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# Final Study Guide MTH 162

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Calculus 2 Chapter 66A Improper Integrals Integrals that are over an in nite interval This means there is no nite area under the curve There are three cases for improper integrals l faoo f x dx Here the interval is a through all values in the 00 direction This can be solved by taking the limit of the integral For most functions the integral as the left most point of the region t approaches in nity is going to one set value that can be calculated by its limit This is the process to solve this is fa fxdx 11mm fa39fxdx limtooFt Fa Where F is the antiderivative of fX 2 If fxdx This can be thought of the same as the previous case but in the opposite direction This is solved by fb fxdx limtoo fbtxdx limtooFt Fa 3 If fxdx Here there is an endless interval in both directions f ofxdx ffoofxdx famfxdx SO f ofxdx limtooFt Fa limtooFt Fa If such an integral as these has a nite limit solvable limit the integral is said to converge at the value of the integral If such an integral as these has a limit that does not exist the integral is said to diverge Calculus 2 Chapter 66B Discontinuous lntegrands Limit approaches a vertical asymptote This means f x will be discontinuous at some point Recall A vertical asymptote is created Where the denominator equals 0 Case 1 Suppose x a is a vertical asymptote for f x Then ffxdx 11mm ft fxdx limtaFb Ft Note that if the limit exists then Ft exists This is integration from a vertical asymptote Case 2 Suppose x b is a vertical asymptote for f x Then f fxdx 11mm f fxdx limtbFt Fa This is integration to a vertical asymptote Case 3 Suppose x c is a vertical asymptote for f x Where a lt c lt b b b Then fa f xdx f xdx fc f xdx limtc fifoMx limtc ftb fxdx Ft Fa Fb Ftl in case three for the integral to converge both limits must exists Calculus 2 Chapter 74 Arc Length arc length s Def1nition The distance along line f x from a to b The arc length is considered smooth if f x is continuous and differentiable and f x is differentiable along a b Derivation of arc length formula Consider that P a lt x0 lt x1 lt xn lt b is a partition of a b Start by taking a line segment from xi1 f xi1 to xi f x0 Using the distance formula we get Li JCxi xi 12 fxi fxi 12 Multiply by xi xi 12 2 Li1W xcxrx xixi 1 Li1fxi fxi 12 Ax xixi 1 By the mean value theorem There exists wt in a b such that f w so i i 1 Li 1 f39W2 Ax Take the integral of this Adds up all the line segments on the interval a b to approximate s b s 1 f w2 dx a There are two cases when calculating arc length Case 1 When differentiating with respect to X yfx an Sb Case 2 When differentiating with respect to y 9901 6133 Sb 3de 1 2 dx Chapter 81 Infinite Sequences Sequence An ordered listing of numbers separated by commons Ex NIH wIN whlw U1IIgt three dots at the end are required since it is an infinite sequence Each of these values can be represented by the term an n is the position in the sequence WIN 1 3 so a1 a2 2613 andsoon The nth term might be represented by an nth term formula 39 n For th1s sequence an 2 n1 This formula describes sequence for every term Ex 1 Write the first 5 terms of sequence Who s nth term is given by n1 SOa21a3a42a a 63 quotn21 12 25 3105 417 5 26 13 3 2 5 3 Th1s 1s wr1tten as 1 3 E E E dontforget the three dots at the end Another way to found the terms in a sequence is through recursion An equation is given relating one term to the nest in the sequence Therefore the previous term is needed to nd the next 2 Write the rst 5 terms of the sequence de ned recursively by a12 an1an4 so a2a146 a36410 a41O414a5144 18 Answer 2 6 10 14 18 3 Write the rst 10 terms of the Fibonacci sequence de ned recursively by a1 1a2 1an an2 an1 a3112 a4123 c1525 c1628 a713 a821 a934 a1055 answer 11235813213455 Factorial Notation If n is apositive integer Then n factorial notated 11 1 x 2 x 3 x x n 1 x 11 It s important to know that 25 25X24gtlt23gtlt22 X21 or 25 25X24 4 Find the value oft quotI 10 x 9 x 8 720 2n2 2n2x2n2 1x2n2 2 2n2 x 2n1 x 2n 5 Slmpllfy 2n 2n 2n 2n2gtlt2n1 4n26n2 Common sequences Sequence Example nth term pos Integers 1234 an 2 11 pos even ints 2468 an 2 211 pos odd ints 1357 an 2 2n 1 pos powers of 2 24816 an 2 2quot pos powers of 2 w 124816 an 2 2quot 1 All int 01122 an 2 W If a1 a2 a3 a4 is an in nite sequence consider limnoo an 4 n E so an Ex consider n1 NIH wIN rhlw I I lim a lim n This means as the terms increase the values added up could approach a number Since polynomials of same degree are in numerator and denominator 1 11mnoo E 1 There is a horlzontal asymptote at y 1 n x Inth1s case can be treated as n1 x1 Basic limits 1 1 1 n 11mnoo Z 0 11mnoo Zn 1 0 11mnoo z n 11mnoo 2 0 Calculus 2 Chapter 82 In nite Series This is the notation for a series 2300 an It means that all the values of an from 0 to in nity will be added up So fo an a1 a2 a3 an1 an 1 1 1 1 1 Ex E127E23T6H 1 Consider the in nite series 291 an 2 a1 a2 a3 an1 an Rules Form sequence of partial sums SPS 1 52 S3 S4Sn Where Sn a1a2 an oo Zn 1 00 Zn 1 0 1 39EX 2111 2n 2111 2 1 2 1 2111 1 2 1 limnooSn limnoo 1 i 1 0 1 2n 1 If sequence of partial sums converges then the series 291 an also converges to S and is referred to the sum of the series If it converges nd the sum 2 If the SPS diverges then the series 291 an diverges 0 1 l l l l l l EXZn1 1234 SPS 51 152 12S3 123 3 11 25 hm oo Sn 1 E g E D1verges Test for Divergence and Convergence Nth test for Divergence Given the series 291 an If limnaoo an i 0 the limit DNE Then Eff an diverges Proof suppose 291 an converges Thus limnoo Sn 2 S is the sum of the series Sn n l an gt Sn Sn 1 an limnoo an 2 limngooSn Sn1 S S 0 The limits are the same since the two are the same other then one term Calculus 2 chapter 83 If the series converges there are 6 other tests that can be used to determine convergencedivergence l Integral Test apply only to series with all positive numbers or all negative numbers Compute the series as an in nite integral since If the limit converges then so does the series and same for divergence 0 1 oo 1 EX 1 Zn1 n24 compute f1 x24 2 lt2 lt2 lt2 2 2 o This integral converges so the series converges but not to the same number dx limtootan 1 2 221 flooidx 1imtoolnx 11m1nt 1n 1 oo This diverges so the series diverges Side note 2le an is the sum of the areas of the rectangles over the line while I100 f x dx is the area under a continuous function so2 1an gt f1 fltxgtdx means the sum of the series doesn t equal the sum of an 2 Basic Comparison test apply only to series with all numbers or all numbers Idea is to compare a series to another that you know converges or diverges Two types of series to compare to l Geometric series a Effzoarnaarar2m raquot n a rst term r common ratio Converges if 1 lt r lt 1 limnnoo rquot 0 Diverges if Irl gt 1 For this series Sn 2 a ar a r2 arquot So rSn arar2ar3arn n ThCI CfOI C7 Sn TSn a arn gt Sn n Now test the limit lim 00 HT L sum Of a n 1 7 1 1 71 EX 1 233204 371 612 r Since 1ltrlt1 Sum 22 2 ziz 5 2 Zai 2 5 2531 for both 1 lt r lt 1 5n 5 E E 3 2 13 Sum 5 5 1 1 2 3 6 2 Pseries 231 nip If p gt 1 the series converges If p S 1 the series diverges In the basic comparison test Compare to 2 bn which is known if it converges or diverges 1 If an lt bn and 2 bn converges then 2 an converges 2 If an gt bn and 2 bn diverges then 2 an diverges 3 If an lt bn and 2 bn diverges then no conclusion 4 If an gt bn and 2 bn converges then no conclusion 3 Limit Comparison Test apply only to series with all numbers or all numbers Suppose limnoo L gt If 0 lt L lt 00 then 1 If 2 bn converges then 2 an diverges 2 If 2 bn diverges then 2 an diverges 4 Ratio Test apply to both series with all numbers or all numbers Note on alternating series 1n1 Altemat1ngtest2an E 2 1 Z n 1 1 1 1 Ser1es ofabsolute values Elanl E 1 E E Z lf 2 an and Elanl converge 2 an is absolutely convergent If 2 an converges and Elanl diverges 2 an is conditionally convergent Ratio test Let 2 an be a series and let 1quot limnoo an 1 If 0 S r lt 1 then 2 an is absolutely convergent 2 If r gt 1 then 2 an is divergent 3 If r 1 then the test fails 5 Root Test apply to both series with all numbers or all numbers Let 2 an be a series and let L limnoo quotJ Ianl 1 If 0 S L lt 1 then 2 an is absolutely convergent 2 If L gt 1 then 2 an is divergent 3 If L 1 then the test fails 6 Alternating Series Test for convergence Tl An alternating series can be expressed as Z 1 Let 2 an be an alternating series If 1 an1 lt Ianl and 2 limnnmlanl 0 Then 2 an is convergent Calculus 2 Study Guide Exam 2 Chapter 58 L hospitals Rule L Hopital s rule Let f x and gx be differentiable functions If 1 limxa f x limxa gx 0 or 2 limx gtafx limx gta9x too Then f x if both limits exist 9 x Certain problems will require that you make use of L Hopital s rule multiple times 00 and g are called indeterminate forms however they are not the only ones 0 oo Indetermmate Forms 6 5 0X00 00 oo 00 1 000 What s important to know is that L Hopital s rule only works for 00 and 3 so you have to change the other indeterminate forms into them For the indeterminate forms 1 00 000 let L lim fxgx lnL lnlim fxgx lim 1n39fxgx lnL lim gx lnfx lim In E lnjx lim lnL 6th e W What ever answer the limit gives is the power of e Ex 1 1 limx0 6quot 3x 10 L x 1 e 3 1n ex3x x 4 so lnL 11mx01nex 3xx 11mx0 9 3x I 4 This however is not the answer according to the equation above The full answer is 64 Chapter 60 Techniques for Integration 1 Division For when there are two polynomials in the fraction and the highest degree of X in the numerator is equal to or greater then that of the bottom Also the equation can not be broken into parts 2 f2x 11 x24 2x2x11 2x2 8 x 3 2x2x11 x 3 f x24 f2 x24 x 3 1 2 3 1 x f2fxz8 fx24 2x lnx 8 tan C 2 Rationalizing Substitution f lett Vx 1 nowputxanddxinterms ofy xt2 1 dx2tdt x2 t2 12 2 2 fm t tht ft 12dt f2t 2dt t3 21 vx 13 2vmc 3 Completing the Square f x2x613 dx If the bottom polynomial was in the form of x2 bx or x2 bx the integral could easily be done To do this Beak the last number up for bottom to be perfect square plus or minus removed number I x1 f x1 x2 6x13 x 324 Obtain the form x2 bx b2 or x2 bx b2 Now use substitution using the completed square wx 3 dwdx and xw3 f x1 fW31 x 324 w24 W 4 l 2 1K fwz4fw24 21nw 42tan 2 1nx 6x 13 2tan 1 C Chapter 61 Integration by Parts The idea is to take one integral and break it up into parts to solve it So recall that f du u And that if v vx and u ux Then 11w 2 dCuv udv vdu dx Know this we can get the integration by parts formula fduv fudv fvdu uvfudvfvdu f udv uv f vdu formula for integration by parts Ex lfsin 1xdx usin 1x du 1xzdx dvdx vx j391d 1f1al 1 1 2quotla l sm x xxsin x x xxsin x x x 2 x ll x2 Now use u substitution but since we already are using a u lets do W substitution w 1 x2 dw 2x 1 1 x sin 1x fx1 x2 dx xsin 1x fw3 dw l E xsin 1x wz C xsin 1x 1 x22 C Tabular Method Don t forget the rules mentioned on the next page regarding this method Can only be used when you have fxquot sinax dx fxquot cosax dx or fxneaquot dx For the previous problem of f x2 sin x dx Step 1 make a quick table Step 2 differentiate u on the left and integrate v on the right until the u gets to zero Step 3 Now pair up the values diagonally so x2 pairs with cosx and so on Step 4 starting at the top line with positive and alternating U V between that and positive go down the rows as seen above x2 sin x 2x cos x 2 sin x Step 5 Now write out the pairings with their corresponding 0 COS 00 Signs sz sinx dx x2 cosx 2x sinx 2 cosx O C x2 cosx 2x sinx 2 c0506 C More rules Need positive integer power of X The second function needs to be easy to integrate many times Despite these the Tabular method may not be the best depending on the situation Cha ter 62 A Tri inte rals There are three types Type 1 f sinm x cosn x dx To solve for this one you either use the identity sinzx coszx 1 or on of the power reducing formulas 1 sm2 x E 1 cos 2x cos2 x E 1 cos 2x To use the power reducing formulas you need to have these special cases a Only f sinm x dx where m is a positive integer b Only f cosn x dx where n is a positive integer c f sinm x cosquot x dx where n and m are positive integers If there is an odd power you need to use the trig identity to solve in terms of either cos x or sin x Type 2 fsecm x tanquot x dx For this one either you use usecx dusecxtanxdx or utanx dusec2x dx to put the integral in terms of either one of these two trig function Make use of the identity 1 tan2 x 2 sec2 x Type 3 f cscm x cotn x dx For this one either you use u cscx du cscxcotxdx or u cotx du csczx dx to put the integral in terms of either one of these two trig function Make use of the identity 1 cot2 x csc2 x Chapter 62 B Trigonometric Substitution There are three instances where trig substitution is used 1 Integral involved Vaz u2 Set u asinH so that 9 sin 1 and du acos 9 d9 Place the known values on a right triangle so that Use SOHCAHTOA to relate to right triangle Hypotenuse a opposite side from angle u Pythagoras theorem gives adjacent side Vaz u2 We can then nd that Vaz u2 a cos 9 With these values we can substitute in the original equation 2 Integral involved Vaz u2 set u atan 9 so that 9 tan 1 and du asec2 9 d9 Place the known values on a right triangle Adjacent side a opposite side from angle u Pythagoras theorem gives Hypotenuse Vaz u2 We can nd that Vaz u2 a sec 9 3 Integral involved Vuz a2 set u asecH so that 9 2 sec 1 and du asecH tanH d9 Place the known values on a right triangle Adjacent side a Hypotenuse u Pythagoras theorem gives Opposite side Vuz a2 We can nd that Vuz a2 a tan 9 In all three cases you try to substitute the original equation with these new equations to make the derivative easier to nd Drawing the triangles for all three cases is also helpful in making sense of each substitution For de nite integrals before substitution set the range of the integral to be in terms of 9 so you don t have to change the nal equation to be back in terms of x Chapter 63 Integration by Partial Fractions x 19 x2 3x 10 For Integrate f Step 1 Factor the denominator of the integral I 96 19 I 96 19 x x2 3x 10 x2x 5 Step 2 Set equation in the integral to equal its partial fractions and multiply entire equation by the original denominator 96 19 A B m EE gt x 19 Ax 5Bx2 If the equation is an improper fraction one would have to divide the numerator by the denominator before setting it equal to its impartial fractions Remember an improper fraction is when the degree of X of the numerator is equal to or greater then that of the denominator Step 3 Solve for A and B 2 methods Method 1 Substitution Set X equal to a number that eliminates only one of the variables x 2 21A 70 A3 x5 140B7 B 2 Substitute them back into the original equation and solve f z f x 25 Iain 6136 Method 2 Equating the coefficients Multiply out variables and combine like terms x 19Ax 5Bx2 gtx 19xAB 5AZB Set the corresponding coefficients equal to each other ABl 19 5AZB Add the functions together to cancel out one of the variables 5ASB5 B2 A3 SA 23 19 7B 2 14 Step 4 Separate the integral by the partial fractions and solve 96 19 3 2 Ide fx2 EdX 31nx2 21nx 5C There are four cases or ways of treating the factor in the denominator when splitting the original function into partial fractions 1 NRFL Nonrepeating Linear Factors The function such as x 2 from above has X to the rst degree and the Whole function its self is to the first degree A x z x2 Only When solVing this case use method 1 to solve for the variables 2 RLF Repeating Linear Factors The function has X to the first degree and the Whole function its self is to the second degree or higher A B A B C x22 x 2 ac 22 39 39 39 39 quot or x 23 x 2 x 22 x23 39 39 39 39 quot 3 NRQF Nonrepeating Linear Factors The function has X to the second degree or higher and the Whole function its self is to the first degree AxB x2 2 x2 2 39 39 39 39 quot 4 RQF Repeating Linear Factors The function has X to the second degree or higher and the Whole function its self is to the second degree or higher AxB DxC x2 22 x2 2 92 22 39 39 39 39 quot Here is a practice example for how you would separate a more compleX fraction into its partial fractions x92x55x7 A B C D ExF GxH xI x3x 3x24x292 x x2 x3 96 3 x24 x29 x292 Calculus 2 Chapter 52 notes Logarithms Letagt0at1 Then ylogax means ayx In other words log a N equals the exponent that is needed to write N as a power of a Graphs ylnxxgt0 f m ylnx JCqu EX 39l 1 log28y can be written as 23 8 soy 4 2 log5 y through the same process y 2 1 3 log42y soy since4 Z Special Bases Base 10 also called common logarithms loglo x which is commonly written without the base to be log x Base 6 also called natural logarithms loge x which is commonly written as In x 1 De nition ofe e limx01 x x 2718 Properties of logs 110ga101n10 310gaax2xlnexx 210gaa 11ne 1 4 alogax xelnx x solongasxgt 0 Let P and Q be positive 510gaPQ logaPlogaQ lnPQ lnPan 610ga5210gaP logaQ lnPQ lnP an 7 logaPN NlogaPlnPN N lnP ChangeofBase formula lnN loga N E Proof y loga N gt ayzN gtlnaylnN gt y In a In N In a Derivative of Natural log Recall for function fX Difference quotient is f xh f x h fxh fx I h i 0 limh0 if it exists is the derivative of fX 1 e limx01 9c Iffx lnx x gt 0 then f x 2 lnhx lnx 1 xh h 1 Proof f x 11mh0 f 11mh0 E lnT 11mh01n1 h h 1 1 Ifm xmh so x h xm i 1 1 1 i 1 lllrr1ln1 mxm rhino 1n1 mmx Elnglrirnm mm Elne E If f x lnlxl then f x i The proof would be the same Now if the chain rule were applied with this I d l alnw udu u Ex 1 i1nsecx2 2x 1nsecx 2 secxtcmx Ztanx dx dx secx d d d 5 2 1n 2 ln3x 2 1n2x 3 3 2 dx 2x3 dx dx 3x2 2x3 6x213x6 3 j xsinamxz 1 coslnx2 1 2quot x21 4 Find 3 1nx2 y2 2 xy 4 xziyz 2x 23 3 x3quot 3 gt 296 230quot x3y yzy xzy 3393 gt2 X39 xzy y3 3633 xyzyr gt 2x xzy y3 ylx3 xyz I 2xx2yy3 y x3xy2 2y 5 Find the equation of the tangent to the line y x2 lnx2 3 at point 20 yr xz 2quot 2x1nx2 3f 2 164ln1 16 x2 3 y 016x 32 Logarithmic Differentiation Step 1 Take the natural log of both sides of the equation Step 2 Simplify expression using the properties of logarithms Step 3 Take the derivative of both sides and isolate y on one side Ex 2 1 Find y y W Step 1 lny lnquot23 44 Step 2 lny 1nx2 3 41n3x 4 1n xi 1nx2 3 4ln3x 4 1 Step 3gt if 393 3 y39 M3319 3 g lt Squot gtlt 31 5 2 Find y39 y x2 1lnx Step l1ny lnx2 1 quot Step 2 lny lnxlnx2 1 Step 31131 lnx 2x 1nx2 1 x21 y x2 11nxlnx x22 1 1nx2 1 Logarithmic Integration lfu ux gt du u x and thenfdju lnIuI C This general example makes use of u substitution The Ln of the absolute value of u is used if u isn t greater than 0 for all values Ex 1 96161 u x2 1 du Zxdx then rewrite the integral to E f i d 1 2 1 2 51nlx 1 C or Elnx 1 C since x2 1 gt 0 for all real numbers 2 u 1 cosx du Sinxdx 61 lnu C ln1 cosx C 3 f35 u3x5 du3dx 2 df ln3x5 C Have to keep absolute value signs since 3x 5 can be negative 2 4 fx de f1 dx x1nx 3C 6x2 5x 11 5 f 2x3 dx When the polynomial of the nominator is equal to or greater than that of the bottom improper fraction one has to divide the 2 functions Here is how long division between two functions is applied in this problem 3x 2 6x2 5x 11 3x2x 3 4x 11 22x 3 remainder 5 The result will be in the format of a polynomial proper fraction So 6x2 5x 11 5 dex f3x2 3dx 2x2x 21n2x 3C 2x Trigonometric Integrals These are important and should be memorized sin x 1 ftanxdx 2f dx throughu substitution 2 f1ncosx C flnlsecxl C C0596 C0596 2 fcotxdx 2f dx lnlsinxl C 1ncscx C Slnx SCC 9639 SEC xtan 9639 SEC2 xseC x tan x 3 f 1 x CSCXtanx f secxtanx u secxtanx du secxtanx seczx dx sec2 xsec x tan x f dx 1nsecx tanxl C secxtanx 4 fcscxdx 1ncscx cotxC Calculus 2 Chapter 53 quot Natural Exponential Function ex i cxp 1 exylnyx Therefore 1 1n ex x For the derivative for ex we use logarithmic differentiation yex lnylnex gt lnyx Take the derivative ofboth sides gt iy 1 gt y y gt y 2 ex d gt quotquot Therefore ex 2 ex This is a little different when the chain rule is applied at du e e dx dx For the integral f ex 2 ex C EX 1 x 1 x 1 x x 1yEe Slnx Ee cosxge Slnx e cosx 1 1 y 5ex cosx ex s1nx ex Slnx ex cosx 5Zex51nx exsmx 2 xey yex xeyy ey yex y ex gt xexy exy yex ey gt I x x x y Izyexey 31068 e ye e gt y xeLex 1 x2xex 2x2xex 3gtyx2exx lnyx1 x2ex gt EV39ZW gt y ym I 2 x x 2x2xex y x e x2ex 4 fexxex 1dx u substitution u 2 996 1 du exdx 3 fexVex1dx fxZdu u C ex15C Calculus 2 Chapter 54 Derivative of regular log functions If you recall the change of base formula loga N 2 11111 1 Then 10 x lnx i 1nx 1 dx ga dx lna lna dx xlna Derivative and integral of number to the power of X 1 101quot Use logarithmic differentiation to solve this dx 1nylnax gt 1nyxlna gt 3 11y 21na This is true since In a is a constant y aquot In a or with the chain rule 1a aud ulna dx dx 2 faxlnadxaxC faxiaxC lna Ex 1 y 10g2 1 1 1 x 5 1 1 x 1 1 1 1 n ln ln ln 2 y ln2 x2 21n2 x2 ln 22 x x ln4 x x2 2 gx 5603 g x 5603 ln5 sinx 3fx22xdxfx2dxf2xdx x32xC 4f33dxf1dxf131 xdx u1 x du 1dx x f3udu2x i31 xC ln3 Calculus 2 Chapter 56 Inverse Trig Functions These functions can be thought of as doing normal trig equations but backwards There will be proofs for the derivatives of sin 1 x tan 1 x and sec 1 sin 1 x Def1nition Derivative Proof Ex 2 tan 1 x Definition Derivative Proof coszy sin2y 1 so 1x y sin 1 x means siny x The range of answers is restricted to be Only need one positive 1 and one negative 4 quadrant 7T V5 77 Ex51n11 and 51n1 2 2 4 d 1 1 d s1n u dx ll u2 du Recall that coszy sin2 y 1 so cos2 y 1 sin2 y y sin 1x gt siny x Therefore cos2 y 1 x2 gt cosy iVl x2 But in quadrants l and 4 range of sinquot1 x cos y is positive I I 1 I 1 Slnyzx gt y Cosy 1 gt y cosy y 1 x2 1 x I ex ex y gt y 1ex2 1182x y tan 1x gt tany x The range of answers is restricted to be Only need one positive 1 and one negative 4 quadrant Ex tan391 1 E 1 d 1u2 du i 1 dx tan u Recall that coszy sinzy 1 gt 1 tan2x seczx cos2 y cos2 y cos2 y y tan 1x gt tany x Therefore sec2 y 1 x2 gt sec2 y 1 x2 In quadrants 1 and 4 range of tan391 x sec x is positive I I 1 1 tanyZx gt y seczy 1 gt y seczx gt y m 3 sec 1 x Def1nition y 2 sec 1 x means secy x The range of is restricted to 0 S y S and TC S y S 3 On1y need one positive 1 and one negative 3 quadrant Derivative i 1 1 dx sec u u2 1du Proof Recall that 1 tan2 y 2 sec2 y gt tany i sec2 y 1 But in quadrants 1 and 3 range of secquot1 x tan y is positive y sec 1x gt secy x 1 1 Therefore secytany y 1 gt 3 secytany xwxz Ex 1 39 L 1 2x 1 1 fx tan 2x f x x Zx21 2 tan 2x 4962 2 tan 2x 1 1 2 y sec lobe2 4 u x2 4 du x2 4 2x xx2 4 1 2 m m21xx2 4y x 3 DemoW 3 y xtan 1x 1nx2 1 xtan 1x 1nx2 1 x 1 2x x x y tan 1x tan1x x21 2 x21 x21 x21 4 x2 x sin 1 y 2 yequot Use implicit differentiation 2x y sin 1 y 2 yequot y ex W sin 1 y xy yex1 yz y ex1 yz y x ex1 yz 1 yzyex sin 1 y 2x W1 y2yexsin1y2x y x ewayz 5 y 2 sec xsec1 x Use logarithmic differentiation y 2 sec xsec1x gt lny sec 1x lnsec x 1 1 1 yy sec x secx secxtanx1nsecx xm y ysec 1xtan x 125 2 sec xsec1 x sec 1x tan x 125 Integrals of Inverse Trig Functions If u ux and a is a constant Where a gt 0 then 2 sin 1 C j du Vazu2 du 1 u 1 jx2a2 atan aC jal J 1 C um asec a Ex dx exdx du 1 MW 26quot 2996 5 szcm x f l z sin 1 C sin 1 C 2 f2x5dx2f 2x f 5 x2 49 x2 49 x249 2 E 1 E gt1nx 49 7tan 7 C 2 3 f2x x11 dx 2 Need to divide the two functions x 4 2 2x2 x 11 2x2 8 Remainder x 3 x24 3 f2x2x11d 2x32 x x24 x x24 x24 x24 X39 2C gt 2x l1nx2 4 Etan 1 2 2 3dx dx 4fx 9962 25 f3x 3x2 52 u3x du3 a5 gt 1 13xC 5sec 5 Calculus 2 Chapter 57 Hyperbolic Functions exe39x ex e39x nhx coshx Slnh tanh 2 2 coshx 1 coshx sechx cschx coth coshx smhx smhx Graphs rush I K 0 39 sinlu f Ill tunh 39 39 1 FIGURE 1 FIGURE 2 lt l f V 39 lnhl U 6 L h 6 Jlt A Hyperbolic Identities cosh2 x sinh2 x 1 since a hyperbolic function is x2 y2 1 2 2 exe x ex e x ezx2e 2x ezx 2e 2x 4 Proo 1 2 2 4 4 4 Derivatives of Hyperbolic Functions d du d du d du 51nhu coshu coshu Slnhu tanhu sech2 u dx dx dx dx dx dx d du d du cschu cschu cothu sechu sechutanhu dx dx dx dx d du cothu csch2 u dx dx Proofs exe x l isinhx l i x x l x x exex 2 2dxe e 2e e 2 coshx d d sinh x cosh x cosh x sinh x sinh x cosh2 x sinh2 x 1 2 2 tanhx sech x dx dx cosh x cosh2 x cosh2 x cosh2 x d coshx sinh2 x cosh2 x 1 3 cothx csch2 x dx sinh x sinh2 x sinh2 x Integrals of Hyperbolic Functions du du fcoshu dusmhuaC fsmhu ducoshuEC 2 du 2 du fsech u dutanhuEC fcsch u cothuEC du du fsechutanhu du sechua fcschucothu du cschua Ex sinhx l y cosh x sinhx sinhx cosh xsinhx cosh x sinh x sinh2 x cosh x smhx Multiply by inverse y coshxsinhx y cosh2 x sinh2 x 2 sinhx coshx coshx sinh x2 e39x x u u x l 2fe Slnhxdx Usetheidentity fe smhxdx fe 2 dx Zfe 1dx gt llexxlexlxC 2 2 4 2 sinhx dx u coshx du Slnhx 3ftanhxdx 2f cosh x 62 lnlul C lnlcoshxl C Calculus 2 chapters 91 92 Parametric equations Equations and curves in the XY plane can be represented with 1 y f x 2 x f y 3 F x y O implicit 4 Parametric Equation x fa Where t is referred to as a parameter y 90 This represents a point on a curve as shown by the below table x t2 For y t3 t x y Point at xy 0 0 00 1 1 1 11 1 1 11 2 4 8 48 2 4 8 48 Ex Graph without using a table x t 1 y t2 Through substitution y x2 easy enough to graph x sin t 2 y Sing t gt 3 x due to the domain ofthe sin function only from 1 to 1 t 3 x 6 gt y x2 domain of et is O lt t y 621 4 Eliminate the parameter find the equation of the graph x3 2t 3 x y23t 2t3x Iquot 3x E 2 E y 223 x 22 2 2x2 96 4 x 4 2 cos9 C0509 T y 1 2sin 9 Smog 3 2 5 square both sides and add them 2 2 1 so x 42y124 This is a circle When the parameter can t be easily done one needs to use differentiation dy Given x fa Then d y y 90 d f dx Ex x t2 t 2 1 F1nd the equatlon of the 11ne of tangent to the curve t3 3t at t 2 a Plug in t to find starting point x 4 y 2 at t 2 Point of tangency 42 dy d3 2 b dx 2t1 Att Z dx 3 3 Slopeoftangent c Find the equation for the tangent y 23x 4 gt y3x 10 Second derivatives x ft dzy d dy dy Givenyzga 5 dx dx Ex d2 x 9 sin 9 1 F1nd y of dx2 y 1 cos9 x 9 sin 9 d y 5M9 so now 39 9 dx 1 cos0 39 L 1 cos0 2 d3quot I 2 2 2 d y E dl 1 c050c050 51n 0 cosQ cos 0 51n 0 cosQ l 1 dx2 d0 1 cost92 1 cost92 1 cost92 1 c050 2 dy 1 d 3 dQ 1 c056 1 1 1 dx2 1 c050 1 c050 1 c050 1 cos 192 Parametric equations and Arc length For calculating arc length we already know two formulas 2 1yfxanSb sf1 dx b d 2 2 xf aSySb sfa 1 dy There is a third using parametric equations x xt 3 313 aStSb 3fo 22dt Ex 1 Find the length of the arc for fl if 31522 914 212 4152 s f01V9t4 4t2dt join912 4dt u 912 4 du 18t s iflxEdu i3u 1 i9t2 41 i 13V13 18 18 0 18 3 0 27 0 27 x9 sin9 y 1 COS 9 is an example of a cycloid Area under a curve Remember b For regular equatlons fa f x dx IS the area under the curve Ex Find the area under one arch of the cycloid x i 9 5m 9 y 1 cos 9 Area f9902nydx y 1 c059 dx 1 cos 9 2 f0 1 cos 6X1 cos 9d9 2 f0 1 2 c059 cos 9d9 f02n1 ZCOSH 1 c0529 d9 foznG ZCOSH c0529 d9 3 1 2 3 9 2 sm9 sm29 X2739 2 37 square umts 2 4 0 2 Calculus 2 Chapters 93 94 Rectangular coordinates x y Polar coordinates 139 Consider Point P x y in rectangular or r 9 in polar Conversion From 1quot 9 to x y c059 2 gt x rcosH sin9 2 gt y rsinH x2 y2 2 From x y to r 9 r W cost9xz y2 sin9 2 tan9 For tangent it could be in multiple quadrants depends on point For 33 the point is in quadrant 1 so 9 g For 33 the point is in quadrant 4 so 9 57 Ex 1 Convert from polar form to rectangular form apoint25n r22 9 5 T l x rc059 2c053 22 1 yrsin925in5n2 T 1 b curve 1quot 3 c059 i c059 2 r 37x 1 2 3x y2 x2 2 3x circle ii r23rcosl9 xrcosl9 x2y23x 2 Convert from rectangular form to polar form a point 1 V3 TWV1 32 tan9 3 92 35 b curve 2x 3y 2 4 2rcosl93rsin94 r 2cosQ351n0 Graphs of polar equations Attached Handout For slope of a tangent in regards to polar equations Given polar curve 1 f 9 What is 2 xrcosl9 y r Sin 9 This is a parametric equation so d d dy a f 6 sin0f0 cos 0 d 81n0rcos0 dx f 0 cos 0 fX sin 9 c030 1 sin 0 Area under a curve region In rectangular form the representation for area under a curve is f f x gxdx Where the area is incremented by Widths of rectangles In polar form the area is incremented by s arc length area 0 sector 6 6 1 4 area of sector 2 gtlt7tr2 r26 radians areaof ClTCle 2n 2n 2 Consider a polar curve 7 f 9 The interval of integration is from 9 a to 9 b R is the total region bounded by r f 9 AA is the area of the incremented sectors of 39 u 39 39 i H the curve u Therefore AA fx2A9 so A f69fx2 d9 EX 1 Find the area of one pedal of r cos59 Need to nd out the range of integration Find Where r 0 When 59 E 9 l 2 10 The rst pedal is then bound in the region 110110 1 2 1 721 1 1 quot10 soA 2 XEfO cos 56MB f0 E1c05106d9 9 1 Osm106 ea 01 2 Find the area of the region bounded by the graphs r 3 2 cos 9 and r 3 sin 9 O a nd their intersection points 3 2cosB3 sin9 tan91 6 and b Find Which is the upper function and Which is the lower function plot some points in the limits of integration V r3 Zsin9 r3 2cosl9 r 0 3 o 1i 0 9 1 W2 3 nz 3 7t 5 n 5 37T2 3 ant2 3 2 1 2n AA farthest sector from pole shortest sector from pole AA 3 2 cos 92A9 3 2 sin 92A9 5 1 T A 22 9 12c0594c0529 3 125in64sin29 Z From here it becomes a long but doable integration problem remember the power reducing formulas when doing this Calculus 2 Chapter 86 87 Representations of Analytical Functions Last chapter we found the interval of convergence of a power series Ex for 25 the interval of convergence is 2 2 This means that for all values of X from 2 to 2 excluding 2 the series of the function will converge For each of the X values the sum of the series will change Therefore power series represent functions of X n so 2 x a function with the same domain 2 2 n2quot xquot x x2 x3 x4 2 39 quot n2quot 2 8 24 64 This looks like a polynomial but with infinitely many forms Ex22213 0xn 1xx2 x3 x4 This is a geometric series where a1 1 and r x So for any X within 11 the series converges to sum ii i 2 xquot for this Domain 1 1 x 1 A function is analytic if It has derivatives of all orders f x f x f x f quot x Any function represented by a power series is analytic lf analytic at X c it has derivatives of all orders at X c Series the represent elementary functions are called either Taylor or Maclaurin series Taylor series A start with fX being analytic at X c f1nd the power series that represents the function and is centered at 0 fx 22 oanx Cquot ao a1x C a2x C2 Determine what an is for each individual term n an 2 fn fc Note 0 1 The nth derivative of function evaluated at c all over n factorial Each evaluation gives an for that single term in the series Ex Suppose fx a0 a1x c a2x c2 a3x c3 Plug in c for all X 0 fc a0 f nf Findf x Oa12a2x c3a3x c2 f1 f C a1 Findf x O Zaz 6a3x c u 142 f C 2612 gt 2 Zlc This is done for every derivative in nitely many to nd series Maclaurin Series B These series are a type of Taylor series where fX is analytical at x 0 Then f x 290 auxquot same as Taylor series but c is equal at 0 EX Verify the Maclaurin series representation of ex fcx ex 2200 anxn a0 196 612962 So since fquotx 2 ex gt f 0 X is always zero making n fquot0 1 n n n Tl We can now derive that e 2 fo0 since 11 2 3 62quot 0u 12x2i2im n 2 3 Remember that the exwill always be 1 when evaluated at X 0 EX Find the Taylor series for In x about X l c 1 fx lnx a0 a1x c a2x c2 a3x c3 f 10 SOf1 1n1 0 T I 0 a0 fxgtfl11gtLI1l12a 1 1 1 f x x 12gtf 1 1 9 a2 f x x33gt f 1 2 gt a3 f4x gt f41 6 gt i a4 First four non zero terms lnx x 1 ax 12 x 13 ix 14 Go further to nd the interval of convergence n1 Find endpoints with ratio test limnoo R 1 in this case Endpoints are at X 0 and X 2 At X 2 series is 1 E g i alternating harmonic converges Therefore say that the it converges to In x evaluated at 2 or In 2 Memorize these Maclaurin Series 11i 1uu2u3 21021quot 1ltult1 u Domain due to both being geometric series mi 2 1 u u2 u3 2O 1quotuquot 1 lt u lt 1 u2 u3 00 uquot 3e 1uZEm Znog allu I u3 u5 u7 0 1nu2n1 4 Slnu u EE 7 39 2n02 n allu u2 u4 u6 0 1nu2n 5cosu 1 ZZ E En02 n allu 6 Binomial series 1 u 1 pu NH 1 391 392 1 lt u lt 1 2 3 If p is a nonnegative integer then it will eventually reach zero and the series has a nite amount of term Ex 1 Express x3 in power series of X 2 Find a0 a1 a2 a3 such that x3 a0 a1x 2 a2x 22 a3x 23 Can t go higher then third term since one can only differentiate x3 three times fx x3 gt f2 8 gt a0 foo 8 I 1 frx 3x2 gt gt a1 f2 1 2 fllx 6x gt frr2 12 gt a2 f2 6 2 III III f3 f x 6 gtf 2 6 gt a3 3 Sox3812x 26x 22x 23 1 2 Find the Maclaurin Series for sin 2x Plug into the series from above 3 5 7 2x 2x 2x 2x 3 5 7 sin 2x 3 Find the Maclaurin Series for V 1 x2 1 lt x lt 1 l fx 2 V1 x2 1 x22 Binomial Series 1 1 1 1 3 gt 1 o 2 W zx 329mm gt 1lx2lx4 ix6 2 8 16 1 32x 4 Find the Maclaurin Series for 32x 1 x f x 1 l 1 Matches one of the models above 3 2 3 3 3 3 3 3 9 27 9 27 81 2quot oo 5 F1nd sum of ser1es Zn0 E This is the 6quot model so sum is 62 Algebraic Operations for Series Let 2200 anxn and 905 210120 bnxn Then within a common interval of convergence 1 f x 906 2306111 bnxquot 2 f x gx 2306111 bnxquot 3 fx x gx 20 cnx39 where Cu 2 aobn albn1 an1b1 anbo Mimics termbyterm multiplication of polynomials but with in nitely many terms Ex 1 Find the Maclaurin Series for sinhx Recall that sinhx a6quot 6quot Find the series for ex and 6quot then subtract them 2 3 4 5 6 7 ex 1xx x x x x x 2 3 4 5 6 7 x2 x3 x4 e x1 x 2 3 4 5 6 7 2x5 2x7 F07m 3 equot equot 02x023il0 3 5 7 3 5 7 Sosinhxl 2x2i2i2im xx x x 2 3 5 7 3 5 7 sin x 2 Represent as a power series x x3 x5 x7 smxx 1 3 5 7 x 3 5 7 Differentiation and integration of power series Let fx a0 alx azx2 a3x3 2220 auxquot Then f quot a1 Zazx 3a3x2 Eff0 auxquot 1 Or 1 2 1 3 1 4 oo 1 n1 ffxdx c aox a1x a2x a3x Zn0 anx 2 3 4 n1 Note If f x 20 auxquot has a nite interval of convergence Then f x 2 fo0 auxquot 1 has same interval but may lose endpoints f x 290 auxquot1 has same interval but may gain endpoints Ex 1 Differentiate the series for sin x x3 x5 x7 fxs1nxx 3 5 7 3x2 5x4 7x6 x2 x4 x6 fx1 31 51 7 m 246m 2 Represent f sinx2 dx as a power series 39 2 2x6 fx 51nx x 3 5 7 I x3 x7 x11 x15 Ismx dx C 73 115 157 3 If f x i 1 lt x lt 1 nd a series representation for f x 1xx2x3x4m sof x12x3x24x3m 4 a Represent f x 1 lt x lt 1 as a series 1 1 x239 1 1x21 x2x4 x6x8 b By integration represent tan 1x as a series 1 1 1 1 1 f tan1xcx x3 x5 x7 x9 1 x2 3 5 7 9 When X 0 c 0 0 show series converges at both endpoints 1Tl1 211 1 Atx1theseriesis1 l3 31Z 3 5 7 9 Converges by Alternating Series Theorem 111 211 1 Atx 1theseriesis 1l ll l 2 3 5 7 9 Converges by Alternating Series Theorem 1 Find the sum ofthe series 1 g E g g as shown in c the series is tan 11 which goes to E e Represent it as a series 1 1 1 1 1 4 4 4 4 1 so n4 4 3 5 7 9 3 5 7 9

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