unit3 who sept24

8/14/2019 Unit3 Who Sept24

1/43

MIT OpenCourseWarehttp://ocw.mit.edu

18.01 Single Variable Calculus

Fall 2006

For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
http://ocw.mit.edu/http://ocw.mit.edu/http://ocw.mit.edu/termshttp://ocw.mit.edu/termshttp://ocw.mit.edu/


2/43

Lecture18 18.01Fall2006

Lecture18: Definite IntegralsIntegralsareusedtocalculatecumulativetotals,averages,areas.

Areaunderacurve: (SeeFigure 1.)1. Divideregionintorectangles2. Addupareaofrectangles3. Take limitasrectanglesbecomethin

a b a b(i) (ii)

Figure1: (i)Areaunderacurve;(ii)sumofareasunderrectanglesExample1. f(x) =x2,a=0,b arbitrary

1. Divide intonintervalsLengthb/n= baseofrectangle

2. Heights: 2

b b1st: x= , height =

n n 22b 2b

2nd: x= , height =n n

Sumofareasofrectangles: 2 2 2 2

b b b 2b b 3b b nb b3+ + + + = (12 + 22 + 32 + +n2)

n n n n n n n n n3

1


3/43


a=0 b

2Figure2: Areaunderf(x) = x above [0, b].Wewillnowestimatethesumusingsome3-dimensionalgeometry.

ConsiderthestaircasepyramidaspicturedinFigure 3.

n = 4

n

Figure3: Staircasepyramid: left(topview)andright(sideview)1st

level: nnbottom,representsvolumen2.2nd level: (n1)(n1),representsvolumne(n1)2),etc.Hence,thetotalvolumeofthestaircasepyramid isn2 + (n1)2 + + 1.

Next,thevolumeofthepyramidisgreaterthanthevolumeofthe innerprism:1 1 1

12 + 22 + +n2 > (base)(height)= n2 n= n3 3 3 3

andlessthanthevolumeoftheouterprism:1 1

12 + 22 + +n2


4/43


Inall,1 1n3 12 + 22 + +n2 1 (n+1)3

= 3


5/43


GeneralPicture

a bci

y=f(x)

Figure5: Onerectangle fromaRiemannSum

Divide intonequalpiecesoflength=x= ban

Pickanyci inthe interval;usef(ci)astheheightoftherectangle

Sumof

areas:

f(c

1

)x+

f(c

2

)x+

+

f(c

n

)x

n

Insummationnotation: f(ci)x calledaRiemannsum.i=1

Definition:n b

lim f(ci)x= f(x)dx calledadefinite integralan

i=1

Thisdefiniteintegralrepresentstheareaunderthecurvey=f(x)above[a, b].

Example3.

(Integrals

applied

to

quantity

besides

area.)

Student

borrows

from

parents.

P =principalindollars,t=timeinyears,r= interestrate(e.g.,6% isr= 0.06/year).Aftertimet,youoweP(1+rt) =P +P rt

Theintegralcanbeusedtorepresentthetotalamountborrowedasfollows. Considerafunctionf(t),theborrowingfunctionindollarsperyear. Forinstance,ifyouborrow$1000/month,thenf(t)=12,000/year. Allowf tovaryovertime.

Sayt= 1/12 year=1 month.ti =i/12 i= 1, ,12.

4


6/43


7/43


Lecture 19: First Fundamental Theorem ofCalculus

Fundamental Theorem of Calculus (FTC 1)Iff(x)iscontinuousandF(x) =f(x),then b

f(x)dx=F(b)F(a)a

Notation: F(x) b =F(x) x=b =F(b)F(a)a x=a

b b3 3 b3 a3x x2; x2dx=Example1. F(x) = F(x) =x =

3 33 , 3a aExample2. Areaunderonehumpofsinx(SeeFigure 1.)

0 sinx dx=cosx

=cos(cos0)=(1)(1)=2

0

1

Figure1: Graphoff(x)=sinxfor0x.

1 16= 1 1

60 = 65dx=x

Example3. x60 0

1


8/43


Intuitive Interpretation of FTC:dx

x(t) isaposition; v(t) =x(t)= isthespeedorrateofchangeofx.dt bv(t)dt=x(b)x(a) (FTC1)

aR.H.S.ishowfarx(t)wentfromtimet=atotimet=b(differencebetweentwoodometerreadings).L.H.S.representsspeedometerreadings.

n

i=1

x(b)x(a) = v(t) cancel each other, whereas an

( ) approximates the sum of distances traveled over times t t tv i

thThe approximation above is accurate if ( ) is close to ( ) on the interval. The interpretationit tv v iof ( ) as an odometer reading is no longer valid if changes sign. Imagine a round trip so thattx v

0. Then the positive and negative velocitiesodometer would measure the total distance not the net distance traveled.Example4.

2

0 sinx dx=cosx2

=cos2(cos0)=0.0

The integral represents the sum of areas under the curve, above the x-axis minus the areas belowthex-axis. (SeeFigure 2.)

+

-

1

2

Figure2: Graphoff(x)=sinx for0x2.

2


9/43


Integralshaveanimportantadditiveproperty(SeeFigure 3.) b c c

f(x)dx+ f(x)dx= f(x)dxa b a

a b c

Figure3: Illustrationoftheadditivepropertyof integralsNewDefinition:

a bf(x)dx= f(x)dx

b aThisdefinitionisusedsothatthefundamentaltheoremisvalidnomatterifa < borb < a. Italsomakes it so that the additive property works for a,b,c in any order, notjust the one pictured inFigure 3.

3


10/43


Estimation: b b

Iff(x)g(x),then f(x)dx g(x)dx(onlyifa < b)a a

xExample5. EstimationofexSince1e forx0,

1 11dx exdx

0 0 1 exdx=ex1 =e1e0 =e1

0 0Thus1e1, or e2.

xExample6. Weshowedearlierthat1+xe . Itfollowsthat 1 1

(1+x)dx exdx=e10 0

1 21x 3

(1+x)dx= x+ =2 20 03 5

Hence,2e1,or,e 2 .

Change of Variable:Iff(x) =g(u(x)),thenwewritedu=u(x)dxand

g(u)du= g(u(x))u(x)dx= f(x)u(x)dx (indefiniteintegrals)Fordefinite integrals:

x2 u2f(x)u(x)dx= g(u)du whereu1 =u(x1), u2 =u(x2)

x1 u1

2 4

Example7. x3 + 2 x2dx1

Letu=x3 + 2. Thendu= 3x2dx = x2dx= du; 3x1 = 1, x2 = 2 = u1 = 13 + 2 = 3, u2 = 23 + 2=10, and

2 10 4 10 4du u5 10535x3 + 2 x2dx= u = =1 3 3 153 15

4


11/43


Lecture20: SecondFundamentalTheoremRecall: FirstFundamentalTheoremofCalculus(FTC1)

Iff iscontinuousandF =f,then bf(x)dx=F(b)F(a)

aWecanalsowritethatas b x=b

f(x)dx= f(x)dxx=aa

Doallcontinuousfunctionshaveantiderivatives? Yes. However...Whataboutafunction likethis?

2ex dx=??

Yes,thisantiderivativeexists. No,itsnotafunctionwevemetbefore: itsanewfunction.Thenewfunctionisdefinedasan integral:

x2

F(x) = et dt0

2

ItwillhavethepropertythatF(x) =ex .sinx1/2Othernewfunctionsincludeantiderivativesofex2, x ex2, ,sin(x2),cos(x2), . . .

xSecondFundamentalTheoremofCalculus(FTC2)

xIfF(x) = f(t)dtandf iscontinuous,then

aF(x) =f(x)

GeometricProofofFTC2: Usethearea interpretation: F(x)equalstheareaunderthecurvebetweenaandx.

F = F(x+ x)F(x)F (base)(height) (x)f(x) (SeeFigure 1.)Fx f(x)F

Hence lim = f(x)x 0 x

But,bythedefinitionofthederivative:F

lim =F(x)x 0 x

1


12/43


x+xxa

F(x)

F

y

Figure1: GeometricProofofFTC2.Therefore,

F(x) =f(x)AnotherwaytoproveFTC2 isasfollows:F 1 x+x xx = x f(t)dt f(t)dta a

1 x+x=

xf(t)dt (which istheaveragevalueoff onthe intervalxtx+ x.)

xAsthe lengthxofthe intervaltendsto0,thisaveragetendstof(x).ProofofFTC1(usingFTC2)

xStart with F = f (we assume that f is continuous). Next, define G(x) = f(t)dt. By FTC2,

aG(x) =f(x). Therefore, (FG) =FG =ff =0. Thus, FG= constant. (Recall weusedtheMeanValueTheoremtoshowthis).Hence,F(x) =G(x) +c. FinallysinceG(a)=0, b

f(t)dt=G(b) =G(b)G(a) = [F(b)c][F(a)c] =F(b)F(a)a

which isFTC1.Remark. IntheprecedingproofGwasadefinite integralandF couldbeanyantiderivative. Letusillustratewiththeexamplef(x)=sinx. Takinga=0 intheproofofFTC1, x x

G(x)= cost dt=sint =sinx andG(0)=0.0 0

2


13/43


If,forexample,F(x)=sinx+21. ThenF(x)=cosxand b

sinx dx=F(b)F(a)=(sinb+21)(sina+21)=sinbsinaaEveryfunctionoftheformF(x) =G(x) +cworks inFTC1.Examplesofnew functionsTheerrorfunction,which isoftenusedinstatisticsandprobability,isdefinedas

22 xerf(x) = et dt

0and lim erf(x) = 1 (SeeFigure 2)

x

Figure2: Graphoftheerror function.Anothernewfunctionofthistype,calledthelogarithmic integral,isdefinedas

x dtLi(x) =

lnt2This functiongivestheapproximatenumberofprimenumbers lessthanx. Acommon encryptiontechnique involvesencodingsensitive information likeyourbankaccountnumbersothat itcanbesent over an insecure communication channel. The message can only be decoded using a secretprimenumber. Toknowhowsafethe secret is, a cryptographer needs to knowroughly how many200-digitprimesthereare. Youcanfindoutbyestimatingthefollowing integral:

10201 dt10200 lnt

Weknowthatln10200 =200ln(10)200(2.3)=460 and ln10201 =201ln(10)462

3


14/43


Wewillapproximatetoonesignificantfigure: lnt500for200t10201.Withallofthatinmind,thenumberof200-digitprimesisroughly1

10201 10201 10200dt dt 1 910200 lnt 10200 500 = 500 1020110200 500 10198

ThereareLOTSof200-digitprimes. Theoddsofsomehackerfindingthe200-digitprimerequiredtobreak intoyourbankaccountnumberareveryveryslim.

AnothersetofnewfunctionsaretheFresnelfunctions,whichariseinoptics:x

C(x) = cos(t2)dt0xS(x) = sin(t2)dt

0Besselfunctionsoftenarise inproblemswithcircularsymmetry: 1

J0(x)= cos(xsin)d2 0

Onthehomework,youareaskedtofindC(x). Thatseasy!C(x)=cos(x2)

x dtWewilluseFTC2todiscussthefunctionL(x)= fromfirstprinciplesnext lecture.

t1

1 Themiddleequality inthisapproximation isaverybasicanduseful fact

bc dx=c(ba)

aThink of this as finding the area of a rectangle with base (ba) and height c. In the computation above, a =

110200, b=10201, c=500

4


15/43


Lecture21: ApplicationstoLogarithmsandGeometry

ApplicationofFTC2toLogarithmsThe integral definition of functions like C(x), S(x)of Fresnelmakesthemnearlyaseasytouse aselementary functions. It is possible to draw their graphs and tabulate values. You are asked tocarry out an example or two of this on your problem set. To get used to using definite integralsandFTC2,wewilldiscussindetailthesimplestintegralthatgivesrisetoarelativelynewfunction,namelythelogarithm.

Recallthat n+1xxndx= +cn+ 1

except when n=1. It follows that the antiderivative of 1/x is not a power, but something else.So letusdefineafunctionL(x)by x dt

L(x) =t1

(Thisfunctionturnsouttobethelogarithm. Butrecallthatourapproachtothelogarithmwasfairlyinvolved. Wefirstanalyzedax,andthendefinedthenumbere,andfinallydefinedthelogarithmas

xtheinversefunctiontoe . Thedirectapproachusingthis integralformulawillbeeasier.)AllthebasicpropertiesofL(x)followdirectlyfrom itsdefinition. NotethatL(x) isdefinedfor

0< x


16/43

Lecture21 18.01Fall2006 ab dt

Tohandle ,makethesubstitutiont=au. Thenta

dt=adu; a


17/43


18/43


First,graphthesefunctionsandfindthecrossingpoints(seeFigure 3).y+ 2 = x=y2

y2y2 = 0(y2)(y+ 1) = 0

Crossingpointsaty=1,2. Plug theseback intofindtheassociatedxvalues,x=1 andx=4.Thusthecurvesmeetat(1,1)and(4,2)(seeFigure 3).Therearetwowaysoffindingtheareabetweenthesetwocurves,ahardwayandaneasyway.HardWay: VerticalSlicesIf we slicetheregion betweenthe twocurvesvertically, weneed toconsider twodifferentregions.

x = y2

y = x 2

(1,1)

(4, 2)

(0, 0)

(0, -2)

dx

2Figure4: The intersectionofx=y andy=x2.Where x >1, the regions lowerbound is the straight line. For x


19/43


x = y2

(1,1)

(4, 2)

y = x 2 ; (x = y +2)(0, 0)

(0, -2)

dy

2Figure5: The intersectionofx=y andy=x2.2. VolumesofsolidsofrevolutionRotatef(x)aboutthex-axis,comingoutofthepage,toget:

f(x)

yrotate an x-y plane section

by 2 radians

x

z

dx

Figure6: Asolidofrevolution: thepurpleslice isrotatedby/4and/2.Wewanttofigureoutthevolumeofasliceofthatsolid. Wecanapproximateeachsliceasa

disk withwidthdx,radiusy,andacross-sectionalareaofy2. Thevolumeofoneslice isthen:dV =y2dx (forasolidofrevolutionaroundthex-axis)

Integratewithrespecttoxtofindthetotalvolumeofthesolidofrevolution.5


20/43


21/43


Lecture22: VolumesbyDisksandShellsDisksandShells

Wewill illustratethe2methodsoffindingvolumethroughanexample.Example1. Awitchscauldron

y

x

2Figure1: y=x rotatedaroundthey-axis.

Method1: Disksy

x

thickness of dy

a

Figure2: VolumebyDisks fortheWitchsCauldronproblem.The area of the disk in Figure 2 is x2. The disk has thickness dy and volume dV = x2dy.

ThevolumeV ofthecauldronisa

V = x2dy (substitute y=x2)0 a 2ay a2

V = ydy= =2 0 20

1


22/43


23/43


Thethinshell/cylinderhasheightax2,circumference2x,andthicknessdx.dV = (ax2)(2x)dx x=a a

V = (ax2)(2x)dx= 2 (axx3)dxx=0 0

2 4 2 2 2x x a a a a a2= 2 a

2 4 0 = 2 2 4 = 2 4 = 2 (sameasbefore)Example2. TheboilingcauldronNow,letsfillthiscauldronwithwater,andlightafireunderittogetthewatertoboil(at100oC).Letssay itsacoldday: thetemperatureoftheairoutsidethecauldronis0oC.Howmuchenergydoesittaketoboilthiswater,i.e. toraisethewaterstemperaturefrom0oCto100oC?Assumethe

y

x

70oC

100oC

Figure5: Theboilingcauldron(y=a=1meter.)temperaturedecreaseslinearlybetweenthetopandthebottom(y=0)ofthecauldron:

T =10030y (degreesCelsius)Usethemethodofdisks,becausethewaterstemperatureisconstantovereachhorizontaldisk. Thetotalheatrequiredis

1H = T(x2)dy (unitsare(degree)(cubicmeters))

01

= (10030y)(y)dy0

1 1= (100y30y2)dy=(50y210y3)

=40(deg.)m30 0

Howmanycaloriesisthat? 3

1cal 100cm#ofcalories =

cm3 deg(40) 1 m =(40)(106)cal=125103kcalThereareabout250kcalsinacandybar,sothereareabout

1#ofcalories = candybar 103500candybars

2So,ittakesabout500candybarsworthofenergytoboilthewater.

3


24/43


R

velocity

Figure6: Flowisfasterinthecenterofthepipe. Itslowssticksattheedges(i.e. theinnersurfaceofthepipe.)Example3. PipeflowPoiseuillewasthefirstpersontostudyfluidflowinpipes(arteries,capillaries). Hefiguredoutthe

velocityprofileforfluidflowinginpipes is:v = c(R2r2)

distancev = speed =time

r

v

cR2

R

v=c(R2-r2)

Figure7: Thevelocityoffluidflowvs. distance fromthecenterofapipeofradiusR.Theflowthroughtheannulus(a.k.aring)is(areaofring)(flowrate)

areaofring=2rdr (SeeFig. 8: circumference2r,thicknessdr)v isanalogoustotheheightoftheshell.

4


25/43


dr

r

Figure8: Cross-sectionofthepipe.

R

R

totalflowthroughpipe = v(2rdr) =c (R2r2)2rdr0 0 R

(R2R2r2 r4R

= 2c rr3)dr= 2c2 4 00

flowthroughpipe = cR42

NoticethattheflowisproportionaltoR4. Thismeanstheresabigadvantagetohavingthickpipes.Example4. DartboardYouaimforthecenteroftheboard,butyouraimsnotalwaysperfect. Yournumberofhits,N,at

2

radiusr isproportionaltoer .2

N =cerThislookslike:

1

2r

y = ce-r2

Figure9:

This

graph

shows

how

likely

you

are

to

hit

the

dart

board

at

some

distance

rfrom

its

center.

Thenumberofhitswithinagivenringwithr1 < r < r2 isr2

2c er (2rdr)

r1

Wewillexaminethisproblemmoreinthenextlecture.

5


26/43


Lecture23: Work,AverageValue,ProbabilityApplicationof IntegrationtoAverageValue

Youalreadyknowhowtotaketheaverageofasetofdiscretenumbers:a1 +a2 a1 +a2 +a3

or2 3

Now,wewanttofindtheaverageofacontinuum.

y=f(x)

a b.

x4

y4.

Figure1: Discreteapproximationtoy=f(x)onaxb.

Average y1 +y2 +...+ynn

wherea=x0 < x1 < xn =b

y0 =f(x0), y1 =f(x1), . . . yn =f(xn)and

n(x) =ba x= bn

a

andThe limitoftheRiemannSumsis

blim(y1 + +yn)b

n

a= f(x)dx

anDividebybatogetthecontinuousaverage

y1 + +yn 1 blim = f(x)dxn n ba a

1


27/43


area = /2

y=1-x2

Figure2:

Average

height

of

the

semicircle.

Example1. Findtheaverageofy=1x2 ontheinterval1x1. (SeeFigure 2)

1 1 1 Averageheight=

2 1x2dx= 2 2 = 41

Example2. Theaverageofaconstant isthesameconstant b1

53dx=53ba a

Example3. Findtheaverageheighty onasemicircle,withrespecttoarclength. (Used notdx.SeeFigure 3)

equal weighting in

different weighting in x

Figure3: Differentweightedaverages.

2


28/43


y = sin

1 1(

cos) 1 2Average = sind= (

cos

(

cos0))==0 0

Example4. Findtheaveragetemperatureofwaterinthewitchescauldronfromlastlecture. (SeeFigure 4).

2m

1m

Figure4: y=x2,rotatedaboutthey-axis.First,recallhowtofindthevolumeofthesolidofrevolutionbydisks.

1 1 y2 1 =

0 2V = (x2)dy= ydy= 20 0Recall that T(y)=10030y and (T(0)=100o;T(1)=70o). The average temperature per unitvolume iscomputedbygivinganimportanceorweightingw(y) =y tothediskatheighty.

1T(y)w(y)dy

01w(y)dy0

Thenumeratoris 1 1 1(10030y)ydy=(500y210y3) =40Tydy=

0 0Thustheaveragetemperatureis:

0

40=80oC

/2Comparethiswiththeaveragetakenwithrespecttoheighty:

1 11Tdy=

1 (10030y)dy=(100y15y2)1

=85oC00 0

Tislinear. LargestT =100oC,smallestT =70oC,andtheaverageofthetwois70+100

=852

3


29/43


30/43


dr

width dr,

circumference 2r

weighting ce-rr

Figure6: Integratingoverrings.

3r1

2Theringhasweight cer (2r)(dr)(seeFigure 6). Theprobabilityofadarthittingyourbrotheris:

12r1 ce

r22rdr6cer22rdr0

Recall that 1 = 5312 is our approximation to the portion of the circumference where the little6

brotherstands. (Note: er2 =e(r2 ) not(er)2 ) b

2 1er2 b 1 1 deb2 2 er2 2=2rerrer ea+dr=

a =2 2 2 draDenominator:

2

2 1 R 1eR2 1e02 1=er er +rdr= =2 2 2 200

(NotethateR2 0asR.)

Figure7: NormalDistribution.

2 213r1 cer 2rdr 13r1 er rdr 1 3r1 226 2r1 = 6 2r1 = er rdr=er 3r1Probability=

cer22rdr 2rdr 3 6er 2r12r10 05


31/43


32/43


VolumebySlices: AnImportantExample

2ComputeQ= ex dx

Figure8: Q=Areaundercurvee(x2).Thisisoneofthemostimportantintegralsinallofcalculus. Itisespeciallyimportantinprobabilityandstatistics. Itsan improperintegral,butdontletthosesscareyou. Inthis integral,theyreactuallyeasiertoworkwiththanfinitenumberswouldbe.

TofindQ,wewillfirstfindavolumeofrevolution,namely,2

V =volumeunderer (r= x2 +y2)Wefindthisvolumebythemethodofshells,whichleadstothesameintegralasinthelastproblem.

2

The shell or cylinder under er at radius r has circumference 2r, thickness dr; (see Figure 9).2ThereforedV =er 2rdr. Intherange0rR,

R2 2 R

=eR2 +er 2rdr=er0

WhenR, eR2 0,0

2

er 2rdr= (sameasinthedartsproblem)V =0

7


33/43


34/43


35/43


36/43


Lecture24:Numerical IntegrationNumerical Integration

Weusenumerical integrationtofindthedefiniteintegralsofexpressionsthat look like: b(abigmess)

aWe also resort to numerical integration when an integral has no elementary antiderivative. Forinstance,there isnoformulafor x 3

cos(t2)dt or ex2dx0 0

Numerical integrationyieldsnumbersratherthananalyticalexpressions.Welltalkaboutthreetechniquesfornumericalintegration: Riemannsums,thetrapezoidalrule,

andSimpsonsrule.1.RiemannSum

a b

Figure1: Riemannsumwith leftendpoints: (y0+y1+. . .+yn1)xHere,

xi xi1 = x(or,xi =xi1 + x)

a=x0 < x1 < x2


37/43


2. TrapezoidalRuleThe

trapezoidal

rule

divides

up

the

area

under

the

function

into

trapezoids,

rather

than

rectangles.

Theareaofatrapezoidistheheighttimestheaverageoftheparallelbases:

base1 + base2 y3 +y4Area=height = x (SeeFigure 2)

2 2

y3

y4

x

Figure2: Area= y3+y4 x2

a b

Figure3: Trapezoidalrule=sumofareasoftrapezoids.

TotalTrapezoidalArea = x y0 +y1 +y1 +y2 +y2 +y3 +...+yn1 +yn2 2 2 2

= xy

20 +y1 +y2 +...+yn1 +y

2n

2


38/43


Note: The trapezoidal rule gives a more symmetric treatment of the two ends (a and b) than aRiemannsumdoestheaverageof leftandrightRiemannsums.

3. SimpsonsRuleThis approach often yields much more accurate results than the trapezoidal rule does. Here, wematchquadratics(i.e. parabolas), insteadofstraightorslanted lines,tothegraph. Thisapproachrequiresanevennumberofintervals.

x x x

y0

y1

y2

x x

Figure4: Areaunderaparabola.

y0 + 4y1 +y2Areaunderparabola=(base)(weightedaverageheight)=(2x)

6Simpsonsrulefornintervals(nmustbeeven!)

1Area=(2x) 6 [(y0 + 4y1 +y2) + (y2 + 4y3 +y4) + (y4 + 4y5 +y6) + + (yn2 + 4yn1 +yn)]Noticethefollowingpattern inthecoefficients:

1 4 11 4 1

1 4 11 4 2 4 2 4 1

3


39/43


0 1 2 3 4

1st chunk 2nd chunk

Figure5: AreagivenbySimpsonsrulefor four intervalsSimpsonsrule:

b xf(x)dx

3(y0 + 4y1 + 2y2 + 4y3 + 2y4 +. . .+ 4yn3 + 2yn2 + 4yn1 +yn)

aThepatternofcoefficientsinparenthesesis:

1 4 1 = sum 61 4 2 4 1 = sum 12

1 4 2 4 2 4 1 = sum 18Todoublecheckplug inf(x) = 1 (neven!).

x x n n 3 (1 + 4 + 2 + 4 + 2 + + 2 + 4 + 1) = 3 1 + 1 + 4 2 + 2 2 1 =nx (neven)

4


40/43


1 1Example1. Evaluate dx using two methods(trapezoidal and Simpsons) of numerical

1 +x20integration.

0 1x x

Figure6: Areaunder(1+

1x2) above [0,1].

x y0 +y1BySimpsonsrule:

2 2 2 2 5 2 2 2 2 5 4

x 1/2 4 1(y0 + 4y1 +y2) = 1 + 4 + = 0.78333...3

Exactanswer:3 5 2

1 1 1 dx=tan1x =tan11tan10 =

4 0 = 4 0.7851 +x2 0Roughlyspeaking,theerror,|SimpsonsExact|, hasorderofmagnitude(x)4.

0

x0 1

1 452

1Bythetrapezoidalrule:

1/(1+x2)

12

+ y2 = (1) + + = + + = 0.7751 1 1 1 4 1 1 1 1 4 1

5


41/43

Exam3Review 18.01Fall2006

Lecture25: Exam3ReviewIntegration

1. Evaluatedefiniteintegrals. Substitution,firstfundamentaltheoremofcalculus(FTC1),(andhints?)

2. FTC2: d x

f(t)dt=f(t)dx a

xIfF(x) = f(t)dt,findthegraphofF,estimateF,andchangevariables.

a3. Riemannsums;trapezoidalandSimpsonsrules.4. Areas,volumes.5. Othercumulativesums: averagevalue,probability,work,etc.Therearetwotypesofvolumeproblems:

1. solidsofrevolution2. other(dobyslices)

Intheseproblems,therewillbesomethingyoucandrawin2D,tobeabletoseewhatsgoingoninthatoneplane.

Insolidofrevolutionproblems,thesolidisformedbyrevolutionaroundthex-axisorthey-axis.You willhavetodecidehowtochopupthe solid: into shellsordisks. Putanother way, youmustdecide whether to integrate withdx ordy. After making that choice, the rest of the procedure issystematicallydetermined. Forexample,considerashaperotatedaround they-axis.

Shells: heighty2y1,circumference2x,thicknessdx Disks(washers): areax2 (orx2

2x2),thicknessdy; integratedy.1Work

Work=Force DistanceWeneedtouseanintegraliftheforce isvariable.

1


42/43


Example1: Pendulum. SeeFigure 1Considerapendulumof lengthL,withmassmatangle. Theverticalforceofgravityismg (g=gravitational

coefficient

on

Earths

surface)

L

mass m

mg

Figure1: Pendulum.In Figure 2, we find the component of gravitational force acting along the pendulums path

F =mgsin.

mg

Figure2: F=mgsin(forcetangenttopathofmotion).

2


43/43


Is it possible to build a perpetual motion machine? Lets think abouta simple pendulum, andhowmuchworkgravityperformsinpullingthependulumfrom0 tothebottomofthependulumsarc.

NoticethatF varies. Thatswhywehavetouseanintegralforthisproblem.

0 0

W = (Force) (Distance)= (mgsin)(Ld)0 0

W =Lmgcos0 =Lmg(cos01)=mg[L(1cos0)]0

InFigure 3,weseethattheworkperformedbygravitymovingthependulumdownadistanceL(1cos) isthesameasifitwentstraightdown.

L

L(1-cos)

Figure3: Effectofgravityonapendulum.Inotherwords,theamountofworkrequireddependsonlyonhowfardownthependulumgoes.

Itdoesnt

matter

what

path

it

takes

to

get

there.

So,

theres

no

free

(energy)

lunch,

no

perpetual

motionmachine.

unit3 who sept24

Documents