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8.5 Rational l Functions
181
systemoftwoequationsintwounknowns.Therearemanywaystoproceed;here’sone: If
7=A+BthenB=7 Aandso 6=3A 2B=3A 2(7 A)=3A 14+2A=5A 14.
ThisiseasytosolveforA:A=8=5,andthenB=7 A=7 8=5=27=5.Thus
Z
7x 6
(x 2)(x+3)
dx=
Z
8
5
1
x 2
+
27
5
1
x+3
dx=
8
5
lnjx 2j+
27
5
lnjx+3j+C:
Theanswertotheoriginalproblemisnow
Z
x
3
(x 2)(x+3)
dx=
Z
x 1dx+
Z
7x 6
(x 2)(x+3)
dx
=
x2
2
x+
8
5
lnjx 2j+
27
5
lnjx+3j+C:
Nowsupposethatx
2
+bx+cdoesn’tfactor. Againwecanuselongdivisiontoensure
thatthenumeratorhasdegreelessthan2,thenwecompletethesquare.
EXAMPLE 8.5.6
Evaluate
Z
x+1
x2+4x+8
dx. The e quadratic denominatordoesnot
factor.Wecouldcompletethesquareanduseatrigonometricsubstitution,butitissimpler
torearrangetheintegrand:
Z
x+1
x2+4x+8
dx=
Z
x+2
x2+4x+8
dx 
Z
1
x2+4x+8
dx:
Therstintegralisaneasysubstitutionproblem,usingu=x
2
+4x+8:
Z
x+2
x2+4x+8
dx=
1
2
Z
du
u
=
1
2
lnjx
2
+4x+8j:
Forthesecondintegralwecompletethesquare:
x
2
+4x+8=(x+2)
2
+4=4
x+2
2
2
+1
!
;
makingtheintegral
1
4
Z
1
x+2
2
2
+1
dx:
Usingu=
x+2
2
weget
1
4
Z
1
x+2
2
2
+1
dx=
1
4
Z
2
u2+1
du=
1
2
arctan
x+2
2
:
Thenalanswerisnow
Z
x+1
x2+4x+8
dx=
1
2
lnjx
2
+4x+8j 
1
2
arctan
x+2
2
+C:
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182
Chapter 8 8 Techniques s ofIntegration
Exercises 8.5.
Findtheantiderivatives.
1.
Z
1
4 x2
dx)
2.
Z
x
4
4 x2
dx)
3.
Z
1
x+10x+25
dx)
4.
Z
x
2
4 x2
dx)
5.
Z
x
4
4+x2
dx)
6.
Z
1
x+10x+29
dx)
7.
Z
x
3
4+x2
dx)
8.
Z
1
x+10x+21
dx)
9.
Z
1
2x x 3
dx)
10.
Z
1
x+3x
dx)
Wehavenowseensomeofthemostgenerallyusefulmethodsfordiscoveringantiderivatives,
andthereareothers. Unfortunately,somefunctionshavenosimpleantiderivatives;insuch
casesifthevalueofadeniteintegralisneededitwillhavetobeapproximated. Wewill
seetwomethodsthatworkreasonablywellandyetarefairlysimple;insomecasesmore
sophisticatedtechniqueswillbeneeded.
Ofcourse,wealreadyknowone waytoapproximateanintegral: if f wethinkof the
integralascomputing anarea, , wecan add upthe e areasofsomerectangles. Whilethis
isquitesimple,it isusuallythecase thata large numberofrectanglesisneededto get
acceptableaccuracy.Asimilarapproachismuchbetter: weapproximatetheareaundera
curveoverasmallintervalastheareaofatrapezoid. Ingure8.6.1weseeanareaunder
acurveapproximatedbyrectanglesandbytrapezoids;itisapparentthatthetrapezoids
giveasubstantiallybetterapproximationoneachsubinterval.
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Figure 8.6.1
Approximatinganareawithrectanglesandwithtrapezoids.
Aswithrectangles,wedividetheintervalintonequalsubintervalsoflengthx. A
typicaltrapezoidispicturedingure8.6.2;ithasarea
f(x
i
)+f(x
i+1
)
2
x. Ifweaddup
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8.6 NumericalIntegration
183
theareasofalltrapezoidsweget
f(x
0
)+f(x
1
)
2
x+
f(x
1
)+f(x
2
)
2
x++
f(x
n 1
)+f(x
n
)
2
x=
f(x
0
)
2
+f(x
1
)+f(x
2
)++f(x
n 1
)+
f(x
n
)
2
x:
ThisisusuallyknownastheTrapezoidRule. Foramodestnumberofsubintervalsthis
isnottoodiculttodowithacalculator;acomputercaneasilydomanysubintervals.
x
i
x
i+1
(x
i
;f(x
i
))
(x
i+1
;f(x
i+1
))
.
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Figure8.6.2
Asingletrapezoid.
Inpractice,anapproximationisusefulonlyifweknowhowaccurateitis;forexample,
wemight needaparticularvalueaccuratetothreedecimalplaces. Whenwecomputea
particularapproximationto anintegral,theerroristhedierence betweenthe approxi-
mationandthetruevalueoftheintegral. Foranyapproximationtechnique,weneedan
errorestimate,avaluethatisguaranteedtobelargerthantheactualerror. IfAisan
approximationandEistheassociatederrorestimate,thenweknowthatthetruevalue
of the integral isbetween A E E and d A+E. In n the case of ourapproximation of the
integral,wewantE=E(x)tobeafunctionofxthatgetssmallrapidlyasxgets
small.Fortunately,formanyfunctions,thereissuchanerrorestimateassociatedwiththe
trapezoidapproximation.
THEOREM 8.6.1
Supposef hasa a secondderivative f00 everywhere onthe interval
[a;b],andjf
00
(x)jM forallxintheinterval. . Withx=(b a)=n,anerrorestimate
forthetrapezoidapproximationis
E(x)=
b a
12
M(x)
2
=
(b a)
3
12n2
M:
Let’sseehowwecanusethis.
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184
Chapter 8 8 Techniques s ofIntegration
EXAMPLE8.6.2 Approximate
Z
1
0
e
x2
dxtotwodecimalplaces. Thesecondderiva-
tiveoff=e
x
2
is(4x
2
2)e
x
2
,anditisnothardtoseethaton[0;1],j(4x
2
2)e
x
2
j2.
Webeginbyestimatingthenumberofsubintervalswearelikelytoneed. Togettwodec-
imalplacesofaccuracy,wewillcertainlyneedE(x)<0:005or
1
12
(2)
1
n2
<0:005
1
6
(200)<n
2
5:77
r
100
3
<n
Withn=6,theerrorestimateisthus1=6
3
<0:0047.Wecomputethetrapezoidapproxi-
mationforsixintervals:
f(0)
2
+f(1=6)+f(2=6)++f(5=6)+
f(1)
2
1
6
0:74512:
Sothe true value of the integralisbetween 0:74512 0:0047 7 =0:74042 and 0:74512+
0:0047=0:74982. Unfortunately,therstroundsto0:74andthesecondroundsto0:75,
sowecan’tbesureofthecorrectvalueintheseconddecimalplace;weneedtopickalarger
n. Asitturnsout,weneedtogoton=12togettwoboundsthatbothroundtothesame
value,whichturnsout to be 0:75. Forcomparison, , using g 12rectanglesto approximate
theareagives0:7727,whichisconsiderablylessaccuratethantheapproximationusingsix
trapezoids.
Inpracticeitgenerallypaystostartbyrequiringbetterthanthemaximumpossible
error;forexample,wemighthaveinitiallyrequiredE(x)<0:001,or
1
12
(2)
1
n2
<0:001
1
6
(1000)<n
2
12:91
r
500
3
<n
Hadweimmediatelytriedn=13thiswouldhavegivenusthedesiredanswer.
Thetrapezoidapproximationworkswell,especiallycomparedtorectangles,because
thetopsofthetrapezoidsformareasonablygoodapproximationtothecurvewhenxis
fairlysmall.Wecanextendthisidea:whatifwetrytoapproximatethecurvemoreclosely,
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8.6 NumericalIntegration
185
byusingsomethingotherthanastraightline? Theobviouscandidateisaparabola: ifwe
canapproximateashortpieceofthecurvewithaparabolawithequationy=ax
2
+bx+c,
wecaneasilycomputetheareaundertheparabola.
There are an innite numberof parabolasthrough anytwo given points, , but only
one throughthreegivenpoints. If f wend aparabola throughthreeconsecutive points
(x
i
;f(x
i
)), (x
i+1
;f(x
i+1
)),(x
i+2
;f(x
i+2
)) on the curve, , it t shouldbe quiteclose to the
curve overthewholeinterval[x
i
;x
i+2
],asingure8.6.3. Ifwe e dividetheinterval[a;b]
intoanevennumberofsubintervals,wecanthenapproximatethecurvebyasequenceof
parabolas,eachcoveringtwoofthesubintervals. Forthistobepractical,wewouldlikea
simpleformulafortheareaunderoneparabola,namely,theparabolathrough(x
i
;f(x
i
)),
(x
i+1
;f(x
i+1
)),and(x
i+2
;f(x
i+2
)).Thatis,weshouldattempttowritedowntheparabola
y=ax2+bx+cthroughthese pointsandthenintegrateit,andhopethattheresult is
fairlysimple. Althoughthealgebrainvolvedismessy,thisturnsouttobepossible. The
algebraiswellwithinthecapabilityofagoodcomputeralgebrasystemlikeSage,sowe
willpresent the resultwithout allof thealgebra;you cansee howto doitinthisSage
worksheet.
Tondtheparabola,wesolvethesethreeequationsfora,b,andc:
f(x
i
)=a(x
i+1
x)
2
+b(x
i+1
x)+c
f(x
i+1
)=a(x
i+1
)
2
+b(x
i+1
)+c
f(x
i+2
)=a(x
i+1
+x)
2
+b(x
i+1
+x)+c
Notsurprisingly,thesolutionsturnout tobequitemessy. Nevertheless,Sagecaneasily
computeandsimplifytheintegraltoget
Z
x
i+1
+x
x
i+1
x
ax
2
+bx+cdx=
x
3
(f(x
i
)+4f(x
i+1
)+f(x
i+2
)):
Nowthesumoftheareasunderallparabolasis
x
3
(f(x
0
)+4f(x
1
)+f(x
2
)+f(x
2
)+4f(x
3
)+f(x
4
)++f(x
n 2
)+4f(x
n 1
)+f(x
n
))=
x
3
(f(x
0
)+4f(x
1
)+2f(x
2
)+4f(x
3
)+2f(x
4
)++2f(x
n 2
)+4f(x
n 1
)+f(x
n
)):
Thisisjustslightlymorecomplicatedthantheformulafortrapezoids;weneedtoremember
thealternating2and4coecients;notethatnmustbeevenforthistomakesense.This
approximationtechniqueisreferredtoasSimpson’sRule.
Aswiththetrapezoidmethod,thisisusefulonlywithanerrorestimate:
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186
Chapter8 TechniquesofIntegration
x
i
x
i+1
x
i+2
(x
i
;f(x
i
))
(x
i+2
;f(x
i+2
))
..
..
..
..
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Figure8.6.3
Aparabola(dashed)approximatingacurve(solid).
THEOREM8.6.3
Supposef hasafourthderivativef(4) everywhere onthe interval
[a;b],andjf
(4)
(x)jM forallxintheinterval.Withx=(b a)=n,anerrorestimate
forSimpson’sapproximationis
E(x)=
b a
180
M(x)
4
=
(b a)
5
180n4
M:
EXAMPLE 8.6.4
Letusagain approximate
Z
1
0
e
x
2
dx totwodecimalplaces. The
fourth derivative of f f =e
x
2
is(16x
2
48x
2
+12)e
x
2
; on [0;1]thisisat t most 12 in
absolutevalue. Webeginbyestimatingthenumberofsubintervalswearelikelytoneed.
Togettwodecimalplacesofaccuracy,wewillcertainlyneedE(x)<0:005,buttaking
acuefromourearlierexample,let’srequireE(x)<0:001:
1
180
(12)
1
n4
<0:001
200
3
<n
4
2:86
[4]
r
200
3
<n
Sowetryn=4,sinceweneedanevennumberofsubintervals. Thentheerrorestimate
is12=180=4
4
<0:0003andtheapproximationis
(f(0)+4f(1=4)+2f(1=2)+4f(3=4)+f(1))
1
34
0:746855:
Sothetruevalueoftheintegralisbetween0:746855 0:0003=0:746555and0:746855+
0:0003=0:7471555,bothofwhichroundto0:75.
8.7 Additional l exercises
187
Exercises8.6.
Inthe following problems, compute the trapezoid and Simpsonapproximations using 4 subin-
tervals, and compute theerror r estimate foreach. (Findingthe e maximumvalues of thesecond
andfourthderivatives canbechallengingforsomeof these; ; youmayuse agraphingcalculator
or computer softwareto estimate the maximum values.) If f youhaveaccess to Sage or similar
software,approximateeachintegraltotwodecimalplaces. Youcanuse e thisSageworksheetto
getstarted.
1.
Z
3
1
xdx)
2.
Z
3
0
x
2
dx)
3.
Z
4
2
x
3
dx)
4.
Z
3
1
1
x
dx)
5.
Z
2
1
1
1+x2
dx)
6.
Z
1
0
x
p
1+xdx)
7.
Z
5
1
x
1+x
dx)
8.
Z
1
0
p
x+1dx)
9.
Z
1
0
p
x+1dx)
10.
Z
4
1
p
1+1=xdx)
11. UsingSimpson’sruleonaparabolaf(x),evenwithjusttwosubintervals,givestheexactvalue
of theintegral,because theparabolasusedtoapproximatef f willbef f itself. . Remarkably,
Simpson’s rulealsocomputes s the e integral of a cubicfunction f(x) = ax
3
+bx
2
+cx+d
exactly.Showthisistruebyshowingthat
Z
x
2
x
0
f(x)dx=
x
2
x
0
32
(f(x
0
)+4f((x
0
+x
2
)=2)+f(x
2
)):
Thisdoesrequireabitofmessyalgebra,soyoumayprefertouseSage.
Theseproblemsrequirethetechniquesofthischapter,andareinnoparticularorder.Some
problemsmaybedoneinmorethanoneway.
1.
Z
(t+4)
3
dt)
2.
Z
t(t
2
9)
3=2
dt)
3.
Z
(e
t
2
+16)te
t
2
dt)
4.
Z
sintcos2tdt)
5.
Z
tantsec
2
tdt)
6.
Z
2t+1
t+t+3
dt)
7.
Z
1
t(t 4)
dt)
8.
Z
1
(25 t2)3=2
dt)
9.
Z
cos3t
p
sin3t
dt)
10.
Z
tsec
2
tdt)
11.
Z
e
t
p
e+1
dt)
12.
Z
cos
4
tdt)
188
Chapter 8 8 Techniques s ofIntegration
13.
Z
1
t+3t
dt)
14.
Z
1
t2
p
1+t2
dt)
15.
Z
sec
2
t
(1+tant)3
dt)
16.
Z
t
3
p
t+1dt)
17.
Z
e
t
sintdt)
18.
Z
(t
3=2
+47)
3
p
tdt)
19.
Z
t
3
(2 t2)5=2
dt)
20.
Z
1
t(9+4t2)
dt)
21.
Z
arctan2t
1+4t2
dt)
22.
Z
t
t+2t 3
dt)
23.
Z
sin
3
tcos
4
tdt)
24.
Z
1
t 6t+9
dt)
25.
Z
1
t(lnt)2
dt)
26.
Z
t(lnt)
2
dt)
27.
Z
t
3
e
t
dt)
28.
Z
t+1
t+t 1
dt)
9
Applications of Integration
Wehaveseenhowintegrationcanbeusedtondanareabetweenacurveandthex-axis.
Withverylittlechangewecanndsomeareasbetweencurves;indeed,theareabetween
a curve andthe x-axismaybe interpretedasthe area between thecurve anda second
\curve"withequationy=0.Inthesimplestofcases,theideaisquiteeasytounderstand.
EXAMPLE9.1.1
Findtheareabelowf(x)= x
2
+4x+3andaboveg(x)= x
3
+
7x
2
10x+5overtheinterval1x2.Ingure9.1.1weshowthetwocurvestogether,
withthedesiredareashaded,thenf alonewiththeareaunderfshaded,andthengalone
withtheareaundergshaded.
x
y
0
1
2
3
0
5
10
x
y
0
1
2
3
0
5
10
x
y
0
1
2
3
0
5
10
Figure 9.1.1
Areabetweencurvesasadierenceofareas.
189
190
Chapter 9 9 ApplicationsofIntegration
It isclearfromthegurethatthe areawewantistheareaunderf f minusthe e area
underg,whichistosay
Z
2
1
f(x)dx 
Z
2
1
g(x)dx=
Z
2
1
f(x) g(x)dx:
Itdoesn’tmatterwhetherwecomputethetwointegralsontheleftandthensubtractor
computethesingleintegralontheright. Inthiscase,thelatterisperhapsabiteasier:
Z
2
1
f(x) g(x)dx=
Z
2
1
x
2
+4x+3 ( x
3
+7x
2
10x+5)dx
=
Z
2
1
x
3
8x
2
+14x 2dx
=
x4
4
8x3
3
+7x
2
2x
2
1
=
16
4
64
3
+28 4 (
1
4
8
3
+7 2)
=23 
56
3
1
4
=
49
12
:
Itisworthexaminingthisproblemabitmore. Wehaveseenonewaytolookatit,
byviewingthedesiredareaasabigareaminusasmallarea,whichleadsnaturallytothe
dierencebetweentwointegrals. Butitisinstructivetoconsiderhowwemightndthe
desiredareadirectly. Wecanapproximatetheareabydividingtheareaintothinsections
andapproximating the area of each section bya rectangle,asindicated in gure 9.1.2.
Theareaofatypicalrectangleisx(f(x
i
) g(x
i
)),sothetotalareaisapproximately
n 1
i=0
(f(x
i
) g(x
i
))x:
Thisisexactlythesortofsumthatturnsintoanintegralinthelimit,namelytheintegral
Z
2
1
f(x) g(x)dx:
Ofcourse,thisistheintegralweactuallycomputedabove,butwehavenowarrivedatit
directlyratherthanasamodicationofthedierencebetweentwootherintegrals.Inthat
exampleitreallydoesn’tmatterwhichapproachwetake,but insomecasesthissecond
approachisbetter.
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