《微积分》课程教学课件（Calculus）06. Applications of Definite Integrals

CALCULUSTHOMAS'CALCULUSARLYTRANSCENDENTAE8.6JAMESSTEWARTIndeterminate Forms andL'Hopital's RuleERMSONAThe meaninglkess expressions which cannot be evaluated in a consistentwayare ndcterminate foms.Suchas0,co0,ooco,o,andTHEOREM6-LHopitat'sRuteSuppose thit f(e) = g(a) 0, that f andg arediffmtiableneninl/ainingandtatg(x)onifThenx)1)1.Indeterminate FormO/0Han suefaning that the limit on the right side of this equa55Esing EHopitaly RaleTofindfGx)aby IHopital's Rule, continue to differentiate fand g,so long as we still get thefonm O/o at r a,Butas soon as one or the other of these derivatives is diffesent from zero at x α we stop differentiating. L'Hopitals Rule does not applywhen either the mumerator or demominator has a finite nonzero limitEXAMPLE1The following limits involve o/0 indeterminate forms, so we apphyr'Hopital's Rule, In, it must be applied repeatedlyEXAMPLE 2 Be careful to apply FHapital's Rule correcty:(a)im3=sinz lim3.c05) 3 c05.x2S营VI+2V11-!-m-f-0Sng,lim i tand(b) limin40EXAMPLE3In this example the one-sided limits are diffrent.V1+x-1 -x/2(1/2)1 + x)-1/2 1/2Solrglimeiateag(0)lim电国信-国#-82xx0+x(/4)(1 + x)-/org mai fonl=m1cOSx402Soll3r2Y雪罪SgNetlimt iefound目

2016/11/15 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 8.6 Indeterminate Forms and L’ Hopital’s Rule ^ 1. Indeterminate Form 0/0 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The meaningless expressions which cannot be evaluated in a consistent way are indeterminate forms. Such as 0/0, ∞·0,∞-∞,00,and 1∞ ( ) ( )' ( ) f x g x Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE4Find the limits of these co/oo forms:asSolutionerator and den(a) The nuinator are discmuousat =w/2,sowe investigate theonesided limits there.To apply IHopitalRule, we canchoose /tobeany openinterval with x = /2 as an endpoint.-, r x 2.IndeterminateFormsseexmtrtimarimar-1 + tan xsecx0/80,00,00-00sec x tan xsecx=, lim,sinx=1lim/I+tanx.limsecxseex_=lseexseexSo limr i+any, and Jima +tans- lim, +tanx1/x一=0002VxVE:A会-EXAMPLE6Fiod the limit ofthisoc-oo formEXAMPLE5Findthe limits ofthese co-0forms(-)(a) m (rsin)(b) J VinxIfxtbenx-nds-1-0SolutionSolutioSimilarly.ifx--o-,then.sin.x-o"and(a) (in)(n)1--VNeither form revealsted oca/asat happensinthelimit.Tofind out,the fraction-1-1/x--1/2/iHipralyRuleThen we apply I'Hopital's Rule to the resultm(-2V) - 0品hnsinx2x9-0.国广Limits that lead to the indeterminate forms , o°, and oo°can sometimes be handled byfirst taking the logarithm of the function, We use I'Hopital's Rule to find the limit of thclogarithm expression and then exponentiate the result to find the original function limitIf lim, In f(x) -- L, thenlim fo) - lin ehroe.3.Indeterminate PowersHere e may be either finite or infinite.目山

2016/11/15 2 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 2. Indeterminate Forms ∞/∞,∞·0,∞-∞ 2 ( /2) ( /2) ( /2) sec sec tan lim lim lim sin 1 x x x 1 tan sec x x x x    x x           ( /2) ( /2) / 2 sec sec sec lim lim lim =1 x x 1 tan 1 tan 1 tan x x x x So and   x x x           Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 3. Indeterminate Powers

EXAMPLEApply I"Hopital's Rule to show that lim,or (1 + x)/ = e.EXAMPLE8Find limx/SolutiomThe limit leads to the indeterminate form 1e, We let f(x) = (1 + x)/ andfind lim,-grIn f(x)SincoThe limit leads to the indeterminate form oo, We let f(t) = x/ and findSolutionlim,In f(x).Sincen Ju)= n(I+)/=Im(1+ )In fo) = In x/w = .'Hopital's Rule now applies to giveI'Hopital Rule givesIn(1 + x)m, o im.一1+x-蓝Tm= ↓ = 1.-1 = 0.Therefore, + - m F ee' =Therefore lin / im /() im eh Ao e° = 1.国AEExercisessint?1613.jm1225. _m(--号)x -121. jimo in(se a)Chapter 6sine4()1n327.limApplicationofDefinite Integrals51. lim x/-ne57.m(t+2)2m)6()66. Jim, sinx*Inx6二广6.1OVERVIEWIn Chapter 5 we saw that a continuous function over a closed interval has definite integral, which is the limit of any Riemann sum for the function. We proved thawe could evaluate definite integnals using the Fundimental Theorem of Calculus. We alsoVolumes Using Cross-Sectionsfound that the area under a curve and the area between two curves could be computed adefinite integrals.白百

2016/11/15 3 Exercises 0 1 6  2 1 ln 3 1 2  1 e  1 2 e 3 e 0 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Chapter 6 Application of Definite Integrals Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 6.1 Volumes Using Cross-Sections

1. The cross-sectionof solid S is the planeregion formed by intersecting S with aplane1.CrossSectionandVolumeofHLuE6.tAmioeSxofCylindrical Solid2. If the cylindrical solid has aknown base area A and height h,thenVolumebase area xheighAhnanetteCtetiedneaTaoinidwHindeandfednwahAAApmosirRda-Ge Sl) hes heigh-4-12.Volumes of Solids ofLopcal sabRevoluton:The Method of SlicingATHEURE63 Atypioat tlissabinthesolidsFIGURE 6.4The solidthin slab inFigure 6.3 is shown enlangod here.liiadby thlecylindnical solid with广eS(xa) beareaA(n)and heighCThe volume , of this cylindrical solid is A(xa) - A.xi, which is approximately the samevolume as that of the slab:Volume ofthe kth slab V, =A(x) A)The volume Vofthe entire solidSis thierefore approximated by the sum ofthese cylindri-cal volumes1-2n-2)Calculating the Volume of a Solid1.Sechheoldypicalss-sectonThis is a Riemann sum for the function 4(x) on [a, b], We expect the approximations from2. Find a formula for A(x), the area of a typical cross-section.these sums to improve as the norm ofthe partition of [a, b] goes to zero. Taking a partitionof [a, b] into n subintervals with [Pf→ 0 gives3,Find the fimits of integration4. negute 4(x)to find the volumega04(x)dsDEFINITIONTheolumeofasotidoCnltegrahross-soctional area A(x)fromxtoxbistheinlegralof.4fromaloV[A)dAAAESRAE

2016/11/15 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 1. Cross Section and Volume of Cylindrical Solid Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1. The cross-section of solid S is the plane region formed by intersecting S with a plane. 2. If the cylindrical solid has a known base area A and height h, then Volume=base area X heigh=Ah. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 2. Volumes of Solids of Revoluton: The Method of Slicing typical slab

EXAMPLE2EXAMPLE1Apymid3mhighhsqurehsethatismonsidTherossAcurdwodgescutfromacircularcylinderofadius3bywoplansOne plane is perpendicular to the axis of the cylinder. The second plane cnrosses the firsectice of thepyramid perpendicular to the altitudex m down from the vertex isa squareplane at a 45*sngle at the center ofthe cylinder.Find the volume of the wedge.X m on a side. Find the volume of the pyramid.2V9-2(t) (heighi)(width) (o)(2V9 P)Selution=.2xV9 2A sketck. We draw the pyramid with its alitude along the x-axis.miThe rectangles run fr=. 3, so we haveE-0Xad iswrtheorigindincludeapicalcossctic[aa) d-L2rV9-Pa(-V9-2)2.4 formula for,d(x). The.cross-section.at x is a square x mcte on a side, so jts area ib--(9-FIGURE 6.6 The Enple 2Ax)=1.stioodperpendicdartofhexatisTeThe limisofigynitiThesuans leon the plais fiomx tox33,cross-setios are rextangke-0+f(omJutagrtatelo findhe volam= 18.A-[=[起-号-9m.ExercisesVolumes by SticingEXAMPLE3Caalienis principlesys that solids withequlaltitudes and idenicalFind the volumes of the solids in Elxercises 110.tionsachithvehsewolue (figure7hisfolscross-sectmedistely from thdefinitioa of vodumese the cross-sectioaal area functios A(x1, The solid lies betseen planes perpeadicselar to the x-axis at 0aadtheiteral a,b]aethesmeforothsolidand r 4., The cross-sectioms perpendicular to the axis on thinterval o ≤ x ≤ 4 are squares whose diagonals rn from theparaholay = Vx to the parabolay VrA) ( de 42x dx =[x] = 16A(x) dx =ameylndFIeuBE6.7CooolmtorcwicThesesolikvealchcaeanlodwithsacksofcoin国五2. The solid lies between planes perpendicular lin the x-axisaExercisesThe molid fie hten plins pependicler m te satiExercises1 and LThe cesons-I and x i.The erogs-sectis perpendicular to thes-aticinular dals shae dindesm fiom the pandbetwen these planese squares whose bases run from the semi-to the panhoa y=2~1circke y - -Vi'o the semicinde y = V1?()=(base)=(2y/1-P=4(1)-4-L4=4r-+-(2) 2((22+),bA(x) =dv] M)dx (1-2+)=↓-g+],-2(1-3+)g普白E

2016/11/15 5 Exercises Exercises Exercises 2 2 2 2 A x base x x ( ) ( ) (2 1 ) 4(1 )      1 2 3 1 1 1 4 16 ( ) 4(1 ) [4 ] 3 3 b a V A x dx x dx x x          

5. The base of a solid is the region betcen the curve y. 2Vsinand the interval [0, w] on the x-axis. The erous-sections perpendicular to the x-axis ana equilalenal triangles with beses running from thes-axis to th4,Thoilie bwppdureExercisesExercisescurvieasshownintheaccompamying figure.and x=1. The cros-sectiones pepedicularto the saxiheAtuecmtheseplanesaresgareswhosediagomalsrunfromthemicircley=V1rtothesemicirclepmV1xVss)2-2)2-bk squres with bases rumning from the x-axis to the curveV=,(=2(1-)=x(n9)-V5an(b) STEP 1) () = (tide) =(2V/in x)(2/in )=4 sln xSTEP 2)a≤0,b=r4ndx=-4=B0+1)2STEP3)VA(x) dx国AEExercises Tho solid les betwem planes perpendietlar to the sratise/3 and x = g/3. The cros-sections perpendicular to thox-axis arcacireulardikx withdiamciersnmning fom thecury tanx to the curvey.sec xIhsquasees uhose bases rin fiom the curvenxtohcurve y = seex3.Solids of Revolution:P(2 x12)(b) STEP 1) A() = (edgn) = (r5xTheDiscMethodSTEP2)a=-55-(2 m2*-1-29)dx=2(V1)4/5STEP 3) V -A(x) ds ()+()(x) dxSTEP.S)V--+(由)(-++(由)(5-号)日EXAMPLE4()RotationAbout thex-axsThe region between the curve y Vx, o ≤ x ≤ 4, and the x-axis irevolved about the x-atis to generate a solid. Find its voltSelutionWe dnw figures shoming the ngion,atypical rndinV3and the getened solid The volume isa[Rn)ParBtetvaThe solid generated by rotuating (orDREandolidof8revohmg)apneregonabouanaxsudutioathlisEEsptanecalledasolidofnerofatioFCURE 6.8:The region (s) ax iolid aoolun(b) inamete4(x) (nadius) [R(x)PNoluebyDiskIXE告人(x)d[R(x)Pd

2016/11/15 6 Exercises 2 2 2 2 ( ) (2 1 ) ( ) 2(1 ) 2 2 diagonal x A x x      1 3 1 2 1 1 8 ( ) 2(1 ) 2 3 3 b a x V A x dx x dx x                 Exercises Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 3. Solids of Revolution: The Disc Method The solid generated by rotating (or revolving) a plane region about an axis in its plane is called a solid of revolution. (1)Rotation About the x-axis

EXAMPLESThecindlePRohEXAMPLEFind the volurcofihesnlenerated by revolving.he tegon bou++y-aAbouttheby y Vr and the lisis y m I,x 4 abouFind ils wolumLine ParallelNE-to the x-axisayio&sTheslenny-V-PaupksSotutisnWe dew figures showing the segion, a typicat radius, and the generated solid(Figure 6.10), The volume i(x) y, w(a?-)m(R(x)P drPsTherefore, the wolu[-[V-1a[[----[[2Vg+1]4thpndindogn国L号-2号+-lrioputEXAMPLE7Find the volume of the solid generated by revolving the region between(3)Rotation Ahouthe y-axsthe y-axis and the curve x = 2/y, I ≤ y ≤ 4, about the yarisTo find the volume ofa solid gemeralod by revolvingaregion berween they-axis andaTcurvex = R(y),e y d, about the y-axis, we use the same method withx replaced byyIn this case, the circular cross-section is(6)4(y) = [mdiusP = [R()P,and the definitionofvolume givesVolume by Disks for Rotatlon About the y-axisSolurtioeWe draw fieurcs sbowine the rcgion. a typisal radius, and the gencratod solidM(y)dy[R(y)Pdy(Figune 6.11). The volume im[R(yIPyRalnkig-Fts1-()0五--4[-4]-3(+jRoutun AlouLbeLine Paralel lo the y-aseEXAMPLE8Find the volume of the solid generated by revolving the region betweentheparabola x y2+ 1 and the line x 3 about the line x =FT-2a/2(ayi)4.SolidsofRevolution:+++aVthe Washer MethodV5We draw figures showing the region, a typical radius, and the generated solidSolution(Figure 6.12). Note that the crosg-sections are perpendicular to the line x = 3 andhave y-coordinates from y V2 to y = V2. The volume ise[RyiPadyy-AV2whear-[2 jfay64eV24+2[44y2 + y'1dyaiealmt5LVSlide 4- 42

2016/11/15 7 (2)Rotation About the Line Parallel to the x-axis (3)Rotation About the y-axis (4)Rotation About the Line Parallel to the y-axis Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 42 4. Solids of Revolution: the Washer Method

()RotationAboutthex-axis()RotationAboutthex-axisVolume by Wiashers for Rotation About the xr-axts(R()P [) y4(x)dionsofthe solid ofteinlFGURE6.11Thoedhnot disks, so the integrCWasher MethodadlestoslighlydiferemfomulaIfthe regioe we revohve to generale a solid does not borderon or cruss the axis ofrevolotion,the solid has a hole in it (Figure 6.13), The cross-sections perpendicular to the axis ofrevolution are washers (the purplish circular surface in Figure 6.13) instead of disks. Thedimensions of a typical washer areOuter radius: R(x)Iner rsdius: (x)The washer's area is() [R(x)P [r(x)P ([R(x)P [)P),lide 4- 43Slide 4- 44EXAMPLE9The region bounded by the curve y = x* + I and the line y = x + 33.Find the limits of integration by finding the x-coordinates ofthe intersection points ofthe curve and fine in Figure 6.14a.is revolved about the x-axis to generate a solid. Find the volume of the solid.x+1=x+3x+x2=0(x +2)(x =1) = 01.2=-2,F=1Limits ofiategrntion4,Evaluate the volume integral.nkervddinlcgratiorHirber(a)Solutionra2TaR-RotfasaroundzatiDraw the region and sketch a line segment across it perpendicular to the axis of revolution (the red segment in Figure 6.14a)+3-（2+）Walses from Stepo22.Find the outer and inner radi of the washer that would be swept out by the line segandtment if it were revolved about the x-axis along with the region.These radi are the distances of the ends of the line segment from the axis of revolu.(86)xSirmelifyalgshrioalytion (Figure 6.14).R(x) = + + 3Outerradius:号.r() =x2+ 1Inner radius:EXAMPLE10The region bounded by the parabols y = and the line y = 2x in thefirst quadrant is revolved about the y-axis to generate a solid. Find the volume of thesolid.(2)RotationAboutthey-axisSolution-f-vThe radii of the washer swept out by theTo find the volume of a solid formed by revolving a region about the y-axis, wuline segment are R(y) = Vy, r(y) = y/2use the same procedure as in Example 9, but integrate with respect to y instead of x.The line and parabola intersect at y. oInthissituationthelinesegment sweeping outatypical washerisperpendiculartothey-axis (the axis of revolution), and the outer and inner radii of the washer are func-and y = 4, so the limits of integration aretions ofy.c = 0 and d = 4. We imtegrate to find ihe volume:(R() [)P))RettimVolume byWashers forRotation About they-axisSbae faendiad=-([-[)-A- ([RO)P-[rOXSin dapetin61--(-)(---domwp catly teliaSlide 4- 47Slide 4.48

2016/11/15 8 Slide 4 - 43 (1)Rotation About the x-axis Slide 4 - 44 (1)Rotation About the x-axis Washer Method Slide 4 - 45 Slide 4 - 46 Slide 4 - 47 (2)Rotation About the y-axis Volume by Washers for Rotation About the y-axis V A y dy R y r y dy d c d c   ( )  ([ ( )] [ ( )] ) 2 2  Slide 4 - 48

17, Ahout the j-axisIAbouttheS-axhnes by the Disk MethodIn Exercises 15-18, find the volume of the solid generated by revolvExercisesExercisesing the shaded region about the given axi“()15, About ther-axis16, About the j-axis+2y-21B()()==4du.ryoouuoy-i.--- [(-V=(-1+a)o=-a+tam og/-1(-++1-0)4--P--[(-α--(++)-+)15. H(x) =y=1 1V[R(x)P dsIR(d)=iexca s R(s) =0ia=0amd bgare the lmlta afiugalinn=+(2-1+喜)-号0-1-00(te x toe xP dx = r n)----oP--[()---r[-1 2[(-0)0]lide 4- 49Slide 4- 50Find the wolumes of the solids generated by revolving the regionsFind che volumes of the solids generaled by revolving the regionboundod by the lines and curves in Exereises 19-28 about the x-axibounded by the lines and curves in Exercises 31-36 about the y-axisExercisesExercises19. y-,y-0, x-220. y=.y=0.X=231. The region enclosed by x = Vsy, x= 0, y =1, y =121y=V9-1=22.Y32, The region enclosed by x = y/, = 0, y = 2(rd19, R(0)=)(R(x1Pdx[R(g)P dy =+yd. Ro)=V-y+Va-]-x-/5y2V5*, g1(-1 =22m. 8(x=-(R(y)P dy =32. R(y) = y3/ μ VRxIPe-[(9-2)d21. (2) = V9-2=4r-(9-xa)a0218=36J8-3u4-51Slide 4-526.21.Slicing withCylindersVolumes UsingCylindrical ShellsHowever, thismethodofslicing issometimesawkwardto apply,as wewillilustrate inoufirst example. To overcome this difficulty, we use the same integral definition for volume,but obtain the area by slicing through the solid in a different waySlide 4- 53Slide 4-54

2016/11/15 9 Slide 4 - 49 Exercises 15. 16. Slide 4 - 50 Exercises 17. 18. Slide 4 - 51 Exercises 19. 20. 21. Slide 4 - 52 Exercises 31. 32. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 53 6.2 Volumes Using Cylindrical Shells Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 54 1. Slicing with Cylinders

.EXAMPLE1The region enclosed by the x-axis and the parabola y = f(x) = 3x x2is revolved about the vertical line x=1 to generate a solid (Figure 6.16). Find the volumeof the solidEach slice is sitting over a subinterval ofthe x-axis of length (width) Axg. Its radius isLi-happroximately (1 + xa),and its height is approximately 3xe x,2. If we unroll the cylin-Asisdra-l*der at x, and flatten it out, it becomes (approximately)a rectangular slab with thickness A.t:*(Figure 6.18).The outer circumference of the kth cylinder is 2r ·radius = 2m(1 + x),thPandthis is the lengthofhe rolled-out rectangularslab. ts volume is approximated by thatFIGURE6.16(a) The graph of the region in Example 1, before revolutionPofa rectangular solid.(b) The solid formed when the region in part (a) is revolved about theaxis of revolution x = 1V, = circumference × height X thickness2m(1 + x) (3.=x) 4xgslide 4-55Slide 4-58Summing together the volumes V, of the individual cylindrical shells.over the interval[0, 3] gives the Riemann sum=22n(m + 1(3x )mTaking the limit as the thickness Ax → 0 and n + 00 gives the volume integral2. The Shell Method 2m(x± + 10(3m x) Ar)y-lim2(x + 1)(3x x) dr2w(3x2 + 3r g x) dx(2 + 3x x) dx2-23+号+-1+-454-Slide 4- s8Suppose the region bounded by the graph of a nonnegative continuous functionWe approximute the volume of the solid Sby summing the volumes of the sbells swept outy=f(x) and thes-axisoer thefiniteclosed interval[a,b] liestotheright the verticalby the n rectangles based on P:line x L (Figure 6.19a)., We assume a ≥ L, so the vertical line may touch the regionbut not pass through it. We generate asolid S by rotating this region about the verticalline L.df iewoladliaetiThe limit of this Riemanm sum as cach AO gives the volume of thenunofrevulutionsolid as a definite integral:-7r(shell radius)(shell height) dt=fun-L)/(x)dt2m(xriaonal Line-Shell Formala for Revolution About a Vertical LineThe volume of the solid generated by revolving the region between the x-axis andLet P be apartition ofthe interval [a,b] by the points athe graph ofa continuous function y f(x) 0, L S a.S x S b,abouta ver<<X=htical line x L isand let c, be the midpoint of the kth subinterval [-1, xa].- We approximate the region inFigure6.19a withangesbdonthispartton[]ypicalappxmating rV- [~2-(sell )(shell)Setangle has height f(c,) and width Ax, = Xg x-, If this rectangle is rotated about the.-2元dTrsatnvertical linex= Ltenasell isswept out,as in Figure 6.19b.Afomula from goometrytells us that the volume of the shell swept out by the rectangle is

2016/11/15 10 Slide 4 - 55 Slide 4 - 56 Slide 4 - 57 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 58 2. The Shell Method Slide 4 - 59 Slide 4 - 60 Horizontal Line 2 d c shell shell dy radius height           