1
2
cos(4):
A somewhat neateralternative tothismethod istochangethe originallimitstomatch
thevariableu. Sinceu=x
2
,whenx=2,u=4,andwhenx=4,u=16. Sowecando
this:
Z
4
2
xsin(x
2
)dx=
Z
16
4
1
2
sinudu=
1
2
(cosu)
16
4
1
2
cos(16)+
1
2
cos(4):
Anincorrect,anddangerous,alternativeissomethinglikethis:
Z
4
2
xsin(x
2
)dx=
Z
4
2
1
2
sinudu=
1
2
cos(u)
4
2
1
2
cos(x
2
)
4
2
1
2
cos(16)+
1
2
cos(4):
Thisisincorrectbecause
Z
4
2
1
2
sinudumeansthatutakesonvaluesbetween2and4,which
iswrong.Itisdangerous,becauseitisveryeasytogettothepoint
1
2
cos(u)
4
2
andforget