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
Cut pages from pdf file - copy, paste, cut PDF pages in C#.net, ASP.NET, MVC, Ajax, WinForms, WPF
Easy to Use C# Code to Extract PDF Pages, Copy Pages from One PDF File and Paste into Others
cut pages from pdf; convert few pages of pdf to word
Cut pages from pdf file - VB.NET PDF Page Extract Library: copy, paste, cut PDF pages in vb.net, ASP.NET, MVC, Ajax, WinForms, WPF
Detailed VB.NET Guide for Extracting Pages from Microsoft PDF Doc
delete pages from pdf reader; extract pages from pdf file online
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. )
C# PDF copy, paste image Library: copy, paste, cut PDF images in
PDF image cutting is similar to image deleting. So, in C# demo code below, we will explain how to cut image from PDF file page by using image deleting API.
delete pages from pdf in reader; copy pdf page to clipboard
VB.NET PDF copy, paste image library: copy, paste, cut PDF images
PDF image cutting is similar to image deleting. So, below example explains how to cut image from PDF file page by using image deleting API.
copy one page of pdf to another pdf; delete page from pdf file online
8
Techniques of Integration
Overthenextfewsectionsweexaminesometechniquesthatarefrequentlysuccessfulwhen
seekingantiderivativesoffunctions. Sometimesthisisasimpleproblem,sinceitwillbe
apparentthatthefunctionyouwish tointegrateisaderivativein somestraightforward
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
C# PDF File & Page Process Library SDK for C#.net, ASP.NET, MVC
Image: Copy, Paste, Cut Image in Page. Link: Edit Redact Text Content. Redact Images. Redact Pages. Annotation & Text. Add Text Box. Drawing Markups. PDF Print. Work
cut pages from pdf reader; cut and paste pdf pages
C# PDF Page Insert Library: insert pages into PDF file in C#.net
Add and Insert Blank Pages to PDF File in C#.NET. This C# demo explains how to insert empty pages to a specific location of current PDF file.
a pdf page cut; cutting pdf pages
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
VB.NET PDF Page Insert Library: insert pages into PDF file in vb.
Moreover, you may use the following VB.NET demo code to insert multiple pages of a PDF file to a PDFDocument object at user-defined position.
delete pages of pdf preview; extract page from pdf online
VB.NET PDF Page Delete Library: remove PDF pages in vb.net, ASP.
Ability to remove consecutive pages from PDF file in VB.NET. Enable specified pages deleting from PDF in Visual Basic .NET class.
extract page from pdf; extract page from pdf reader
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
likeordinaryarithmeticgivesthecorrect answer,evenifsomethingmorecomplicatedis
C# PDF File Split Library: Split, seperate PDF into multiple files
note, PDF file will be divided from the previous page of your defined page number which starts from 0. For example, your original PDF file contains 4 pages.
extract pages from pdf; export pages from pdf online
C# PDF Page Delete Library: remove PDF pages in C#.net, ASP.NET
Ability to remove a range of pages from PDF file. Description: Delete consecutive pages from the input PDF file starting at specified position. Parameters:
copy web pages to pdf; extract pdf pages acrobat
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
correctanswer. Anotherpossibility,forexample,is: : Sincedu=dx=2x,dx=du=2x,and
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,
andnisapositiveinteger.Weletu=ax+bsodu=adxordx=du=a. Then
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
a6=0.Againweletu=ax+bsodu=adxordx=du=a.Then
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
168
Chapter 8 8 Techniques s ofIntegration
tosubstitutex
2
backinforu,thusgettingtheincorrectanswer 
1
2
cos(4)+
1
2
cos(2). A
somewhatclumsy,butacceptable,alternativeissomethinglikethis:
Z
4
2
xsin(x
2
)dx=
Z
x=4
x=2
1
2
sinudu= 
1
2
cos(u)
x=4
x=2
1
2
cos(x
2
)
4
2
cos(16)
2
+
cos(4)
2
:
EXAMPLE8.1.4
Evaluate
Z
1=2
1=4
cos(t)
sin
2
(t)
dt. Letu=sin(t)sodu=cos(t)dtor
du==cos(t)dt.Wechangethelimitstosin(=4)=
p
2=2andsin(=2)=1. Then
Z
1=2
1=4
cos(t)
sin
2
(t)
dt=
Z
1
p
2=2
1
1
u2
du=
Z
1
p
2=2
1
u
2
du=
1
u
1
1
1
p
2=2
1
+
p
2
:
Exercises 8.1.
Findtheantiderivativesorevaluatethedeniteintegralineachproblem.
1.
Z
(1 t)
9
dt)
2.
Z
(x
2
+1)
2
dx)
3.
Z
x(x
2
+1)
100
dx)
4.
Z
1
3
p
1 5t
dt)
5.
Z
sin
3
xcosxdx)
6.
Z
x
p
100 x2dx)
7.
Z
x
2
p
1 x3
dx)
8.
Z
cos(t)cos
sin(t)
dt)
9.
Z
sinx
cos3x
dx)
10.
Z
tanxdx)
11.
Z
0
sin
5
(3x)cos(3x)dx)
12.
Z
sec
2
xtanxdx)
13.
Z
p
=2
0
xsec
2
(x
2
)tan(x
2
)dx)
14.
Z
sin(tanx)
cos2x
dx)
15.
Z
4
3
1
(3x 7)2
dx)
16.
Z
=6
0
(cos
2
x sin
2
x)dx)
17.
Z
6x
(x 7)1=9
dx)
18.
Z
1
1
(2x
3
1)(x
4
2x)
6
dx)
19.
Z
1
1
sin
7
xdx)
20.
Z
f(x)f
0
(x)dx)
8.2 Powers s ofsineandcosine
169
Functionsconsistingofproductsofthesineandcosinecanbeintegratedbyusingsubsti-
tutionandtrigonometricidentities. Thesecansometimesbetedious,butthetechniqueis
straightforward. Someexampleswillsucetoexplaintheapproach.
EXAMPLE8.2.1
Evaluate
Z
sin
5
xdx. Rewritethefunction:
Z
sin
5
xdx=
Z
sinxsin
4
xdx=
Z
sinx(sin
2
x)
2
dx=
Z
sinx(1 cos
2
x)
2
dx:
Nowuseu=cosx,du= sinxdx:
Z
sinx(1 cos
2
x)
2
dx=
Z
(1 u
2
)
2
du
=
Z
(1 2u
2
+u
4
)du
= u+
2
3
u
3
1
5
u
5
+C
= cosx+
2
3
cos
3
1
5
cos
5
x+C:
EXAMPLE 8.2.2
Evaluate
Z
sin
6
xdx. Usesin
2
x=(1 cos(2x))=2to o rewritethe
function:
Z
sin
6
xdx=
Z
(sin
2
x)
3
dx=
Z
(1 cos2x)
3
8
dx
=
1
8
Z
1 3cos2x+3cos
2
2x cos
3
2xdx:
Nowwehavefourintegralstoevaluate:
Z
1dx=x
and
Z
3cos2xdx= 
3
2
sin2x
170
Chapter 8 8 Techniques s ofIntegration
areeasy.Thecos
3
2xintegralislikethepreviousexample:
Z
cos
3
2xdx=
Z
cos2xcos
2
2xdx
=
Z
cos2x(1 sin
2
2x)dx
=
Z
1
2
(1 u
2
)du
1
2
u3
3
1
2
sin2x 
sin
3
2x
3
:
Andnallyweuseanothertrigonometricidentity,cos
2
x=(1+cos(2x))=2:
Z
3cos
2
2xdx=3
Z
1+cos4x
2
dx=
3
2
x+
sin4x
4
:
Soatlonglastweget
Z
sin
6
xdx=
x
8
3
16
sin2x 
1
16
sin2x 
sin
3
2x
3
+
3
16
x+
sin4x
4
+C:
EXAMPLE8.2.3 Evaluate
Z
sin
2
xcos
2
xdx.Usetheformulassin
2
x=(1 cos(2x))=2
andcos
2
x=(1+cos(2x))=2toget:
Z
sin
2
xcos
2
xdx=
Z
1 cos(2x)
2
1+cos(2x)
2
dx:
Theremainderisleftasanexercise.
Exercises 8.2.
Findtheantiderivatives.
1.
Z
sin
2
xdx)
2.
Z
sin
3
xdx)
3.
Z
sin
4
xdx)
4.
Z
cos
2
xsin
3
xdx)
5.
Z
cos
3
xdx)
6.
Z
sin
2
xcos
2
xdx)
7.
Z
cos
3
xsin
2
xdx)
8.
Z
sinx(cosx)
3=2
dx)
9.
Z
sec
2
xcsc
2
xdx)
10.
Z
tan
3
xsecxdx)
Documents you may be interested
Documents you may be interested