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# Calculus MAT 021B

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This 467 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 021B at University of California - Davis taught by Qinglan Xia in Fall. Since its upload, it has received 20 views. For similar materials see /class/187344/mat-021b-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15

54 fundamental theorem of calculus Calculus Calculus consists of two parts 0 Differential Calculus Math2 1A This part comes from the study of Calculus Calculus consists of two parts 0 Differential Calculus Math2 1A This part comes from the study of velocity problem Calculus Calculus consists of two parts 0 Differential Calculus Math2 1A This part comes from the study of velocity problem slope problem Calculus Calculus consists of two parts 0 Differential Calculus Math2 1A This part comes from the study of velocity problem slope problem rate of change etc Calculus Calculus consists of two parts 0 Differential Calculus Math2 1A This part comes from the study of velocity problem slope problem rate of change etc o Integral Calculus Math21B Calculus Calculus consists of two parts 0 Differential Calculus Math2 1A This part comes from the study of velocity problem slope problem rate of change etc o Integral Calculus Math21B This part comes from the study of area problem Calculus Calculus consists of two parts 0 Differential Calculus Math2 1A This part comes from the study of velocity problem slope problem rate of change etc o Integral Calculus Math21B This part comes from the study of area problem 0 Today we will see that they are closely related by a very important theo rem The Fundamental Theorem of Calculus Mean Value Theorem Theorem If f is continuous on a b then at some point C in a b b nabianmmx Sometimes we write it as b AfWWw for some 3 in a b Proof min f 56 S f 56 S max f 56 Abmm fxdx g Ab xwx g bmaxfxdx b a minfxb a fxdx maxfxb a a b minfltxgt g b i aa fxdx g maxfltxgt Since f is continuous by the intermediate value theorem there exists some 3 in a b so that b N b i a mezzo Suppose f is continuous on a b For any x in a b consider the function J ftdt a What is F Fx h F 1 x lino h h hm f5 flttgtdt ffflttgtdt h gt0 h 1 h gt 1 gimo f ck Where ck is some point between x and x h f 56 That is d m g flttgtdt fltccgtu The fundamental theorem of calculus Part I TheoremIf f is continuous on a b then f ofquot f tdt is continuous on a b and di erentiable on a b and its derivative is f F ltcgt 36mm fltccgt NowF f so fax f tdt is an antiderivative of f Thus suppose G is an antiderivative of f then 5U ftdt Cx o a for some constant 0 Let x a a ftdt Ca C thus C Ga a Therefore I ftdt Cx Ca In particular let x b b ftdt 05 Ca In particular let x b b ftdt Cb Ca a The fundamental theorem of calculus Part II Theoremf f is continuous on a b and F is any aniiderivaiive of f on la bl b a fxdx Fltbgt Fa Sometimes we write it as b Aimmmm3 Remark o Roughly Part I says x d g f f bd FFb adx la That is inte ration and differentiation are inverse o erators 9 while Part 11 says Remark o Roughly Part I says x d g f f bd FFb adx la That is inte ration and differentiation are inverse o erators 9 while Part H says Remark Part H says that the de nite integral b a fltccgtdcc W M for any antideriVatiVe F of f Remark o Roughly Part I says x d g f f bd FFb ad a That is integration and differentiation are inverse operators while Part H says Remark Part 11 says that the de nite integral b a fltccgtdcc W M for any antiderivative F of f Thus to evaluate a de nite integral of a continuous function we only need to nd its antiderivative Without having to calculate limits of Riemann sums This observation saves a lot of energy 0 x 05 2 d xgt 2 3 3x 21H 2 e g8 32 2 32 2 2lne g 2ln1 2 2 3 Part1 d x ff is also very powerful offsmtdt Part1 d x ff is also very powerful offsmtdt d J sin tdt Smx dx a 2 of ffet 1dt Part1 d x ff is also very powerful offsmtdt d J sin tdt Smx dx a 2 of ffet 1dt d x 2 2 d a a d 5 1 J 7 U W jm may i 5 1 dd J 3 2tdt 39 d 5 1 dt 61 1 3 8275 d 1 dt d 5 31 82 1 382 r u d x 1 dt 61 1 5t4 2 d m 1 4dt dIE 1 5 13 2 Lety ff dt nd Let u 302 then y flu dt Thus dy 1 du 5 U4 dy Known Unknown dm d21 dt dIE 1 5t4 2 Lety ff Ldt nd 5t4 2 U 1 Let u 10 then y f1 mdt Thus dy 1 du 5 U4 dy Known Unknown dm By the chain rule any 1 n 1 236 2w dx du dx 5u4 dw 5x8 5x8 Thus d 2 1 2x 00 dxl 5t4 5x c Find the formula for d 9515 a Where g is a differentiable function and f is integrable c Find the formula for d 9515 a where g is a differentiable function and f is integrable Let y 990 ftdt nd Let u then y f5 ftdt Thus dy fful dy dy Known Unknown c Find the formula for d 9515 t where g is a differentiable function and f is integrable Layj fmnmd Let u then y f5 ftdt Thus a g I f U d d Known Unknown By the chain rule dy dy dU du I a du d M d fltgltxgtgt g lt55 Thus d 956 fW IfM d d 913 dx Mm d W a flttgtdt film flttgtd ltgltxgtgtg ltxgt flthltxgtgth ltxgt d 4 1 tdt Cm 135L 2 2 l e o 2 2 132 1 3x 6x 2 132 o Practise 53 The De nite Integral Last time we de ned Riemann sums as follows y M my mm rz Fm FIGURE 59 The rccmngles approximalc the legxon between the graph loe mamaquot J A and the xraxis Let P 10951 79 be any partition of a7 17 and let Ck be a point in the kth subinterval 9516479510 the corresponding Riemann sum is de ned by Sp Z ak Axki k1 This is called a Riemann Sum for f on 11717 It approximates the signed area of the region R Limits nt RJemann Sums To increase accuracy of approximation we let the norm HP of partitions approaches zero If the values of all Riemann sums approach a limiting value I as the norm of partitions approach zero we call this value I the de nite integral of f DEFINITION The De nite Integral as a Limit of Riemann Sums Let L 39 4 4 39 1 39ab I is the de nite integral of f over 11 2 and ihai I is his limit of the Rismann sums Ziacl i ifthe following condiiion is sniisried Given any number s gt 0 here is a corresponding number a gt 0 such man for every partilion P m x xquot oru b with HPH lt 5 and any choice of q in In xi we have icil AM it lt 5 i1 Notation the De nite Inleng The symbol for the number I is I I I I The function is the integrand Upper hum of uilcgmlion he variable of inlegml lon b Inlegral Slgquot a When you nd the value oflhe inlegml ynu have cvalualed he integraL Lower limit of inlegraliun lnmginl ul 1min ii In I Riemann Integrable When the limit exists we say f is Riemann integrable over ab and the limit value is called the de nite integral of f over a b In this case we write as n b HIlji r n Z fckAxk fxdx A0 k1 a Question Which functions are Riemann integrable ie for which functions the de nite integral always exists 0 Continuous functions are always integrable 0 Increasing functions are always integrable o Decreasing functions are always integrable o Piecewise continuous functions are always integrable A Nonintegrable Function Suppose fxl if x is a rational number and f 0 if x is an irrational number Then this function is not Riemann integrable over 0 1 This function f is called the Dirichlet function Why Let P be some partition of 0 1 We may construct Riemann sum SP Z f0kA95k k1 0 Choice 1 pick 0 from rational numbers then fckAxk Z 1 Am l k1 k1 0 Choice 2picllt ck from irrational numbers then TL fckAxk Z 0 Axk 0 k1 k1 So the limit depends on the choices of ck the function f is not integrable Properties of De nite Integrals f and g are integrable the de nite integral satis es the following rules mate 5 Rules satis ed by de nite integrals u 1 l oamoInlegimian 39dx 7 1mm rlleumn I n 2 2m mdllllllle thl xdxo lxrviln39wimmu a h 3 CansmmMuupte kJtlx k frL unxummi a n Y xdx IAhx4Y l7 b h 39h 4 Sum undDi mate x agmm Xzbrt gxdx a u 1 n c a s Additivily mmer mm fwd I h r 6 MtlvainIImtuulily 1r has maximum value max and minimum value min f on 11 b then A minf39bllSXdx5 max ba b n Dumitmlinn fx Zglx on a b fxdx 2 gmdx r39 n x 2 01m 11 b gt fxdr 2 o Minn m a 2m Widm Interval 1b Crmxlanl Milli153 c Sum mm hk r dx quot1m gmm The area over a palm is 0 Shown for I 2 Areas add nlax miquot n o a I oh b d Addiriviryjbr liqMile inlegmls39 c MaxMin Inequury f Dominalion hm4 mm fxwx min b r u s fh xhlx x 2 g1xonab 39quot h u s quotinfb in quothmth 7 gum FIGURE 511 Example Suppose 2 5 2 fxdx 37 fxdx 717 and hxdx 2 1 2 1 Find ffzmy 3hxdx De nition of area for nonnegative integrable functions De nition fy f is nonnegative and integrable over a closed interval cg b then the area under the curve y f over cg b is the de nite integral of f from a to b b A Example 0 Let c be any constant then b cdm 7 0 Suppose I gt 0 What is f xdx 0 Suppose I lt 0 What is f xdx o What is xdx o What is fob xgdx Answer SubdiVide 0 b into n subintervals of equal width Pick ck to be the rightside endpoint kAx Then kb 2b ZfckA95 2 k1 k1 b3 n 2 Ezk k1 i Enn12n1 i n3 6 as n approaches in nity Therefore b 3 b 2 acdx 0 3 b 3 3 b a mgdx a 3 Average Value of A Function Recall that when f is a continuous nonnegative function on a b then the average of f is o In general we have area of the region under the graph of f aw b a b i fa fxdx i b a In general we have De nition ff is integrable on a b then its average value on a b also called its mean value is 12 mm mm Example Find the average value of M v4 an on 12 52 Notation and Limits of Finite Sums The sigma notation Z Often we need to write sums with a large number of terms a1a2an We may write it in a compact form using sigma notation TL Zaka1a2an k1 Here a Z the summation symbol stands for sum Greek letter sigma k the index of summation It tells us where the sum begins at the number below the Z symbol and where it ends at the number above 2 o ak a formula for the kth term The index k ends at i n Hquot H The summation symbol 2 a I I V 1 g Greek 1 er Sigma I 49k Is a formula for u ill lLF L 2 The imdm If ElliJle at k E I Example What do the following sigma notations stand for ism k W sin 3 sin 4 sin 5 H 2 2 2 2 5 k 39Zk0k1 5 k 012 3 4 5 Emit t titgta k0 100 1m Zml 1m 11 11 1 39 m 2 3 4 100 ml Any letter can be used to denote the index but typically we use ij k or m n The index may start at any integers but often it starts With l or 0 Also note that signs are alternating here 5 E33333333x5 k1 In general Express the following sums in sigma notation 13233343536373 7 13233343536373ZZk3 k1 flt2gtflt3gtflt4gt flt234gt 234 flt2gtflt3gtflt4gtflt234gtZfltngt 712 0357911 Example 3579 ll These are sums ofodd numbers which may be expressed as 2k l or 2k 1 So it may be written as 39 222 27 1 or 311 2i 1 0r 2 2739 3 They all represent the same sum 3 5 7 9 ll So the same sum may be expressed in different forms Note 1 1 23 33 43 Me Me II MFbx NDI H NDI H OOIH IH V A w CA3 00 CAD 4 00 V In general we have Algebraic rules for nite sums In general we have a n n n Zltakbkgt ZakF25 k1 k1 k1 b n n n 2 bkgt Zak Zbk k1 k1 k1 c For any constant value 3 independent of the index k 71 71 Z 0 0 Z 6 61 k1 In particular Example k1 quot 1 1 n n k1 n 2 713811171 1 71 713811171 1 191 Example Find the sum of i 1 91 k k 1 which is 1 12 23 nn1 k 1 n 1 Z using partial fraction k1 1 1 I 1 1 1 1 2 2 3 n n 1 1 1 n n1 n1 Example Calculate k1 n n n 52k Zmzk3 191 191 k1 Notethat n 1 Zk123nnltn2gt 191 Indeed let S 221 k Then S123ltn 1n nn 121 28 n1n1n1n1 nnl Therefore 221 k m There are other formulas 6 n n l 2 2 You may prove them using mathematical induction 2 n k2 nltnlgtlt2nlgt 121 M 3 P w 3 x So nn12n1 n2 71 12 T 6 2 15 2n 13n23n 3712 n14 Limits of nite sums Recall that n n 1 k1kkl n1 Note if we let n gt 00 then we have TL 1 1 kltklgt n1 gtoonl Thatis 11 1 1 12 23 nnl The limit of nite approximations to an area Example Find the area of the region We rst approximate the area and then take the limit Step 1 we subdivide the interval 0 1 into n equal Width subintervals 01 12 n ln 7n 7 n7 7 n 7n 1 z39 Here A2 was for each 2 Step 2 We approximate using the rule of the rightside endpoint 1 1 2 1 1 fff n n n n n n n k 1 f a a k 1 I Z 5 5 k1 n n Step 3 Simplify the expression 3 fk A n k1n2 1714 An 311 1 n2 n2 2 n4 2 n 1 n 12 2n 4712 Now let n gt oo ie the number of subintervals approaches 00 and the also subinterval width approaches 0 we have 1 1 1 A n gt 2 4 4 In fact if we use any other nite sum approximation it will also convergent to the same value For this reason we de ne the area of the region to be the limiting value Riemann Sum introduced by German mathematician Bernhard Riemann Suppose f is an arbitrary function on a b gt FIGURE 58 A lypicxl continuous function 11 overa clasedinlerval 41 b Here f may be positive or negative Also f is not required to be continuous here We want to introduce the notion of a Riemann sum which approximates the signed area of the function f This notion underlines the theory of the de nite integral studied in the next section Upperlower sum approximations etc are just some special Riemann sums Partition of a b xn1 between a and b satisfying Choosen 1 points x1x2 altx1ltx2ltltxn1ltb Usually we denote a x0 and b xn So ax0ltx1ltx2ltltxn1ltxnb The set P 07 17 75371 17 71 is called a partition of a b I n I39 1 l f ru 25539 171 I1 The partition P divides a b into n closed subintervals 5607 561 561 562 7 5672 1711672 Norm of a partition Foreachk l2 Aitk LEk k1 F IE gt xg l l I l I H I Xquot quot IJ I A Let maXAxk k l2 n n the Width ofthe k th interval xk1 wk is It I I f I Tn I er 7quot b be the Width of the longest subintervals called the norm of the partition P Example P 2 22 25 285 3 gives a partition of 2 3 Am 22 2 02 Awg 03 ALE3 035 Am 015 The norm ofthe partition P is 035 Let P 0 1 scn be a partition of a b JIA39J FIGURE 59 The remangles approximale the region between the graph orme runcxion v x and me xraxis For each k 1 2 n let ck be some point in the kth subinterval 33k1 33k and we consider the product f Ck 39 Ak It represents the signed area in the sense that o If ak gt 0 ak Ark means the area ofthe kth rectangle o If f ck g 0 f ck Ark means the negative of the area of the rectangle Then we sum all these products to get 71 SP Z fltCkgt ASEg kl This is called a Riemann Sum for f on a b It approximates the signed area of the region R How to increase accuracy 0 How about let n approaches in nity This does not necessarily increase accuracy 0 The right answer is let the norm P gt 0 If maxAxk k l 2 n is small then all ofthe subintervals have a small Width 0 If all subintervals have equal Widththen the common Width Abe a n and Ta In this case proaches in nity P approaches 0 is equivalent to n ap We hope as P gt O Riemann sums S p Will have a limit But does the limit always exist Important Factzlf f is continuous then as P gt 0 all Riemann sum Sp will approach a single limiting value We will de ne this value to be the area of the region 5 3 The De nite Integral Review Riemann Sums Last time we de ned Riemann sums as follows y m rm rm km recianglc Cg ls FIGURE 59 The recmngles approximate me region between lhe graph oflhe Function 139 x and me raxis Let P 330 1 scn be any partition of a b and let ck be a point in the kth subinterval k1 ask the corresponding Riemann sum is de ned by TL SP Z 010 39 AM kl This is called a Riemann Sum for f on a b It approximates the signed area of the region R Limits of Riemann Sums To increase accuracy of approximation we let the norm HP of partitions approaches zero If the values of all Riemann sums approach a limiting value I TL lim 2 fckAak I M H0 H as the norm of partitions approach zero we call this value Ithe de nite inte gral of f lim S llPllno P DEFINITION The De nite Integral as a Limil oi Riemann Sums e A n e a J is me de nite integral or f over in b and dial 1 is me limit of the Riemann sums 22l cl Axl irllie following condilion is satis ed iven any number 5 gt a mere is acorresponding number a gt 0 such mm for every parlilinnr xuxx em 7 with llrll lt Sand any choice of cl in 11 in we have Iiick Axk 7 l lt 5 i1 Notation the De nite Integral The symbol for the number I is I I I The Function is the integrand Upper lll39l39lll 0F integration if quotb r is the variable of integration Integral Sign fx dx 9 When you nEl the value aner limit at integralwn x j r v Uflhe miegml you have Integral 0Hquot mm 1 to h rquot evaluated 1 3 mtegmla Riemann Integrable When the limit exists we say f is Riemann integrable over ab and the limit value is called the de nite integral of f over a b In this case we write as n b Wy 0fltckgtAk Ia 50de Riemann Integrable When the limit exists we say f is Riemann integrable over ab and the limit value is called the de nite integral of f over a b In this case we write as n b Wy l t0fltckgtAk Ia 50de Question Which functions are Riemann integrable ie for which func tions the de nite integral always exists Riemann Integrable When the limit exists we say f is Riemann integrable over ab and the limit value is called the de nite integral of f over a b In this case we write as n b Wy l t0fltckgtAk Ia 50de Question Which functions are Riemann integrable ie for which func tions the de nite integral always exists 0 Continuous functions are always integrable Riemann Integrable When the limit exists we say f is Riemann integrable over ab and the limit value is called the de nite integral of f over a b In this case we write as n b Wy l t0fltckgtAk Ia 50de Question Which functions are Riemann integrable ie for which func tions the de nite integral always exists 0 Continuous functions are always integrable 0 Increasing functions are always integrable Riemann Integrable When the limit exists we say f is Riemann integrable over ab and the limit value is called the de nite integral of f over a b In this case we write as n b Wy l t0fltckgtAk Ia flt50gtd50 Question Which functions are Riemann integrable ie for which func tions the de nite integral always exists a Continuous functions are always integrable 0 Increasing functions are always integrable Decreasing functions are always integrable Riemann Integrable When the limit exists we say f is Riemann integrable over ab and the limit value is called the de nite integral of f over a b In this case we write as n b Wy l t0fltckgtAk Ia flt50gtd50 Question Which functions are Riemann integrable ie for which func tions the de nite integral always exists a Continuous functions are always integrable 0 Increasing functions are always integrable Decreasing functions are always integrable Piecewise continuous functions are always integrable A Nonintegrable Function Suppose f xl if x is a rational number and f 0 if x is an irrational number Then this function is not Riemann integrable over 0 1 This function f is called the Dirichlet function A Nonintegrable Function Suppose f xl if x is a rational number and f 0 if x is an irrational number Then this function is not Riemann integrable over 0 1 This function f is called the Dirichlet function Why Let P be some partition of 0 1 We may construct Riemann sum SP Z 13chka 1571 a Choice 1 pick ck from rational numbers then 71 71 Z fltCkgtAIEk 1 Aitk 1 191 k1 Choice 2pick ck from irrational numbers then TL TL Zflt0kgtA50k 0 ka 0 k1 k1 So the limit depends on the choices of ck the function f is not integrable Properties of De nite Integrals When f and g are integrable the de nite integral satis es the following rules TAELE 53 Rules satis ed by de nite integrals n Iv 1 Order39ofbllegmlian Xdx 7 1mm lmmmun I a 2 2m Widhlmervul xdx0 lsallk39muwn u 1 a 3i CunsmmMulnplc kx1lx k fxdx xmxmmm a n h b xdx 71004 is I h h 4 Sum nntlDifemnce x igxdx xdri gxdr h 39c 39c 5r AddiIiVin xdx flxdx fxdx u a 6 Mumm Inequality If has maximum value max and minimum value min on 21 b then minf b r a s mm s maxfb 7 11L b h 7 Duminan39urt x 2 gx on 4 b MM 2 gxgtdx 39Iv x 2 01m 4 b mm 2 o Wilma nl Zem mam Interval Z 1km u The area over a poim is u b Canxlun Mulliplc39 h h k x 41 Shownrm 2 max min 1 1 x c Sum n I z x gxdxf11lxSYrlx Areas add d Addmviwfar de nite inlcgr39als39 mm v 4x 39 xm FIGURE 511 a a b e MaxMin Incqualilv min I bans um gnawbra r Domination x 2 gx on 4 1 A I max gum Example Suppose 2 5 5 fxd 3 fxdx 1 and hxdx 2 1 2 1 Find 15 2fltxgt 3hltxgtdx Example Suppose 2 5 5 fxd 3 fxdx 1 and hxdx 2 1 2 1 Find 15 2fx 3hltxgtdx Solution ff fxdx 3 1 2 5 5 5 12fx 3hxdx21fxdx 3A hdx22 32 2 De nition of area for nonnegative integrable functions De nition If y f is n0nnegative ana integrable over a closed interval a b then the area under the curve y f over a b is the de nite integral offfrom a to b b A fxdx Example 0 Let c be any constant then b edx Cl De nition of area for nonnegative integrable functions De nition If y f is n0nnegative ana integrable over a closed interval a b then the area under the curve y f over a b is the de nite integral offfrom a to b b A fxdx Example 0 Let c be any constant then b edx Cl Abedx Cb a Answer Suppose I gt 03 What 0 Suppose b gt 0 What is fob rdx Answer The integral from 0 to b is the area of the triangle So b b2 xdx 0 2 0 Suppose b lt 0 What is fob xdx 0 Suppose b gt 0 What is fob rdx Answer The integral from 0 to b is the area of the triangle So b b2 xdx 0 2 0 Suppose b lt 0 What is fob xdx Answer Obxdx b0 0 de b0ltd 3 5 ltbgt 0 Suppose b gt 0 What is fob rdx Answer The integral from 0 to b is the area of the triangle So b b2 xdx 0 2 0 Suppose b lt 0 What is fob xdx Answer Obxdx b0 0 de b0ltd 3 5 ltbgt Therefore fob xdx g for any b What is fab xdx 0 Suppose b gt 0 what is fob rdx Answer The integral from 0 to b is the area of the triangle So b b2 xdx 0 2 0 Suppose b lt 0 what is fob xdx Answer Obxdx b0 0 de b0ltd 3 5 ltbgt Therefore fob xdx g for any b What is fab xdx Answer b b a 2 2 2 2 xdxxdx xdb a b a a 0 0 2 2 2 o What is fob My What is fobedx Answer Subdivide 0 b into n subintervals of equal Width Pick ck to be the rightside endpoint kAx Then Enema 3 61 91 b3 n 2 Z 773quot k1 63 Mn 12n 1 as n approaches in nity Therefore b b3 x2dx 0 3 o In general we have b b a 3 3 dex 2dx dex b a a 0 0 3 Average Value of A Function Recall that when f is a continuous nonnegative function on a b then the average of f is Average Value of A Function Recall that when f is a continuous nonnegative function on a b then the average of f is area of the region under the graph of f cwltfgt b a b fa f xdx b a Average Value of A Function Recall that when f is a continuous nonnegative function on a b then the average of f is area of the region under the graph of f cwltfgt b a b fa f xdx b a In general we have De nitionIf f is integrable on a b then its average value on a b also called its mean value is b cwltfgt b i a melee Example Find the average value of fltxgt 4 2 on 2 2 Example Find the average value of fltxgt 4 2 on 2 2 Solution Note that the graph of y V4 x2 is the upper semicircle of radius 2 centered at the origin Example Find the average value of fltxgt 4 2 on 2 2 Solution Note that the graph of y V4 x2 is the upper semicircle of radius 2 centered at the origin Because f is continuous and nonnegative the area is also the value of the integral of f from 2 to 2 2 V4 2dx 27T 2 Example Find the average value of fltxgt 4 2 on 2 2 Solution Note that the graph of y V4 x2 is the upper semicircle of radius 2 centered at the origin Because f is continuous and nonnegative the area is also the value of the integral of f from 2 to 2 2 V4 2dx 27T 2 Therefore the average value of f is 2 cwltfgt 1 24 x2d 2 2 2 If an integrand involves the terms a2 x2 a2x2 or Vx2 a2 for a gt O we would like to use the trigonometric substitution Case I v a2 x2 If an integrand involves the terms a2 x2 a2x2 or Vx2 a2 for a gt O we would like to use the trigonometric substitution Case 1 Va2 x2 Letx asinQWhere g 6 3 then ml 9 1 a2 x2 a2 a2 sin2 6 2 a2 l sin2 6 quotV HE I 2 2 x in a oos 6aoos6 Example Evaluate de Let x 2 sin 6 then dx 2 COS 6d6 Thus may W2 COS 6d6 M2 COS 6d6 2 COS 6 2 COS 6d6 11008266 41 l 320829d6 26sm26C NOW we need to change 6 back to variable w LI 6 Sl 2 4 x2 a 0086 7i 394 6Mwm5 Tmm tVqx 26 2n6aE6 C39 2Mm 2jE EEEC 2 2 2Mm V4 xz0 Example Evaluate Example Evaluate 9 2 xd LI Let x Ssin6 where g 6 g Then dw 3COS 6d6 and 2 W3C086d6 32 1 u2dubysett1ngucose u 32 32 du u l n1 3 u 1 3 1 C 2nu 1 3 3086 1 1 C 2n00861 3 COSQ D2 gt2 3 6 1 C 5 1 COS 2fl sin26 1 6 3COSQ31I1 COS 0 st 3 9 2 x9 x231n C x Example 3 Find the area enclosed by the ellipse Example 3 Find the area enclosed by the ellipse x2 y2 gEr Solution In the rst quadrant the curve is b ya2 x2where0 x a a Thus a A 4 SVa2 x2dx O a Va2 x2dx a O Let x a sin 6 then 4 1 A b2 a2 a281n26a0086d6 a 0 4 b2a2cos26d6 a 0 4ab2Wd9 0 sin 26 81 71Gb Case 11 Va2 2 Case H Va2 512 3 Letx atan6 with lt 6 lt then mugT2 a2se026 a se06 8606 a J a tan 9 Example Evaluate 1 dx V9x2 Let x 3 tan 6 then dx 3 sec2 6d6 and 3 sec2 6 1 dx d6 V9x2 V99tan26 38e026 1nsec tan6 0 V9x2 x ln9x2x ln3C 1 2 9x2dx ln9 xC Therefore 111 2 a2 Letxase06036 ltgor7r 6lt377TThen Vx2 a2 a2sec26 a2 atan6 atanQ 3 41 3 a ruse Example Evaluate 4 V x2 4 dx 2 LE Letxase60 6 ltgor7r 6lt377TThen Vx2 a2 a2sec26 a2 atan6 atanQ 3 45 3 a ruSECH Example Evaluate 4 QEZ 4 dx 2 LE Let x 2860 6 then dx QSeCQtan 6d6 Whenx 2 1e 28e06 2 6 O Whenx 4 1e 28606 4 6 Thus rm 2 dx x I 1 2 3 L642 sec 6 tan 6d6 0 QSeCQ QtaHQQSeCQtaHQdQ 0 28606 2tan26d6 032sec26 1d6 2 tan6 6 2 g m gw oogt1 A oogt1 gt1 Example Evaluate 3 3 T x d LE 0 4x2 932 Example Evaluate 3 3 T x 32 dx 0 4x2 9 Note that 1amp2 9 32 V 4x2 93 22 32 Let 2x Btan 6 ie x gtan 6 then 3 dcc gsec2 6d6 V4x2 9tan269 388C6 Whenx O gtan Oand6 O Whenx 37 gtaHQ SO 6 arctan Thus and Slw Slwalwcx 47W 0 36 3 32dx oogt1 tan 63 3 3 sec 63 item 6 sec 6 tan26lt 0 86026 U 8802 6d6 2 3 oogt1 3 2 3 16 E 3 3 tan 6d6 0 8606 tan 6 sec 6 d6 2u2 1 u2 1 u du by setting u sec 6 WW x 2 3 16 Q u1 Example Find the volume of the solid generated by revolving about the x aXis the region bounded by the curve y the x axis and the lines x O and x 2 A x24 Example Find the volume of the solid generated by revolving about the x aXis the region bounded by the curve y the x axis and the lines x O and x 2 Solution By the disk method the volume is 2 2 2 d1 V 0 7TRx dx l67r0 ltx24gt2 A x24 Example Find the volume of the solid generated by revolving about the x aXis the region bounded by the curve y the x axis and the lines x O and x 2 Solution By the disk method the volume is 2 2 2 d1 V 0 7TRx dx l67r0 ltx24gt2 To evaluate the integral we set x 2tan6 dx 2seo2 6d6 A x24 x244tan2644tan26 l 4seo26 L Integration of Rational Functions 5 x Fractions AIM Evaluate an integral of a rational function 58 Where P and g are polynomials Example Evaluate 6x4 17x3 122 3x 5 dx 6x2 17x 12 AIM Evaluate an integral of a rational function 58 Where P and g are polynomials Example Evaluate 6x4 17x3 122 3x 5 dx 6x2 17x 12 Indeed one can express the integrand as the sum of a simpler fractions 6x4l7x312x2 3x 5 1 3 2 x 6x2l7x12 2x3 34 Thus 6x4 17x3 12x2 3x 5 dx 6x2 17x 12 1 3 2 d 2x3 355 5 1 x3 1n2x3 1n3x4 0 Thus 6x4 17x3 12x2 3x 5 dx 6x2 17x 12 1 3 2 d 2x3 355 5 l x3 iln2x3 ln3x4 0 The method for rewriting rational functions as a sum of simpler functions is called the method of partial fractions Thus 6x4 17x3 12x2 3x 5 dx 6x2 17x 12 1 3 2 d 2x3 355 5 l x3 iln2x3 ln3x4 0 The method for rewriting rational functions as a sum of simpler functions is called the method of partial fractions Question How to rewrite a ration function as a sum of simpler functions a Step 1 Use the method of long division to express PCB 9ltgt Mtgtfltgt such that deg f lt deg g Then Pltwgtgltxgt Mltwgtfltxgt x fltxgt gltxgt gltxgt W gm HIE Here M is a polynomial and m is a proper rational function a Step 1 Use the method of long division to express PCB 9ltgt Mtgtfltgt such that deg f lt deg g Then pm gltcMltxfltxgt ZMltgt g 56 g 6 1 Here M is a polynomial and m is a proper rational function That is a rational function can be eXpressed as the sum of a polynomial and a proper rational function a Step 1 Use the method of long division to express PCB 9ltgt Mlt56gtfltgt such that deg f lt deg g Then PCB 9ltgt Mlt56gtflt56gt M gltxgt gltxgt 5 Here M is a polynomial and is a proper rational function That is a rational function can be eXpressed as the sum of a polynomial and a proper rational function Example dx x35x2 2xl x l 3 52 2 1 xx x x26x4 x l x l 35x2 2l dzr x l 5 26x4 dx x l l x33x24x51nx l0 a Step 2 Find the factors of g In theory any polynomial with real coef cients can be Written as a product of real linear factors x 7 and real irreducible quadratic factors x2 px g In practice the factors may be hard to nd e g 6x4 17x3 12x2 x2 2x 3 3x 4 0 Step 3 Find partial fractions of the proper rational function fx WV 1 P ctlmd Il39 Pill39lial lil39a illm rglxl l39mpcrl Lern e r hes linear factor orgm Suppose that r e r is the highest power ofx r that divides gx Then to this factor assign the sum urine in partial fractions A 7 7 7 r 9 r Do this for each distinci linear factor OngI Lent2 px 1 be a quadralic factor ufgr Suppose Ihal x2 px q is the highest power ofthis factor that divides gx Then to this factor C i S x assign the sum ofthe ll partial in lion 811 C 81x x3prq r3ixiq2 Kr Cquot r2 px q 39 linear factors with real coefficients fractions Clear the resulting equation of fractions and arrange the terms in decreasing powers 0 x equations for the undetermined coefficients Example Suppose g has the following forms nd partial fractions of a proper rational function x f ltxgt M gt g 56gt With degf lt degg l g x 23 then A B C Rltgt lt2gtx 22x 23 2 g x x 2 x 292 x 4 then Example Suppose g has the following forms nd partial fractions of a proper rational function fgt M gt g 56gt With degf lt degg l g x 23 then A B C Rltgt lt2gtx 22x 23 2 g x x 2 x 292 x 4 then A B C D RWmpe x 4 3 g x 32 x 22 then Example Suppose g has the following forms nd partial fractions of a proper rational function fgt M gt g 56gt With degf lt degg l g x 23 then A B C Rltgt lt2gtx 22x 23 2 g x x 2 x 292 x 4 then A B C D RWmpe x 4 3 g x 32 x 22 then A B C D cc 3 w 32 cc2 x22 4 g 2 3x 1002 then Rltxgt A B 7 D Rltxgtlt 3gtx 322x22 491 x2 3x1002then AxB CxD R 2 E 3 100 2 33 1002 59w2 5x 502Cr 3fithen A B C D cc 3 w 32 cc2 x22 4 g x2 3x 1002 then Rltxgt Ax B Ox D Rx 2 IE2 3 100 2 33 100 5 g x2 5 502 x 33 then AxB CxD E F G Rltxgt 2 2 2 3 550 x25x50 3 x B x B How to nd these undetermined coef cients Emmp a 64 5 6 Example LC x2 5x 6 x25x6 c2c3 we can decompose the integrand into the form of x A B x256 2x3 Since so Ax3Bx2 1 Example LC x2 5x 6 x25x6 c2c3 we can decompose the integrand into the form of x A B x256 2x3 Since so Ax3Bx2 1 Now let s determine the values of the constants Method 1 Equate the coef cients for each power of x xAx3ABx2BABx3A2B Therefore 1 AB 0 3A2B so 33A3B 2B3BB B3A 2 Therefore 1 A B 0 3A 2B so 33A3B 2B3BB B 3 A 2 Method 2 at 1 let x 2 then 2A 23B 22 A let x 3 then Therefore Thus w d x25x6 23d w x2 x3 21nx231nx30 Example 2x3 2x 1 x x2 12 The form of partial fraction decomposition is Thus w d x25x6 23d w x2 x3 21nx231nx30 Example 2x3 2x 1 x x2 12 The form of partial fraction decomposition is 2x32x1 A BxC DxE 2 2 2 x21 1 x21 Multiplying byx x2 12 we have 2x32x1 Ax212BxCxx21 DxEx Ax42x21Bx4xCc3cDc2Ec ABx40x32ABDx2CExA Equate the coef cients for each power of x AB0 02 2ABDO CE2 A1 Therefore A1B 1C2D 1E0 Thus 2x3 2x 1 2d x x2 1 1 x2 x Cm x21 x212 Thus 2x3 2x 1 2d x x2 1 1 x2 x 2 2cm 1 x21 1 dx1nxC x x2dx x21 2 d d 21x21x 1 7111LE2 1 1 2arctanxC 2d x2 1 1 d39 7bysettmgux21 u 2 1 C 271 1 0 2x21 Therefore 2x32x1 xx212 1 1 In x 1nx212aretanx C ll 2 2x21 Example x2 3x 3 2dx x x 2 The form of partial fraction is Example x2 3x 3 2dx x x 2 The form of partial fraction is x2 3 3 A B C 2 2 cc2 56 56 cc2 Multiply by x x 22 on both side we have x23x3Altx22Bxx2Cx Example x2 3x 3 2dx x x 2 The form of partial fraction is x2 3 3 A B C 2 2 cc2 56 56 cc2 Multiply by x x 22 on both side we have x23x3Altx22Bxx2Cx Letx O we have 34A Letx 2 we have 4 63 20 Letx l we have lA B C Thus 3 1 1 A B C 4 4 2 Therefore x23x3 xltx22 3 l 1 Z 4 2 dx cc cc2 lt2gt2 3 1 1 1 1 2 C 41ax41q 2lt2gt Heaviside Method Suppose is a proper rational function and 9CD CU r1gtltfv r2gtltCU rngt is a product of n distinct linear factors Then there is a quick way to expand x into partial fractions 956 Note that fltscgt fltccgt A1 A2 An gltxgt x r1x r2x m x r1x r2 LE T n Multiplying x 7 1 on both sides flt56gtlt50 7 1gt lt T1gtlt T2gt lt mgt A1 A2 Ai An x rl x rl x rg x n x m Thatis flt56gt SC 7 2gt lt50 7 z gt lt56 mgt A1x 7 1 A AZ An x r x ri x m Let x 7 1 we have A1 Heavisidl Merllnd I Wme the quanem wuh gx mmed x gm x mu r x 2 um W Lu I r 7 r2 w A39 lt 7 m 7 fl392 quot1 quot2 7 mm 7 r quot39r2 7 r I39u Aquot 7 n mm r2r r 39 3 Write 1h211ur1ialfmcxian expansion afxgx as iltx 2 m ltx rltx n Example In one of the previous examples x x A B x25x6x2x3c2 cc3 By the Heaviside method x 2 A 7 x3 2 2 3 LE B i 3 2 3 32 Example 4x25x3 dx x32x2 2 x2x2 x2 xZ lx2x llx2 Therefore 4 5x3 4 5x3 14 3 C x32x2 x 2 x 1x1x2 x 1x1x2 We use the Heaviside method to calculate coef cients 4 5x3 453122 ampMxmx 23 6 4 5x3 4 531 w 1Mxm34 24 4 5x3 m 4039 C 3 x 1r1 x2 3lt 1 3 Therefore the integral 4x25x3 d 2 1 3 do x32x2 x 2 x l 1 x2 21nx 1 1n131nx2C x 12x23 x1 0 ln 55 integrals and the substitution Given a function f how to nd its inde nite integral ie the family of antiderivatives of f Given a function f how to nd its inde nite integral ie the family of antiderivatives of f If f is clearly recognized as derivatives of some function F then fxd C Example 0 f xndx Given a function f how to nd its inde nite integral ie the family of antiderivatives of f If f is clearly recognized as derivatives of some function F then fxd C Example ofxndx Ifn y l nl0 1 1 xndx n Given a function f how to nd its inde nite integral ie the family of antiderivatives of f If f is clearly recognized as derivatives of some function F then fxd C Example ofxndx Ifn y l nl0 1 1 xndx n l dxlnxC w IF O J S m 36d LE 7 a sindzr A a sin xdw sinxdx CosxC o fCOS xdx COSEdE sinx C ofemdx a sin xdw sinxdx oosxC o foosxdx oosxdx sinx C ofemdx edxe0 How about more complicated functions a sin xdw sinxdx COSEC o foosxdx oosxdx sinx C ofemdx edxe0 How about more complicated functions Need some integration techniques Example 7 2 2136i J Example 2 2tet dz Solution Let u 252 then du 2tdt 2 2 2tet dteudueu0et 0 Example 2tet2dt Solution Let u 252 then du 2tdt 2tet2dt eudu e C 872 C 2 2 Check et C 2238 j 39939 This technique is called the substitution rule The Power Rule in Integral form Suppose u is any differentiable functionand n is any number different from 1 Then the Chain rule tells us that i unl ndu dx unl undu C n l for any differentiable function u and n y l n lgt u Therefore The Power Rule in Integral form Suppose u is any differentiable functionand n is any number different from 1 Then the Chain rule tells us that i lt unl n du dx unl undu C n l for any differentiable function u and n y 1 For example What is W5 4x99 5x4 4dx n lgt u Therefore The Power Rule in Integral form Suppose u is any differentiable functionand n is any number different from 1 Then the Chain rule tells us that l ltun gtZZUHQE dx n 1 dx unl undu 0 n1 for any differentiable function u and n y 1 For example What is W5 4x99 5x4 4dx Therefore x5 4x100 0 100 x5 4x99 5x4 4dx Theorem The substitution rule If f is continuous on an interval I and u 9x is di erentiable Then fltgltxgtgtg ltxgtdx mu Proof Suppose F is an antiderivative of f then 17 ltgltfvgtgt F lt9ltgt 9 f lt9ltgtgtglltgt gt Thus is antiderivative of fg Therefore Remark To evaluate f hxdx Where may be a complicated function a If possible nd suitable functions f and 9 such that hltfvgtdfv f ltgltfvgtgtg ltfvgtdfv Typically f is a relatively simple function and you know how to evaluate f f While 9x is some simplecomplicated term 0 Substitute u gltxgt and du g xdx to express hltscgtdcc fltgltscgtgtg ltxgtdcc fltugtdu Evaluate f f udu with respect to variable u 0 Replace u by 9x in the nal result Practice makes perfect You need practice on lots of problems to gradually build up your ability on solving these problems Emmp a o f sm5 4W9 Example 0 f sinlt56 4d6 Let u 56 4 and du 5d6 then d 1 1 sm56 4d6 sin a Cosu C ECOSlt56 4 C 5 d 39 f Twig Example 0 f sinlt56 4d6 Let u 56 4 and du 5d6 then d 1 1 sm56 4d6 sm U Cosu C ECOSlt56 4 C 5 d 39 f Twig 5 Let u 3x 5 and du de then d 1 1 1 1 d 1 1 35 guu 3nuC 3n3x5C O Sin W W dw f Let u and du dw then SHde28mudu QCOSUC 2 OSEC 0 f C082 xdx f Let u and du dw then SHde28mudu QCOSUC 2 OSEC 0 f C082 xdx 1 2 2 C082ddg81n4x0 o f sin2 xd f Let u and du dw then SHde28mudu QCOSUC 2 OSEC 0 f C082 xdx 1 2 2 C082ddg81n4x0 o f sin2 xd 1 COS 2x x sin 2x 2 sm x x 2 x 2 4 or 2 2 sm2dx1 Cos2xdxx 3Sm4xCE Sm xC ofdw 2 oflndex Let u lnx and then du So 1 2 21 d deZguduu2owho SE LE 2 56305756 fw e O o fx2e3dx Let u x3 then du 3x2dx So 3 x2 x dx 6 d3 6 33 of d1 ee Note that dx ewdx e6 21 Let u em then du emdx So d du arctanu C arctan ex C ee u21 ltgt O f tam mix o f tan xdx sin x tan xdx dm COS x Let u COS x then 39 d tanxdxSmdx Zu lnuC lncosx0 COS LE sz2 f 22113 o f tan xdx tanxdx Let u COS x then Smx tanxdx szz f 22113 Letu Z2 1 22dz z2113 dx COS LU 13 3u2du u sin x dm COS LU d Zu lnuC lncosx0 then U3 Z2 1 and 3u2du 22dz Thus 3 3 3udu 5152 C z2 123 C of dd J a22 dIE 1 1 1 W x a2 x2 2a ax a x 1 Zlt1na lna C 39 f 126122 dIE 1 1 1 W x a2 x2 2a ax a x 1 Zlt1na 1na x 0 f afo Let x cm then dx adu So dx adu 1 du 1 t 0 1 t xC arcanu arcan a2 x2 a2 2112 a 1 U2 a a a f 1 x1 f d1 1V1 Letu xx 1 then U2 1 x and 2udu dx So dx 2udu 1xx1 1u 1 2 1 lt 1ugtdu 2u 1nu1C 21 1n11C O 3185wa f COSJ x COS LU dx COS LU COS2 LE COS LE 26 sin x U 9 Where u sin x H H 2 H g Q HI D U 1n C u 1Sm H ml le H E 1 sinx 4 f CIA412 1d Let u V4x2 1 then U2 4x2 1 Therefore 2udu Sxdx That is udu 4mm 1 dx 4x dcc xx42 1 4x2 4x2 1 udu U2 1u du u21 arctan u C arctan V 4x2 1 C O 7 dd J a In 51 lnln 51 54 The fundamental theorem of calculus Calculus Calculus consists of two parts 0 Differential Calculus Math21A This part comes from the study of velocity problem slope problem rate of change etc o Integral Calculus MachlB This part comes from the study of area problem 0 Today we will see that they are closely related by a very important theorem The Fundamental Theorem of Calculus Mean Value Theorem Theorem If f is continuous on L1 b then at some point c in cg b b N bid mom Sometimes we write it as b mm 1 b a for some 0 in a7 b Proof minim s M s max M abrnin fxdx S fxdx g maxfxdx b minmb a fltxgtdxsmaxfltxgtltb agt bfa bfmdrr max rr Since f is continuous by the intermediate value theorem there exists some 0 in m b so that b N L molar b a min M 3 Suppose f is continuous on m b For any x in m b consider the function What is Fx Fm 115 FxhL Fx zh z hm fa ftdt fa ftdt hgt0 h zh i i W illirr ck where ck is some point between 05 and x h r That is d I t dt H d f gt M The fundamental theorem of calculus Part 1 Theorem If f is continuous on a b then ftdt is continuous on a b and di erentiable on a b and its derivative is fxr d x F t dt x d f0 ftrr NowFx at so f ftdt is an antiderivative of Thus suppose C05 is an antiderivative of f then Wuomo for some constant C Let x a Wuqwamm0 qw Therefore UuGm Gmy In particular let ac b a UmuG Gmy The fundamental theorem of calculus Pan H a Theorem If f is continuous on a b and F is any antiderivutive off on a b then b lmeF F Sometimes we write it as Remark 0 Roughly Part1 says d I g f 7 f b d FF 1d I That is integration and differentiation are inverse operators while Part H says Remark Part H says that the de nite integral b molar Fltbgt M for any antiderivative F of f Thus to evaluate a de nite integral of a continuous function we only need to nd its antiderivative without having to calculate limits of Riemann sums This observation saves a lot of energy Example 0 f05 xgdx 7 1 5 dx i 0 f0 fag i 1mm gm Part1 d I ff is also very powerful d I i o E fa s1ntdt 7 A I t21 7 0 d1 fa 6 dt 7 d 5 1 i o E I 362dt 7 O 2 d I 1 dt 7 dx 1 5 t4 d 91 roach where g is a differentiable function and f is integrable Let y ffw ftdt nd 3 5 Let u 905 then 3 fauftdt Thus Find the formula for dy u Known j Z Unknown 3 3 By the chain rule dy i dy du i du i i im imm m Thus 91 ftdt mar gltrrgt i 8 ftdt 7 A d 4 1 i U E f1312 2e dt quot o Practise C d we rquot tan3t2dt d3 1sinz 82 Integration by Parts Example Evaluate Example Evaluate xcosxdx xcosxdc sinxldx It seems not that easy to solve it But ml sin xdx Note that is easy to evaluate Example Evaluate xcosxdx xcosxdc sinxldx It seems not that easy to solve it But xsin xdx is easy to evaluate Is there any relation between f x sin x dx and f x sin xdx Note that Example Evaluate xcosxdx xcosxdc sinxldx It seems not that easy to solve it But xsin xdx is easy to evaluate Is there any relation between f x sin x dx and f x sin xdx Indeed by the product rule Note that x sin x xsinx x sin x Therefore cosxdx xsinxldx xsinxl xlsinxdx xsinx sindx xsinx cosx C The technique used here is called integration by parts In general the product rule says ltf ltccgt g ltxgtgt f ltxgtg ltxgt f ltccgt 9 lt56 Thus fxgd f xgxdx ogsmite femow omowmynmdowx Therefore fomwwmfomm 3 omewm This is called the integrationbyparts formula Let u f and v g then du fl dr and do g dx udvuv vdu Therefore Exalmp a Find f 1m mica Example Find f xln xdx First try It xdv lndx du dx 2 Not so clear here Example Find f xln xdx First try It xdv lndx du dxv Not so clear here Second try u lnxdvxdx 2 du diandvx x 2 2 2 x x dx 1 xlndx 2 nx 2 x 2 2 2 LI LI LI LU Example Find f In mix Example Find f xln xdx First try It xdv lndx du dxv Not so clear here Second try u lnxdv xdx 2 dud andv x 2 2 2 x x dx 1 1 xnxdx 2nx 2 x 2 2 2 x x x x Example Find f In xdx u 1nxdvdx dx du vyc x lnxdx xlnx xldx x xlnx xC So u 1nxdvdx d du xvx x So 1 lnxdx xlnx x dx x xlnx C Now if we want we may use it to evaluate f x In xdx again lnxdx xdlnx x xltx1n x xlnx xdx 2 21 2 1 d0 Therefore 2 Qxlnxdx x2lnx 30 Thus I 2 2 xlnxdxln C To evaluate a complicated integral using the method of integration by parts one needs to express F weir udv Key part Find suitable to and do if possible To evaluate a complicated integral Fxdx using the method of integration by parts one needs to express F xdx odo Key part Find suitable to and do if possible Some general hints on how to make the correct choice of o and do 1 Choose to f to be a function that becomes simpler after differentia tion or at least does not become more complicated 2 Choose do to be the complicated part but still be something you can inte grate 3 f odo should be simpler or at least not more complicated than f odo Emmp a f1 acemdw J0 Example 1 xemdx 0 u xdv emdx a Let du dxve By the integration by parts formula xexdx xex exdx 538 8 C x 1em C Terefore Therefore A useful order ln x polynomial ex sin x or COS x When the integrand is the product of them pick the later one into the do term Therefore A useful order ln x polynomial ex sin x or COS x When the integrand is the product of them pick the later one into the do term Example ex 08 xdx Therefore A useful order ln x polynomial ex sin x or COS x When the integrand is the product of them pick the later one into the do term ex 08 xdx Example Let u e dv oosxdx du ewdmm sinx So ex COS xdx ex sinx smxexdx e sin xdx ex COS x ex COS xdx So ex COS xdx ex sinx smxexdx e sin xdx ex COS x ex COS xdx Therefore ex eosxd exsmx exeosx ex eosxdx 2exeosxd exsmerexeostrC em sineosx exeosxdx C Examph f 2 Sm 5mm Example f x2 sin xdx Let So 2 sin xdx u x2 do sin xdx du 2mm 21 cosx x2d cos x x2 cos x cos xd x2 x2 cos x 2 cos xdx x2 cos x 2xd sin x x2 cos x 2 sin x sin xd 2x x2cosx2xsinx2cosx0 Example Use integrationbyparts to derive the reduction formula xnexdx xnex nxTL lewdx Here we reduce the power of the variable in the integrand Example Use integrationbyparts to derive the reduction formula xnexdx xnex nxTL lexdx Here we reduce the power of the variable in the integrand e g x4exdx x4ex 4 x3exdx x4e 4 fem 3x2exdx x4ex 4x3 12 ex Qxexdx x4ex 4x3 12x28m 24 xex exdx x4ex 4x3 12x2e 24xex 248 C Emmp a 2 353g dac Example 2 x3e dx 2 u x3dvex dx du 3x2dxv First try Example 2 x3e dx 2 u x3dvex dx du 3x2dxv First try Second try to ex2 do x3dx 2 4 du 2xex dxv I It becomes more complicated than the original one Example 2 x3e dx 2 u x3dvex dx du 3x2dxv First try Second try to ex2 do x3dx 2 4 du 2xex dxv I It becomes more complicated than the original one Third try So 2 x3 x dx du 2 x2 dv xex dx x2 2xdx v 58 1 2 1 2 x2Eex 2x exdx 2 2 8m xem dx wl pwl p Third try 2 2611 xex dx 2 du 2xdx v 58 2 1 2 1 2 x3 xdxx2 ex 2x exdx 2 2 2 gem xem dx 2 2 3 U So x 12 1 2 C e 28 2 Fourth try We rst use the substitution 2 x2 dt 2dx 2 x 1 2 eC Then 2 t 1 x3 x dx etdt tetdt 2 2 Then we use the integration by parts 2 1 x3 x dx Etetdt 1 M t 2 e 1 1 Elie Eetdt 1 2lief thrC 1 1 x2 1ex 0 Emmp a SEC mm J Example 8603 xdx 8803 xdx sec xd tan x secxtanx secxtanx sec x tan x 8803 xdx S C mix tan xd sec x tan x sec x tan x dx sec tan2 xdx seCtan SGCLE 8602 w 1 dx Therefore Thus 2 see3 mix 8803 xdx NIH seextanxseedx seextanx 1nseex 1321an C seextanx 1nseex tanxl C Emmp a Example 71 The Logarithm De ned as an Integral What is 5 ln 2 We have seen and used the logarithmic function In 2 and exponential function 5 many times But What are they How didwe describe 5 Intuitive and informal I 5 is some constant 5 27182821828 I 5 When nis an integer eg 52 5 5 etc I 5 With 7 being arational number eg 5i 52 I When 2 is irrational the precise meaning of 5 is not so clear Then In 2 is de ned as the inverse of 5 Also We claimed Without proof that 5 5 and lnz In this chapter We Will give a rigorous approach to the de nitions and properties of these functions and We study aWide range of applied problems in Which they play a role De nition of the Natural Logarithm Function DEFINITION The Natural Lngarithm Function 1 lnX7dlv xgt0 To understand this inction We Want to study its I domain I range I graph monotonicity concavity etc I derivative I inverse I other properties Domain of int 2 gt U The function is not de ned for 2 g 0 I iroltrlt1umnni inn 1 gm he negsmvc 0mm ma Iixgtnhennr 41 l gives quotus am FIGURE 71 The gmph cry lnxand its V gt 039 relation to the functan v 1xr The graph ofthc Iogan ihm rises above Lhe r u sx moves from to the righL Ind I falls below the axis a5 moves from I 0 the left By using rectangles to obtain nite approximations ofthe area under the graph ofy and over the interval between 1 and 2 We can approximate the values ofthe inctio n IABLE 71 Typical 2place values of n x v In x 0 unde ned 005 10 05 069 I 0 2 069 3 110 4 139 10 230 DEFINITION The Number 2 1116 1 That is 5 is the point on the xaxis for Which the area under the graph ofy and above the interval 1 5 equals 1 5 2718281828 The Derivative of y ln 05 By the fundamental theorem of calculus d d I1 1 l dt h 0 dxnx dxlt1tgt xwenxgt If u is differentiable and positive then by the Chain Rule d l du i u ltmultxgtux Ei ux eg d l l a nbx abwhenbxgt0 In particular if b l and 05 lt 0 then d l ln x Whenx lt 0 x dx Thus dlllgt1 h 0 1195 w en a dx x 7 Thisgives l dxlnx0 a and MDU ultxd 1 c 63 cos x dx 695 s1nx Example Find Example Find sec 2 d2 In Sec 2 Lanz Farniiiar algebraic properties of in 1 l lnbx lnh 1 2 lug In a qu 3 in flux 4 lnx r In x rany raiienni number Sketch ofproof Let u z bad then lnuy7lnu1 Tlny 1n 71nl7 many 1nl7rlnz The graph and range of h r 0 1n 2 is an increasing function because 0 the graph of in z is concave down because lure 27 o linngg0O In a 00 because 0 lim0 Ina 00 because Thus the range of In a is 00 00 The graph of y In 33 looks like The graph of yogx The inverse function of In 1 and the Number 6 The function In a is an increasing function of a With 0 domain 000 0 range 00 00 So it has an inverse 1n1x with c domain 00 00 I range 000 DEFINITION The Natural Exponential Function For every real numberx equot Irr39 x xpx The graph ofy hfl z 5 is the graph of 1112 re ected across the line y z FIGURE 7 The graphs cry lnxand 1 cxpx The number e is Inquot I exp 11 E Innn Equations for lquot and In 39 Ema unigt0 in a x allx Example Find d lnsm m1 E5 Example Find 1DltEwtacosaw The Derivative and Integral of 5 gas Baa y Moreover We have 5 dz a O Ifu is any differentiable function of 2 then by the Chain rule 51 u udu EB 5 E Thus 5 du 51W 0 Example Suppose x arisinZOwZ Fl 63 Lengths of plane curves Given a curve how to calculate its length 0 a line segment on the xiaxis a b Length b 7 a o A line segment from a point lexz39yz 131961791 t0 1329627 92 P1lxx39y1 Length 961 9622 91 92 o A circle Length 27W G I39 0 An arc Length 9r That is 27W o How about 2 2 a b o A cardioid o The trajectory of the stone that you A thrown o Trajectory of your dog These curves may be represented by a parametric curve f t 905 You may record the position of your dog at time Let C be a curve given parametrically by the equations xflttgtay9lttgtaa t b The length of the curve may be approxi mated by the sum of the lengths of polyg onal path L a Z P 1PZ Each P 1PZ equals to We it 1W ltglttigt gm 0 FIGURE 624 The curve C de ned parametrically by the equations 3 fir and y gm a 5 t 5 b The length of the curve from A to B is approximated by the sum of the lengths of the pnlygnnai path straight line segments starting at A 301th to P and 50 cm ending at B P Now assume f g are continuously differentiable they have continuous derivatives on a b By the mean value theorem flit flit i fllttfgtAt 7 and 90 QW i glltt gtAtz for some 15 25 in 134 13 Therefore LmeWWmew Zb a xf lttgt2 gllttgt2dt as P a 0 Thus the length of the curve C is L b tf WZ gllttgt2dt b ltj jgt2 ltj gt2dt Example Find the length of the parametric curve reosty 7 sth fOIOStSQ 39 Example Find the length of the parametric curve reosty 7 sth fOIOStSQ 39 27T 27T L 7 sin 02 7 COS t2dt V 7 2dt 27W 0 0 Example Find the length of the parametric curve m rcosty 7 sth f0r0 t 27r 27T 27T L 7 sin 02 7 cos t2dt V 7 2dt 27W 0 0 Note that the curve is simply a circle here Example Find the length of the parametric curve m rcosty 7 sth f0r0 t 27r 27T 27T L 7 sin 02 7 cos t2dt V 7 2dt 27W 0 0 Note that the curve is simply a circle here How about if 0 g t g 4 Example Find the length of the parametric curve m rccsty 7 sth f0r0 t 27r 27T 27T L 7 sin 02 7 cos t2dt V 7 2dt 27W 0 0 Note that the curve is simply a circle here How about if 0 g t g 4 47139 47F L 7 sin 02 7 cos t2dt V 7 2dt 471397 O O Example Find the length of the parametric curve m rccsty 7 sth f0r0 t 27r 27T 27T L 7 sin 02 7 cos t2dt V 7 2dt 27W 0 0 Note that the curve is simply a circle here How about if 0 g t g 4 47139 47F L 7 sin 02 7 cos t2dt V 7 2dt 471397 O O The circle is traveled twice this time Example Find the length of the parametric curve x 7130822234 TSi Zt forOgtgg Example Find the length of the parametric curve x 7130822234 rsin2t for 0 g t g dx dt d d il 27 sin 75 COS t 7 sin2t 27 COS t sint 7 sinlt2tgt d w2 7 2 sin22t 7 2 sin22t r sinlt2t Therefore 30822 7 L 0 xir sm2tdt r r What does the curve look like Example Find the length of the parametric curve x 7130822234 rsin2t for 0 g t g dx dt d d il 27 sin 75 COS t 7 sin2t 27 COS t sint 7 sinlt2tgt d w2 7 2 sin22t 7 2 sin22t r sinlt2t Therefore 30822 7 L 0 xir s1n2tdt r r What does the curve look like Aline segmentxy rwithx 2 0y 2 O I 239353 I y sin DEIEE 39 Example Find the length of the parametric curve C5 E If x eos3ty sin3t forO g t g 27T FIGURE 626 The astroici in Example 2 r ms 1 y sin 0 5 il39 5 2 Example Find the length of the parametric curve x eos3ty sin3t forO g t g 27T FIGURE 626 The aatroid in Example 2 d d i 3 0082t sin I d y 381n2teost dt dx d E y2 9 30842 sin2 I 9 sin4t C082 t dt 9 9 sm2 2 C082 tlteos2t sin2 I Z sin22t 27T 9 27T 3 L sin22tdt sin2tdt 3 g Cos2t g 40 ismlt2tgtdt 6 Th 6 Therefore Length of the graph of y f Given a continuously differentiable function y f x a g x g b we can assign x t as a parameter Then the graph of f is de ned parametrically by xayf with a g t g b In this case dx 2 d3 2 2 2 lt3 W 1flttgtl 1fltrvgtl Therefore Llbtlf x2dAbtl2dx WW How about the length of the graph of x g Length of the graph of y f Given a continuously differentiable function y f x a g x g b we can assign x t as a parameter Then the graph of f is de ned parametrically by xayf with a g t g b In this case dx 2 d3 2 2 2 lt3 W 1flttgtl 1fltrvgtl Therefore LULlef deLAlegde WW How about the length of the graph of x g L ZRHdWWLMy b lt3 My a a dy Example Find the length of the curve 3 x Vet 16h 0 WithO g l Example Find the length of the curve 3 x V e7j MI 0 with 0 g y g 1 dx We 1 dy So 9 1 d 1 y L0 td y21dyO Ve ydy2 12E 1 2 Example Find the length of the curve y x3 0 g x g 1 2 Example Find the length of the curve y xi 0 g x g 1 First approach Here dy 2 1 dx 3 and is not differentiable at x O 2 Moreover L fol V l 3xdx 2 Example Find the length of the curve y 0 g x g 1 First approach Here dy 2 1 dx 3 and is not differentiable at x O 1 42 Moreover L f0 1 9x 3 dx Second approach Here comes a better approach Rewrite the equation as 3 xy20 y 1 Then dx 3 a 1 m L yldy 0 4 Thus Let u y 1 then 13 I 4 8 133 L 1 f1 gdu 27ltlt4gt gt The short differential Formula Recall that Formally we may write div ltj gt2dt ltZ fgt2 ltdtgt2 Cl ltdtgt2 de2dy2 b L xdx2dy2 a Thus Moreover we may set d5 xdeeryZ and treat d5 as the differential of arc length A Thus With this in mind a simple way to recall the formula is Are Length d5 With this in mind a simple way to recall the formula is Are Length d5 eg lfy then d5 V dx2 dy2 dx2 f dx2 3 l f 2dx Thus L Lde Abbi1 f x2dx Example Find the length of the curve 1 y lt x 8 1 With0 x 2 Example Find the length of the curve 1 y lt x 8 1 with 0 g x g 2 1 J 1 J dy lt dx 8 dx lt 8 dx 1 1 1 d82 d2 gt2d2 1821 2 2xgtd2 lt L l L gtdx2 So 1 d5 yew edx 105 Polar coordinates Given a point P in the plane how to represent it mathematically o Cartesian coordinates O U x axis U y axis 139 a Polar coordinate O U polar axis j quotquot 0 Usually the polar axis is drawn horizontally to the right and corresponds to the positive x axis in Cartesian coordinates De nition of Polar Coordinates For any point P in the plane let o 7 be the directed distance from O to P pin a a 6 be the directed angle from the polar axis r Origin pole to the llne OP g o 6 is positive when measured counter 0 Initial 1 3 39 clockwise FIGURE 1035 To define polar 9 is negative when measured Clockwise coordinates for the plane we start with an origin called the pole and an initial ray 0 the angle 6 1s usually measured 1n radl ans Then the point P is represented by the ordered pair 7 6 and 736 are called polar coordinates of P Note that if 7 0 then 0 6 always represent the pole O for any 6 Negative 7 There are occasions When we Wish to allow r to be negative That is Why we use directed distance in de ning PU 9 Also it is convenient to agree that in 9 represents the same point as r 9 71 That is in 9 and r 9 lie on the same line at the same distance M from 0 but on opposite sides of 0 ms quot6 FIE URE 103 7 Polnr coordinates can have negative rvalues Example Plot each of the following polarform points A 4 g 8 47 g 0 s a 7D 3 gt E 4 F 4 G 4 Example Plot each of the following polarform points 7T 7T 7T 7T A477B477 7 7D 7 3 3 C 3 6 3 6 E 431 F 4ZG 41 2 2 2 A X 3 2 i g D x x 1r x l y 1 x l l o l y D l l v r l l y r O y l v 7 I 1 27 V72 C 3 222 B eh EFG 4 l l a Note that 7 6 7 6 27r397 6 2w 7 6 7r all represents the same point Or in general for any integer n 7 639 27m and 7 6 2n 1 7T all represent the same point P 7 6 Therefore polar representation is not unique Example Find all the polar coordinates of the point P 2 7T 6 0 Note that 7 7 t9 7 9 2700 9 7 27139 77 9 7r all represents the same point Or in general for any integer n 7 7 9 27m and 77 9 2n 1 7r all represent the same point P 7 7 9 Therefore polar representation is not unique Example Find all the polar coordinates of the point 132 7r6 or r 2 the complete list of angles is 7r677r6i 2713977r6 i 471397 H 72 For T 2 the complete list of angle is 57r6 57r6 i 2713757176 i 47139 The corresponding coordinate pairs of P are rS b 7139 57139 FIGURE 1035 Thepman27r6hasIn n elymany 2 7 2771717 and 2 if 2771717 7 7 7 6 6 polar Coordinate pairs Example 1 form 07i17i2 Relation between polar and Cartesian Coordinates When we use both polar and cartesian coordinates in a plane we place a the pole 0 be the same point as Raw 5 the origin 0 PM Pins othe polar axis be the positive ammoquot r m origin 9 zugra LE aXIS39 J I Initial ray Then any point P in the plane can be represented by 7 6 in polar form or x y in cartesian form FIGURE 1041 The usual way to relate polar and Cartesian coordinates a To change from polar 7 6 to Cartesian x we use xroosQ y rsinQ Example Convert the point 2 from polar to Cartesian coordinates x rcos6 y rsinQ a To change from polar 7 6 to Cartesian x we use Example Convert the point 2 from polar to Cartesian coordinates 7T 1 2 2 1 LU COS3 2 3 y s1n3 2 So the point is l in Cartesian coordinate a To change from Cartesian x y to polar 7 6 we use arotan ifx y 0 6 if r 0 y gt 0 if x 034 lt 0 7 ixxZerZ Note the sign of 7 is chosen to ensure that the point P lie in the appropriate quadrant a sign is if P is in quadrant I or IV 039 sign is ifP is in quadrant H or H Example Convert the point from Cartesian to polar coordi nates tan6l so6 2 5 52 52 52 2 2 2 2 5 7 xy lt 2 gt2 4 4 so 7 i5 Since the point is quadrant III 7 5 Therefore the polar coordinate of the point is 5 Example What curve is represented by 073 so 6 g 2 5 5 2 52 52 2 2 2 2 3 5 7 x y lt 2 gt 2 4 4 so 7 i5 Since the point is quadrant Ill 7 5 Therefore the polar coordinate of the point is 5 Example What curve is represented by a 7 3 It is a circle With center 0 and radius 3 In Cartesian coordinates it is x2 32 9 o7 3 E so6 6 2 2 2 2 5 3 5 5 5 722y2lt gt 3 52 2 2 4 4 so 7 i5 Since the point is quadrant Ill 7 5 Therefore the polar coordinate of the point is 5 Example What curve is represented by a 7 3 It is a circle With center 0 and radius 3 In Cartesian coordinates it is x2 32 9 o 7 3 This is the same circle as 7 3 06 so61 2 5 5 2 52 52 2 2 2 2 5 7 xy lt2 gt2 4 4 so 7 i5 Since the point is quadrant Ill 7 5 Therefore the polar coordinate of the point is 5 Ch Example What curve is represented by a 7 3 It is a circle With center 0 and radius 3 In Cartesian coordinates it is x2 32 9 o 7 3 This is the same circle as 7 3 7T 6 I It is a straight line Whose cartesian equation is y x Example Graph the sets of points Whose polar coordinates satisfy the fol lowing conditions ol r 2and 6 7T Example Graph the sets of points Whose polar coordinates satisfy the fol lowing conditions oi r Zand ggegw a 2 rlt3and6 or 06 0 g 6 3 no restriction on 7 Example Find a polar equation for the circle 02 y 22 4 Example Find a polar equation for the circle 02 y 22 4 x2y24y4 4 x2 y2 4y 0 7 2 47 sin 6 0 7 Oor7 4sin6 As 7 0 correspond the pole O which is also on the curve 7 4 sin 6 the circle is represented by the polar equation 7 4sin6 Example Find an equivalent Cartesian equation for the following polar equations o7 cos65 Example Find a polar equation for the circle 02 y 22 4 x2y24y4 4 x2 y2 4y 0 7 2 47 sin 6 0 7 Oor7 4sin6 As 7 0 correspond the pole O which is also on the curve 7 4 sin 6 the circle is represented by the polar equation 7 4sin6 Example Find an equivalent Cartesian equation for the following polar equations o7 cos65 This is simply the vertical line x 5 o 7 2 67 sin 6 This is simply the vertical line x 5 o 7 2 67 sin 6 This gives x2 32 6y A circle r 30089 28m9 This is simply the vertical line x 5 o 7 2 67 sin 6 This gives x2 32 6y x2 y 32 9 A circle 4 r 30089 28m9 en Tlt3COS 281n6 37 0086 27 sin6 3x 2y This is the line Sir19 This is the line 07 21ISint9 In Cartesian coordinates r2 27 1 sin 6 332 32 27 23 2y 12 1 27 2 2 2 lt y 11gtx2y2 2 This is a Cardioid meaning heart In general 7 a1ioos6 or a1isin6 represents a cardioid o 7 cos 29 This is the line 07 21ISint9 In Cartesian coordinates r2 27 1 sin 6 332 32 27 23 2y 12 1 27 2 2 2 lt y 11gtx2y2 2 This is a Cardioid meaning heart In general 7 a1ioos6 or a1isin6 represents a cardioid o 7 cos 29 This is a fourleaved Rose In Cartesian co ordinates cos 29 cos2 9 sin2 9 r3 T2 cos2 6 T2 sin2 6 T3 2 x2 32 Therefore 62 Volume by cylindrical shells Motivation Recall that the volume of a solid is given by b V Axdx EL 3 o Solids of revolution the Disk Method b V WRx2dx EL 0 Solids of revolution the Washer ethod What happens for the following region rotates about the line x L xL What happens for the following region rotates about the line x L xL Trouble May not easy or impossible to nd x y and x What happens for the following region rotates about the line x L xL Trouble May not easy or impossible to nd x y and x Answer Need a new method SHELL Method Vertical axis of revolulion Vertical axis of revolution 63 Rectangle heighl E ick FIGURE 620 When the region shown in a is rewle about the vertical line I L a solid is produced which can 3933 sliced into cylindrical Sh lls A typical shell is shown in b V 8 Z Vk k What is Vk then A omerclmumlmm 21r radius Radius thickness 7 5 micknexs 0mm cimum ucncc 21r radius FIGURE 51 Imagine cultng and unmlling a cylindrical shell to get 1 Hal nearly rectangular solid Vk m 271 shell radius shell height Az Thus let gt O we have b V 27rlt shell radius shell height dx a Thus let gt O we have b V 27rlt shell radius shell height dx a HOW to nd shell heightradius Thus let a O we have b V 27r shell radius shell height da CL How to nd shell heightradius Draw the region and sketch a line segment across it parallel to the aXis of revolution Then Shell heightthe segment s height Shell radiusthe distance from the aXis of revolution to the segment Here a is the thickness variable Shell radius 2 a L and height f b V 2 27733 Lfada Example The region enclosed by the curve y xltl x on 0 l and the x aXis is revolved about the y aXis to generate a solid Find the volume of the solid Example The region enclosed by the curve y xltl x on 0 l and the x aXis is revolved about the y aXis to generate a solid Find the volume of the solid Here thickness variable is x from O to l shell height y xltl x and shell radius x 2 Example The region enclosed by the curve y on 07139 and the x aXis is revolved about the y axis to generate a solid Find the volume of the solid 2 Example The region enclosed by the curve 3 SID 5 3 on 0 7t and the IE m aXis is revolved about the y aXis to generate a solid Find the volume of the solid Here thickness variable is x from O to 7T sin2 a shell height y shell radius 2 a 7T V 27T shellradius shellheight dx 0 7T 2 Sl x 27Tx dx 0 LE 7T07T1 COS2d 7T2 Example The disk x2 y2 a2 is revolved about the vertical line x b With b gt a gt O to generate a torus Find the volume of the solid Example The disk x2 y2 a2 is revolved about the vertical line x b With b gt a gt O to generate a torus Find the volume of the solid Here thickness variable is x shell height 2y 2 a2 x2 and shell radius b x a V 27T shellradius shellheightdx a a 27Tb x 2 a2 2d a a a 27r 26V a2 x2dx 2x a2 x2dx a a Note that 2x a2 x2 is an odd function on the symmetric interval a a so a 2x a2 dex 0 a Thus L V 27T 2b a2 x2d a 2 27v 2b 27r2a2b Example The region bounded by the curve 3 the x aXis and the line a 4 is revolved about a the y aXis b the m aXis c the line 3 4 to generate a solid Find the volume of the corresponding solid for each case Case 21 Revolving about the yiaXis Shall radius y gig Shel radius lmerva of inlegmlinn a b FIGURE 521 2Th I Q n In I Mmru n 39 lhThr hrll swept out by the vzriical sagmaul in pan 3 with a width Ax Here shell height xE shell radius x Thus 4 2 V 27m dx 27T552g 0 Thus 4 2 2 V 27m dx 27T55 E r 0 Or Using the washer method 2 v 0 WW y22dy m2 Case b Revolving about the xiaxis Shell heighl y Shell mdins m muRE 522 2TI region shell dlmensinns and inlcrval cfilllegl39miun in Example 339 b The shell swept out by the horizontal segmem in pan 3 wim s widm All Here shell height 4 e y and shell radius y V O 27ry 4 7 y2dy 87L Or Using the disk method 42 4 7T x27r V fo d 287T Case 0 Revolving about the line y 4 Here shell height 4 32 and shell radius 4 y 2 1047 V 0 27Tlt4 y 4 y2dy Or Using the washer method 4 104 v WW 4 dx 0 Summary of the Shell Methad Regardless of the position ofllle axis of revolution horizontal or vertical ihe Sleps for implcmcnting the shcll mclhod are Ihcsc 1 Draw he region and sketch 1 line xegmerll across it parallel to the axis of rcvolulion Lube the scgmcnt39s height or length shell height and disiancc from the axis ofrcvoiuliml shell radius 2 Fl39rldlhe limits ofiluegralioll for the Il39lickness variable 3 Integrate the producl 217 shell radius shell height with respect to lhe thickness variable x ur y lo Find the volume Chapter 8 Techniques of Integration 81 Basic Integration Formulas mm 31 Basic integration formulas 1 zl14uC 2 fkdulmC anynumherk 3dudududu 1 4 quot117quot C u zI 5 T muc 6 sin1tlu cosu C n 1 7 cosudu 511111 C s seczudu lanu c 9 csczudu icon c 11 0 secxtlnnudu secu C cscu colutlu cscu C 12 jlanxadxl ln cosM c Inlsecul c 13 cotudu1nsinu c ln cscuj c 14 equotdu L39 C 15 aquot1u quotc Inquot 11gt0a l6 sinhudu 12051111 C 17 coshudu sinth C du u 18 7sn39iC VET quot in 1 u 19 2 2Ztan39zC 1114 1 20 f7iselt 11 V111 12 u du 1 u 21 fith 7 C agt0 Va2 M1 a d quot 6 For more complicated integrals if possible we use techniques to rewrite them into standard forms Basic techniques of integration 1 Substitution by setting to Example Evaluate For more complicated integrals if possible we use techniques to rewrite them into standard forms Basic techniques of integration 1 Substitution fltgltscgtgtg ltxgtdx fltugtdu by setting to Example Evaluate 2 1 ex 1llmlt2x dx x Let u x2 lnx du 2m 2 l 2 e 1n2x dxeudueu0 ex HuerC x Practice Practice de V1 41n2 Let u 2km then du arcsinu C arcsinlt21n x C de du le 4ln2x Vl U2 2 Completing the square Example Evaluate dx 24x x2 2 Completing the square Example Evaluate dx 24x x2 24x x2 x24 466 x 22 Thus by setting to x 2 du 6 u7 2 arcsmi C arcsinx C xE xE Practise 4 2am jg 352 6931 Practise 4 de gxl m40 4 de 4 de gxl MHamp0 2x 21 2 arctan 39L 2arctan1 arctan 1 21 7T 3 Using trigonometric identities Basic trigonometric identities 2 2 sin xcos x1 sec2x tan2x 1 2 1 cos2x cos x 2 1 2 sin2 x C2Slt x sinlt2xgt 2 cos x sin x Example 0 V1 cos tdt 7T 0 V1 COS tdt 7T 0 t 2 C082 dt 7T 2 0 t COS dt W 2 0 t xicos dt 7T 2 t 2 sm 07T 0 N 2 4 Using long division to reduce an improper fraction 2x2 7x4 dc x2 Example 4 Using long division to reduce an improper fraction Example 2x2 7x 4 dx x 2 2 2 7 4 6 x2 26 2 11 x x2dx x2 11x261nx20 5 multiply by a form of 1 Example sec xdx 5 multiply by a form of 1 Example sec xdx sec xdx sec x tan x SGCLE dx sec x tan x 8802 x see x tan x dx secxtanx lnsecxtanx 0 6 Separating fractions Example x2md x 2xxx 1 6 Separating fractions Example x2x 16m 2xxx 1 x2x 1dx 2xxx 1 1 1d x 2xx 1 x xx 11nxC 1 2 8 dx 0142 Practise 6 Separating fractions Example x2x 16m 2xxx 1 x2x 1dx 2xxx 1 1 1d x 2xx 1 x xx 11nxC 1 2 8 dx 0142 Practise 1 2 2 E 8 dwd 014x2 0142 1 du 2du 01u2 1 u arctan 1n Z 1112 4 Example secltx2 5dx 1 2 2 E 8 dwd 014x2 0142 1 du 2du 01u2 1 u arctan 1n Z 1112 4 Example secltx2 5dx Let u x2 5 then du 2xdx x sec2 5dx 1 d QS CU u 1 1nsecutanu 0 1 1n secltx2 5 tan2 5 C Example 823 1 Let u x2 5 then du 2xdx x sec2 5dx 1 d 28ecu u 1 1nsecutanu 0 1 1n secltx2 5 tan2 5 C Example 34 823 1 Let u V8231 1 U2 823 1 SO 2udu 28236134 2u2 1dy 823 1 udu U2 Du du U2 1 arctanu C arctan 231 1 C Example 823 1 udu U2 Du du U2 1 arctanu C arctan 231 1 C Example LetuE U2 so 2ududx d x Qudu uZ u 2du u l 21nu 1C 21n 1C lnxdx x 4x1n2 x Example LetuE U2 so 2ududx d x Qudu uZ u 2du u l 21nu 1C 21n 1C lnxdx x 4x1n2 x Example Let u In x then du 87 Numerical integration AIM Given a function f 05 we want to evaluate the de nite integral b f dx a So far we have learned many integration techniques However in some situations it is impossible to nd the exact value Situation 1 There are many integrals of functions that do not have elementary antiderivatives For instance sin dac These integrals look easy but it can be proved that there is no way to express these integrals as nite combination of elementary functions Situation 2 When a function is determined from a scienti c experiment through instrument readings or collected data there may be no formula for the function In both cases we need to nd approximate values for de nite integrals How to nd approximate values A 0 mm One method we already know approximate by Riemann Sums b n f f dag N f ck Aack 1 k1 02 09 EUR 59 The rectangles approximate the region between the graph ofthe mde y x and thexv39dxls However Riemann Sums approximation is sometime not very ef cient We want better approximations The Trapezoidal Rule v ftx Trapezoid area 0 2 39 4 k gt1 In I Ax FIGURE 810 The Trapezoidal Rule approximates short stretches of the curve y x with line segments To approximate the integral of f from a to b we add the areas of the trapezoids made by joining the ends of the segments to the wax is The area is then approximated by adding the areas of all the trapezoids T gym a Ar gym a Ar 39 quot ltywyt1gtm ltyt1ytgtm Areas y1y2 Mm 72 A2 790 2 91 2 42 297271 97 l hv l rapvmidul Run V Tn approxlmate f flx dx use 7 Ax T7 7 yo 2M Zyz ZyH y The y s are the Values of at xhe partition poinls 7 u ALI a An x a n lAxx 7 3m where Ax b e V le Use the 39Il39apezoidal Rule With n 8 to estima Exarnp te f1 22212 Compare the estimate With the exact value Solution Partition 1 2 into 8 subintervals of equal length Use the Trapezoidal Rule We have The exact value of the integral is 3 2 x 2 8 l d if 777772r3333 x x 1 3 3 TheErroris 299 77 1 128 37384 The percentage error is 00011 011 EXAMPLE 2 Averaging Temperatures An observer measures he oulside lemperature every hour from noon until midnighl recording me temperatures in me following table Time N 1 2 3 4 S 6 7 8 9 10 II M Temp 63 65 66 68 70 69 8 68 65 64 62 58 55 What was the average temperature for Ihe 124mm period Error Estimates for the Trapezoidal rule A result from advanced calculus says that if f N is continuous on m b then b b 7 a fltxgtdz T e 12 f cM2 for some number 6 between a and 17 Thus as Ax approaches zero the error de ned by b 7 a Er 7 12 f cAI2 approaches zero as the square of AI Note that if lf xl g M on mil then if f cAz2l ibl z f ltcgtlt 7 gt2i 2 lEr bia bia 12 n IllW 39Ir I AIIIII39 n Ihr lln Impounda Rlllu If f is cominuous an M is any upper bound for I116 values of quot on 21 b then mu mar ET in the trapezoidal appmximation onhc imegral of from a o b few steps satis es the inequality Mb v a3 l E S lrl 2W2 Example How many subdivisions should be used in the Trapezoidal Rule to approximate 2 ln2 ldx 1m with an error whose absolute value is less than 104 Simpson s Rule Approxunation LISng parabolas Here We require n be an even number y FIGU RE 11 we find the shaded area to be 315 By integrating from II to I 33900 1 4quot yz onal xH xquot b FIGURE Elk Simpson39s Rule approximates shon stretches nfme curve with pmbolas yoAhLBhc 31C y2Ah2BhC yoy22Ah220 hAa2BasCdas Area Ah320h h 10122020h 10124C 3 E 3 h g 10 y24m The area is then approximated by adding the areas under all the parabolas b h h NEW Z gltyo 4y1 yz y2 4 y4 h ltyn72 4yn71 31 h glty0 4y1 Zyz 4 2y4 39 272 4yn71 31 Note the pattern of the coef cienm in the above rule 1424242 4241i Simpsle Ruli To approximate f m it use S ya 4y1 Zyz 4y ZyH 4ynil yquot The y s are the values of j at the partition poinls an a 017 1Axx b x The numbern is even and Ax b 7 un Error Estimates 0 Simpson s Rule A result from advanced calculus says that if the fourth derivative f 4 is continuous on b 7 a a ram s 7 km flt4gtltcgtltmgt4 m b then for some number 6 between a and 17 Thus as Ax approaches zero the error de ned by Es 11 flt4gtltcgtltmgt4 approaches zero as the fou1th power of Ax Note that if lf4xl g M on a7 bthen 7 W W flt4gtltcgtltmgt4l bia l w l E v Q I39lu lirrnl39 Iislinlale lm39 Sllupmn s Rule If f is continuous and M is any upper bound for the values of 39l on u b than me error E in Ihc Simpson s Rule approximation oflhe inmng of f from a to b for n steps sansrics the incqual39ry Ml mi 180114 IESI S Example Use Simpson s Rule withn 4 to approximate 32 5I4d1 Example Use Simpson s Rule with n 8 to estimate 12 12511 Compare the estimate with the exact value Solution PaItition l7 2 into 8 subintervals of equal length l l1l llisl lgliglglglalzl lyxl1l lElHlZlTl1lHl4l Review for Midterm 2 Information for Midterm 2 a When This Friday November 142008 on lecture time 12 10 100pm o What Contents of Chapter 6 section 71 and section 72 This is a closed booknotesfriends exam No Calculator is allowed Types of problems a Part I 10 Multiplechoice problems Totally 70 points 7 points for each problem Setup only Typically nd the right formula to calculate volume work etc No evaluation is needed here Also no partial credits here a Part 11 Here are some simple calculations You need to provide details for your work Partial credits will be given 10 points Find volumeworksurface areacentroid etc 10 points Logarithmexponential functions 10 points Exponential growthdecay A previous exam for the last year has been posted on the course webpage You are encouraged to take it rst and then compare your answers with the provided solutions Helpful resources a This Review a Previous Exam posted on the course webpage o Homework In particular PAPER HOMEWORK o Lecture Notes a Textbook 0 Online learning materials Volume p Crossrscclmn Rm x wi harm Am FIGURE 5 A cross566mm of the solid S formed by imexsccxing s with a plane Iquot perpendicular m me was hmugl1 the poi in me inlenal u b V a Axdx Where Aac is the area of a crosssection Volume for solids of revoluIion 0 Disk method 7 V 71Rz2dz a Here the crosssection is a disk of radius o Washer method 7 v mm 7 m dz a 0 Shell method I V 27T shell radiusshell heightda CL Length of curves The general formula for calculating length of curves is Lzds Where d8 is the arc length Parametric curve x f I y 0y b y L V1 d a dx x b dx L 2nw atHwgt Moments center of mass centroid The general formula for calculating center of mass is moment xmw Center of mass mass a center of mass of an one dimensional rod with density 6 fab x6xdx fab 6xdx centroid of a plate With a constant density 6 Example Find the centroid of a thin plate covering the region bounded above by y Rltwgt and below by y Mr over the interval a b Where Rx gt x for each x E a b 1 3 Recall that b My fa idm a M 1 dm 1 N fa ydm y M 1 dm So we need to nd i g and dm for a typical strip For a typical vertical strip the center of mass i g is Rx 2 ix andjg dm 6 dA 6R rltxgtgtdx Thus M dm lb 6R rltxgtgtdx 6A MxQdm6bwdx I My fdm 61 rltxgtgtdx Therefore the centroid i g is given by If xltRltxgt rltxgtgtdx max rltxgtgtdx j fl Rx27rx2d jam 7 wow Area of surfaces of revolution g m o molmion FIGURE 551 The arm ofth surface swept on by revolving an AB about he axis shown here is f b 2W 4 Th exact expression depends on he formulas for p and d5 The general formula for calculating length of curves is S 27rpds Here 3 is the radius from the axis of revolution to d5 d5 d12 ch2 is the arc length Revolution about the asaxis 17 d2 dy S QWili V a Eydt ll Ify fa 2 0 then b S 27Tfa 1faz2da Similar results for revolution about the yiaXis 1 dz dy S 2 1 72 izdt 1 Wm dt dt Ifa gy 2 0 then b S 27rgy 9y21dy Work 9 Constant force along a line 39gt d Work force distance 9 Variable force F along a line Fx gt a b I I W F weir a In particular for a spring the force is given by F kx FIGURE 660 Li iug the bucket in FIGURE Example 4 WW1 5 661 The olive oil and ank in e i W Z Wk where Wk equals to the weight of the kth slab distance from the kth slab Fluid pressures and Forces 0 Constant depth 0 Variable Depth 17 F wy0yL2dy Surface level of fluid A min plate F whA or F whA Where h is the depth of the plate s centroid and A is the area of the plate Use this formula When the centroid and the area of the plate is obvious A y 1 sum UI mud I Surface level of fluid Submerged vertical i 39 plate Stri b 39 39 39 39 f d ptll E centroid depth 3 quot A in Plate centroid a x 4 LEN 3 7 FIGURE 668 The Force against one side Strap length at level 3 ofthe platens w h plate area LogarithmExponential function 0 Some basic formulas derivative integration etc alogax x 10ga ax x Exponential growthdecay Chapter 6 Applications of de nite integrals 61 Volume by slicing and Rotation about an Axis Announcements 0 Midterm I This Friday in Class 0 Paper Homework 3 will not be collected But you still need to work on them Solutions will be provided tomorrow online Motivation Given a solid object how to calculate its volume For simple solids we may have geometric formulas o The volume of a cylinder is 39 The V01ume Of a Cylindrical V A h sol1d w1th arb1trary base is o The volume for a sphere is lltighl I 4 3 V 777 PIAquot region whose m wlisl lmed on region 3 am we know use area x mm All The V01ume for a cone FIGURE 6 2 The volume ofu cylindrical solid is always de ned V 1 A h to be in base area times is Liei l Question Given a more complicated solid how to calculate its volume 0 Slice the solid into sum of smaller plecesi oEach small piece may be approx1 mated by a cylinder V V 16 7 Vk m AxkAxk Approximming cylinder has all m H on mm ha hcigm Ark wk rm Plane 1 x 1 Th cylinder39s MM is m region Km wim am Ami FIGURE 63 A typical Ihin slab in he solid S As a result given a partition P107I17quot397In of 11 the volume of the solid S can be approximated by a Riemann sum V m 2 AxkAxk k Here Aac denotes the area of the crosssection at ac pl Crossrsecuan Rm ll WI lnren A s l 39 1 b A crosssuction of the solid 5 formed by inlersecling s will a plum Iquot perpendlcular lo lhe raxls lluough he poinl l in me illlerval 11 bl To increase accuracy of approximation we let the norm gt 0 V Axdx Whenever Aac is integrable on 1117 DEFINITION V0 ume The volume ofa solid arknown integrable crms scclional area Ax from x a to x b is the integral ofA from a to bl I Ax dx To apply the formula in the de nition to calculate the volume of take the following steps Calculating the Volume of a Solid 1 Skach the xylid and a typical crasssection 2 Find aformulufbr Au me area nfa typical crosssection 3 ind the limits oimegmlion 4 Integrate Ax using me Fundamental Theorem a solid Example Suppose Am A for a positive constant A then the volume is h V AdzAh 0 Same newsman es al every level FIGURE 66 Cavalieriiv Principle These solids luv Ihe same volume which can be illustrated with slacks of coins Example 2 Example Volume of a coni cal solid Altzgt 52 a AW 7 a2 A Thus 39 1495 Axz a a a a V Azdzi xzdz AO O In particular for a cone For a pyramid Typvcm crussr ecnon Solids of Revolution The Disk Method The solid generated by rotating a plane region about an axis in its plane is called a solid of revolution Observe that the crosssection area AW is the area of a disk of ra dius Rx the distance of the planar region s boundary from the axis of revolution Thus the area 1496 WWW V Axdx 7rRx2dz mun 53 The region a and solid of revolution h in Example 4 Example Find the volume of the solid generated by ro tating the region between the curve y and the xiaXis over 04 about the xaXis 4 2 16 V 7rEdx7r 87r 0 Example Find the volume of the solid generated by rotating the region between the curve y sin x and the x axis over 0 2 about the xaxis Example Find the volume of the solid generated by rotating the region between the curve y sin x and the x axis over 0 2 about the xaxis V 27T 7T sin2 xdx 0 27T 1 2 7T cos rdx 0 2 7T sinlt2xgt 2 2i 12 1lo glt27T 7T2 Example Rotate the ellipse 12 y2 t along the yaXis and get a solid What is the volume of the solid 1 Example Rotate the ellipse 12 y2 t along the yaXis and get a solid What is the volume of the solid 1 Here for each y Therefore b 27Ta2ltb 37169 In particular if a b then we get a sphere Whose volume is 4 2 4 3 V 7Ta CL 7TCL 3 3 Example Find the volume of the solid generated by revolving the region bounded by y and the lines y 17 x 4 about the line y 1 fl ai FIGURE 610 The region a and solid ofrevulu on h in Example 639 Example Find the volume of the solid generated by revolving the region bounded by y and the lines y 17 x 4 about the line y l W 100 FIGURE 610 The region a and solid ofrevulmion h in Example 639 The volume is 7 4 2 7 4W 967 2x77 V717TRxdxi1 f ud 7 7T 6 i Solid of Revolution The Washer Method If the region we revolve to generate a solid does not border on or cross the axis of revolution the solid has a hole in it In this case the crosssections perpendicular to the aXis of revolution are washers instead of disks A typical washer has outer radius Rz and inner radius Wmhcr FIGU RE 61 3 39 39 jg Alx dx leads m a slighlly different formula 139 nli llt m m ml The area of the washer is Solid of Revolution The Washer Method If the region we revolve to generate a solid does not border on or cross the axis of revolution the solid has a hole in it In this case the crosssections perpendicular to the aXis of revolution are washers instead of disks A typical washer has outer radius Rz and inner radius Washer FIGURE 613 39 39 n nii llt vlv um vnl jg Alx dx leads m a slighlly different formula The area of the washer is Am WM2 7 mo Therefore the volume of the solid of revolution is V lb Altxgtdx 7T AbRx2 7 gt2ldx This method is called the washer method Example The region bounded by the curve y x2 l and the line y x 3 is revolved about the xaXis to generate a solid Find the volume of the solid 56 glfiubstition and area between curves Last time we knew that for inde nite integral we have the substitution rule flt9lt56gtgt9 lt56gtd56 fltugtdu ifwe let u How about de nite integral Last time we knew that for inde nite integral we have the substitution rule L wwwxfWWu ifwe let u 9x How about de nite integral Answer We have a similar one b 95 fmmwmm5 WWW 1 9a Why Let F be any antideriVatiVe of f Then F is an antideriVatiVe of fg So by the Fundamental theorem of Calculus Remark There are two methods to evaluate f fg xdx 0 apply the substitution rule for de nite integral b I 91 fltgltxgtgtg ltccgtdac fltugtdu a W a or apply the substitution rule for inde nite integral fgxglxdx fudu some function ofx 0 Then substitute x a x b in the end Which way is better Depends on the particular problem Advantage for the rst one don t need to replace u by Advantage for the second one don t need to calculate endpoints for each substitution step Example Evaluate fof Example Evaluate OW x2331 3dx Method 1 We apply the substitution rule for de nite integral by letting 2 u x l 2wd sdu u23u8 lt22 1 9 0 x2113 1 1513 2 1 2 2 Example Evaluate OW x2331 3dx Method 1 We apply the substitution rule for de nite integral by letting 2 u x l 2wd sdu u23u8 lt221gt9 0 x2l13 mil3 2 1 2 2 Method 2We apply the substitution rule for inde nite integral by letting 2 u x l 21 dU 3 23 3 2 23 1 ltx21gt13dx u13 2U 0 2x 0 Therefore V7 255 3 3 3 9 2 23x 23 23 1 1 0 x2113d 256 gt ICED 2lt8 gt 2 3 2 d3 64 Exampm 62 m d1 a In a ln n 51 Let u In x then 4 Example 2 e4 d Example fa Wallace Letulnxthen fez 56 f d 62 1nln1 gt 2 um 1n4dt 1n2 t 751 4 l tlt132 1 4 I1 gt2 nltn gt nlt1 2gt m 1 gtln2 ln2 Example What is 2 sinz5 230317 2 Seems not that easy to evaluate But indeed the value is simply 0 Why Here the integrand gz sinz5 53 is an odd function on a symmetric interval 2 2 Let f be a function on a symmetric interval a a o f is an odd function if f 513 f o f is an even function if f 513 3 sin 515 a gz so g is odd De nite integrals of symmetric functions Theorem Let f be continuous on a symmetric interval a a 01f f is even then a a fxdx 2 fd a 0 01f f is odd then a f xd 0 a Example f32sinx5 x3dx 0 because sin x5 x3 is an odd function on a symmetric interval 2 2 Proof of the theorem Assume f is odd then a 0 a afxdx afxdxO fxdx But flt gtd anodes a fltugtdu by using substitution u x 0 a f since f is an odd function 0 a 0 fudu fltcgtdc 0afudu Oa fxd 0 Therefore Example Example 5 1001 5 sin3 2x tan x 1dx 5 x7Tl 4 4 5 3 1001 E5 5 sin 2x tan x1dx0 1dx10 5 74 lt4 gt 5 1001 7T4 5 Thls IS because Sl 3 2x tan 4 x IS an odd functlon on a symmetrlc interval 5 5 Emmp a Example ltIc 4gtdcc Note that f 4 is an even function on a symmetric interval So x 4dx 203x 4dx 203x 4dx 2 2i mg Example ltIc 4gtdcc Note that f 4 is an even function on a symmetric interval So x 4dx 203x 4dx 203x 4dx 2 2i 454 9 2 129 24 15 Note f 1 a xdx O as x is an odd function on a symmetric interval Example ltIc 4gtdcc Note that f 4 is an even function on a symmetric interval So x 4dx 203x 4dx 203x 4dx 2 2i 454 9 2 129 24 15 Note f 1 a xdx O as x is an odd function on a symmetric interval 04 xdx 8 y 0 because 0 4 is not a symmetric interval Area between curves 0 If f 2 O is integrable on a b then the area between the graph of f and the XaXis is b 39 Suppose f 9 2 95 gt 0 then the area between f and 9513 is f my bgzdz ll 3 ii I E 3 E 8 o How about f 2 9513 Without the condition that 9513 Z O o How about f 2 9513 Without the condition that Z O Note that we may always pick a large constant M so that fmM2mmM20 Thus the area is b b a ltfltv Mgt ltgltxgt Mgt1dx W gltxgtdc CL De nition1ff and g are continuous function with f 2 gltxgt on a b then the area of the region between the curves y f and y gltxgt from a to b is b A fltxgt gltxgtdx De nitionf f and g are eontinuousfunetion with f 2 gz 0n 6119 then the area of the region between the curves y f and y gz from a to b is b A W gltxgtda What happens if we do not require f 2 955 De nitionf f and g are eontinuousfunetion with f 2 gz 0n 6119 then the area of the region between the curves y f and y gz from a to b is b 44Z 9Wmo What happens if we do not require f 2 955 The area between curves is 14 met mewm Example Find the area of the region enclosed by the parabola y 2 52 and the line y z Example Find the area of the region enclosed by the parabola y 2 52 and the line y z Step 1 Find the intersection points of these two curves That is we solve simultaneously for 13 IE1 1E2 2 So the intersection points are P1 1 1 and 1322 2 On 1 2 the curve y 2 x2 is above the line y x Therefore Step 2 Find the area 2 Area 2 x2 xdx 1 2 2 x2dx 1 3 LI LI 2 i290 g i 34 8 1 1 4 2 amp 31t f3g 6 32 4lt Example Find the area of the region in the rst quadrant that is bounded above by y and below by the xaxis and y 52 14 Intersection points are 0 DAVE O and 4 2 So the area is Method 1 fom df4 z2 14dz M 4 4 d 2 1405 Method 2 or method 3 integration with respect to y 2 O y14 y2dy In general for regions like we may nd the area between two curves by integrating with respect to y instead of 13 d AI fy 9ydy Example Find the area of the region in the rst quadrant that is bounded above by the XaXis and below by y 2132 and 1 y 3 Intersection points are 0 O 3 O and 1 2 The area is Method 1 Integration With respect to 13 Example Find the area of the region in the rst quadrant that is bounded above by the XaXis and below by y 2132 and 1 y 3 Intersection points are 0 O 3 O and 1 2 The area is Method 1 Integration With respect to 13 1 3 o 1 3 Method 2 Integration With respect to y f02lt3 ygt dy 88 Improper integrals Recall that in the study of de nite integral f f z dz we assume that a o the interval 11 is nite o the function f is bounded Such integrals are said to be proper How about the area under the following curves FIGURE 817 Are Ihc areas under these in nile curves finile This leads to improper integrals Improper integr s 0 Type I the interval is in nite 1700 70071 or 70000 0 Type II the function f is unbounded Idea improper integrals are calculated as limits Type I Integral over in nite intervals 1 If f at is continuous on ooo then ffltxgtdxaggrgo fltegtde 2 If f at is continuous on 700 27 then 5 mode 1irn bfxdx Lt asset em o mwf em In each caseifthe 1irnitis nitewe saythe improper inniti 39 r r inte a1f the 1irnit fails to exist the improper integral diverges Note that ff f 00030 converges ifand only ifboth f mode and jg mode converge wdj 0 592 3 Iff at is continuous on 700 cc then where o is any real number Example Evaluate Important Example For what values of p does the integral converge Direct Comparison Test Let f and g be continuous on cg 00 with for all x Z a Then 1 Leo f dx converges if jam 9 dag converges 2 ff 9 dag diverges if f dag diverges Remark Usually compare with the important example 11 17 for some p Example dx 2 05V 1 952 How about dm 7 2 mm 1 Limit Comparison Test Suppose f and g are two nonnegative functions on a 00 and if im M 11 g 05 L0ltLltoo Then 00 00 f dx and g dx both converge or both diverge Example Does the integral dx 2 mm converge ExamplezDoes the integral 0 5 t t 2 3 i 4 xlogx s1nltan e 5d 0 converge Type II Improper integral integrals of unbounded function 1mm Type n Impmper Imegrals Integrals offunctious that become infinite at a point within the interval ofixlle gralicn are improper integrals my 11 1 If x is continuous on a b and is discontinuous at a then i i mm limA fxd u 9quot r 2 lf x is continuous on 1 o and is discontinuous at b then h i fwd clinkf fxdx If x is discontinuous at c where n lt clt b and cominmus m as c U c bs Ihcn bxdx xldi hfxdx y E 1 quotHEEI al In P K l l msp 39 lh l 39 quot 39 r r tlwinw gral diverges 67 Fluid Pressures and Forces Fluid Pressure Note that darns are thicker at the bottom than at the top Why Because the pressure against them increases With de th FIGURE 664 Tu wilhsland Ihc increasing pressure dams an built thickens they go down In general We have The PmsuruDepm Enumiun In a uid that is standing still the pressure p at depth 1 is the uid39s weight dcnsity w times 2 p wk 1 Weightdens A uid39s Weighldcnsily is its weight per unit volume Typical values lbl1 are Gasoline 42 Here the constant w is the uids Weight density Mercury 849 391 645 Molasses 100 Olive oil 57 Seawater 64 ler 624 Fluid Force How to calculate the uid force Recall that pressure is force per unit area Suppose a thin plate is horizontally submerged in water Surface level of fluid The L A rm h L L P w h Thus the uid force on a constant depth surface is A quot39quot39 PM F pressure area whA Note also that hA gives volume so whA gives the weight of the body above the plate Example A cup is lled with some water Find the force on the base base area lZ ftAZ How about the force on side This leads to variable depth The variable depth formula Suppose athin plate is vertically submerged in the uid FIGURE 656 The force excrtcd by a uid against one side ofu thin at hurizo slrip is nboul A sirip length a level F pressure gtlt area Sui fac 39 of n d FZFC Zw depth ofthekth strip Ak e 2 Was 7 yeLyAyc a w by0 7yLydy mal w X snip depth X IyAy l39lw Illk gl al for Fluid For39L eAgainsl n 1 ml Fhll Plate iii 39J 39b39 w ul mi y 1110 y r on ihe y axis Lei My be the length oflhe horizontal strip meas ure may in an cu W quot ened by me uid againsl one side of the plate is F r w strip depth 10W On each strip depth is constant and thus pressure is constant Thus Example Find the force exerted by the water against the side of the cup 1ft base area H2 IVE ExampleA at plate is submerged vertically in a y392X 0 swimming pool Find the force exerted by the wa ter against one side of the plate y2x Example A at parabolic plate is submerged ver tically in a swimming pool Find the force exerted by the water against one side of the plate ExampleA at disk is submerged vertically in wa ter Find the force exerted by the water against one side of the plate Fluid ches and Centmids ccmmld dcpm we know the location of the centroid of a submerged at vertical plate we can take a shortcut to nd the force against one side ofthe plate FIGURE 668 The force againsl one side ofhe plate 5 M W plak area F mommy w moment ofthe region about the suxface level line w depth ofplate s centroid area ofthe plate wEA Therefore F wEA ExampleA at rectangle is submerged vertically in water Find the force exerted by the water against one side of the plate Example A at isosceles right triangular plate with base 6ft and height 3 ft is submerged verti cally base up 2 ft below the surface of a swim ming water Find the force exerted by the water against one side of the plate Pool surface at 1 ft J39 IUTI 39J r J E 5 Depth 5 y 3917 If m 39mmmn J ft FIGU RE 667 To f ind the force on one side of the submerged plate in Example 1 we can use a coordinate system like the one here 87 Numerical integration Paper Homework Due next Monday in class Online Homework Due next Monday early morning 200am Happy Thanksgiving AIM Given a function f we want to evaluate the de nite integral Abfw ix So far we have learned many integration techniques However in some situations it is impossible to nd the exact value Situation 1 There are many integrals of functions that do not have elemen tary antiderivatives For instance f sin dex f l x4dx f d f cos ex dx f emzdm f dx flnltln x dx fdx These integrals look easy but it can be proved that there is no way to express these integrals as nite combination of elementary functions Situation 2 When a function is determined from a scienti c experiment through instrument readings or collected data there may be no formula for the function In both cases we need to nd approximate values for de nite integrals It How to nd approximate values One method we already know ap proximate by Riemann Sums l x I W kn mclnngle TL b afltxgtdxZfltcngtAas k 1 FIGURE 59 The recangles approximale he region between Ihe graph ofme function y x and the xraxis However Riemann Sums approximation is sometimes not very ef cient We want better approximations The Trapezoidal Rule 1vquot f I Trapezoid area gm yam yr 2 lt gta In x9 39 x xii n FIGURE 810 The Trapezoidal Rule appruximates shun stretches 9f the curva y fix with line segments Tm appmximate the integral of f from a to b we add the areas of the trapezoids made by joining the ends uf the segments tn the x axis The area is then approximated by adding the areas of all the trapezoids T aye in aye mm 39 ltyne2 ynei x ynei WM A 110 y1 112 ynei gym E 2 y02y12y2quot392yn71yn I lie 39I rapvzoidal Rule To approximate f x dx use T yo Zyl 2y 2le y The y s are the Values of at the partition points In mxl a Am a 2m new a n1AXI b where Ax I 7 nn Example Use the Trapezoidal Rule with n 8 to estimate ff x2dcc Com pare the estimate with the exact value Example Use the Trapezoidal Rule with n 8 to estimate ff x2dw Com pare the estimate with the exact value Solution Partition 1 2 into 8 subintervals of equal length x 125 7 2 8 218 2815 w2181 12191692254 y mmmzmwm

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