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Review Chater 5 Ch 4 Most often used de nite Integrals 1 f kfzdz kffzdz 2 zndz if n 31 71 5 dz In my 3 sinz dz 7 cos f cosz dz sinzZ 51 The Area of a Plane Region 1 Region bounded by z axis A Aabove 173315 l Abelow 1753st l above 1753st below 3st 2 Region bounded by y axis A mm 7gltygtdy right y7axis left y7axis 3 Region between two curves First7 decide vertical slice dz or horizontal slice 17 vertical slice A fup fdownxdx or d horizontal slice A grighty 7 gleampydy 52 53 Volumes of Solids Disks or Washers Shells 1 Revolving region from one function Method of Disks revolving about z axis 7 V 7T fz2dz fz radius of cross section disk revolving about y axis 7 V 7Tde gy2dy 2 Revolving region from two functions First7 decide vertical slice dz or horizontal slice Case 1 the chosen slice is perpendicular to the revolving axis Washers revolving about z axis 7 V 7T fu2 7 fdz2dz ad revolving about y axis 7 V 7T 97y2 7 gly2dy Case 2 the chosen slice is parallel to the revolving axis Shells b revolving about y axis 7 V 27T zfuz 7 fdzdz z distance from the revolution axizs7 fuz 7 fdz height of slice d revolving about z axis 7 V 27T 979 919519 1 54 Length of a Plane Curve 1 Length of Curves tb L We g t2dt ta 117 1 f z2dx yd 1 912026111 110 2 Area of a Surface of Revolution about s axis b A 27T fz 1 f z2dx 55 Works 1 Springs kz k spring constant x change of length from the natural length 17 W kxdx a b distance from the natural length at the beginning and end respectively A D V Pumping Water 17 W 6 AyDydy where I a y coordinate of the bottom of the water b y coordinate of the top of the water 6 weight density of water given in the problem Ay area of horizontal cross section Dy distance which the cross section will be lifted up to 56 Center of Mass 1 Discrete Case a line total moments M xlml ann 2 mm xi distance from the fulcrum mi size of mass total mass m 7 M center z 7 m b plane total moments My Z zimi Mm total mass m yiim 7m center E 7 y m 2 Continuous Case a line b b M 6dz 6x density7 m 6xdx 047 M 1 center z 7 Math166 Section T Spring 2009 Review Chapter 9 Part I 91 In nite Sequences 1 A sequence a17a27a37 is an ordered list of real numbers We denote a sequence by 11012013 7 by anfL1 or simply by an 2 We say that the sequence an converges to a limit L7 or lim an L if all but nite number Hoe of terms of sequence are very close to the nite number L If not7 we say that the sequence diverges 3 To determine the convergence of sequences7 a L Hopital s ruleconsidering a continuous function am Only when an has or form as n a 00 b Alternating sequence7 H1n1an lim 71 Han converges if and only if lim an 0 Hoe Hoe c Squeeze Theorem If an and are two sequences that converge to the same limit L7 and if bn is a sequence such that an an bn on then bn also converges to L For examples7 78 g g g 71 g lH i l 2quot7 n 92 In nite Series 00 V L 1 An in nite series is a sum in the form 2 ak and the nth partial sum is de ned by S Z ak k1 k1 If the sequence of partial sums Sn converges to a limit S7 then the series also converges to S If not7 then the series diverges 0 Zak lim Sn Hoe k1 2 Geometric Series 00 Zarkil a M arz mg converges to la only when 71lt r lt1 k l 77 3 Harmonic Series 171 1 1 1 d1 ggi g1 1verges g V Collapsing Series 00 2ak 7 1111 lim a1 7 an since the nth partial sum is Hoe k1 Sn a1 a2 a2 a3 as a4 an iam1gt a1 iam1 5 N th Term Test for divergence 00 If lim an 31 0 or diverges then the series Zan diverges However lim an 0 does NOT naoo 4 00 n1 guarantee that the series converges for example harmonic series 93 Positive Series Integral Test 1 V lntegral test Let 2 an be a series of positive terms and suppose am is continuous positive and nonincreas 00 ing on 1 00 Then the in nite series 2a converges if and only if the improper integral n1 00 am dz converges 1 2 p series 1 Z J converges if and only ifp gt 1 n1 94 Positive Series Other Tests 1 V Ordinary Comparison Test Let 2 an 2 1 be series of positive terms and suppose 0 g an lt I Then a lf 2 bn converges so does 2 an b If 2a diverges so does 1 c For other cases the test is inconclusiveuse other tests 2 Limit Comparison Test Apply this test if an is a rational expression in Let 2 an 2 on be series of positive terms and suppose lim 7 L Then Hoe n a If 0 lt L lt 00 then 2 an and Z on converge or diverge together b If L 0 and Z on converges then 2 an also converges Remark In many cases L will be either 1 or O 00 n n 1 Exam le For a series take and b 7 7 the uotient of 7p gn237 an 712 7L n2 n7 q largest degree terms in numerator and denominator Then nd L as follows L lim 7 71700 n lim 2 3 1 by l Hopital s rule twice Hence 2 an and Z on converge or diverge together 714700 77 by Limit comparison test Since we know that Eb Z harmonic series diverges we conclude that 2 an also diverges Ratio Test Apply this test if an involves 71 r or OJ V a Let 2 an be a series of positive terms and suppose lim 1 p Then Hoe an a If p lt 1 the series converges b If p gt 1 or p 00 the series diverges c If p 1 the test is inconclusiveuse other tests 0 2 1 2 y Example For a series 2 n7 rst nd p as follows p lim an lim M 7 1 nl n7ltxgt an n7ltxgt n 1l n l 1 2 1 2 lim 1 W lim 1 1 7 0 lt 1 Hence the series converges by 7171 71 71 717100 71 71 Ratio test 95 Alternating Series 271 1an a1 7 a2 a3 7 a4 1 Alternating series test If an gt an gt 0 and lim an 0 then 271 1an converges Hoe V 2 Two kinds of convergence of 2 un a 211 converges absolutely if 2 converges b Zun converges conditionally if Zun converges but 2 diverges OJ V g V 1 1 E 1 71 7 b 1 tl b t 71 17 d t 11 xamp e n11 converges a so u e y7 u n converges con 1 iona y Absolute Convergence Test for 2 un lf 2 converges7 then 2 un converges Absolute Ratio Test for 2 un If un involves 71 r or Wall 11ml Suppose lim p Then Hoe a If p lt 17 then 2 un converges absolutelyhence converges b If p gt 1 or p 007 then Zun diverges c If p 17 the test is inconclusiveuse other tests M 00 2 1 2 l 711 n 1Z 1 1 2 lim 7 lim 7 1 7 0 lt 1 Hence the series converges Hoe n1 nacoltn1 n n2 absolutely by Absolute Ratio test Review Chapter 10 104 Parametric Representation of Curves in the Plane 1 Parametric Representation of Curves m ft and y gt7 where t in I Usually7 I is a closed interval ab 2 Calculus for curves de ned parametrically Let x ft and y gt be a parametric equation such that f and g are continuously differentiable with f t 31 0 on a lt t lt b Then7 dy 7 dydt dz dzdt39 105 Polar Coordinate System P 719 1 Polar 736 i Cartesian Ly z 7 cost97 y rsin6 2 Cartesian Ly Polar 730 T2 2 127 ta n03 1 3 Symmetries 7 7 t9 730 27139 77 7 6 7T 76 is symmetric to 7736 about the orgin 7 7 9 is symmetric to 7quot7 70 about the x axis 7 9 is symmetric to 7r 70 7quot7T 7 9 about the y axis 106 Graphs of Polar Equations Use your graphing calculator 1 Polar equations and Graphs a Circle 7 2acos t9 7 2asin6 b Cardioid r a i acos t9 7 a i asin0 c Limacon without inner loop a gt b gt 0 r a i bcos t9 7 a i bsin6 d Limacon with inner loop 0 lt a lt b r a i bcos t9 7 a i bsin6 e Rose 7 acosn0 r asinn6 71 odd n leaves7 71 even 271 leaves 2 lntersections in polar coordinate system 7 ft9 and r 90 a Draw the graph and check the locations of intersections including the origin points b Solve ft9 90 c If the sollutions are at the same position7 pick only one 107 Calculus in Polar Coordinates 1 1 Area A 5 f62d6 a Draw thew graph b Find 04 and 6 Check the starting position with 6 0 If the graph passes through the origin point7 then slove ft9 0 C If possible7 use the symmetry Review Chater 9 Part II 96 Power Series 1 Power Series in x has the form 0 n 2 Eanx ia0a1xagz n0 2 Convergence Set the set on which a power series converges 3 types First use the Absolute Ratio test then use other tests at the end points a 0 single point Radius of Convergence0 b 7R7R7 plus possibly one or both end points intervals Radius of ConvergenceR C R the real line Radius of Convergenceoo 97 Operations on Power Series We can do operations on Power series as we do on polynomials i7 gtlt7 derivative7 integrate7 composite 98 Taylor and Maclaurin Series 1 Taylor Series based at a m w weewfyhvam 2 Maclaurin Series Taylor Series based at a 0 f f i g u 2 3 Important Maclaurin Series a m x2 x3 b6l 3 5 3E5lim 2 4 dCOSlEEi 1zx2z371ltxlt1 17 0 sinx zi 99 Taylor Approximation to a Function 1 Taylor7s Formula with Remainder based at a f a n we NO f ltagtltz e a To 7 a mac n a Pm faf azafTa of order 71 based at a and used for approkirnation n1 a 7 can where 0 lies between z and a7 is used for error 71 bound x 7 a is called Taylor polynomial Review Chater 9 Part I 91 In nite Sequences 1 De nition of Sequence A sequence 11612 is an ordered arrangement of real numbers We denote a sequence by 11612 7 by Lngtff17 or simply by an 2 If lim am L7 then lim an L 92 In nite Series Zak n th partial sum Sn a1 an k1 00 1 If limyH00 Sn S lt 007 then Zak converges to S 2 Geometric Series k1 00 Z arkil a or arz converges to ghly when 71lt 7 lt1 1 3 Harmonic Series diverges k 1 4 Collapsing Series 2ak 7 ak1gt lim a1 7 an17 k1 since the n th partial sum Snai 020203quot39anan101 iamr 5 n th Term Test for devergence lf lim ak 71 07 then the series E ak diverges kaoo k1 00 However7 lim ak 0 might not imply that E ak converges kaoo k1 93 lntegral Test 1 lntegral Test 00 Z ak converges iff amdm converges7 where n is a positive integer 1 2 p series Test 1 2 E converges iff p gt 1 If p S 17 then it diverges k1 1 94 Positive Series other tests 1 Ratio Test Apply this test if an involves n1 7 or an1 p Then7 a n Assume that an 2 0 and lim Hoe a p lt 1 i Zan converges b p gt 1 or p 00 i Eon diverges c p 1 i The test is inconclusive Use other tests 2 Limit Comparison Test Apply this test if an is a rational expression in Assume an 2 07 on 2 07 and lim 7 L Then7 a If 0 lt L lt 007 then 2 an and Z on converge or diverge together b If L 0 and Z on converges7 then Eon converges 3 Ordinary Comparison Test Assume 0 S on S on for n 2 N Then7 a lf 2 on converges7 so does 2 on b lf 2 an diverges7 so does 2 on 95 Alternating Series 271 1an a1 7 a2 a3 7 a4 1 Alternating Series Test for 271 1an If an gt an gt 0 and lim an 07 then 271 1an converges 2 Two kinds of convergence of Zn where foo lt un lt 00 a Zun converges absolutely if E converges b Zun converges conditionally if Zun converges but 2 diverges 3 Absolute Convergence Test for Zun lf 2 converges7 then Zun converges absolutely 4 Absolute Ratio Test for Zun If on involves n1 7 or un1 Th U l p en a p lt 1 i Zun converges absolutely b p gt 1 or p 00 i Zun diverges c p 1 i The test is inconclusive Use other tests Assume that lim 1 Hoe Math166 Section T Spring 2009 Review Chapter 7 71 Integration by Substitution Standard Integral Formspage 383 kdukuC ur1 7 C 71 71 776 lnlulCr7l urdu sinu du icosuC seczu dutanuC e due C U a duLC ilna cosu du sinuC secutanudusecu0 tanudu7lnlcosul0 csc2u du7cotuC cscucotuduicscu0 cotudulnlsinul0 du aziu 2 arcsin C a2bu2 iarctan C 1 Memorize the Standard Integral Forms 2 Integration by Substitution gltzgtg lt m dm by substituting u 9m7 so du g mdz Good candidates for a substitute u are in general the following 0 Inside the square root 0 The exponent o The denominator 0 Inside the parenthesis 72 Integration by Parts 1 Integration by parts Inde nite case u du M C Froaw 07 De nite case b 1 1b uzvm dm 71buzvm dm Strategy Given a function as an integrand try to split it into two functions so that one of them can be identi ed as the derivative of something Then you can use the above rule 2 The LIATE rule A rule of thumb for choosing u and 1 is to be u by whichever function comes rst in this list L logarithmic functions ln m logs m inverse trigonometric functions arcsin m arctan z algebraic functions 1 zmz trigonometric functions sin m tan m EHgtH exponential functions em 5m am 73 Some Trigonometric Integrals 1 Type 1 sinnmdz fcosnmdz and Type 2 sinmzcosnzdm a If some function have odd degrees then Step 1 Factor out sinz or cosz from the function with odd degree If all functions have odd degree then from the function with positive minimum degree Step 2 Apply sin2 z cos2 x 1 to the function where the factor came from Step 3 Use a substitution u cosz or sinz which is not the factor b If all functions have even degrees then use Half angle Identity 2 17cos2z 2 1cos2z s1nx cosxf 2 2 Type 3 sin mm cos nz dm f sin mm sin nz dm f cos mm cos nz dz Use Product identities i 1 i i sin mm cos nz 5 s1nm nz s1nm 7 nz i i 1 sin mm s1n nz 75 cosm nz 7 cosm 7 nz 1 cos mm cos nz 5 cosm nz cosm 7 nz 3 Type 4 ftannzdz7 fcotnzdz ftan z dz l fcot z dz 1 Factor out tan2 z 1 Factor out cot2 z 2 Use tan2zsec2z71 2 Use cot2zcsc2z71 3 Make a substitution 3 Make a substitution utanz7 dusec2z dz ucotz7 du7csc2zdz 4 Type 5 tanm z sec z dz7 f cotm z csc z dz f tanm z sec z dz l f cotm z csc z dz m odd7 71 any 1 Factor out sec ztanz 1 Factor out cscz cot z 2 Use tan2zsec2z71 2 Use cot2zcsc2z71 3 Make a substitution 3 Make a substitution usecz7 dusecztanzdz ucscz7 du7csczcotz dz m any7 71 even 1 Factor out sec2 z 1 Factor out cscz z 2 Use sec2ztan2z1 2 Use csc2zcot2z1 3 Make a substitution 3 Make a substitution utanz7 dusec2z dz ucotz7 du7csc2zdz 74 Rationalizing Substitutions 1 Integral involving 7 dz 1 b u dzb gt u dzbnu 1duddz 2 Integral involving v d2 7 z2 zdsint gt dzdcostdt7 deiz2dcost7 for iggtgg 3 lntegral involving d2 z2 7r 7r zdtant gt dzdsec2tdt7 x112 l z2dsect7 for 7 lttlt 4 lntegral involving z2 7 d2 7r zdsect gt dzdsecttantdt7 szidzdtant7 for0 tlt 75 Integration of Rational functions Step 1 If the degree of numerator is at least that of denominator7 then get a proper rational function x qz by long division Step 2 Factor Step 3 Do the partial fraction decompositionPFD o For each factor of the form am bk7 i i i am I am b2 am bk o For each factor of the form azz bx c 7 Blm Cl Bgm 02 Bmm Cm amz bm 0 am bm 32 am bm cm Step 4 Integrate term by term Math166 Section T Spring 2009 Review Chapter 8 81 amp 82 Indeterminate forms 0 1 6 g forms use the L Hopital s ruleseveral times if needed 00 9 or g then lim M lim f o oo H gltzgt H gm H W 96 0 2 0 00 oo 7 00 forms transform problems to 6 or E then use the L Hopital s rule 00 3 00 000 1 forms consider their logarithm then nd the limit of log of the given expression using the L Hopital s rule then nd the limit of the given expression that is y the given expression gt lim lny gt lim y lim eh y mac mac mac Example lim z This is of the form 00 Now we proceed as follows Let y mm then ma lny ln mm mlnz We rst nd the limit for lny limJr lny limJr zlnm which is in the ma0 ma0 form 0 00 Write it as lim E which is in the form 9 By L Hopital s Rule this limit is 0 12 l 40 I 00 Then passing the limit lim y lim eh y ellmw otlny 50 1 za0 za0 4 Determinate forms 0 g 000 ioo 000oooooo 000oooooo 000000 0 0 83 Improper Integrals with In nite Limits These are de nite integrals with unbounded intervals 1 b b 00 b dm lim dm and dm lim dm foo aaioo a a 174 a If the limit exists and is nite we say that the integral converges otherwise it diverges co 0 co 0 b 100 fltzgt dz 7 00 fltzgt dz fltzgt dz 7 am fltzgt dz b11330 fltzgt 0 co 00 If both x dx and x dx converge then x dx converges otherwise the 0 700 OO integral diverges 84 Improper Integrals with In nite integrands These are de nite integrals with unbounded integrands 1 Integrand that is unbounded at an end point a unbounded at the right end point ie lim x 00 or 7 00 x7 b t dx lim dx a t7vb a If the limit exists and is nite we say that the integral converges otherwise it diverges b unbounded at the left end point ie limJr x 00 or 7 00 x7 17 b dx lim dx a t7va t If the limit exists and is nite we say that the integral converges otherwise it diverges 2 Integrand that is unbounded at an interior point ie for some 0 between a and 1 lim x 00 or 7 00 x70 gymzap10mdzf zgtdiggingpadztgrgf zmz If both limits exist and are nite we say that the integral converges otherwise it diverges Calculus I Review De nition 1 The derivative of the function f m h 7 M h 1 f m hlglo which equals the slope of the tangent line to the graph of f at m We denote the derivative of f by quotz7 Dm m7 or fz De nition 2 The de nite integral 1 1171095 195 hm i fiwmi WHO H which equals the signed area between the graph ofy and the miaris from m a to m b The relation between the derivatives7 the de nite integrals7 and the antiderivatives is given by the Fundamental theorem of Calculus Theorem 1 Fundamental theorem of Calculus WW2 Hz dm where F is any antideriuatiue of f on a7 1 FUD Fa To nd derivatives and antiderivatives7 we use the following rules The Power rule mr1 70 in 71 imrrmril rd T1 7g lnlmlC ifr7l Trigonometric functions sinmcosm cosmdz sinmC tanzseczm seczmdz tanzC dimsecmsecztanm secmtanmdz seczC coszisinm sinmdz 7coszC 1 t 7 2 2 d 7 t C Wcoziicscm cscz ziicomi csczicscmcotz cscmcotmdz 7csczC Exponential and Logarithmic functions dLiemem emdmem0 d m m m am Ea alna a dmm0 ilnxl ldmlnz0 dm m m Exercise 1 75 5d dm m o x3 x3 dm 1 1 1 39 a d d o i gdm dm 1 o isinm sinzdz dm 1 o icosz cosmdm dm d 1 1 0 E5 5 dm 1 75m 51d dm m Other important rules for derivatives The Product rule The Quotient rule The Chain rule ltf9mgt f 9xg x Exercise d 2 0 E s1nz d i o isinzcosm dm 1 i 75m s1n z ln z dm isinm dm m 1 5m if dz lnz d o Esin2m d 305122 0 E5 sinkos 2x Things to remember 0 Derivatives and integrals can both be split up when adding and subtracting For example7 DMZ sin 35 DMZ Dm sin 35 z2sinzdz m2dz sinmdz o However7 you can not split up when multiplying and dividing For example7 DMZ sin 35 7g Dmz2Dmsin z m2sinmdz imzdzsinmdz The only exception to this is constants you can always factor a number out of any derivative or integral xermule THE V F WC W C mwFEJULKO muerQDCDU estoio 6 ab 39 4 3 yc m 317 o ij 3 Anagram imcx Cs m6 fmd it Q m a C 0 C46 gamma no mwcam E wwcawupiog i 4 by u CF h m f in 66ng an W3 0wa v ri r5div Q1 46ng 00quot Qwwaiy 5 WW Mg 5 Q5 966 3 1 361 me 30 inU NFLMwB S m vaw 15 RSURWW L amp 300 SOC P Q Cng i F FJQCIQ Xw Aw H03 56 guide Jewxi 0 Lam 4f 0969 W mec3msx taswL 0 WW wwwmw mm W Rakxckumag Sem m 9 x v 14 OY CESDVQW 6w r2 v 39 1 2 1 W3 gm 5 rm ver 07F 16 W12 kng W3 chva 9 av can 0 ch CU 0 WVka M 8m I 5 Math166 Section T Spring 2009 Review Chapter 5 Review Integrals b 1 b m dz Fm a Fan 7 M 2 fwltzgtdzkabfltzgtdz 3 Ab z mdzbezdxi1bgxdx 4 Await I l7 l7 l7 5 sinmdm icosm coszdz sinz 04 a 04 I m71 r 51 The Area of a Plane Region 1 Region bounded by m axis A Aabove m axis Abelow m axis JCW dm 7 m dm above m ax1s below m ax1s 2 Region between two curves 7 A a fabove fbelow dm 5253 Volumes of Solids 1 General formula If the area of the cross section Am or Ay is known7 then VAzdz or Aydy 2 Solids of revolutionDisks or Washers Shells Remember the following rules7 b 0 Volume area of Disk Washer or Shell 2 Area of Shell 27rrh I 0 Area of Disk wrz Area of Washer 7rrbig2 7 7rrsma11 First decide integrating with respect to z dz or integrating with respect to y a For integrating wrt m consider a vertical slice then revolve it i If we have a shellrevolving axis is perpendicular to z axis then nd the radius of base r and the height h of the shell as functions of m Then use the rules ii If we have a disk or a washerrevolving axis is parallel to z axis then nd the radius r of the disk or the radii Tbig and r 1 of the washer as functions of z Then use the rules sm al b For integrating wrt y consider a horizontal slice then revolve it i If we have a shellrevolving axis is perpendicular to y axis then nd the radius of base r and the height h of the shell as functions of y Then use the rules ii If we have a disk or a washerrevolving axis is parallel to y axis then nd the radius r of the disk or the radii Tbig and rsmau of the washer as functions of y Then use the rules 54 Lengths of a Plane Curve 1 Length of curves 7 S d8 1 d 2 d 2 ds dt foracurve by parametric equations where 2 1 dm for a curve by y f 2 1 dy for a curve by z 9y b A 27rde 04 2 Area of a surface of revolution 55 Works 1 Work Force x Distance WZbFzdz 2 Springs km where k is a spring constant it may be given if not we have to determine and z is the change of length from the natural length m2 W dm 1 3 Pumping the water First place a vertical cross section on the coordinate system Then integrate with respect to y by the method of slice approximate and integrate or use the following 17 W6 Altygtdltygt dy where a y coordinate of the bottom of the water I y coordinate of the top of the water 6 density of water Ay area of horizontal cross section at y dy distance to which the cross section will be lifted 56 Center of Mass 1 Discrete Case a Line V L Total moment M zlml znmn Zimi i1 V L Total mass mm1mnzmi i1 Center of Mass i M m b Plane 7L 7L Total moments My Zzimh Mm Emmi i1 i1 7L Total mass m Eml i1 M M Center of Mass z 7 3 J m m 2 Continuous Case a Line varying density b Total moment M z6z dz 7 1 Total mass m 6z dz 1 Center of Mass z 7 m b Plane constant density 6 ie homogeneous Total moments I My dz7 17 z z Mm Wmsz dz 7 Total mass m 6fz 7 gz dz 1 My Mm m Centroid z 7 y m 3 Pappus Theorem If a region R7 lying on one side of a line in its plane7 is revolved about that line7 then the volume of the resulting solid equals to the area of R multiplied by the distance traveled by its centroid That is7 V 27rdA7 where d is the distance between the revolving axis and the centroid7 and A is the area of the region R Review Chater 7 71 Integration by Substitution I Memorize the Standard Integral Forms on page 383 2 Integration by Substitution rumpac dz fudu M o Fltgltzgtgt 0 72 Integration by parts 2 choice of and im LIATE iz 73 Some Trigonometric Integrals I If some function have odd degrees7 then Step I actor out sinx or cosm depending on which one has lower degree Step 2 Apply sin2 cos2 x 1 to the remaining term to have only one trigono metric function which is different from the factored out Step 3 u cos or u sin 2 If all functions have even degrees7 then use Half angle Identity licos2z Icos2z sin2 7 coszx 2 2 74 Rationalizing Substitutions I Integrands involving quotaz b u 7 ax b 2 Trigonometric substitutions Va 7 2 i m asint Va x2 i z atant 2 7 a2 i z asect 75 Integration of Rational functions N 96 DW Step 1 Get a proper rational function f qx by long division Step 2 Factor Step 3 Do the partial fraction decomposition Step 4 Multiply both sides by D and solve for the unknown constants Step 5 Integrate term by term Review Chater 8 817 82 Indeterminate Forms 1 L7Hopitals Rule hm L95 hm fz if H2 996 H2 9 96 f 0 90 2 lim fz9m Exponential Type 0000071 0 has 6 or g form Step 1 Take logarithm lnfx9w g lnfz Step 2 Compute lim g lnf Oz In 60 hm lnfx9w Step 3 hm fx9 60 83 Improper Integrals In nite Limits of Integration 17 b 1 fxd hm fxdz 7 b 2 mm blim mm 00 0 oo 3 mm fzdx fxdz foo foo 0 If both right integrals converge7 then fzd converges Otherwise7 it di verges 84 Improper Integrals In nite Integrands 1 Let f be continuous on the half open interval ab and suppose that lirlni fx 00 or foo Then7 bfzdz Jig atfzdx 1 If this limit exists and is nite7 we say that the integral converges Otherwise7 we say that the integral diverges 2 If f is continuous except at z c and lim f ioo7 then fltzgtdz cf96d96 brew where a lt c lt b