Calculus and Analytic Geometry
Calculus and Analytic Geometry MATH 222
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VII The Integral In this chapter We de ne the integral of a function on some interval a b The most c mmon interpretation of the integral is in terms of the area under the graph of the given function so that is Where We begin 50 Area under 3 Graph a I Let f be a function Which is de ned on some interval a g x g b and assume it is positive ie assume that its graph lies above the x axis How large is the area of the region caught between the x axis the graph ofy and the vertical lines y a and y 1 One can try to compute this area by approximating the region With many thin rect angles To do this you choose a partition of the interval ab ie you pick numbers 1ltquot39ltInWith axoltx1 ltI2ltltxn71 ltxnb These numbers split the interval a 1 into n subintervals io lly lilyizly 7 linilyxnl Whose lengths are Ax1x17co AM 122 761 Ax xnixnil In each interval We choose a point Ck ie in the rst interval We choose 0 Cl 1 in the second interval We choose x1 g 02 g m and in the last interval We choose some number M71 g CT g in See gure 29 We then de ne n rectangles the base of the kth rectangle is the interval xkihxk on the xaxis While its height is here k can be any integer from 1 to The area of the kth rectangle is of course the product of its height and Width ie its area is fckAck Adding these We see that the total area of the rectangles is 44 R fclAx1 f82AI2 fcnAxT This kind of sum is called a Riemann sum If the partition is su 39iciently ne then one Would expect this sum ie the total area of all rectangles to be a good approximation of the area of the region under the graph Replacing the partition by a ner partition With more division points should improve the approximation So you Would expect the area to be the limit of Riemannsums like R as the partition becomes ner and ner77 A precise formulation of the de nition goes like this S n to n w 3 5 n m n n 4 FIGURE 29 Forming a Riemann sum FIGURE 30 Re ning the partition De nition 501 If f is afunction de ned on an interval a7b then we say that b modes 1 ie the integral of fx from x a to 1 equals I iffor every E gt 0 one can nd a 6 gt O sue t at f01A1 f02AI2 fcnAn 7 I lt 5 holds for every partition all of whose intervals have length Axk lt 6 51 When f changes its sign If the function f is not necessarily positive everywhere in the interval a g x g b then we still de ne the integral in exactly the same way as a limit of Riemann sums whose mesh size becomes smaller and smaller However the interpretation of the integral as the area of the region between the graph and the xaxis77 has a twist to it FIGURE 31 lllustrating a Riemann sum for a function whose sign changes Let f be some function on an interval a g x g b and form the Riemann sum R fC1AI1 f82AI2 fcnAxn that goes with some partition and some choice of ck When f can be positive or negative then the terms in the Riemann sum can also be positive or negative lf gt 0 then the quantity fckAxk is the area of the corresponding rectangle but if lt 0 then fckAxk is a negative number namely minus the area of the corresponding rectangle The Riemann sum is therefore the area of the rectangles above the xaxis minus the area below the axis and above the graph Taking the limit over ner and ner partitions we conclude that b d area above the xaxis below the graph a x x minus the area below the xaxis above the graph 52 The Fundamental Theorem of Calculus De nition 521 A function F is called an antiderivative of f on the interval one has Fx for all x with a lt I lt b a bl if For instance c2 is an antiderivative of x but so is Cx c2 2008 111 Theorem 522 If f is afunction whose integral f exists and ifF is an anti derivative of f on the interval ab then one has 2 45 mm Fb 7 Fa a proof Was giVen in lecture Because of this theorem the expression on the right appears so often that Various abbreviations have been invented We Will abbreviate 521 Terminology In the integral dx the numbers a and b are called the bounds of the integral the function Which is being integrated is called the integrand and the Variable x is integration variable The integration Variable is a dummy variable If you systematically replace it With another Variable the resulting integral Will still be the same For instance7 1 2 1 3 1 1 dag 75 lame E7 o and if you replace x by 4 you still get 1 2 3 1 1 1 Ar dad w lube a Another Way to appreciate that the integration Variable is a dummy Variable is to look at the Fundamental Theorem again 2 fa dag Fb 7 Fa The right hand side tells you that the Value of the integral depends on a and b and does not anything to do With the Variable x Exercises 52 7 Find an antideriVatiVe F for each of the following functions Finding antideriVatiVes inVolVes a fair amount of guess Work but With experience it gets easier to guess antiderivatives a we 7 221 1 b we 7 17321 c we 7 x2 7 21 11 d we 7 x4 7 x2 ltegtfltxgt71x lt1gtfltxgt7 g we 7 e h we 7 i 2x 1 lt1 we 7 e 1 we 7 2 x e 7 e7 1 k we 7 2 lt1 we 1 2 e70 e 1 m 2 n 7 W 0 we 7 sum p we 7 13 Cl i 003 r fx cos 2x 5 sinc 7 7r3 t sinx sin21c u we 7 smltxgt v we 7 2xlt1 W In each of the following exercises you should compute the area of the indicated region and also of the smallest enclosing rectangle with horizontal and vertical sides Before computing anything draw the region 522 7 The region between the vertical lines x O and x 1 and between the xaxis and the graph of y x3 523 7 The region between the vertical lines x O and x 1 and between the xaxis and the graph of y in here 71 gt O draw for n 12 3 4 524 7 The region above the graph of y below the line y 2 and between the vertical lines x O x 4 525 7 The region above the xaxis and below the graph of x2 7 is 526 7 The region above the xaxis and below the graph of 4x2 7 4 527 7 The region above the xaxis and below the graph of 1 7 4 528 7 The region above the xaxis below the graph of sinx and between x O and x 7r 529 7 The region above the xaxis below the graph of 11 2 a curve known as Maria Agnesi s witch and between x 0 an 7 1 5210 7 The region between the graph of y 1x and the xaxis and between x a and x I here 0 lt a lt b are constants eg Choose a 1 and b if you have something against either letter a or b 5211 7 The region above the xaxis and below the graph of 5212 7 Compute 1 V17 xzdx 0 Without nd an antideriVatiVe for 1 7 x2 you can nd such an antideriVatiVe but it s not easy This integral is the area of some region Which region is it and What is that area 5213 7 Compute 12 1 1 a 17x2dx b Hindi c i27xidx O 71 71 Without nding antideriVatiVes 53 The indefinite integral The fundamental theorem tells us that in order to compute the integral of some function f oVer an interVal a b you should rst nd an antideriVatiVe F of In practice much of the effort required to nd an integral goes into nding the antideriVatiVe In order to simplify the computation of the integral 46 b modes M 7 M the following notation is commonly used for the antideriVatiVe 47 Fa mm For instance c2 dI gag sin5x dx7 cos5x etc The integral Which appears here does not have the integration bounds a and b It is called an inde nite integral as opposed to the integral in 46 Which is called a de nite integral You use the inde nite integral if you expect the computation of the antideriVatiVe to be a lengthy a air and you do not Want to Write the integration bounds a and a the time It is important to distinguish between the tWo kinds of integrals Here is a list of differences TNDEFINITE INTEGRAL DEFINITE INTEGRAL is a function of x f is a number I Was de ned in terms of Rie By de nition is any function mann sums and can be interpreted as ofx whose derivative is area under the graph of y x at least When gt O x is not a dummy Variable for example x is a dummy Variable for example Qxdx x2C and f2tdt t2C are 01 Qxdx 1 and 01 Qtdt 1 so functions of dilfferent Variables so they I1 Qxdx fol Qtdt are not equal 531 You can always check the answer Suppose you want to nd an antideriVatiVe of a giVen function and after a long and messy computation which you don t really trust you get an answer You can then throw away the dubious computation and differentiate the you had found If F c turns out to be equal to then your is indeed an antideriVatiVe and your computation isn t important anymore For example suppose that we want to nd I lnx do My cousin Louie says it might be x lnx 7 x Let s see if he s right d7xlnx7x x1lnx71lnx Who knows how Louie thought of this7 but it doesn t matter he s right We now know that flnxdx xlnx 7 x C 532 About 0 Let be a function de ned on some interVal a g x g b If is an antideriVatiVe of on this interval7 then for any constant C the function FxC will also be an antideriVatiVe of So one giVen function has many different antideriVatiVes7 obtained by adding different constants to one giVen antideriVatiVe Theorem 531 If F1x and F2x are antidem39vatz39ves of the same function on some interval a g x g b then there is a constant C such that F1x F2x C Proof Consider the difference C13 F1c 7 Then Gx 7 7 07 so that C13 must be const nt Hence F1c 7 F2x C for some constant It follows that there is some ambiguity in the notation f do Two functions F1x and F2x can both equal without equaling each other When this happens they F1 and F2 differ by a constant This can sometimes lead to confusing situations eg you can check that 2sinxcoscdx sin2 x 2sinxcoscdx 7cos2x are both correct Just differentiate the two functions sin2 x and 7cos2x These two answers look different until you realize that because of the trig identity sin2 x cos2 x 1 they really only differ by a constant sin2 x 7 cos2 x 1 To avoid this kind of confusion we will from now on never forget to include the arbitrary constant 0 in our answer when we compu e an 7He took math 222 and learned to integrate by parts 533 Standard Integrals Here is a list of the standard integrals everyone should know fxdac Fx 0 n n1 x dxn10 foralln7 71 l dx ln 0 Note the absolute values x ewdxew0 aaC dx a 0 don t memorize use a e 1m ln 1 sincdx 7cosxC coscdx sinx C tanc dx 7 ln l cos xl 0 Note the absolute values dc 7 arctanxC 1 x2 7 1 7 dx arcsinx C 1 7 x2 The following integral is also useful but not as important as the ones above dx 1 1sinx 7r 7r cosx7 ln17sinx0for 7 ltIlt All of these integrals should be familiar from the differentiation rules we have learned so far7 except for for the integrals of tanx and of col You can check those by differentiation using ln lna 7 lnb simpli es things a bit 54 Properties of the Integral Just as we had a list of properties for the limits and derivatives of sums and products of functions the integral has similar properties Su pose we have two functions and gx with antiderivatives and C13 respectively Then we know that 173 Gm M was me gm in words F G is en antiderivative of f g which we can write as 48 fltxgtgltxgtdx fIdI gltxgtde Similarly cFx ems cfx implies that 49 efede cfxdx if c is a constant These properties imply analogous properties for the de nite integral For any pair of functions on an interval 11 one has b b b so W gmidx 7 mm gmdx a a a and for any function f and constant 0 one has b b 51 cfxdx c a a De nite integrals have one other property for which there is no analog in inde nite integrals if you split the interval of integration into two parts then the integral oVer the whole is the sum of the integrals oVer the parts The following theorem says it more precisely Theorem 541 Given a lt c lt b and afunctz39on on the interval 11 then b c b 52 mm 7 mm mm Proof Let F be an antideriVatiVe of Then C mm 7 Fc 7 Fa and b mm 7 Fb 7 Fa so that b modes 7 M 7 M 7 M 7 Fltcgt Fltcgt 7 M fxdxbexdx ar we have always assumed theat a lt b in all inde nitie integrals f The fundamental theorern suggests that when I lt a we should de ne the integral as b a 53 mm 7 Fb 7 Fa 7 7Fa 7 Fb 7 72 mm For instance 0 1 xdx7 xdx7 1 o 55 The definite integral as a function of its integration bounds I tZdt 0 What does I depend on To see this you calculate the integral and you nd ig wii iiaila I73t73x 307313 E Consider the expression So the integral depends on x It does not depend on t since t is a dummy Variable77 see 52 1 where we already discussed this point In this way you can use integrals to de ne new functions For instance we could de ne z tzdt O 117 Which Would be a roundabout Way of de ning the function 13 x33 Again since t is a dummy variable We can replace it by any other variable We like Thus Ixa2da de nes the same function namely x The previous example does not de ne a new function x33 An example of a new function de ned by an integral is the errorfunction from statistics It is given by e 2 a 72 54 erfc Clf 7 e dt V7 0 so erfx is the area of the shaded region in gure 32 The integral in 54 cannot be com I FIGURE 32 De nition of the Error function puted in terms of the standard functions square and higher roots sine cosine exponential and logarithms Since the integral in 54 occurs very often in statistics in relation With the socalled normal distribution it has been given a name namely erfc How do you di erentiate afunction that is de ned by an integral The answer is simple for if F then the fundamental theorem says that f N dt Fmi Fm and therefore 0 dt 996 7 F00 F m fa d an a ft dt me A similar calculation gives you d b a ft dt 7M So What is the derivative of the error function We have d 2 7 2 f i 7 t dt er 9 dacl 0 e l 2 d 7 tz if dt 7 dx 0 e 2 2 75 7 56 Method of substitution The chain rule says that W F ltGltxgtgt we so that F Gac G e do FGe 0 561 Example Consider the function 2x sinx2 3 It does not appear in the list of standard antideriVatiVes we know by heart But we do notice8 that 2x x2 3 So let s call C13 x2 37 and icosu then 7cosx2 3 and dF G sinx2 3 22 FGac Gw so that 55 2c sinc2 3 dx 7 cosc2 3 C 562 Leibniz notation for substitution The most transparent way of computing an integral by substitution is by following Leibniz and introduce new Variables Thus to do the integral fGIG IdI where F u7 we introduce the substitution u Cx and agree to write du dGx Gx dx Then we get fGIG I dx fu du M 0 At the end of the integration we must remember that u really stands for C13 so that faea ede Fol 0 FGe 0 As an example let s do the integral 55 using Leibniz7 notation We want to nd 2c sinx2 3 dx and decide to substitute z x2 3 the substitution Variable doesn t always have to be called Then we compute dz dc2 3 2x dx and sinx2 3 sinz so that 2x sinx2 3 dx sinzdz icosz 0 Finally we get rid of the substitution Variable z7 and we nd 2xsinc2 3 dx 7cosc2 3 C 3 You will start notioing things like this after doing several examples When we do integrals in this calculus class7 we always get rid of the substitution Variable because it is a Variable we inVented7 and which does not appear in the original problem But if you are doing an integral which appears in some longer discussion of a reallife or reallab situation then it may be that the substitution Variable actually has a meaning eg the effective stoichiometric modality of CQF selfinhibition in which case you may want to skip the last step and leaVe the integral in terms of the meaningful substitution Variable 563 Substitution for de nite integrals For de nite integrals the chain rule d FGI F GIG I fGIG I imp lies I fGIG IdI FltGltbgtgtiFltGlta which you can also write as b 56 fGIG IdI fltugtdm 564 Example of substitution in a de nite integ39ral Let s compute 1 x O 712 dx using the substitution u C13 1 x2 Since du 2x dx the associated inde nite integral is 1 1 1 W 333 E z d W AM A 2 To nd the de nite integral you must compute the new integration bounds C0 and C1 see equation 56 If x runs between x O and x 1 then u C13 1 x2 runs between u 1 0 1 and u 1 12 2 so the de nite integral we must compute is 1 2 L 1 l 57 O 1I2 dx 7 21 H du7 which is in our list of memorable integrals So we nd 1 x 21 2 O Wdu1dulnu1 ln2 Sometimes the integrals in 57 are written as 1 2 L 1 l A012dx7241udu to emphasize and remind yourself to which Variable the bounds in the integral refer Exercises 56 7 Compute these derivatives d an a a 0 d 0 dx C E S 1 x2 d sins di 6 EA 17x2 1 9 dt 1 b lnzdz I an 25 d d1 52ds I an t2 f endx 562 7 Compute the second derivative of the error function How many in ection points does the graph of the error function have Compute some of the following integrals 566 7 6x5 7 22 4 7 7x 3x 75 46 7wdc 564 7 xaaxxaaaC axdc m7W7W 7 W 76ew1dx 566 726 65 4 5677 3x75dc 72 4 568 7 5de hm 1 4 569 7 265 1 1 4 5611 7 265 u 1 1 5611 7 1722732662 0 2 5612 7 5562 7 4x 3 dx 1 O 5616 7 554 7 612 14dy 73 1 5614 7 59725535d5 O 4 5615 7 dx O 1 5616 7 23762 0 a 56177 7 dt 7 1 t2 t4 2 67 2 t t dt 1 t4 2 2 56197 I 162 1 5618 7 W 5620 5621 5622 5625 5626 O 5627 7 x13dx 71 72 4 7 5628 7 27162 5 x 1 E 2 5629 7 mm 1 1 M11556 7 75EE dx 1 O 3 I 71 2 2 5632 7 dx 56 5 7 x 7 x dx 1 g 4 711 7r3 2 5633 7 sintdt 56 6 7 x 7leldc 7r4 7 1 7r2 2 5634 7 cos 2sin d 56 7 72242710 12 o 7 o 7r2 2 5635 7 cos sin2 d9 56 8 7 where o 7 0 7r 4 4 lt 56 367 tanx x5 if07lt17 27f3cosx x 1f1 x 2 quot2 cotx quot 5637 7 d 7 A Sinx x 5649 Hr where Wm x if77r x 0 dag fI sinx7 if0ltx 7n 5650 7 Compute 2 I 221x23dx o in two diHerent ways 00 RH en La 0 Hh x JV H gs ls 1 N 6 x a Ebrpand 1717x237 multiply with 2x and 5641 1 85 dx integrate each term 93 b Use the substitution u 1 x2 5642 8 2t dt 5651 7 Compute 7E 2 n 56 437 Edi In2c1c dx 7 752 I 3 565271f xx7122and 112 56 447 MAW 13mm f 7 2 5653 7 Sketch the graph of the curVe y 1 and determine the area of the region enclosed by the curve the xaxis and the lines x 07 x 4 5654 7 Find the area under the curVe y 6x 4 and above the xaxis between x O and x 2 Draw a sketch of the curVe 5655 7 Graph the curVe y 2x17 x2 x 6 01 and nd the area enclosed between the curVe and the xaxis Don t evaluate the integral7 but compare with the area under the graph of y 17 N 5656 7 Determine the area under the curVe y xa2 7 x2 and between the lines x O and x a 5657 7 Graph the curVe y 2x9 7 x2 and determine the area enclosed between the curVe and the xaxis 5658 7 Graph the area between the curVe y2 4x and the line x 3 Find the area of this region 5659 7 Find the area bounded by the curVe y 4 7 x2 and the lines y O and y 3 5660 7 Find the area enclosed between the curVe y sin 2x 0 g x g 7r4 and the axes 5661 7 Find the area enclosed between the curVe y cos 2x 0 g x g 7r4 and the axes 122 M 7 Graph y2 1 x and nd the area enclosed by the curVe and the line x 2 M 7 Find the area of the region bounded by the parabola yz 4x and the line y 216 w 7 Find the area bounded by the curVe y x2 7 x and the line x 2y M 7 Find the area bounded by the curVe x2 4y and the line x 4y 7 2 M 7 Calculate the area of the region bounded by the parabolas y x2 and x y2 5667 7 Find the area of the region included between the parabola yz x and the line x y 2 5668 7 Find the area of the region bounded by the curVes y and y x 5669 7 Here are the graphs of a function x its derivative and an antideriVar tiVe of Unfortunately the graphs are not lap beled Identify which graph is which Erplain your answer 5670 7 Below is the graph of a function y H06 wyggggg V a 96 Which among the following statements true7 b Fa Fc 7 TrueFalse Reason 0 Fb O 7 TrueFalse Reason d Fb gt Fc 7 TrueFalse Reason e The graph of y has two in ection points7 TrueFalse Reason Use a substitution to evaluate the following integrals 2 5671 7 MHZ 7 1 1u 3111216 5678 7 d 7 10032x x 7r3 5679 7 sin2 Cos d 7r4 3 1 5681 7 dr 2 rlnr 5681 i dx 1coszx 5682 i dx 1slnx 1 M7 zvlizzdz o 2 5684 7 In dx 7 1 I 2 56857 12 210d 50 3 5686 7 sin70032p4 do 2 5687 7 ae 2 da 5688 7 dt 7 t2 VI Exponentials and Logarithms In this chapter we rst recall some facts about exponentials xy with I gt O and y arbitrary they should be familiar from algebra or precalculus What is new is perhaps the de nition of xy when y is not a fraction eg 234 is the 4th root of the third power of 2 323 but what is 2 Then we ask what is the derivative of a 7 The answer leads us to the famous number 6 R 2718 281828 459 045 235 360 287 471352 662 497 757 247 093 699 95 Finally we compute the deriVatiVe of loga x and we look at things that grow exponential y77 42 Exponents Here we go oVer the de nition of xy when x and y are arbitrary real numbers with x gt For any real number x and any positiVe integer n 1 2 3 one de nes n times n x x x m and if x 7 O T 1 x 7 7 in One de nes x0 1 for any x 7 0 To de ne xpq for a general fraction g one must assume that the number x is positiVe One then de nes 34 xpq q xp This does not tell us how to de ne N is the exponent a is not a fraction One can de ne in for irrational numbers a by taking limits For example to de ne 2V2 we look at the sequence of numbers you get by truncating the decimal expansion of 2 ie 14 141 1414 111 a214 a3141m a41414m Each an is a fraction so that we know what 2 is eg 2amp4 move 2 Our de nition of 2 2 then is 2V2 lim 2 new ie we de ne 2V2 as the limit of the sequence of numbers 2 19214 10921417 100921414 See table 2 Here one ought to proVe that this limit exists and that its Value does not depend on the particular choice of numbers an tending to a We will not go into these details in this course It is shown in precalculus texts that the exponential functions satisfy the following properties 35 Nib NH i Nib y Iab x 2 1 7 14000000000 263901582154 6 14100000000 2657371628193 14140000000 2664 749650184 14142000000 2665119088532 14142100000 2665137561794 14142130000 2665143103798 14142135000 2665144027466 TABLE 2 Approximating 2V2 Note that as x gets closer to 2 the quantity 2 0 appears to converge to some number This limit is our de nition of 2V2 rovided a and b are fractions One can show that these properties still hold if a and b are real numbers not necessarily fractions Again7 we won t go through the proofs here Now instead of considering in as a function of x we can pick a positive number a and consider the function a This function is de ned for all real numbers x as long as the base a is positive 421 The trouble with powers of negative numbers The cube root of a negative number is well de ned For instance 37 72 because 72 78 In View of the de nition 34 of xp 7 we can write this as ear3 W W 72 But there is a problem since you would think that 78 6 78 However our de nition 34 tells us that vs W W 2 Another example 74 2 74 is not de ned but7 even though 74 374 416 2 is de ned There are two ways out of this mess 1 avoid taking fractional powers of negative numbers 2 when you compute xpq rst simplify the fraction by removing common divisors of p an 1 The safest is just not to take fractional powers of negative numbers Given that fractional powers of negative numbers cause all these headaches it is not surprising that we didn t try to de ne in for negative x if a is irrational For example 787r is not de ned6 6There is a de nition of 73quot which uses complex numbers You will see this next semester if you take math 222 43 Logarithms Brie y y loga x is the inverse function to y a This means that by de nition 1110ng ltgt xay In other words loga x is the answer to the question for which number y does one have y The number loga x is called the logarithm with base a of L In this de nition both a and x must be positive For instance 238 212 2 1 so log2 8 3 log2 log2 71 Also log273 doesn t exist because there is no number y for which 2y 73 2y is always positive and log3 2 doesn t exist either because y log3 2 would have to be some real number which satis es 73W 2 and we don t take noninteger powers of negative numbers 44 Properties of logarithms In general one has 1 loga a x and 1 Oh x There is a subtle difference between these formulas the rst one holds for all real numbers x but the second only holds for I gt 0 since loga x doesn t make sense for x g 0 Again one nds the following formulas in precalculus texts loga Iy loga I loga 1 lost 3 loga I 10ga 21 36 loga xy y loga x logb x 1 Oga x logb a They follow from 35 45 Graphs of exponential functions and logarithms Figure 27 shows the graphs of some exponential functions y a with different values of a and gure 28 shows the graphs of y log2 x y log3 x logl2 x log13c and y log10 x Can you tell which is which Yes you can From algebraprecalc recall If a gt 1 then a is an increasing function and l If 0 lt a lt 1 then a is a decreasing function l In other words for a gt 1 it follows from 1 lt M that a lt a if 0 lt a lt 1 then 1 lt 2 implies a 1 gt a 3 2 1 U 1 2 3 FIGURE 27 The graphs of y 2 23 12 01 and y 415 FIGURE 28 Graphs of some logarithms 46 The derivative of a and the definition of e To begin7 We try to di erentiate the function y 2 de 2xAa 7 2w So if We assume that the limit 2 7 1 A1330 Ax 0 exists then We have 12 37 dI 2171 for smaller and smaller values of A13 Which leads you to suspect that the limit actually exists and that C R 0693 4 in fact prove that the limit exists but We Will not do this here Once We know 37 We can compute the derivative of a for any other positive number a To do this We Write a 210g a and ence 02 On your calculator you can compute 7 One can aw 210g2 aw Qxlogz a By the chain rule We therefore get dax de logg a dx 7 d x C Zamogw dx log2a x C log2 12w3910g2 a C log2 a a So the derivative of a is just some constant times a the constant being Clog2 a This is essentially our formula for the derivative of a but one can make the formula look nicer y introducing a special number namely We de ne 2M 7 1 e 210 Where 0 lim 7 AacHO Ax One has 6 x 2718 281818 459 This number is special because if you set a C then ClogzaClogzeClog221CC i1 and therefore the derivative of the function y e is 38 Read that again the function e is its oWn derivative The logarithm With base 6 is called the Natural Logarithm and is Written lnx logE x Thus We have 39 elmC x ln 6 x Where the second formula holds for all real numbers x but the rst one only makes sense for x gt O For any positive number a We have a 6 and also a an aclna 76 By the chain rule you then get 40 47 Derivatives of Logarithms Since the natural logarithm is the inverse function of e70 We can nd its derivative by implicit differentiation Here is the computation Which you should do yourself The function loga I satis es awe x Differentiate both sides and use the chain rule on the left lnaafwfI 1 Then solve for to get 1 f W Finally We remember that afa I Which gives us the derivative of a da dI I ln a In particular the natural logarithm has a very simple derivative namely since lne 1 We have 41 48 Limits involving exponentials and logarithms Theorem 481 Let r be any real number Then Zfa gt 1 lim Ilaix O I MX 7 lim 1 0 w ioo ax This theorem says that any exponential will beat any power of I as x gt 00 For instance as x gt 00 both I1000 and 1001 0 go to in nity but 1000 1 7 win 100m so in the long run for very large I 1001 Will be much larger than 1000 07 Proof when a e We Want to shoW limac gt0 Ile 0 To do this consider the function Il1e lts derivative is dT167I I 7 f as 7 7 Therefore lt O for I gt r 1 ie is decreasing for I gt r 1 It follows that lt fr 1 for all I gt r 1 ie ITTleiw lt r 1T1e7l1 for I gt T 1 Divide by I abbreviate A r 1T1e7l1 and We get r 1IT 7 IT1e T 17 IIle7x Tiw OltIe lt forallIgtr1 I The Sandwich Theorem implies that limac gt0 Ile 0 Which is What We had promised to shoW D Here are some related limits r as agt1 lim L 00 DNE w oo x l m gt 0 lim O w oo xm mgt0 limoxmlnx0 aca The second limit says that even though lnx becomes in nitely large as x gt 00 it is always much less than any power am with m gt O rea To prove it you set x et and then t sm which leads to l t t 1 lim if an lim 7 t Sm 7 lim 30 x nx x taco emt m taco 55 The third limit follows from the second by substituting x 1y and using lni 7 ln x 49 Exponential growth and decay A quantity X which depends on time t is said to grow or decay exponentially if it is given by 42 Xt Xoe The constant X0 is the value of Xt at time t O sometimes called the initial value of X 77 The derivative of an exponentially growing quantity ie its rate of change with time is given by X t X0 leekt so that dX t dt ln words for an exponentially growing quantity the rate of change is always proportional to the quantity itself The proportionality constant is k and is sometimes called the relative growth rate77 43 kXt This property of exponential functions completely describes them by which I mean that any function which satis es 43 automatically satis es 42 To see that this is true suppose you have a function Xt for which X t kXt holds at all times t Then dXte m 7 def dXt 7 dt 7X dt dt 6 7kXte7m X tekt X t 7 kXte m 0 It follows that Xte kt does not depend on t At t 0 one has Xtekt XoeO X0 and therefore we have Xte XO for all t Multiply with ekt and we end up with Xt Xoe 491 Half time and doubling time If Xt Xoekt then one has Xt T XOeWkT Xoek e ekTXt ln words7 after time T goes by an exponentially growing decaying quantity changes by a factor 6 If k gt 07 so that the quantity is actually growing7 then one calls ln 2 k the doubling time for X because Xt changes by a factor ekT 61 2 2 eVery T time units Xt doubles eVery T time units If k lt 0 then Xt is decaying and one calls T the half life because Xt is reduced by a factor ekT e 1n2 eVery T time units 492 Determining X0 and k The general exponential growthdecay function 42 contains only two constants7 X0 and k and if you know the Values of Xt at two different times then you can compute these constants Suppose that you know X1 Xt1 and X2 Xt2 Then we have XoeM1 X1 and X2 Xoekt2 in which t1t2 X1X2 are giVen and k and X0 are unknown One rst nds C from m1 226 ekm 111 kt717t2 which implies k lnX17lnX2 t1 7 t2 Once you have computed k you can nd X0 frorn XO 5 X2 em em both expressions should giVe the same result Exercises 49 7 Sketch the graphs of the following functions a y e b y 6 c y e 67230 d y ea 7 4e 6 0 26 e y W 0 y W s y 906 h 2 ew4 i y 9626 j y em 7 z Hint for some of these if you have to solVe something like 64 7 363 e 0 07 then call u e 7 and you get a polynomial equation for w namely 104 7 3103 w O 492 7 Sketch the graphs of the following functions a y71n b yang cyxlnx dy 0ltIltoox7 1 e yln2 ltxgtogt f 1112 ltxgtogt g yamE m lt1 h y1n1x2gt i yln273x2 zgt2 j ylncosx m ltg 493 7 The function 5in plays a central in statistics and its graph is called the bell curve because of its shape Sketch the graph of 494 7 Sketch the part of the graph of the function 1 we with x gt 0 Find the limits fx gilt xquot and 1320 where n can be any positiVe integer hint substitute x 7 495 7 A damped oscillation is a function of the form 67 cos bx or 67M sinbx where a and b are constants Sketch the graph of 6 sin 10x ie nd zeroes7 local max and rnins7 in ection points and draw with pencil on paper the piece of the graph with 0 g x g 27V 496 7 Find the in ection points on the graph of 1 x lnx I gt O 497 7 a If x is large which is bigger 2 or 27 b The graphs of x2 and gc 2 intersect at x 2 since 22 22 How many more intersections do these graphs have with 700 lt I lt 00 i 498 7 Find the following lirnits 5x 1 e 7 x2 a 23207 b Jinsom 2w e 7 x C 2 73x 72w d 2 7x 7 7x2 4w e lirn L f lim 54 70700 ew1 x7w6x ac W h ln1 x 7 lnx lnx 4 lt1 JLHSQW J 133961 lnx ln k 1 7 1 1 7 W winzo m 0 x lnz 9 9 7 Find the tenth deriVatiVe of 67 A 9 10 7 For which real number x is 2 7 3 the largest A o m dz dz Xxx dz 7 dz 7 and dz 912 About logarithmic dz erentiation 911 7 Find Hint x 67 3 Let y x 12c 34x 56 and u lny Find dudz Hint Use the fact that ln converts multiplication to addition before you differentiate It will simplify the calculation b Check that the derivative of lnuz is the logarithmic derivative of the function u as de ned in the exercises following 25 chapter 4 4913 7 After 3 days a sample of radon222 decayed to 58 of its original amount a What is the half life of radon222 b How long would it take the sample to decay to 10 of its original amount 4914 7 Polonium210 has a half life of 140 days a If a sample has a mass of 200 mg nd a formula for the mass that remains after t days b Find the mass after 100 days c When will the mass be reduced to 10 mg d Sketch the graph of the mass as a function of time 915 7 Current agricultural experts believe that the world s farms can feed about 10 billion people The 1950 world population was 2517 billion and the 1992 world population was 54 billion When can we expect to run out of food 4916 7 The Archer Daniel Midlands company runs two ads on Sunday mornings One says that when this baby is old enough to vote the world will have one billion new mouths to feed77 and the other says in thirty six years7 the world will have to set eight billion places at the table77 What does ADM think the population of the world is at present How fast does ADM think the population is increasing Use units of billions of people so you can write 8 instead of 8 000 000 000 Hint 36 2 X 18 4917 7 The population of California grows exponentially at an instantaneous rate of 2 per year The population of California on January 1 2000 was 20000000 3 Write a formula for the population Nt of California t years after January 1 2000 b Each Californian consumes pizzas at the rate of 70 pizzas per year At what rate is California consuming pizzas t years after 1990 c How many pizzas were consumed in California from January 1 2005 to January 1 2009 4918 7 The population of the country of Slobia grows exponentially a If its population in the year 1980 was 1980000 and its population in the year 1990 was 1990000 what is its population in the year 2000 b How long will it take the population to double Your answer may be expressed in terms of exponentials and natural logarithms 4919 7 The hyperbolic functions are de ned by e equot e equot sinhx sinhx7 coshz 7 tanhx 2 2 coshx a Prove the following identities cosh2 x 7 sinh2 x 1 cosh 2x cosh2 x sinh2 x sinh2x 2 sinh x coshx b Show that d sinh x x dcoshx 7 coshx7 7 7 slnhx dx 0 Sketch the graphs of the three hyperbolic functions dtanh x 7 1 cosh2 x MATH 222 SECOND SEMESTER CALCULUS Spring 2009 Math 222 2nd Semester Calculus Lecture notes version 15Spring 2009 This is a self contained set of lecture notes for Math 222 The notes were written by Sigurd Angenent starting from an extensive collection of notes and problems compiled by Joel Robbin The LATEX files as well as the XPIG and OCTAVE files which were used to produce these notes are available at the following web site www math wisc eduquotangenent FreeLectureNotes They are meant to be freely available for noncommercial use in the sense that free soft ware is free More precisely Copyright c 2006 si urd B An enent 39 39 39 A to copy 439 A 4 this document under the terms of the GNU Free Documentation License Version 12 or any later version published v m over Texts and no BackJCover TextsVAcopy of quot 39 39 quot 39 39 quotGNU Fr 1quot quotm ta quot Licensequot CONTENTS Methods of Integration 6 The indefinite integral 6 2 You can always check the answer 6 3 About C 7 4 Standard Integrals 8 5 Method of substitution 8 6 The double angle trick 9 7 Integration by Parts 10 8 Reduction Formulas 11 9 Partial Fraction Expansion 14 91 Reduce to a proper rational function 14 92 Partial Fraction Expansion The Easy Case 15 93 Partial Fraction Expansion The General Case 17 10 PROBLEMS 18 Basic Integrals 18 Basic Substitutions 19 Review of the Inverse Trigonometric Functions 20 Integration by Parts and Reduction Formulae 21 Integration of Rational Functions 22 Completing the square 22 Miscellaneous and Mixed Integrals 23 Taylor s Formula and In nite Series 25 11 Taylor Polynomials 25 12 Examples 26 13 Some special Taylor polynomials 29 14 The Remainder Term 29 15 Lagrange s Formula for the Remainder Term 31 16 The limit as x gt O keeping 71 fixed 32 161 Littleoh 32 162 Computations with Taylor polynomials 35 163 Differentiating Taylor polynomials 38 17 The limit n gt 00 keeping x fixed 39 171 Sequences and their limits 39 172 Convergence of Taylor Series 18 Leibniz formulas for In 2 and 714 19 Some proofs 191 Proof of Lagrange s formula 192 Proof of Theorem 166 20 PROBLEMS Taylor s formula Lagrange s formula for the remainder Littleoh and manipulating Taylor polynomials Limits of Sequences Convergence of Taylor Series Approximating integrals Complex Numbers and the Complex Exponential 21 Complex numbers 22 Argument and Absolute Value 23 Geometry of Arithmetic 24 Applications in Trigonometry 24 1 Unit length complex numbers 24 2 The Addition Formulas for Sine 2S Cosine 243 De Moine s formula 25 Calculus of complex Valued functions 26 The Complex Exponential Function 27 Complex solutions of polynomial equations 271 Quadratic equations 27 2 Complex roots of a number 28 Other handy things you can do with complex numbers 28 1 Partial fractions 282 Certain trigonometric and exponential integrals 283 Complex amplitudes 29 PROBLEMS Computing and Drawing Complex Numbers The Complex Exponential Real and Complex Solutions of Algebraic Equations Calculus of Complex Valued Functions Differential Equations What is a Difqu 31 First Order Separable Equations 32 First Order Linear Equations 321 The Integrating Factor 322 Variation of constants for 1st order equations 33 Dynamical Systems and Determinism 34 Higher order equations 35 Constant Coefficient Linear Homogeneous Equations 351 Differential operators 352 The superposition principle 35 3 The characteristic polynomial 354 Complex roots and repeated roots 36 Inhomogeneous Linear Equations 37 Variation of Constants 371 Undetermined Coefficients 38 Applications of Second Order Linear Equations 381 Spring with a weight 382 The pendulum equation 383 The effect of friction 384 Electric circuits 39 PROBLEMS General Questions Separation of Variables Linear Homogeneous Linear Inhomogeneous Applications Vectors 40 Introduction to vectors 401 Basic arithmetic of vectors 402 Algebraic properties of vector addition and multiplication 403 Geometric description of vectors 404 Geometric interpretation of vector addition and multiplication 41 Parametric equations for lines and planes 411 Parametric equations for planes in spacequot 42 Vector Bases 421 The Standard Basis Vectors 422 A Basis of Vectors in generalquot 43 Dot Product 431 Algebraic properties of the dot product 432 The diagonals of a parallelogram 433 The dot product and the angle between two vectors 434 Orthogonal projection of one vector onto another 435 Defining equations of lines 436 Distance to a line 437 Defining equation of a plane 44 Cross Product 441 Algebraic definition of the cross product 442 Algebraic properties of the cross product 443 The triple product and determinants 444 Geometric description of the cross product 45 A few applications of the cross product 451 Area of a parallelogram 452 Finding the normal to a plane 453 Volume of a parallelepiped 46 Notation 47 PROBLEMS Computing and drawing vectors Parametric Equations for a Line Orthogonal decomposition of one vector with respect to another The Dot Product The Cross Product Vector Functions and Parametrized Curves 48 Parametric Curves 49 Examples of parametrized curves 50 The derivative of a vector function 51 Higher derivatives and product rules 110 110 110 113 114 52 Interpretation of 32 t as the velocity vector 5 Acceleration and Force 54 Tangents and the unit tangent vector 55 Sketching a parametric curve 56 Length of a curve 57 58 L0 The arclength function Graphs in Cartesian and in Polar Coordinates 59 PROBLEMS Sketching Pararnetrized Curves Product rules Curve sketching using the tangent vector Lengths of curves Answels and Hints GNU Free Docurnentation License APPLICABILITY AND DEFINITIONS VERBATIM COPYING COPYING IN QUANTITY MODIFICATIONS COMBINING DOCUMENTS COLLECTIONS OF DOCUMENTS AGGREGATION WITH INDEPENDENT WORKS TRANSLATION TERMINATION 10 FUTURE REVISIONS OF THIS LICENSE 11 RELICENSING WNSI P FWNE o 6 Methods of Integration 1 The indefinite integral We recall some facts about integration from first semester calculus De nition 11 Afunction y Px is called on antiderivative ofonotherfunction y fx if Px fxfor all x 4 12 Example P1x x2 is an antiderivative offx 2x P206 x2 2004 is also an antiderivative of fx 2x Ct sin2t 1 is an antiderivative of gt cos2t 1 gt The Fundamental Theorem of Calculus states that if a function y fx is continuous on an interval a g x g b then there always exists an antiderivative Px off and one has 1 rm dx ma 7 PM The best way of computing an integral is often to find an antiderivative P of the given function f and then to use the Fundamental Theorem 1 How you go about nding on antiderivative P for some given function f is the subject of this chapter The following notation is commonly used for antiderivates 2 x fxdx The integral which appears here does not have the integration bounds a and b It is called an inde nite integral as opposed to the integral in 1 which is called a de nite integral It s important to distinguish between the two kinds of integrals Here is a list of differences lNDEFINITE INTEGRAL DEFINITE INTEGRAL f fxdx is a function of x f bfxdx is a number By definition ffxdx is onyfunction of f bfxdx was defined in terms of Rie x whose derivative is fx mann sums and can be interpreted as area under the graph of y fx at least when fx gt O x is not a dummy variable for example x is a dummy variable for example 2xdx x2 C and f 2tdt t2 C are k12de 1 and 012m 1 so functions of diffferent variables so they fol Zxdx fol tht are not equa 2 You can always check the answer Suppose you want to find an antiderivative of a given function fx and after a long and messy computation which you don t really trust you get an answer P You can then throw away the dubious computation and differentiate the Px you had found If Px turns out to be equal to fx then your Px is indeed an antiderivative and your computation isn t important anymore 4 21 Example Suppose we want to find f ln x dx My cousin Bruce says it might be Px x lnx 7 x Let s see if he s right d 1 axlnx7x 7x1lnx71 ilnx Who knows how Bruce thought of thisl but he s right We now know that f In xdx x lnx 7 x C gt 3 About C Let fx be a function defined on some interval a g x g b If Px is an antiderivative of fx on this interval then for any constant C the function Rx Px C will also be an antiderivative of f So one given function fx has many different antiderivatives obtained by adding different constants to one given antiderivative Theorem 31 If P1x and P206 are antiderivatives of the same function f x on some interval a g x g b then there is a constant C such that P1x P206 C Proof Considerthe difference Cx P1x 7 P206 Then Gx P x 7 Fax fx 7 fx 0 so that Cx must be constant Hence P1x 7 P206 C for some constant B It follows that there is some ambiguity in the notation f fx dx Two functions P106 and P206 can both equal f fx dx without equaling each other When this happens they P1 and P2 differ by a constant This can sometimes lead to confusing situations eg you can check that 2sinxcosxdx sinzx 2sinxcosxdx 7cos2x are both correct Just differentiate the two functions sin2 x and 7 cos2 x These two an swers look different until you realize that because of the trig identity sin2 x cos2 x 1 they really only differ by a constant sin2 x 7 cos2 x 1 To avoid this kind of confusion we will from now on never forget to include the arbitrary constant C in our answer when we compute an antiderivative 1He integrated by parts 4 Standard Integrals Here is a list of the standard derivatives and hence the standard integrals everyone should know fxdx Px c xn1 xquotdxn1C foralln7 71 1dx1nxlc x sinxdx7cosxC cosxdxsinxC tanxdx 7lncosxC 1 1x2 dx arctanxC 1 7T idxarcsinxC 77arccosxC 17x2 2 dx 1 1sinx 7T 7T cosx7 n17sinxc for73ltxlt3 All of these integrals are farniliar from first sernester calculus like Math 221 except for the last one You can check the last one by differentiation using In na 7 lnb simplifies things a bit 5 Method of substitution The chain rule says that Lg 7 New Gm so that r cx G xdx PGx 0 4 51 Example Consider the function fx 2x sinx2 3 It does not appear in the list of standard integrals we know by heart But we do notice2 that 2x 062 3 So let s call Cx 32 3 and H14 7 cos u then PGx 7 cosx2 3 and dFGx 2 T ismx 3 E 7 fX mom 01 sothat 2x sinx2 3 dx 7 cosx2 3 C 2 11 39 39 39 quot1 ma n r A 39 I I The most transparent way of computing an integral by substitution is by introducing new Variables Thus to do the integral fGxG xdx where fu Pu we introduce the substitution u C36 and agree to write du dGx Gx dx Then we get fcxc x dx fudu P04 c At the end of the integration we must remember that u really stands for Cx so that fcxc xdx r04 c Hem c For definite integrals this implies AbfGxG xdx RM 7 New which you can also write as GO 3 ffltcltxgtgtdltxgtdx mom GUI 4 52 Substitution in a de nite integral As an example we compute 1 del 0 1x2 using the substitution u Cx 1 x2 Since du 2x dx the associated inde nite integra 1s 1 1 72xdxidu 3 x J v u T du To find the definite integral you must compute the new integration bounds GO and G1 see equation 3 lfx runs between x O and x 1 then u Cx 1 362 runs between 14 1 1 and u 1 12 2 so the definite integral we must compute is 1 x 21 76117 1 A1x2x 2114 which is in our list of memorable integrals So we find 1 x 21 2 7 1 7 1 1 0 1x2 dx 21 udu 2lnu1 zln2 6 The double angle trick If an integral contains sin2 x or cos2 x then you can remove the squares by using the double angle formulas from trigonometry Recall that cos2 a 7 sin2 a cos 2a and cos2 a sin2 a 1 Adding these two equations gives 2 cos a cosZa 1 1 E
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