MATH 222 Second Semester Calculus. Fall 2015

Transcription

1 MATH Second Semester Clculus Fll 5 Typeset:August, 5

2 Mth nd Semester Clculus Lecture notes version. (Fll 5) This is self contined set of lecture notes for Mth. The notes were written by Sigurd Angenent, s prt ot the MIU clculus project. Some problems were contributed by A.Miller. The L A TEX files, s well s the Inkscpe nd Octve files which were used to produce these notes re vilble t the following web site ngenent/miu-clculus They re ment to be freely vilble for non-commercil use, in the sense tht free softwre is free. More precisely: Copyright (c) Sigurd B. Angenent. Permission is grnted to copy, distribute nd/or modify this document under the terms of the GNU Free Documenttion License, Version. or ny lter version published by the Free Softwre Foundtion; with no Invrint Sections, no Front-Cover Texts, nd no Bck-Cover Texts. A copy of the license is included in the section entitled GNU Free Documenttion License.

3 Contents Chpter. Methods of Integrtion 9. Definite nd indefinite integrls 9.. Definition 9. Problems 3. First trick: using the double ngle formuls 3.. The double ngle formuls 4. Problems 5. Integrtion by Prts The product rule nd integrtion by prts An Exmple Integrting by prts once Another exmple Exmple Repeted Integrtion by Prts Another exmple of repeted integrtion by prts Exmple sometimes the fctor G (x) is invisible An exmple where we get the originl integrl bck 6 6. Reduction Formuls First exmple of reduction formul Reduction formul requiring two prtil integrtions A reduction formul where you hve to solve for I n A reduction formul tht will come in hndy lter 9 7. Problems 8. Prtil Frction Expnsion 8.. Reduce to proper rtionl function 8.. Exmple 8.3. Prtil Frction Expnsion: The Esy Cse 8.4. Previous Exmple continued Prtil Frction Expnsion: The Generl Cse Exmple A complicted exmple After the prtil frction decomposition 5 9. Problems 6. Substitutions for integrls contining the expression x + bx + c 7.. Integrls involving x + b 8.. Integrls contining x 9.3. Integrls contining x 3.4. Integrls contining x or + x 3. Rtionl substitution for integrls contining x or + x 3.. The functions U(t) nd V (t) 3. Simplifying x + bx + c by completing the squre 35.. Exmple 35.. Exmple Exmple Problems Chpter summry Mixed Integrtion Problems 38 Chpter. Proper nd Improper Integrls 4 3

4 4 CONTENTS. Typicl exmples of improper integrls 4.. Integrl on n unbounded intervl 4.. Second exmple on n unbounded intervl 4.3. An improper integrl on finite intervl A doubly improper integrl Another doubly improper integrl 44. Summry: how to compute n improper integrl 44.. How to compute n improper integrl on n unbounded intervl 44.. How to compute n improper integrl of n unbounded function Doubly improper integrls More exmples Are under n exponentil Improper integrls involving x p Problems Estimting improper integrls Improper integrls of positive functions Comprison Theorem for Improper Integrls The Til Theorem Problems 55 Chpter 3. First order differentil Equtions 57. Wht is Differentil Eqution? 57. Two bsic exmples 57.. Equtions where the RHS does not contin y 57.. The exponentil growth exmple Summry First Order Seprble Equtions Solution method for seprble equtions A sng: Exmple Exmple: the sng in ction 6 4. Problems 6 5. First Order Liner Equtions The Integrting Fctor An exmple 6 6. Problems Direction Fields Euler s method The ide behind the method Setting up the computtion Problems 67. Applictions of Differentil Equtions 67.. Exmple: crbon dting 67.. Exmple: dting leky bucket Het trnsfer Mixing problems 7. Problems 7 Chpter 4. Tylor s Formul 73. Tylor Polynomils 73.. Definition 73.. Theorem 73. Exmples 74.. Tylor polynomils of order zero nd one 74.. Exmple: Compute the Tylor polynomils of degree, nd of f(x) = e x t =, nd plot them Exmple: Find the Tylor polynomils of f(x) = sin x 76

5 CONTENTS 5.4. Exmple: compute the Tylor polynomils of degree two nd three of f(x) = + x + x + x 3 t = Some specil Tylor polynomils Problems The Reminder Term Definition Exmple An unusul exmple, in which there is simple formul for R nf(x) Another unusul, but importnt exmple where we cn compute R nf(x) 8 6. Lgrnge s Formul for the Reminder Term Theorem Estimte of reminder term How to compute e in few deciml plces Error in the pproximtion sin x x Problems The limit s x, keeping n fixed Little-oh Theorem Definition Exmple: prove one of these little-oh rules Cn we see tht x 3 = o(x ) by looking t the grphs of these functions? Exmple: Little-oh rithmetic is little funny Computtions with Tylor polynomils Theorem How NOT to compute the Tylor polynomil of degree of f(x) = /( + x ) A much esier pproch to finding the Tylor polynomil of ny degree of f(x) = /( + x ) Exmple of multipliction of Tylor polynomils Tylor s formul nd Fiboncci numbers More bout the Fiboncci numbers 9 9. Problems 9. Differentiting nd Integrting Tylor polynomils 9.. Theorem 9.. Exmple: Tylor polynomil of ( x) Exmple: Tylor polynomils of rctn x 93. Problems on Integrls nd Tylor Expnsions 93. Proof of Theorem Lemm Proof of Lgrnge s formul for the reminder 95 Chpter 5. Sequences nd Series 97. Introduction 97.. A different point of view on Tylor expnsions 97.. Some sums with infinitely mny terms 97. Sequences 99.. Exmples of sequences 99.. Definition Exmple: lim n n = Exmple: lim n n = if <.5. Theorem.6. Sndwich theorem.7. Theorem.8. Exmple.9. Limits of rtionl functions.. Exmple. Appliction of the Sndwich theorem... Exmple: fctoril bets ny exponentil 3. Problems on Limits of Sequences 4. Series

6 6 CONTENTS 4.. Definitions 4.. Exmples Properties of series Theorem 4 5. Convergence of Tylor Series The Geometric series converges for < x < Convergence of the exponentil Tylor series The dy tht ll Chemistry stood still 6 6. Problems on Convergence of Tylor Series 7 7. Leibniz formuls for ln nd π/ Problems 9 Chpter 6. Vectors. Introduction to vectors.. Definition.. Bsic rithmetic of vectors.3. Some GOOD exmples..4. Two very, very, BAD exmples..5. Algebric properties of vector ddition nd multipliction.6. Exmple 3. Geometric description of vectors 3.. Definition 4.. Exmple 5.3. Exmple 5.4. Geometric interprettion of vector ddition nd multipliction 5.5. Exmple 6 3. Prmetric equtions for lines nd plnes Exmple Midpoint of line segment Vector Bses The Stndrd Bsis Vectors A Bsis of Vectors (in generl)* Definition Definition 9 5. Dot Product Definition Algebric properties of the dot product Exmple The digonls of prllelogrm Theorem 5.6. The dot product nd the ngle between two vectors 5.7. Theorem 5.8. Orthogonl projection of one vector onto nother 5.9. Defining equtions of lines 5.. Line through one point nd perpendiculr to nother line 5.. Distnce to line Defining eqution of plne Exmple Exmple continued Where does the line through the points B(,, ) nd C(,, ) intersect the plne P from exmple 5.3? 6 6. Cross Product Algebric definition of the cross product Definition Exmple Exmple Algebric properties of the cross product Wys to compute the cross product 8

7 CONTENTS Exmple The triple product nd determinnts Definition Theorem Geometric description of the cross product Theorem Theorem 3 7. A few pplictions of the cross product Are of prllelogrm Exmple Finding the norml to plne Exmple Volume of prllelepiped Nottion 34 Common buses of nottion tht should be voided Problems Computing nd drwing vectors 36. Problems Prmetric equtions for line 38. Problems Orthogonl decomposition of one vector with respect to nother 38. Problems The dot product Problems The cross product 39 Chpter 7. Answers, Solutions, nd Hints 4 Chpter 8. GNU Free Documenttion License 59

8 8 CONTENTS f(x) = df (x) f(x) = F (x) + C (n + )x n = n+ x n = xn+ n + + C n d ln x = x = ln x + C x bsolute vlues!! e x = dex sin x = d cos x cos x = d sin x d ln cos x tn x = e x = e x + C sin x = cos x + C cos x = sin x + C tn x = ln cos x + C bsolute vlues!! + x = d rctn x = d rcsin x x f(x) + g(x) = cf(x) = df (x) + G(x) d cf (x) F dg = df G df G = rctn x + C + x = rcsin x + C x {f(x) + g(x)} = F (x) + G(x) + C cf(x) = cf (x) + C F G = F G F G To find derivtives nd integrls involving x insted of e x use = e ln, nd thus x = e x ln, to rewrite ll exponentils s e.... The following integrl is lso useful, but not s importnt s the ones bove: cos x = + sin x ln + C for cos x. sin x Tble. The list of the stndrd integrls everyone should know

9 CHAPTER Methods of Integrtion The bsic question tht this chpter ddresses is how to compute integrls, i.e. Given function y = f(x) how do we find function y = F (x) whose derivtive is F (x) = f(x)? The simplest solution to this problem is to look it up on the Internet. Any integrl tht we compute in this chpter cn be found by typing it into the following web pge: Other similr websites exist, nd more extensive softwre pckges re vilble. It is therefore nturl to sk why should we lern how to do these integrls? The question hs t lest two nswers. First, there re certin bsic integrls tht show up frequently nd tht re reltively esy to do (once we know the trick), but tht re not included in first semester clculus course for lck of time. Knowing these integrls is useful in the sme wy tht knowing things like + 3 = 5 sves us from lot of unnecessry clcultor use. f(t) Input signl f(t) Electronic Circuit Output signl t g(t) = e τ t f(τ) dτ g(t) t t The second reson is tht we often re not relly interested in specific integrls, but in generl fcts bout integrls. For exmple, the output g(t) of n electric circuit (or mechnicl system, or biochemicl system, etc.) is often given by some integrl involving the input f(t). The methods of integrtion tht we will see in this chpter give us the tools we need to understnd why some integrl gives the right nswer to given electric circuits problem, no mtter wht the input f(t) is.. Definite nd indefinite integrls We recll some fcts bout integrtion from first semester clculus... Definition. A function y = F (x) is clled n ntiderivtive of nother function y = f(x) if F (x) = f(x) for ll x. For instnce, F (x) = x is n ntiderivtive of f(x) = x, nd so is G(x) = x +. The Fundmentl Theorem of Clculus sttes tht if function y = f(x) is continuous on n intervl x b, then there lwys exists n ntiderivtive F (x) of f, 9

10 . METHODS OF INTEGRATION Indefinite integrl Definite integrl f(x) is function of x. b f(x) is number. By definition f(x) is ny function F (x) whose derivtive is f(x). If F (x) is n ntiderivtive of f(x), then so is F (x) + C. Therefore f(x) = F (x) + C; n indefinite integrl contins constnt ( +C ). x is not dummy vrible, for exmple, x = x + C nd tdt = t + C re functions of different vribles, so they re not equl. (See Problem..) b f(x) ws defined in terms of Riemnn sums nd cn be interpreted s re under the grph of y = f(x) when f(x). b f(x) is one uniquely defined number; n indefinite integrl does not contin n rbitrry constnt. x is dummy vrible, for exmple, x =, nd tdt =, so x = tdt. Whether we use x or t the integrl mkes no difference. Tble. Importnt differences between definite nd indefinite integrls nd one hs () b f(x) = F (b) F (). For exmple, if f(x) = x, then F (x) = x is n ntiderivtive for f(x), nd thus b x = F (b) F () = b. The best wy of computing n integrl is often to find n ntiderivtive F of the given function f, nd then to use the Fundmentl Theorem (). How to go bout finding n ntiderivtive F for some given function f is the subject of this chpter. The following nottion is commonly used for ntiderivtives: () F (x) = f(x). The integrl tht ppers here does not hve the integrtion bounds nd b. It is clled n indefinite integrl, s opposed to the integrl in () which is clled definite integrl. It is importnt to distinguish between the two kinds of integrls. Tble lists the min differences.

11 3. FIRST TRICK: USING THE DOUBLE ANGLE FORMULAS. Problems. Compute the following integrls: () A = x, (b) B = t dt, (c) C = x dt, (d) I = xt dt, (e) J = xt.. One of the following three integrls is not the sme s the other two: A = B = C = x, t dt, x dt. Which one? Explin your nswer. 3. Which of the following inequlities re true? () 4 4 ( x ) > (b) ( x )dt > (c) ( + x ) > 4. One of the following sttements is correct. Which one, nd why? x () t dt = 3 x3. (b) t dt = 3 x3. (c) t dt = 3 x3 + C. 3. First trick: using the double ngle formuls The first method of integrtion we see in this chpter uses trigonometric identities to rewrite functions in form tht is esier to integrte. This prticulr trick is useful in certin integrls involving trigonometric functions nd while these integrls show up frequently, the double ngle trick is not generl method for integrtion. 3.. The double ngle formuls. The simplest of the trigonometric identities re the double ngle formuls. These cn be used to simplify integrls contining either sin x or cos x. Recll tht cos α sin α = cos α nd cos α + sin α =, Adding these two equtions gives while subtrcting them gives cos α = (cos α + ) sin α = ( cos α). These re the two double ngle formuls tht we will use Exmple. The following integrl shows up in mny contexts, so it is worth knowing: cos x = ( + cos x) = {x + } sin x + C = x + sin x + C. 4

12 . METHODS OF INTEGRATION Since sin x = sin x cos x this result cn lso be written s cos x = x + sin x cos x + C A more complicted exmple. If we need to find I = cos 4 x then we cn use the double ngle trick once to rewrite cos x s ( + cos x), which results in { I = cos 4 x = ( + cos x)} ( = + cos x + cos x ). 4 The first two terms re esily integrted, nd now tht we know the double ngle trick we lso cn do the third term. We find cos x = ( ) x + cos 4x = + sin 4x + C. 8 Going bck to the integrl I we get ( + cos x + cos x ) I = 4 = x sin x + (x sin 4x) + C = 3x 8 + sin x + sin 4x + C Exmple without the double ngle trick. The integrl J = cos 3 x looks very much like the two previous exmples, but there is very different trick tht will give us the nswer. Nmely, substitute u = sin x. Then du = cos x, nd cos x = sin x = u, so tht J = cos x cos x = ( sin x) cos x = ( u ) du = u 3 u3 + C = sin x 3 sin3 x + C. In summry, the double ngle formuls re useful for certin integrls involving powers of sin( ) or cos( ), but not ll. In ddition to the double ngle identities there re other trigonometric identities tht cn be used to find certin integrls. See the exercises. 4. Problems

13 4. PROBLEMS 3 Compute the following integrls using the double ngle formuls if necessry:. ( + sin θ) dθ.. (cos θ + sin θ) dθ. 3. Find sin x cos x (hint: use the other double ngle formul sin α = sin α cos α.) 4. cos 5 θ dθ 5. Find ( sin θ + cos θ) dθ The double ngle formuls re specil cses of the following trig identities: sin A sin B = cos(a B) cos(a + B) cos A cos B = cos(a B) + cos(a + B) sin A cos B = sin(a + B) + sin(a B) Use these identities to compute the following integrls. 6. sin x sin x. The input signl for given electronic circuit is function of time V in(t). The output signl is given by V out(t) = t sin(t s)v in(s) ds. Find V out(t) if V in(t) = sin(t) where > is some constnt. 3. The lternting electric voltge coming out of socket in ny Americn living room is sid to be Volts nd 5Herz (or 6, depending on where you re). This mens tht the voltge is function of time of the form V (t) = A sin(π t T ) where T = sec is how long one oscilltion tkes (if the frequency is 5 Herz, then 5 there re 5 oscilltions per second), nd A is the mplitude (the lrgest voltge during ny oscilltion). V(t) T T 3T 4T 5T 6T A=mplitude t 7. π sin 3x sin x (sin θ cos 3θ ) dθ. π/ π ( sin θ + sin 4θ ) dθ. sin kx sin mx where k nd m re constnt positive integers. Simplify your nswer! (creful: fter working out your solution, check if you didn t divide by zero nywhere.). Let be positive constnt nd I = π/ () Find I if. sin(θ) cos(θ) dθ. (b) Find I if =. (Don t divide by zero.) V RMS The Volts tht is specified is not the mplitude A of the oscilltion, but insted it refers to the Root Men Squre of the voltge. By definition the R.M.S. of the oscillting voltge V (t) is T = V (t) T dt. (it is the squre root of the men of the squre of V (t)). Compute the mplitude A.

14 4. METHODS OF INTEGRATION 5. Integrtion by Prts While the double ngle trick is just tht, (useful) trick, the method of integrtion by prts is very generl nd ppers in mny different forms. It is the integrtion counterprt of the product rule for differentition. 5.. The product rule nd integrtion by prts. Recll tht the product rule sys tht df (x)g(x) df (x) = G(x) + F (x)dg(x) nd therefore, fter rerrnging terms, F (x) dg(x) = df (x)g(x) df (x) G(x). If we integrte both sides we get the formul for integrtion by prts F (x) dg(x) df (x) = F (x)g(x) G(x). Note tht the effect of integrtion by prts is to integrte one prt of the function (G (x) got replced by G(x)) nd to differentite the other prt (F (x) got replced by F (x)). For ny given integrl there re mny wys of choosing F nd G, nd it not lwys esy to see wht the best choice is. 5.. An Exmple Integrting by prts once. Consider the problem of finding I = xe x. We cn use integrtion by prts s follows: }{{} x }{{} e x = }{{} x }{{} e x F (x) G (x) F (x) G(x) }{{}}{{} = xe x e x + C. e x G(x) F (x) Observe tht in this exmple e x ws esy to integrte, while the fctor x becomes n esier function when you differentite it. This is the usul stte of ffirs when integrtion by prts works: differentiting one of the fctors (F (x)) should simplify the integrl, while integrting the other (G (x)) should not complicte things (too much) Another exmple. Wht is x sin x? d( cos x) Since sin x = we cn integrte by prts }{{} x sin }{{} x = }{{} x ( cos x) }{{}}{{} F (x) G (x) F (x) G(x) F (x) ( cos x) = x cos x + sin x + C. }{{} G(x)

15 5. INTEGRATION BY PARTS Exmple Repeted Integrtion by Prts. Let s try to compute I = x e x by integrting by prts. Since e x = d ex one hs (3) }{{} x }{{} e x = x ex e x x = x e x e x x. F (x) G (x) To do the integrl on the left we hve to integrte by prts gin: e x x = ex }{{} x }{{} ex }{{}. = F (x) }{{} xex e x = xex 4 ex +C. F (x) G(x) G(x) Combining this with (3) we get x e x = x e x xex + 4 ex C (Be creful with ll the minus signs tht pper when integrting by prts.) 5.5. Another exmple of repeted integrtion by prts. The sme procedure s in the previous exmple will work whenever we hve to integrte P (x)e x where P (x) is ny polynomil, nd is constnt. Every time we integrte by prts, we get this P (x) }{{}}{{} e x = P (x) ex e x P (x) F (x) G (x) = P (x)ex P (x)e x. We hve replced the integrl P (x)e x with the integrl P (x)e x. This is the sme kind of integrl, but it is little esier since the degree of the derivtive P (x) is less thn the degree of P (x) Exmple sometimes the fctor G (x) is invisible. Here is how we cn get the ntiderivtive of ln x by integrting by prts: ln x = ln x }{{} }{{} F (x) G (x) = ln x x x x = x ln x = x ln x x + C. We cn do P (x) ln x in the sme wy if P (x) is ny polynomil. For instnce, to compute (z + z) ln z dz

16 6. METHODS OF INTEGRATION we integrte by prts: (z + z) }{{}}{{} ln z G (z) F (z) dz = ( ( 3 z3 + z) ln z 3 z3 + z) z dz = ( ( 3 z3 + z) ln z 3 z + z) dz = ( 3 z3 + z) ln z 9 z3 4 z + C An exmple where we get the originl integrl bck. It cn hppen tht fter integrting by prts few times the integrl we get is the sme s the one we strted with. When this hppens we hve found n eqution for the integrl, which we cn then try to solve. The stndrd exmple in which this hppens is the integrl I = e x sin x. We integrte by prts twice: }{{} e x sin }{{ x } = e x sin x }{{} e x } cos {{ x } F (x) G(x) F (x) G (x) = e x sin x e x cos x = e x sin x e x cos x e x sin x = e x sin x e x cos x 4 e x sin x. Note tht the lst integrl here is exctly I gin. Therefore the integrl I stisfies We solve this eqution for I, with result I = e x sin x e x cos x 4I. 5I = e x sin x e x cos x = I = 5( e x sin x e x cos x ). Since I is n indefinite integrl we still hve to dd the rbitrry constnt: I = 5( e x sin x e x cos x ) + C. 6. Reduction Formuls We hve seen tht we cn compute integrls by integrting by prts, nd tht we sometimes hve to integrte by prts more thn once to get the nswer. There re integrls where we hve to integrte by prts not once, not twice, but n-times before the nswer shows up. To do such integrls it is useful to crefully describe wht hppens ech time we integrte by prts before we do the ctul integrtions. The formul tht describes wht hppens fter one prtil integrtion is clled reduction formul. All this is best explined by n exmple.

17 6. REDUCTION FORMULAS First exmple of reduction formul. Consider the integrl I n = x n e x, (n =,,, 3,...) or, in other words, consider ll the integrls I = e x, I = xe x, I = x e x, I 3 = x 3 e x,... nd so on. We will consider ll these integrls t the sme time. Integrtion by prts in I n gives us I n = }{{} x n }{{} e x F (x) G (x) = x n ex nx n ex = xn e x n x n e x. We hven t computed the integrl, nd in fct the integrl tht we still hve to do is of the sme kind s the one we strted with (integrl of x n e x insted of x n e x ). Wht we hve derived is the following reduction formul I n = xn e x n I n, which holds for ll n. For n = we do not need the reduction formul to find the integrl. We hve I = e x = ex + C. When n the reduction formul tells us tht we hve to compute I n if we wnt to find I n. The point of reduction formul is tht the sme formul lso pplies to I n, nd I n, etc., so tht fter repeted ppliction of the formul we end up with I, i.e., n integrl we know. For exmple, if we wnt to compute x 3 e x we use the reduction formul three times: I 3 = x3 e x 3 I = x3 e x 3 { x e x } I = x3 e x 3 { x e x ( xex )} I Insert the known integrl I = ex + C nd simplify the other terms nd we get x 3 e x = x3 e x 3 x e x xex 6 4 ex + C.

18 8. METHODS OF INTEGRATION 6.. Reduction formul requiring two prtil integrtions. Consider S n = x n sin x. Then for n one hs S n = x n cos x + n x n cos x = x n cos x + nx n sin x n(n ) x n sin x. Thus we find the reduction formul S n = x n cos x + nx n sin x n(n )S n. Ech time we use this reduction, the exponent n drops by, so in the end we get either S or S, depending on whether we strted with n odd or even n. These two integrls re S = sin x = cos x + C S = x sin x = x cos x + sin x + C. (Integrte by prts once to find S.) As n exmple of how to use the reduction formuls for S n let s try to compute S 4 : x 4 sin x = S 4 = x 4 cos x + 4x 3 sin x 4 3S = x 4 cos x + 4x 3 sin x 4 3 { x cos x + x sin x S } At this point we use S = sin x = cos x + C, nd we combine like terms. This results in x 4 sin x = x 4 cos x + 4x 3 sin x 4 3 { x cos x + x sin x ( cos x) } + C = ( x 4 + x 4 ) cos x + ( 4x 3 + 4x ) sin x + C A reduction formul where you hve to solve for I n. We try to compute I n = (sin x) n by reduction formul. Integrting by prts we get I n = (sin x) n sin x = (sin x) n cos x ( cos x)(n )(sin x) n cos x = (sin x) n cos x + (n ) (sin x) n cos x.

19 6. REDUCTION FORMULAS 9 We now use cos x = sin x, which gives {sin I n = (sin x) n cos x + (n ) n x sin n x } = (sin x) n cos x + (n )I n (n )I n. We cn think of this s n eqution for I n, which, when we solve it tells us ni n = (sin x) n cos x + (n )I n nd thus implies (4) I n = n sinn x cos x + n n Since we know the integrls I = (sin x) = = x + C I n. nd I = sin x = cos x + C the reduction formul (4) llows us to clculte I n for ny n A reduction formul tht will come in hndy lter. In the next section we will see how the integrl of ny rtionl function cn be trnsformed into integrls of esier functions, the most difficult of which turns out to be I n = ( + x ) n. When n = this is stndrd integrl, nmely I = = rctn x + C. + x When n > integrtion by prts gives us reduction formul. Here s the computtion: I n = ( + x ) n x = ( + x ) n x ( n) ( + x ) n x x = ( + x ) n + n x ( + x ) n+ Apply to get x ( + x ) n+ = ( + x ) ( + x ) n+ = ( + x ) n ( + x ) n+ x { } ( + x = ) n+ ( + x ) n ( + x ) n+ Our integrtion by prts therefore told us tht x I n = ( + x ) n + n( ) I n I n+, = I n I n+.

20 . METHODS OF INTEGRATION which we cn solve for I n+. We find the reduction formul I n+ = x n ( + x ) n + n n I n. As n exmple of how we cn use it, we strt with I = rctn x + C, nd conclude tht ( + x ) = I = I + = x ( + x ) + I = x + x + rctn x + C. Apply the reduction formul gin, now with n =, nd we get ( + x ) 3 = I 3 = I + = x ( + x ) + I { } = x 4 ( + x ) + 3 x 4 + x + rctn x = x 4 ( + x ) + 3 x 8 + x rctn x + C. 7. Problems. Evlute x n ln x where n.. Assume nd b re constnts, nd compute e x sin bx. [Hint: Integrte by prts twice; you cn ssume tht b.] 3. Evlute e x cos bx where, b. 4. Prove the formul x n e x = x n e x n nd use it to evlute x e x. x n e x 5. Use 6.3 to evlute sin x. Show tht the nswer is the sme s the nswer you get using the hlf ngle formul. 6. Use the reduction formul in 6.3 to compute π/ sin 4 x. 7. In this problem you ll look t the numbers A n = π sin n x. () Check tht A = π nd A =. (b) Use the reduction formul in 6.3 to compute A 5, A 6, nd A 7. (c) Explin why A n < A n is true for ll n =,, 3, 4,... (Hint: Interpret the integrls A n s re under the grph, nd check tht (sin x) n (sin x) n for ll x.) (d) Bsed on your vlues for A 5, A 6, nd A 7 find two frctions nd b such tht < π < b. 8. Prove the formul cos n x = n sin x cosn x + n cos n x, n

21 for n, nd use it to evlute π/4 9. Prove the formul x m (ln x) n = cos 4 x. 8. PARTIAL FRACTION EXPANSION x m+ (ln x) n m + n x m (ln x) n, m + for m, nd use it to evlute the following integrls:. ln x.. (ln x) x 3 (ln x) 3. Evlute x ln x by nother method. [Hint: the solution is short!] 4. For ny integer n > derive the formul tn n x = tnn x n tn n x Using this, find π/4 tn 5 x. Use the reduction formul from exmple 6.4 to compute these integrls: esy.] ( + x ) 3 ( + x ) 4 x ( + x ) 4 [Hint: x/( + x ) n is + x ( + x ) ( 49 + x ) 3.. The reduction formul from exmple 6.4 is vlid for ll n. In prticulr, n does not hve to be n integer, nd it does not hve to be positive. Find reltion between + x nd by setting + x n =.. Apply integrtion by prts to x nd dv =. This gives us, Let u = x du = nd v = x. x x = ( x )(x) x x Simplifying x = + x nd subtrcting the integrl from both sides gives us =. How cn this be? 8. Prtil Frction Expnsion By definition, rtionl function is quotient ( rtio) of polynomils, f(x) = P (x) Q(x) = p nx n + p n x n + + p x + p q d x d + q d x d + + q x + q. Such rtionl functions cn lwys be integrted, nd the trick tht llows you to do this is clled prtil frction expnsion. The whole procedure consists of severl steps tht re explined in this section. The procedure itself hs nothing to do with integrtion: it s just wy of rewriting rtionl functions. It is in fct useful in other situtions, such s finding Tylor expnsions (see Chpter 4) nd computing inverse Lplce trnsforms (see Mth 39.) 8.. Reduce to proper rtionl function. A proper rtionl function is rtionl function P (x)/q(x) where the degree of P (x) is strictly less thn the degree of Q(x). The method of prtil frctions only pplies to proper rtionl functions. Fortuntely there s n dditionl trick for deling with rtionl functions tht re not proper.

22 . METHODS OF INTEGRATION If P/Q isn t proper, i.e. if degree(p ) degree(q), then you divide P by Q, with result P (x) R(x) = S(x) + Q(x) Q(x) where S(x) is the quotient, nd R(x) is the reminder fter division. In prctice you would do long division to find S(x) nd R(x). 8.. Exmple. Consider the rtionl function f(x) = x3 x + x. Here the numertor hs degree 3 which is more thn the degree of the denomintor (which is ). The function f(x) is therefore not proper rtionl function. To pply the method of prtil frctions we must first do division with reminder. One hs x = S(x) so tht x x 3 x+ x 3 x x+ = R(x) = x + x + x f(x) = x3 x + x When we integrte we get x 3 { x + x = x + x + x = x + } x + x. The rtionl function tht we still hve to integrte, nmely x+ x, is proper: its numertor hs lower degree thn its denomintor Prtil Frction Expnsion: The Esy Cse. To compute the prtil frction expnsion of proper rtionl function P (x)/q(x) you must fctor the denomintor Q(x). Fctoring the denomintor is problem s difficult s finding ll of its roots; in Mth we shll only do problems where the denomintor is lredy fctored into liner nd qudrtic fctors, or where this fctoriztion is esy to find. In the esiest prtil frctions problems, ll the roots of Q(x) re rel numbers nd distinct, so the denomintor is fctored into distinct liner fctors, sy P (x) Q(x) = P (x) (x )(x ) (x n ). To integrte this function we find constnts A, A,..., A n so tht P (x) Q(x) = A + A + + A n. (#) x x x n Then the integrl is P (x) Q(x) = A ln x + A ln x + + A n ln x n + C. One wy to find the coefficients A i in (#) is clled the method of equting coefficients. In this method we multiply both sides of (#) with Q(x) = (x ) (x n ).

23 8. PARTIAL FRACTION EXPANSION 3 The result is polynomil of degree n on both sides. Equting the coefficients of these polynomil gives system of n liner equtions for A,..., A n. You get the A i by solving tht system of equtions. Another much fster wy to find the coefficients A i is the Heviside trick. Multiply eqution (#) by x i nd then plug in x = i. On the right you re left with A i so A i = P (x)(x i) Q(x) = x=i P ( i ) ( i ) ( i i )( i i+ ) ( i n ) Previous Exmple continued. To integrte x + x we fctor the denomintor, x = (x )(x + ). The prtil frction expnsion of x + x then is x + (5) x = x + (x )(x + ) = A x + B x +. Multiply with (x )(x + ) to get x + = A(x + ) + B(x ) = (A + B)x + (A B). The functions of x on the left nd right re equl only if the coefficient of x nd the constnt term re equl. In other words we must hve A + B = nd A B =. These re two liner equtions for two unknowns A nd B, which we now proceed to solve. Adding both equtions gives A =, so tht A = ; from the first eqution one then finds B = A = 3. So x + x = / x 3/ x +. Insted, we could lso use the Heviside trick: multiply (5) with x to get x + x + = A + B x x + Tke the limit x nd you find + + = A, i.e. A =. Similrly, fter multiplying (5) with x + one gets nd letting x you find s before. x + x = Ax + x + B, B = ( ) + ( ) = 3, Nmed fter Oliver Heviside, physicist nd electricl engineer in the lte 9 th nd erly th century. More properly, you should tke the limit x i. The problem here is tht eqution (#) hs x i in the denomintor, so tht it does not hold for x = i. Therefore you cnnot set x equl to i in ny eqution derived from (#). But you cn tke the limit x i, which in prctice is just s good.

24 4. METHODS OF INTEGRATION Either wy, the integrl is now esily found, nmely, x 3 x + x = x x + + x = x + { / x 3/ x + } = x + ln x 3 ln x + + C Prtil Frction Expnsion: The Generl Cse. When the denomintor Q(x) contins repeted fctors or qudrtic fctors (or both) the prtil frction decomposition is more complicted. In the most generl cse the denomintor Q(x) cn be fctored in the form (6) Q(x) = (x ) k (x n ) kn (x + b x + c ) l (x + b m x + c m ) lm Here we ssume tht the fctors x,..., x n re ll different, nd we lso ssume tht the fctors x + b x + c,..., x + b m x + c m re ll different. It is theorem from dvnced lgebr tht you cn lwys write the rtionl function P (x)/q(x) s sum of terms like this (7) P (x) Q(x) = + A (x i ) k + + Bx + C (x + b j x + c j ) l + How did this sum come bout? For ech liner fctor (x ) k in the denomintor (6) you get terms A x + A (x ) + + A k (x ) k in the decomposition. There re s mny terms s the exponent of the liner fctor tht generted them. For ech qudrtic fctor (x + bx + c) l you get terms B x + C x + bx + c + B x + C (x + bx + c) + + B mx + C m (x + bx + c) l. Agin, there re s mny terms s the exponent l with which the qudrtic fctor ppers in the denomintor (6). In generl, you find the constnts A..., B... nd C... by the method of equting coefficients. Unfortuntely, in the presence of qudrtic fctors or repeted liner fctors the Heviside trick does not give the whole nswer; we relly hve to use the method of equting coefficients. The workings of this method re best explined in n exmple Exmple. Find the prtil frction decomposition of nd compute f(x) = x + x (x + ) I = x + x (x + ).

25 8. PARTIAL FRACTION EXPANSION 5 The degree of the denomintor x (x + ) is four, so our prtil frction decomposition must lso contin four undetermined constnts. The expnsion should be of the form x + x (x + ) = A x + B x + Cx + D x +. To find the coefficients A, B, C, D we multiply both sides with x ( + x ), x + = Ax(x + ) + B(x + ) + x (Cx + D) x + = (A + C)x 3 + (B + D)x + Ax + B x 3 + x + x + = (A + C)x 3 + (B + D)x + Ax + B Compring terms with the sme power of x we find tht A + C =, B + D =, A =, B =. These re four equtions for four unknowns. Fortuntely for us they re not very difficult in this exmple. We find A =, B =, C = A =, nd D = B =, whence The integrl is therefore f(x) = x + x (x + ) = x x +. I = x + x (x + ) = rctn x + C. x 8.7. A complicted exmple. Find the integrl x + 3 x (x + )(x + ) 3. The procedure is exctly the sme s in the previous exmple. We hve to expnd the integrnd in prtil frctions: (8) x + 3 x (x + )(x + ) 3 = A x + A x + A 3 x + + B x + C x + + B x + C (x + ) + B 3x + C 3 (x + ) 3. Note tht the degree of the denomintor x (x + )(x + ) 3 is = 9, nd lso tht the prtil frction decomposition hs nine undetermined constnts A, A, A 3, B, C, B, C, B 3, C 3. After multiplying both sides of (8) with the denomintor x (x + )(x + ) 3, expnding everything, nd then equting coefficients of powers of x on both sides, we get system of nine liner equtions in these nine unknowns. The finl step in finding the prtil frction decomposition is to solve those liner equtions. A computer progrm like Mple or Mthemtic cn do this esily, but it is lot of work to do it by hnd After the prtil frction decomposition. Once we hve the prtil frction decomposition (8) we still hve to integrte the terms tht ppered. The first three terms re of the form A(x ) p nd they re esy to integrte: A = A ln x + C x

26 6. METHODS OF INTEGRATION nd A (x ) p = A ( p)(x ) p + C if p >. The next, fourth term in (8) cn be written s B x + C x + = B x x + + C x + = B ln(x + ) + C rctn x + K, where K is the integrtion constnt (normlly +C but there re so mny other C s in this problem tht we chose different letter, just for this once.) While these integrls re lredy not very simple, the integrls Bx + C (x with p > + bx + c) p which cn pper re prticulrly unplesnt. If we relly must compute one of these, then we should first complete the squre in the denomintor so tht the integrl tkes the form Ax + B ((x + b) + ) p. After the chnge of vribles u = x + b nd fctoring out constnts we re left with the integrls du u du (u + ) p nd (u + ) p. The reduction formul from exmple 6.4 then llows us to compute this integrl. An lterntive pproch is to use complex numbers. If we llow complex numbers then the qudrtic fctors x + bx + c cn be fctored, nd our prtil frction expnsion only contins terms of the form A/(x ) p, lthough A nd cn now be complex numbers. The integrls re then esy, but the nswer hs complex numbers in it, nd rewriting the nswer in terms of rel numbers gin cn be quite involved. In this course we will void complex numbers nd therefore we will not explin this ny further. 9. Problems. Express ech of the following rtionl functions s polynomil plus proper rtionl function. (See 8. for definitions.) () x 3 x 3 4 (b) x3 + x x 3 4 (c) x3 x x 5 x 3 4 (d) x3 x. Compute the following integrls by completing the squre: () x + 6x + 8, (b) (c) x + 6x +, 5x + x Use the method of equting coefficients to find numbers A, B, C such tht x + 3 x(x + )(x ) = A x + B x + + nd then evlute the integrl x + 3 x(x + )(x ). C x

27 . SUBSTITUTIONS FOR INTEGRALS CONTAINING THE EXPRESSION x + bx + c 7 4. Do the previous problem using the Heviside trick. x Find the integrl x (x ). 6. Simplicio hd to integrte 4x (x 3)(x + ). He set 4x (x 3)(x + ) = A x 3 + B x +. Using the Heviside trick he then found A = 4x x 3 =, x= nd B = 4x x + = 9, x=3 which leds him to conclude tht 4x (x 3)(x + ) = x x +. To double check he now sets x = which leds to = 3 + 9???? Wht went wrong? Evlute the following integrls: x 4 x + x 3 x 4 + x 5 x x 5 x x 3 x x + x 3x + x + x 3x + e 3x e 4x e x + e x e x e x + e x + + e x x(x + ) x(x + ) x (x ) (x )(x )(x 3) x + (x )(x )(x 3) x 3 + (x )(x )(x 3) 4. () Compute positive number. where h is x(x h) (b) Wht hppens to your nswer to () when h? (c) Compute x.. Substitutions for integrls contining the expression x + bx + c The min method for finding ntiderivtives tht we sw in Mth is the method of substitution. This method will only let us compute n integrl if we hppen to guess the right substitution, nd guessing the right substitution is often not esy. If the integrl contins the squre root of liner or qudrtic function, then there re number of substitutions tht re known to help. Integrls with x + b: substitute x + b = u with u >. See.. Integrls with x + bx + c: first complete the squre to reduce the integrl to one contining one of the following three forms u, u, u +.

28 8. METHODS OF INTEGRATION Then, depending on which of these three cses presents itself, you choose n pproprite substitution. There re severl options: trigonometric substitution; this works well in some cses, but often you end up with n integrl contining trigonometric functions tht is still not esy (see. nd.4.). use hyperbolic functions; the hyperbolic sine nd hyperbolic cosine sometimes let you hndle cses where trig substitutions do not help. ( rtionl substitution (see ) using the two functions U(t) = ) t+t nd V (t) = ( ) t t... Integrls involving x + b. If n integrl contins the squre root of liner function, i.e. x + b then you cn remove this squre root by substituting u = x + b.... Exmple. To compute I = x x + 3 we substitute u = x + 3. Then x = (u 3) so tht = u du, nd hence I = (u 3) }{{}}{{} u }{{} u du x x+3 (u 4 3u ) du = = { 5 u5 u 3} + C. To write the ntiderivtive in terms of the originl vrible you substitute u = x + 3 gin, which leds to I = (x + 3)5/ (x + 3)3/ + C. A comment: setting u = x + b is usully the best choice, but sometimes other choices lso work. You, the reder, might wnt to try this sme exmple substituting v = x + 3 insted of the substitution we used bove. You should of course get the sme nswer.... Another exmple. Compute I = + + x.

29 . SUBSTITUTIONS FOR INTEGRALS CONTAINING THE EXPRESSION x + bx + c 9 Agin we substitute u = x +, or, u = x +. We get I = + u = x + so u du = + x u du A rtionl function: we know = + u wht to do. ( ) = du + u = u ln( + u) + C = x + ln ( + x + ) + C. Note tht u = x + is positive, so tht + x + >, nd so tht we do not need bsolute vlue signs in ln( + u)... Integrls contining x. If n integrl contins the expression x then this expression cn be removed t the expense of introducing trigonometric functions. Sometimes (but not lwys) the resulting integrl is esier. The substitution tht removes the x is x = sin θ.... Exmple. To compute note tht I = ( x ) 3/ ( x ) = 3/ ( x ) x, so we hve n integrl involving x. We set x = sin θ, nd thus = cos θ dθ. We get cos θ dθ I = ( sin θ). 3/ Use sin θ = cos θ nd you get ( sin θ) 3/ = ( cos θ ) 3/ = cos θ 3. We were forced to include the bsolute vlues here becuse of the possibility tht cos θ might be negtive. However it turns out tht cos θ > in our sitution since, in the originl integrl I the vrible x must lie between nd +: hence, if we set x = sin θ, then we my ssume tht π < θ < π. For those θ one hs cos θ >, nd therefore we cn write ( sin θ) 3/ = cos 3 θ. After substitution our integrl thus becomes cos θ dθ I = cos 3 θ = dθ cos = tn θ + C. θ To express the ntiderivtive in terms of the originl vrible we use x x = sin θ = tn θ =. x θ x x x = sin θ x = cos θ. The finl result is I = ( x ) 3/ = x x + C.

30 3. METHODS OF INTEGRATION... Exmple: sometimes you don t hve to do trig substitution. The following integrl is very similr to the one from the previous exmple: x Ĩ = ( ). x 3/ The only difference is n extr x in the numertor. To compute this integrl you cn substitute u = x, in which cse du = x. Thus we find x ( ) = x 3/ du u = 3/ u / u 3/ du = ( /) + C = + C u = + C. x.3. Integrls contining x. If n integrl contins the expression x for some positive number, then this cn be removed by substituting either x = sin θ or x = cos θ. Since in the integrl we must hve < x <, we only need vlues of θ in the intervl ( π, π ). Thus we substitute x = sin θ, π < θ < π. For these vlues of θ we hve cos θ >, nd hence x = cos θ. 3 θ 9 x x.3.. Exmple. To find J = 9 x we substitute x = 3 sin θ. Since θ rnges between π nd + π we hve cos θ > nd thus 9 x = 9 9 sin θ = 3 sin θ = 3 cos θ = 3 cos θ = 3 cos θ. We lso hve = 3 cos θ dθ, which then leds to J = 3 cos θ 3 cos θ dθ = 9 cos θ dθ. This exmple shows tht the integrl we get fter trigonometric substitution is not lwys esy nd my still require more tricks to be computed. For this prticulr integrl we use the double ngle trick. Just s in 3 we find x = 3 sin θ 9 x = 3 cos θ. J = 9 cos θ dθ = 9 ( θ + sin θ) + C. The lst step is to undo the substitution x = 3 sin θ. There re severl strtegies: one pproch is to get rid of the double ngles gin nd write ll trigonometric expressions in terms of sin θ nd cos θ. Since θ rnges between π nd + π we hve x = 3 sin θ θ = rcsin x 3,

31 . SUBSTITUTIONS FOR INTEGRALS CONTAINING THE EXPRESSION x + bx + c 3 To substitute θ = rcsin( ) in sin θ we need double ngle formul, sin θ = sin θ cos θ = x 9 x 3 = 3 9 x 9 x. We get 9 x = 9 θ + 9 sin θ cos θ + C. = 9 rcsin x 3 + x 9 x + C..4. Integrls contining x or + x. There re trigonometric substitutions tht will remove either x or + x from n integrl. In both cses they come from the identities ( ) ( ) (9) = tn θ + or = tn θ. cos θ cos θ You cn remember these identities either by drwing right tringle with ngle θ nd with bse of length, or else by dividing both sides of the equtions = sin θ + cos θ or cos θ = sin θ by cos θ..4.. Exmple turn the integrl 4 x 4 into trigonometric integrl. Since x 4 = x we substitute cos θ θ tn θ x = tn θ + x = / cos θ y = / cos θ y = tn θ x = cos θ, which then leds to x 4 = 4 tn θ = tn θ. In this lst step we hve to be creful with the sign of the squre root: since < x < 4 in our integrl, we cn ssume tht < θ < π nd thus tht tn θ >. Therefore tn θ = tn θ insted of tn θ. The substitution x = cos θ lso implies tht = sin θ cos θ dθ. y = x 4 Are =? 4 Figure. Wht is the re of the shded region under the hyperbol? We first try to compute it using trigonometric substitution (.4.), nd then using rtionl substitution involving the U nd V functions (..). The nswer turns out to be 4 3 ln ( + 3 ).

32 3. METHODS OF INTEGRATION We finlly lso consider the integrtion bounds: nd Therefore we hve 4 x = = = = cos θ = = θ =, cos θ x = 4 = cos θ = 4 = cos θ = = θ = π 3. x 4 = π/3 tn θ sin θ π/3 cos θ dθ = 4 sin θ cos 3 θ dθ. This integrl is still not esy: it cn be done by integrtion by prts, nd you hve to know the ntiderivtive of / cos θ.. Rtionl substitution for integrls contining x or + x.. The functions U(t) nd V (t). Insted of using trigonometric substitution one cn lso use the following identity to get rid of either x or x +. The identity is reltion between two functions U nd V of new vrible t defined by () U(t) = These stisfy ( t + ) (, V (t) = t t ). t () U = V +, which one cn verify by direct substitution of the definitions () of U(t) nd V (t). To undo the substitution it is useful to note tht if U nd V re given by (), then () t = U + V, t = U V.... Exmple.4. gin. Here we compute the integrl A = 4 x 4 using the rtionl substitution (). Since the integrl contins the expression x 4 = x we substitute x = U(t). Using U = + V we then hve x 4 = 4U(t) 4 = U(t) = V (t). When we use the substitution x = U(t) we should lwys ssume tht t. Under tht ssumption we hve V (t) (see Figure ) nd therefore x 4 = V (t). To summrize, we hve (3) x = U(t), x 4 = V (t).

33 . RATIONAL SUBSTITUTION FOR INTEGRALS CONTAINING x OR + x 33 U(t) = ( t + ) t t/ V (t) = ( t ) t t U = ( t + ) t V = ( t ) t t = U + V t = U V U V = Figure. The functions U(t) nd V (t) We cn now do the indefinite integrl: x 4 = V (t) }{{} x 4 = = U (t) dt }{{} ( t t ) ( t t + t 3 ) dt = t ln t t + C ( ) t dt To finish the computtion we still hve to convert bck to the originl x vrible, nd substitute the integrtion bounds. The most strightforwrd pproch is to substitute t = U + V, nd then remember the reltions (3) between U, V, nd x. Using these reltions the middle term in the integrl we just found becomes { x ( ln t = ln(u + V ) = ln + x) }.

34 34. METHODS OF INTEGRATION We cn sve ourselves some work by tking the other two terms together nd fctoring them s follows t t = ( t ( ) ) (4) b = ( + b)( b) t = ( t + )( t ) t + t t t = x = ( x x) ( ) t t = ( x ) = x x 4. So we find x 4 = x x 4 ln { x ( + x Hence, substituting the integrtion bounds x = nd x = 4, we get A = 4 x 4 [ x x ( = x 4 ln{ + x) }] x=4 x= = 4 ( ) 6 4 ln + 3 = 4 3 ln ( + 3 ). ) } + C. ( the terms with x = vnish )... An exmple with + x. There re severl wys to compute I = + x nd unfortuntely none of them re very simple. The simplest solution is to void finding the integrl nd look it up in tble, such s Tble. But how were the integrls in tht tble found? One pproch is to use the sme pir of functions U(t) nd V (t) from (). Since U = + V the substitution x = V (t) llows us to tke the squre root of + x, nmely, x = V (t) = + x = U(t). Also, = V (t)dt = ( + t ) dt, nd thus we hve I = + x }{{} =U(t) }{{} dv (t) ( )( ) = t + + dt t t = (t + 4 t + ) dt t 3 = {t 4 + ln t } + C t = ( t ) + 8 t ln t + C.

35 . SIMPLIFYING x + bx + c BY COMPLETING THE SQUARE 35 At this point we hve done the integrl, but we should still rewrite the result in terms of the originl vrible x. We could use the sme lgebr s in (4), but this is not the only possible pproch. Insted we could lso use the reltions (), i.e. t = U + V nd t = U V These imply nd conclude t =(U + V ) =U + UV + V t =(U V ) =U UV + V t t = = 4UV I = + x = 8 ( t t ) + ln t + C = UV + ln(u + V ) + C = x + x + ln( x + + x ) + C.. Simplifying x + bx + c by completing the squre Any integrl involving n expression of the form x + bx + c cn be reduced by mens of substitution to one contining one of the three forms u, u, or u +. We cn chieve this reduction by completing the squre of the qudrtic expression under the squre root. Once the more complicted squre root x + bx + c hs been simplified to ±u ±, we cn use either trigonometric substitution, or the rtionl substitution from the previous section. In some cses the end result is one of the integrls listed in Tble : du u = rcsin u u du = u u + rcsin u du + u = ln( u + + u ) + u du = u + u + ln( u + + u ) du u = ln( u + u ) u du = u u ln( u + u ) (ll integrls +C ) Tble. Useful integrls. Except for the first one these should not be memorized. Here re three exmples. The problems hve more exmples... Exmple. Compute I = 6x x.

36 36. METHODS OF INTEGRATION Notice tht since this integrl contins squre root the vrible x my not be llowed to hve ll vlues. In fct, the quntity 6x x = x(6 x) under the squre root hs to be positive so x must lie between x = nd x = 6. We now complete the squre: 6x x = ( x 6x ) = ( x 6x ) = [ (x 3) 9 ] [ (x 3) = 9 9 [( x 3 = 9 3 At this point we decide to substitute which leds to 6x x = u = x 3, 3 ] ) ]. 9 ( u ) = 9 ( u ) = 3 u, x = 3u + 3, = 3 du. Applying this chnge of vrible to the integrl we get = 6x x 3du 3 u =.. Exmple. Compute du x 3 = rcsin u + C = rcsin + C. u 3 I = 4x + 8x + 8. We gin complete the squre in the qudrtic expression under the squre root: 4x + 8x + 8 = 4 ( x + x + ) = 4 { (x + ) + }. Thus we substitute u = x +, which implies du =, fter which we find 4x I = + 8x + 8 = u (x + ) + = + du. This lst integrl is in tble, so we hve I = u u + + ln ( u + u + ) + C = (x + ) { (x + ) + + ln x + + } (x + ) + + C..3. Exmple. Compute: We first complete the squre I = x 4x 5. x 4x 5 = x 4x = (x ) 9 u form { ( x ) } = 9 3 u form

37 4. CHAPTER SUMMARY 37 This prompts us to substitute u = x, du = 3 3, i.e. = 3 du. We get { ( x ) } I = 9 = 3 u u 3 3 du = 9 du. Using the integrls in Tble nd then undoing the substitution we find I = x 4x 5 = 9 u u 9 ln( u + u ) + C = 9 x ( x ) { 9 x ( 3 3 ln x ) } + + C 3 3 { = (x ) (x ) 9 9 ln 3 x + } (x ) 9 + C = (x ) x 4x ln{ x + x 4x + 5 } 9 ln 3 + C = (x ) x 4x ln{ x + x 4x + 5 } + C 3. Problems In ny of these integrls, is positive constnt.. x x + x x x Evlute these integrls: x 4x 4 / / 3/ 4 x 4 x x + x + x x + x + x 3x + 6x + 6 3x + 6x x +, x Chpter summry There re severl methods for finding the ntiderivtive of function. Ech of these methods llow us to trnsform given integrl into nother, hopefully simpler, integrl. Here re the methods tht were presented in this chpter, in the order in which they ppered: () Double ngle formuls nd other trig identities: some integrls cn be simplified by using trigonometric identity. This is not generl method, nd only works for certin very specific integrls. Since these integrls do come up with some frequency it is worth knowing the double ngle trick nd its vritions.

38 38. METHODS OF INTEGRATION () Integrtion by prts: very generl formul; repeted integrtion by prts is done using reduction formuls. (3) Prtil Frction Decomposition: method tht llows us to integrte ny rtionl function. (4) Trigonometric nd Rtionl Substitution: specific group of substitutions tht cn be used to simplify integrls contining the expression x + bx + c. 5. Mixed Integrtion Problems One of the chllenges in integrting function is to recognize which of the methods we know will be most useful so here is n unsorted list of integrls for prctice. Evlute these integrls: /3 /4 4 3 x sin x x cos x x x x x x x x x + x + 7 x x + x + 7 x 4 (x 36) / x 4 x 36 x 4 36 x x + x 4 x x + 3 x 4 x (x 3) / e x (x + cos(x)) (e x + ln(x)) (x + 5) x + 5x hint: x + 5 = /u x x + x + 7 x x + x + 7 x x + x + 7 x x + x + 7 3x + x x 3 x 4 x 4 6 x (x ) 3 4 (x ) 3 (x + ) 6 x 4x e e 3 e e x + x + 3 x ln x x ln(x + ) x ln x x(ln x) 3 rctn( x) x(cos x)

39 5. MIXED INTEGRATION PROBLEMS π + cos(6w) dw Hint: + cos α = cos α. + sin(x) 34. Find x(x )(x )(x 3) nd (x 3 + ) x(x )(x )(x 3) 35. Compute x 3 + x + x + (Hint: to fctor the denomintor begin with +x+x +x 3 = (+x)+x (+x) =...) 36. [Group Problem] You don t lwys hve to find the ntiderivtive to find definite integrl. This problem gives you two exmples of how you cn void finding the ntiderivtive. () To find I = π/ sin x sin x + cos x you use the substitution u = π/ x. The new integrl you get must of course be equl to the integrl I you strted with, so if you dd the old nd new integrls you get I. If you ctully do this you will see tht the sum of the old nd new integrls is very esy to compute. (b) Use your nswer from () to compute x + x. (c) Use the sme trick to find π/ sin x 37. The Astroid. Drw the curve whose eqution is x 3 + y 3 = 3, where is positive constnt. The curve you get is clled the Astroid. Compute the re bounded by this curve. 38. The Bow-Tie Grph. Drw the curve given by the eqution y = x 4 x 6. Compute the re bounded by this curve. 39. The Fn-Tiled Fish. Drw the curve given by the eqution ( ) y = x x. + x Find the re enclosed by the loop. (Hint: fter finding which integrl you need to compute, substitute x = sin θ to do the integrl.) 4. Find the re of the region bounded by the curves x =, y =, y = x ln x 4. Find the volume of the solid of revolution obtined by rotting round the x xis the region bounded by the lines x = 5, x =, y =, nd the curve x y = x How to find the integrl of f(x) = cos x. Note tht cos x = cos x cos x = cos x sin x, nd pply the substitution s = sin x followed by prtil frction decomposition to compute cos x. Clculus bloopers As you ll see, the following computtions cn t be right; but where did they go wrong? 43. Here is filed computtion of the re of this region: y y = x- Clerly the combined re of the two tringles should be. Now let s try to get this nswer by integrtion. x

40 4. METHODS OF INTEGRATION Consider x. Let f(x) = x so tht { x if x f(x) = x if x < Define F (x) = { x x if x x x if x < Then since F is n ntiderivtive of f we hve by the Fundmentl Theorem of Clculus: x = f(x) = F () F () ( ) = =. ) ( But this integrl cnnot be zero, f(x) is positive except t one point. How cn this be? 44. It turns out tht the re enclosed by circle is zero?! According to the ncient formul for the re of disc, the re of the following hlf-disc is π/. y = (-x ) / - y x We cn lso compute this re by mens of n integrl, nmely Are = Substitute u = x so: x u = x, x = u = ( u), = ( )( u) ( ) du. Hence x = u ( )( u) ( ) du. Now tke the definite integrl from x = to x = nd note tht u = when x = nd u = lso when x =, so x = ( ) u ( u) ( ) du = The lst being zero since ( nything ) =. But the integrl on the left is equl to hlf the re of the unit disc. Therefore hlf disc hs zero re, nd whole disc should hve twice s much re: still zero! How cn this be?

41 CHAPTER Proper nd Improper Integrls All the definite integrls tht we hve seen so fr were of the form I = b f(x), where nd b re finite numbers, nd where the integrnd (the function f(x)) is nice on the intervl x b, i.e. the function f(x) does not become infinite nywhere in the intervl. There re mny situtions where one would like to compute n integrl tht fils one of these conditions; i.e. integrls where or b is not finite, or where the integrnd f(x) becomes infinite somewhere in the intervl x b (usully t n endpoint). Such integrls re clled improper integrls. If we think of integrls s res of regions in the plne, then improper integrls usully refer to res of infinitely lrge regions so tht some cre must be tken in interpreting them. The forml definition of the integrl s limit of Riemnn sums cnnot be used since it ssumes both tht the integrtion bounds nd b re finite, nd tht the integrnd f(x) is bounded. Improper integrls hve to be defined on cse by cse bsis. The next section shows the usul wys in which this is done.. Typicl exmples of improper integrls.. Integrl on n unbounded intervl. Consider the integrl A = x 3. This integrl hs new feture tht we hve not delt with before, nmely, one of the integrtion bounds is rther thn finite number. The interprettion of this integrl is tht it is the re of the region under the grph of y = /x 3, with < x <. y = x 3 A = x 3 Becuse the integrl goes ll the wy to x = the region whose re it represents stretches infinitely fr to the right. Could such n infinitely wide region still hve finite re? And if it is, cn we compute it? To compute the integrl I tht hs the in its integrtion bounds, we first replce the integrl by one tht is more fmilir, nmely A M = M 4 x 3,

42 4. PROPER AND IMPROPER INTEGRALS where M > is some finite number. This integrl represents the re of finite region, nmely ll points between the grph nd the x-xis, nd with x M. y = x 3 A M = M x 3 M We know how to compute this integrl: A M = M [ x 3 = ] M x = M +. The re we find depends on M. The lrger we choose M, the lrger the region is nd the lrger the re should be. If we let M then the region under the grph between x = nd x = M will expnd nd eventully fill up the whole region between grph nd x-xis, nd to the right of x =. Thus the re should be A = lim A M = M lim M M [ x 3 = lim M M + ] =. We conclude tht the infinitely lrge region between the grph of y = /x 3 nd the x-xis tht lies to the right of the line x = hs finite re, nd tht this re is exctly!.. Second exmple on n unbounded intervl. The following integrl is very similr to the one we just did: A = The integrl represents the re of the region tht lies to the right of the line x =, nd is cught between the x-xis nd the hyperbol y = /x. A M M x. A y=/x As in the previous exmple the region extends infinitely fr to the right while t the sme time becoming nrrower nd nrrower. To see wht its re is we gin look t the truncted region tht contins only those points between the grph nd the x-xis, nd for which x M. This re is A M = M x = [ ln x ] M = ln M ln = ln M. The re of the whole region with x < is the limit A = lim A M = lim ln M = +. M M So we see tht the re under the hyperbol is not finite!

43 . TYPICAL EXAMPLES OF IMPROPER INTEGRALS An improper integrl on finite intervl. In this third exmple we consider the integrl I =. x The integrtion bounds for this integrl re nd so they re finite, but the integrnd becomes infinite t one end of the integrtion intervl: lim x x = +. The region whose re the integrl I represents does not extend infinitely fr to the left or the right, but in this exmple it extends infinitely fr upwrd. To compute this re we gin truncte the region by looking t ll points with x for some constnt <, nd compute I = The integrl I is then the limit of I, i.e. x = lim We see tht the re is finite. = [ rcsin x ] = rcsin. x x = lim rcsin = rcsin = π..4. A doubly improper integrl. Let us try to compute I = + x. This exmple hs new feture, nmely, both integrtion limits re infinite. To compute this integrl we replce them by finite numbers, i.e. we compute + x = lim = lim = lim lim b lim b b + x rctn(b) rctn() rctn b lim rctn b = π ( π ) = π. A different wy of getting the sme exmple is to replce nd in the integrl by nd nd then let. The only difference with our previous pproch is tht we now use one vrible () insted of two ( nd b). The computtion goes s follows: + x = lim = lim + x ( ) rctn() rctn( ) = π ( π ) = π. In this exmple we got the sme nswer using either pproch. This is not lwys the cse, s the next exmple shows. y = x

44 44. PROPER AND IMPROPER INTEGRALS Another doubly improper integrl. Suppose we try to compute the integrl I = x. The shorter pproch where we replce both ± by ± would be x = lim x = lim ( ) =. On the other hnd, if we tke the longer pproch, where we replce nd by two different constnts, then we get this x = lim lim b = lim lim b = lim b b b [ x ] lim x At this point we see tht both limits lim b b / = nd lim / = do not exist. The result we therefore get is x =. Since is not number we find tht the improper integrl does not exist. We conclude tht for some improper integrls different wys of computing them cn give different results. This mens tht we hve to be more precise nd specify which definition of the integrl we use. The next section lists the definitions tht re commonly used.. x is the re bove minus the re below the xis.. Summry: how to compute n improper integrl.. How to compute n improper integrl on n unbounded intervl. By definition the improper integrl is given by f(x) (5) f(x) = This is how the integrl in. ws computed. M lim f(x). M.. How to compute n improper integrl of n unbounded function. If the integrtion intervl is bounded, but if the integrnd becomes infinite t one of the endpoints, sy t x =, then we define (6) b f(x) = lim s b s f(x).

45 3. MORE EXAMPLES Doubly improper integrls. If the integrtion intervl is the whole rel line, i.e. if we need to compute the integrl I = f(x), then we must replce both integrtion bound by finite numbers nd then let those finite numbers go to ±. Thus we would first hve to compute n ntiderivtive of f, F (x) = f(x). The Fundmentl Theorem of Clculus then implies I,b = b f(x) = F (b) F () nd we set I = lim F (b) lim F (). b Note tht ccording to this definition we hve to compute both limits seprtely. The exmple in Section.5 shows tht it relly is necessry to do this, nd tht computing cn give different results. In generl, if n integrl f(x) = lim F () F ( ) b f(x) is improper t both ends we must replce them by c, d with < c < d < b nd compute the limit For instnce, b lim f(x) = lim c lim d b d d = lim lim x c d c 3. More exmples c f(x). x Are under n exponentil. Let be some positive constnt nd consider the grph of y = e x for x >. How much re is there between the grph nd the x-xis with x >? (See Figure for the cses = nd =.) The nswer is given by the y = e x y = e x Figure. Wht is the re under the grph of y = e x? Wht frction of the region under the grph of y = e x lies under the grph of y = e x?

46 46. PROPER AND IMPROPER INTEGRALS improper integrl A = = lim M = lim M e x M = lim M =. e x [ e x] M ( ) e M > so lim M e M = We see tht the re under the grph is finite, nd tht it is given by /. In prticulr the re under the grph of y = e x is exctly, while the re under the grph of y = e x is exctly hlf tht (i.e. /). 3.. Improper integrls involving x p. The exmples in. nd. re specil cses of the following integrl I = x p, where p > is some constnt. We lredy know tht the integrl is if p = 3 (.), nd lso tht the integrl is infinite (does not exist) if p = (.). We cn compute it in the sme wy s before, (7) I = lim M = lim M M = p lim M x p [ p x p] M ( ) M p. ssume p At this point we hve to find out wht hppens to M p s M. This depends on the sign of the exponent p. If this exponent is positive, then M p is positive power of M nd therefore becomes infinite s M. On the other hnd, if the exponent is negtive then M p is negtive power of M so tht it goes to zero s M. To summrize: If < p < then p > so tht lim M M p = ; if p > then p <, nd lim M p = M If we pply this to (7) then we find tht if < p then nd if p > then lim M =. M p x p =, x p = p.

47 4. PROBLEMS 47 Q Q R P R P O Figure. Geometric construction of the Versier. The figure shows circle of dimeter tht touches the x-xis exctly t the origin. For ny point Q on the line y = we cn define new point P by intersecting the line segment OQ with the circle, nd requiring P to hve the sme x-coordinte s Q nd the sme y-coordinte s R. The Versier is the curve trced out by the point P s we lets the point Q slide long the line y =. See problem Problems Compute the following improper integrls nd drw the region whose re they represent: / ( + x). (x ) 3 3 x. / x. xe x. x x + x. x x 3 + x. x. x x. x + x. (suggestion: x = u ) ( x + ). x e x xe x xe x e x 6. The grph of the function y = /(+x ) is clled the Versier (see Figure ). Compute the re of the shded region between the Versier nd the circle in Figure. 7. Let be positive constnt. () Drw the grph of y = xe x, x < (the function hs one mximum: where is it?) (b) Compute I = does I depend on? xe x. How (c) Compute lim I. Interpret your nswer using the grphs of y = x x tht you drew in prt (i). 8. e x. How would you define this integrl (one of the integrtion bounds is rther thn + )? 9. () e x sin πx where, b re positive constnts. (You hve done the integrl before see Ch I, 7, Problem 7..) (b) As in the previous problems, drw the region whose re the integrl from () represents. Note tht the function in the integrl for this problem is not positive everywhere. How does this ffect your nswer?

48 48. PROPER AND IMPROPER INTEGRALS f(x) g(x) y = g(x) y = f(x) Figure 3. Compring improper integrls. Here f nd g re positive functions tht stisfy f(x) g(x) for ll x. If g(x) is finite, then so is f(x). Conversely, if f(x) is not finite, then g(x) cnnot be finite either. y = g(x) y = f(x) y = f(x) Figure 4. In this figure f(x) is gin positive, but g(x) is bounded by f(x) g(x) f(x). The sme conclusion still holds, nmely, if f(x) exists, then so does g(x). Cn you tell from your drwing if the integrl is positive?. (A wy to visulize tht x = ) () Show tht for ny > one hs x = ln ; in prticulr, this integrl is the sme for ll >. (b) Show tht the re under the grph of y = /x between x = nd x = n is n ln by splitting the region into: the prt where x, the prt where x 4, the prt where 4 x 8,. the prt where n x n. (c) Explin how the nswer to (b) implies tht the integrl /x does not exist.. The re under the grph of y = /x with x < is infinite. Compute the volume of the funnel-shped solid you get by revolving this region round the x-xis. Is the volume of this funnel finite or infinite? 5. Estimting improper integrls Sometimes it is just not possible to compute n improper integrl becuse we simply cnnot find n ntiderivtive for the integrnd. When this hppens we cn still try to

49 5. ESTIMATING IMPROPER INTEGRALS 49 estimte the integrl by compring it with esier integrls, nd even if we cnnot compute the integrl we cn still try to nswer the question does the integrl exist, i.e. Does lim M M f(x) exist? In this section we will see number of exmples of improper integrls tht much esier to estimte thn to compute. Throughout there re three principles tht we shll use: Integrl of positive function. If the function f(x) stisfies f(x) for ll x then either the integrl f(x) exists, or else it is infinite, by which we men lim M M f(x) =. If the function f(x) is not lwys positive then the bove limit cn fil to exist by oscillting without ever becoming infinite. For n exmple see 5..; see lso 5. for further explntion. Comprison with esier integrls. If y = f(x) nd y = g(x) re functions defined for x, nd if g(x) f(x) for ll x, then g(x) f(x). In prticulr, f(x) exists = g(x) does not exist = g(x) exists, f(x) does not exist. Red 5. for more detils nd exmples. Only the til mtters. If f(x) is continuous function for ll x, then for ny b we hve f(x) exists Furthermore, for ny b we hve (8) f(x) = b b f(x) + f(x) exists. b f(x). This is further explined in Theorem 5.3 nd the exmples following it. 5.. Improper integrls of positive functions. Suppose tht the function f(x) is defined for ll x, nd tht f(x) for ll x. To see if the improper integrl f(x) exists we hve to figure out wht hppens to I M = M f(x) s M. Since we re ssuming tht f(x) the integrl I M represents the re under the grph of f up to x = M. As we let M increse this region expnds, nd thus the integrl I M increses. So, s M there re two possibilities: either I M remins finite nd converges to finite number, or I M becomes infinitely lrge. The following theorem summrizes this useful observtion:

50 5. PROPER AND IMPROPER INTEGRALS 5... Theorem. If f(x) for ll x then either the integrl I = exists, (i.e. I is finite number), or else I = f(x) f(x) =, i.e. lim M M f(x) = Exmple - integrl to infinity of the cosine. To illustrte wht Theorem 5.. sys, let s consider n improper integrl of function tht is not lwys positive. For instnce, consider I = cos x. The function in the integrl is f(x) = cos x, nd this function is clerly not lwys positive. When we try to compute this integrl we get I = lim M M cos x = [ ] M lim sin x = lim sin M. M x= M This limit does not exist s sin M oscilltes up nd down between nd + s M. On the other hnd, since sin M stys between nd +, we cnnot sy tht lim M M cos x = +. Theorem 5.. tells us tht if we integrte positive function then this kind of oscilltory behvior cnnot occur. 5.. Comprison Theorem for Improper Integrls. Suppose f(x) nd g(x) re functions tht re defined for x <, nd suppose tht g(x) f(x) for ll x, i.e. f(x) g(x) f(x) for ll x. If the improper integrl f(x) exists then the improper integrl g(x) lso exists, nd one hs g(x) f(x). This theorem is used in two wys: it cn be used to verify tht some improper integrl exists without ctully computing the integrl, nd it cn lso be used to estimte n improper integrl Exmple. Consider the improper integrl I = + x 3. The function in the integrl is rtionl function so in principle we know how to compute the integrl. It turns out the computtion is not very esy. If we don t relly need to know the exct vlue of the integrl, but only wnt rough estimte of the integrl, then we could compre the integrl with n esier integrl. To decide which simpler integrl we should use s comprison, we reson s follows. Since only the til mtters, we consider the integrnd +x 3 for lrge x. When x is very

51 5. ESTIMATING IMPROPER INTEGRALS 5 lrge x 3 will be much lrger thn, so tht we my guess tht we cn ignore the in the denomintor + x 3 : (9) + x 3 x 3 s x. This suggests tht we my be ble to compre the given integrl with the integrl x 3. We know from our very first exmple in this chpter (.) tht this lst integrl is finite (we found tht it is ). Therefore we cn guess tht our integrl I lso is finite. Now let s try to use the Comprison Theorem 5. to get certinty by proving tht the integrl I does indeed exist. We wnt to show tht the integrl +x exists, so we 3 choose g(x) = + x 3, nd thus f(x) = x 3. We cn compre these functions s follows: ( ) divide both sides first it follows from: x 3 + x 3 for ll x by x 3 nd then by + x 3 tht: + x 3 x 3 for x This tells us tht + x 3 x 3 =. Therefore we hve found tht the integrl I does indeed exist, nd tht it is no more thn. We cn go beyond this nd try to find lower bound (insted of sying tht I is no more thn we try to sy tht it is t lest s lrge s some other number.) Here is one wy of doing tht: + x 3 x 3 + x 3 for ll x = + x 3 x 3 for ll x = x 3 + x 3 for x ( ) divide both sides first by x 3 nd then by + x 3 This implies tht x 3 + x 3. The first integrl here is hlf the integrl we computed in., so we get In summry, we hve found tht x 3. + x 3.

52 5. PROPER AND IMPROPER INTEGRALS 5... Second exmple. Does the integrl I = x x + exist? Since the function we re integrting is positive, we could lso sk is the integrl finite? As in the previous exmple we look t the behvior of the integrnd s x. For lrge x we cn ssume tht x is much lrger thn x, so tht it would be resonble to ignore the in the denomintor x +. If we do tht thn we find tht x x + x x = x. If this were correct, then we would end up compring the integrl I with the simpler integrl x. We know this ltter integrl is not finite (it ws our second exmple, see.) nd therefore we guess tht the integrl I probbly lso is not finite. To give sound rgument we will use the Comprison Theorem. Our gol is to show tht the integrl I = f(x), with f(x) = x + x is not finite. To do this we hve to find function g(x) such tht g(x) is smller thn f(x) (so tht f will be lrger thn g(x)), g(x) is esy to integrte, nd the integrl of g(x) is not finite. The first nd lst point together imply tht I = f(x) g(x) =, which is wht we re trying to show. To complete the resoning we hve to find the esy-to-integrte function g(x). Bsed on wht we hve done bove our first guess would be g(x) = x, but this does not work, since x x + < x x = x. So with this choice of g(x) we get g(x) > f(x) insted of g(x) < f(x). One wy to simplify f(x) nd get smller function is to remember tht by incresing the denomintor in frction we decrese the frction. Thus, for x > we hve f(x) = x x + > So we let g(x) = x. Then we find I = =. x x + x = x x + x x x = x.

53 5. ESTIMATING IMPROPER INTEGRALS The Til Theorem. If y = f(x) is continuous function for x, nd if b > then f(x) exists if nd only if b f(x) exists Moreover, if these integrls exist, then (8) holds: f(x) = b f(x) + b Proof. For ny finite M one hs M b M f(x) = f(x) + f(x) b The theorem follows by tking the limit for M on both sides of this eqution. The following two exmples show how one cn use this fct Exmple. Does the integrl I = x 3 + exist? The integrnd here is the sme s in 5.. where we found tht x 3 + <, f(x). nd in prticulr is finite. Since the function f(x) = /(x 3 + ) is continuous for x we my conclude tht I = x 3 + = x x 3 + so tht I is finite. If we wnt n upper estimte for the integrl we cn use the fct tht we lredy know tht the integrl from to is not more thn nd estimte the integrl from x = to x =. For x we hve x 3 + = x 3 + This implies x 3 + =, nd hence x 3 + = x x = The re under the bell curve. The bell curve which plys centrl role in probbility nd sttistics, is the grph of the function n(x) = e x /b, where nd b re positive constnts whose vlues will not be importnt in this exmple. In fct, to simplify this exmple we will choose them to be = nd b = so tht we re deling with the function n(x) = e x. The question now is, is the re under the bell curve finite, nd cn we estimte it? In terms of improper integrls, we wnt to know if the integrl A = e x

54 54. PROPER AND IMPROPER INTEGRALS n(x) = n( x) so the bell curve is symmetric exists. We write the integrl s the sum of two integrls e x = e x + e x Since the function n(x) = e x is even these two integrls re equl, so if we cn show tht one of them exists, the other will lso exist. When x is lrge x is much lrger thn x so tht e x will be smller thn e x ; since e x is function we know how to integrte it seems like good ide to compre the bell curve with the grph of e x. To mke the comprison (nd to use the Comprison Theorem) we first check if e x e x relly is true. We find tht this is true if x, nmely: x = x x = x x = e x e x. We cn therefore use e x to estimte the integrl of e x for x : e x e x = lim [ e x] M M Between x = nd x = we hve e x, so nd therefore e x = e x e x + ( = lim e M + e ) = e. M =, e x + e. Since the bell curve is symmetric, we get the sme estimtes for the integrl over < x : e x + e. In the end we hve shown tht the re under the bell curve is finite, nd tht it is bounded by A = e x + e. With quite bit more work, nd using n indirect rgument one cn ctully compute the integrl (without finding n ntiderivtive of e x ). The true vlue turns out to be A = e x = π. A = e x y = e x y = e x Figure 5. Estimting the re under the bell curve.

55 6. PROBLEMS 55 For comprison, π nd e 6. Problems Which of the following inequlities re true for ll x > (you get to choose ):. x > x? Answer: True. If x > then x > x nd therefore <. So the inequlity is true if x x x > where we choose = x < x? x x + x > x? x x3 > x /? x 3x + (x x) 3 < x 4? Which of the following inequlities re true for ll x with < x < (you get to choose gin, s long s > ): 3 6. x > 3 x? Answer: Flse. If < x < then x < x nd therefore > 3, which implies > 3. x x x x So, more precisely, the inequlity is flse if < x < where we choose = x x + x < x? x x < x? x + x < x? 4 x + x < x? x x < 4 x? For ech of the following improper integrls drw the grph of the integrnd, nd decide if the integrl exists. If it does try to find n estimte for how lrge the integrl is π u ( u + ) du u 3 ( u + ) du u 4 ( u + ) du sin x x sin(x ) x (wht hppens t x =?) 7. [Group Problem] (Fresnel integrls from optics.) For bckground on Fresnel integrls red the Wikipedi rticle on the subject. Consider the function F (x) = sin x x. () Compute lim x F (x) nd lim x F (x). (b) Show tht F (x) = cos x (c) Show tht the integrl sin x ex- x ists. (d) Show tht cos x = sin x x. sin x x. (e) True or Flse: if for some function f the improper integrl f(x) exists then it must be true tht lim x f(x) =? 8. [Group Problem] Suppose p(x) = e x, where > is constnt. () Show tht p(x) <. (hint: you cn compute the integrl.) (b) Compute xp(x) p(x)

56 56. PROPER AND IMPROPER INTEGRALS by integrting by prts in the integrl in the numertor. (c) Compute x p(x) p(x) by integrting by prts twice in the integrl in the numertor. 9. [Group Problem] Suppose p(x) = e x /σ where σ > is constnt. (σ is the Greek lower cse s, pronounced s sigm. ) () Show tht p(x) <. (hint: there is no ntiderivtive of e x /σ, so don t look for it. Insted follow exmple 5.3..) (b) Compute xp(x) p(x). Hint: there is n esy ntiderivtive for the integrl in the numertor. Do tht one first. (c) Compute x p(x) p(x) by integrting by prts in the integrl in the numertor.. [Group Problem] The re under the bell curve is -fctoril In this problem we look t new function F (n). For ny number n we would like to define F (n) = t n e t dt. The integrl tht defines F (n) is improper so it is not utomticlly cler tht F (n) exists. () Show tht the integrl exists if n. (b) Compute F (), F (), nd F (). (c) Use integrtion by prts to show tht F (n + ) = (n + )F (n) Holds for ny n. Use this to Compute F (). (d) If you set n = in the reltion F (n + ) = (n + )F (n) then you get F () = F ( ) =. This should contrdict the vlue for F () you found in (b). Wht is wrong? (e) Show tht the integrl for F ( ) exists, nd use substitution to show tht F ( ) = e u du. How does this justify the sttement t the beginning of the problem?

57 CHAPTER 3 First order differentil Equtions. Wht is Differentil Eqution? A differentil eqution is n eqution involving n unknown function nd its derivtives. A generl differentil eqution cn contin second derivtives nd higher derivtives of the unknown function, but in this course we will only consider differentil equtions tht contin first order derivtives. So the most generl differentil eqution we will look t is of the form dy () = f(x, y). Here f(x, y) is ny expression tht involves both quntities x nd y. For instnce, if f(x, y) = x + y then the differentil eqution represented by () is dy = x + y. In this eqution x is vrible, while y is function of x. Differentil equtions pper in mny science nd engineering problems, s we will see in the section on pplictions. But first let s think of differentil eqution s purely mthemticl question: which functions y = y(x) stisfy the eqution ()? It turns out tht there is no generl method tht will lwys give us the nswer to this question, but there re methods tht work for certin specil kinds of differentil equtions. To explin ll this, this chpter is divided into the following prts: Some bsic exmples tht give us clue s to wht the solution to generl differentil eqution will look like. Two specil kinds of differentil eqution ( seprble nd liner ), nd how to solve them. How to visulize the solution of differentil eqution (using direction fields ) nd how to compute the solution with computer using Euler s method. Applictions: number of exmples of how differentil equtions come up nd wht their solutions men.. Two bsic exmples.. Equtions where the RHS does not contin y. Which functions y = y(x) stisfy dy () = sin x? This is differentil eqution of the form () where the function f tht describes the Right Hnd Side is given by f(x, y) = sin x. In this exmple the function f does not depend on the unknown function y. Becuse of this the differentil eqution relly sks 57

58 58 3. FIRST ORDER DIFFERENTIAL EQUATIONS which functions of x hve sin x s derivtive? In other words, which functions re the ntiderivtive of sin x? We know the nswer, nmely y = sin x = cos x + C where C is n rbitrry constnt. This is the solution to the differentil eqution (). This exmple shows us tht there is not just one solution, but tht there re mny solutions. The expression tht describes ll solutions to the differentil eqution () is clled the generl solution. It contins n unknown constnt C tht is llowed to hve rbitrry vlues. To give mening to the constnt C we cn observe tht when x = we hve So the constnt C is nothing but y() = cos + C = + C. C = y() +. For instnce, the solution of () tht lso stisfies y() = 4 hs C = 4 + = 5, nd thus is given by y(x) = cos x + 5. We hve found tht there re mny solutions to the differentil eqution () (becuse of the undetermined constnt C), but s soon s we prescribe the vlue of the solution for one vlue of x, such s x =, then there is exctly one solution (becuse we cn compute the constnt C.).. The exponentil growth exmple. Which functions equl their own derivtive, i.e. which functions stisfy dy = y? Everyone knows t lest one exmple, nmely y = e x. But there re more solutions: the function y = lso is its own derivtive. From the section on exponentil growth in mth we know ll solutions to dy = y. They re given by y(x) = Ce x, where C cn be n rbitrry number. If we know the solution y(x) for some vlue of x, such s x =, then we cn find C by setting x = : y() = C. Agin we see tht insted of there being one solution, the generl solution contins n rbitrry constnt C..3. Summry. The two exmples tht we hve just seen show us tht for certin differentil equtions there re mny solutions, the formul for the generl solution contins n undetermined constnt C, the undetermined constnt C becomes determined once we specify the vlue of the solution y t one prticulr vlue of x. It turns out tht these fetures re found in lmost ll differentil equtions of the form (). In the next two sections we will see methods for computing the generl solution to two frequently occurring kinds of differentil eqution, the seprble equtions, nd the liner equtions.

59 3. FIRST ORDER SEPARABLE EQUATIONS First Order Seprble Equtions By definition seprble differentil eqution is diffeq of the form () y dy (x) = F (x)g(y(x)), or = F (x)g(y). Thus the function f(x, y) on the right hnd side in () hs the specil form For exmple, the differentil eqution f(x, y) = F (x) G(y). dy = sin(x)( + y ) is seprble, nd one hs F (x) = sin x nd G(y) = + y. On the other hnd, the differentil eqution dy = x + y is not seprble. 3.. Solution method for seprble equtions. To solve this eqution divide by G(y(x)) to get (3) dy G(y(x)) = F (x). Next find function H(y) whose derivtive with respect to y is (4) H (y) = ( ) dy solution: H(y) = G(y) G(y). Then the chin rule implies tht the left hnd side in (3) cn be written s dy G(y(x)) = H (y(x)) dy = dh(y(x)). Thus (3) is equivlent with dh(y(x)) = F (x). In words: H(y(x)) is n ntiderivtive of F (x), which mens we cn find H(y(x)) by integrting F (x): (5) H(y(x)) = F (x) + C. Once we hve found the integrl of F (x) this gives us y(x) in implicit form: the eqution (5) gives us y(x) s n implicit function of x. To get y(x) itself we must solve the eqution (5) for y(x). A quick wy of orgnizing the clcultion goes like this: To solve dy = F (x)g(y) we first seprte the vribles, dy = F (x), G(y) nd then integrte, dy G(y) = F (x).

60 6 3. FIRST ORDER DIFFERENTIAL EQUATIONS The result is n implicit eqution for the solution y with one undetermined integrtion constnt. Determining the constnt. The solution we get from the bove procedure contins n rbitrry constnt C. If the vlue of the solution is specified t some given x, i.e. if y(x ) is known then we cn express C in terms of y(x ) by using (5). 3.. A sng: We hve to divide by G(y) which is problemtic when G(y) =. This hs s consequence tht in ddition to the solutions we found with the bove procedure, there re t lest few more solutions: the zeroes of G(y) (see Exmple 3.4 below). In ddition to the zeroes of G(y) there sometimes cn be more solutions, s we will see in Exmple. on Leky Bucket Dting Exmple. We solve dz dt = ( + z ) cos t. Seprte vribles nd integrte dz + z = cos t dt, to get rctn z = sin t + C. Finlly solve for z nd we find the generl solution z(t) = tn ( sin(t) + C ) Exmple: the sng in ction. If we pply the method to y (x) = y, we get y(x) = e x+c. No mtter how we choose C we never get the function y(x) =, even though y(x) = stisfies the eqution. This is becuse here G(y) = y, nd G(y) vnishes for y =. 4. Problems For ech of the following differentil equtions - find the generl solution, - indicte which, if ny, solutions were lost while seprting vribles, - find the solution tht stisfies the indicted initil vlues dy = xy, y() =. dy + x cos y =, y() = π. 3 dy + + x + y =, y() = A. 4. y dy + x3 =, y() = A dy + y =, y() = A. dy + + y =, y() = A. dy + x =, y() =. y 8. [Group Problem] Let P (t) be the size of colony of bcteri in Petri dish in some experiment. Assume tht the size of the colony chnges ccording to the so-clled logistic eqution: dp dt = P ( P ), 5

61 5. FIRST ORDER LINEAR EQUATIONS 6 Assume lso tht in the beginning of the experiment the popultion size is P =. () Find the generl solution to the differentil eqution. (b) Find the solution tht stisfies the given initil conditions. (c) How long does it tke the popultion to rech size P = 5? (d) How long does it tke the popultion to rech size P = 9? (e) Wht vlue does P hve when dp is dt the lrgest (hint: you do not need to solve the differentil eqution this question hs very short nswer.) (6) 5. First Order Liner Equtions Differentil equtions of the form eqution re clled first order liner. dy + (x)y = k(x) 5.. The Integrting Fctor. Liner equtions cn lwys be solved by multiplying both sides of the eqution with specilly chosen function clled the integrting fctor. It is defined by (7) A(x) = (x), m(x) = e A(x). Here m(x) is the integrting fctor. It looks like we just pulled this definition of A(x) nd m(x) out of ht. The exmple in 5. shows nother wy of finding the integrting fctor, but for now let s go on with these two functions. Multiply the eqution (6) by the integrting fctor m(x) to get m(x) dy + (x)m(x)y = m(x)k(x). By the chin rule the integrting fctor stisfies Therefore one hs dm(x) = d ea(x) dm(x)y = A (x) }{{} =(x) e A(x) = (x)m(x). }{{} =m(x) = m(x) dy + (x)m(x)y { dy } = m(x) + (x)y = m(x)k(x). Integrting nd then dividing by the integrting fctor gives the solution y = ( ) m(x)k(x) + C. m(x) In this derivtion we hve to divide by m(x), but since m(x) = e A(x) nd since exponentils never vnish we know tht m(x), so we cn lwys divide by m(x).

62 6 3. FIRST ORDER DIFFERENTIAL EQUATIONS 5.. An exmple. Find the generl solution to the differentil eqution Then find the solution tht stisfies dy = y + x. (8) y() =. Solution. We first write the eqution in the stndrd liner form dy (9) y = x, nd then multiply with the integrting fctor m(x). We could of course memorize the formuls (7) tht led to the integrting fctor, but sfer pproch is to remember the following procedure, which will lwys give us the integrting fctor. Assuming tht m(x) is s yet unknown we multiply the differentil eqution (9) with m, (3) m(x) dy m(x)y = m(x)x. If m(x) is such tht (3) m(x) = dm(x), then eqution (3) implies m(x) dy + dm(x) y = m(x)x. The expression on the left is exctly wht comes out of the product rule this is the point of multiplying with m(x) nd then insisting on (3). So, if m(x) stisfies (3), then the differentil eqution for y is equivlent with dm(x)y = m(x)x. We cn integrte this eqution, m(x)y = m(x)x, nd thus find the solution (3) y(x) = m(x) m(x)x. All we hve to do is find the integrting fctor m. This fctor cn be ny function tht stisfies (3). Eqution (3) is differentil eqution for m, but it is seprble, nd we cn esily solve it: dm = m dm = ln m = x + C. m Since we only need one integrting fctor m we re not interested in finding ll solutions of (3), nd therefore we cn choose the constnt C. The simplest choice is C =, which leds to ln m = x m = e x m = ±e x. Agin, we only need one integrting fctor, so we my choose the ± sign: the simplest choice for m here is m(x) = e x.

63 6. PROBLEMS 63 With this choice of integrting fctor we cn now complete the clcultion tht led to (3). The solution to the differentil eqution is y(x) = m(x)x m(x) = e x e x x (integrte by prts) = e x{ } e x x e x + C = x + Ce x. This is the generl solution. To find the solution tht stisfies not just the differentil eqution, but lso the initil condition (8), i.e. y() =, we compute y() for the generl solution, y() = + Ce = 3 + Ce. The requirement y() = then tells us tht C = 3e. The solution of the differentil eqution tht stisfies the prescribed initil condition is therefore y(x) = x + 3e x. 6. Problems. In exmple 5. we needed function m(x) tht stisfies (3). The function m(x) = stisfies tht eqution. Why did we not choose m(x) =?. Why cn t we simplify the computtion t the end in exmple 5. by cnceling the two fctors m(x) s follows: y(x) = m(x) = x = x + C? m(x) x For ech of the following differentil equtions - specify the differentil eqution tht the integrting fctor stisfies, - find one integrting fctor, - find the generl solution, - find the solution tht stisfies the specified initil conditions. In these problems K nd N re constnts. 3. dy = y + x, y() =. 4. dy = y + x, y() =. 5. dy + y + ex =. 6. dy (cos x)y = esin x, y() = A. 7. dy = y + e x, y() =. 8. dy = y tn x +, y() =. 9. dy = y tn x +, y() =.. cos x dy = N y y() =.. x dy = y + x, y() =.. dy = xy + x3, y() =. 3. dy = y + sin x, y() =. 4. dy = Ky + sin x, y() =. 5. dy + x y =, y() = dy + ( + 3x )y =, y() =.

64 64 3. FIRST ORDER DIFFERENTIAL EQUATIONS 7. Direction Fields We cn visulize differentil eqution by drwing the corresponding direction field. Consider differentil eqution dy = f(x, y) where f is some given function. The differentil eqution tells us tht if we know point (x, y ) on the grph of solution y = y(x) of the differentil eqution, then we lso know the slope of the grph t tht point. We cn drw the tngent line to the grph of the solution: Slope m = f(x, y ) (x, y ) A solution Tngent line to the solution t (x, y ) If we hve not yet solved the differentil eqution then we don t know ny points on the solution. In this cse we cn smple the xy-plne, compute f(x, y) t ll our smple points, nd drw the tngent lines solution would hve if it pssed through one of the smple points. In Figure this is done for the differentil eqution dy = y + sin πx. The direction field on the left in Figure gives us n ide of wht the solutions should look like. Whenever solution psses through one of the smple points the slope of its grph is given by the line segment drwn t tht smple point. It is cler from quick look t Figure tht drwing direction field involves computing f(x, y) (i.e. y + sin πx in our exmple) for mny, mny points (x, y). This kind of repetitive computtion is better Figure. Direction field for dy = y + sin πx on the left, nd the sme direction field with solutions to the differentil eqution.

65 8. EULER S METHOD 65 done by computer, nd n internet serch quickly leds to number of websites tht produce direction fields for our fvorite differentil eqution. In prticulr the ODE pge t the Virtul Mth Museum of UC Irvine, drws both direction fields nd pproximtions to solutions found using Euler s method, to which we now turn. 8. Euler s method 8.. The ide behind the method. Consider gin the differentil eqution (), dy = f(x, y). Suppose we know one point on solution, i.e. suppose we know tht given solution to this eqution stisfies y = y when x = x, i.e. (33) y(x ) = y. The differentil eqution then tells us wht the derivtive of the solution is t x = x, nmely, y (x ) = f(x, y ). The definition derivtive sys tht so tht we hve y (x ) = lim h y(x + h) y(x ) h y(x + h) y(x ) (34) lim = f(x, y ). h h Keep in mind tht the right hnd side is wht we get by substituting the x nd y vlues tht we know for the solution in f. So if we know x nd y then we cn lso compute f(x, y ). If we don t know the solution then we cnnot compute the left hnd side in (34), but, following Euler, we cn mke n pproximtion. If insted of letting h we choose smll number h >, then we my ssume tht y(x + h) y(x ) (35) f(x, y ). h Here mens pproximtely equl, which is vguely defined concept. It mens tht the difference between the two quntities in (35) is smll nd we will not worry too much bout the error in (35) (such issues re ddressed in more dvnced courses on Numericl Anlysis; e.g. Mth 54 t UW Mdison). In the pproximte eqution (35) ll quntities re known except y(x + h). After solving (35) for y(x + h) we find y(x + h) y + hf(x, y ). Euler s ide (see Figure ) ws to forget tht this is n pproximtion, nd declre tht we now know new point on the grph of the solution, nmely x = x + h, y = y + hf(x, y ). Assuming tht our solution stisfies y(x ) = y exctly (rther thn just pproximtely), we cn pply the sme procedure nd get nother new point on the grph of our solution: x = x + h, y = y + hf(x, y ).

66 66 3. FIRST ORDER DIFFERENTIAL EQUATIONS y tngent y solution? x x = x + h Figure. Approximting solution with Euler s method. On the left: one step of Euler s method. Knowing one point (x, y ) on the grph does not immeditely tell us wht the solution is, but it does tell us wht the tngent to the grph t (x, y ) is. We cn then guess tht the point on the tngent with x = x + h nd y = y + f(x, y )h lmost lies on the grph. On the right: repeting the procedure gives us sequence of points tht pproximte the solution. Reducing the step size h should give better pproximtions. By repeting this procedure we cn generte whole list of points (x, y ), (x, y ), (x, y ), (x 3, y 3 ), etc... tht lie pproximtely on the grph of the solution pssing through the given initil point (x, y ). 8.. Setting up the computtion. If we re given x strt nd y(x strt ) = y strt, nd if we wnt to find y(x end ) for some x end > x strt, then we cn decide to pproximte y(x end ) by pplying n steps of Euler s method. Since ech step dvnces x by n mount h, ppliction of Euler s method will dvnce x by nh fter n steps. Thus we wnt nh = x end x strt, i.e. h = x end x strt. n Strting with the known vlue of y t x strt we then pply Euler s method n times by filling out tble of the following form: x k y k m k = f(x k, y k ) y k+ = y k + m k h x = x strt y strt m y x = x + h y m y x = x + h y m y 3 x 3 = x + 3h y 3 m 3 y 4.. x n = x + (n )h y n m n y n x n = x end y n The procedure is very mechnicl nd repetitive, nd is best progrmmed on computer.

67 . APPLICATIONS OF DIFFERENTIAL EQUATIONS 67 Once the tble hs been computed the vlues in the first two columns cn be used to grph the pproximtion to the rel solution. 9. Problems. Let y(t) be the solution of dy dt = t y, y(.) =.. We wnt to compute y(3.). () Find n exct formul for y(3.) by solving the eqution (the eqution is both seprble nd liner, so we hve t lest two methods for finding the solution.) (b) If we use Euler s method with step size h =., then how mny steps do we hve to tke to pproximte y(3.)? Compute the pproximtion with h =.. (c) Find the pproximtions if h =., h = /3, nd h =.5. Orgnize the computtions following the tble in 8.. (d) Compre the results from the computtions in (b) nd (c) with the true vlue of y(3.) from prt ().. The function y(x) = e x is the solution to dy = y, y() =. () Approximte e = y() by using Euler s method first with step size h =, then with h = /, nd then with h = /3. Wht re the pproximtions you find for y() in ech cse? (b) Look for pttern in the nswers to (). (c) Find formul for the result of pplying Euler s method n times with step size h = n. 3. Use Euler s method to pproximte the solution to dy = y, y() =, nd, in prticulr to find y(). Use vrious step sizes. How smll do you hve to mke the step size before the nswer seems resonble? Explin.. Applictions of Differentil Equtions Differentil equtions re very often used to describe how some object or system chnges or evolves in time. If the object or system is simple enough then its stte is completely determined by one number (sy y) tht chnges with time t. A differentil eqution for the system tells us how the system chnges in time, by specifying the rte of chnge of the stte y(t) of the system. This rte of chnge cn depend on time, nd it cn depend on the stte of the system, i.e. on y(t). This dependence cn be expressed s n eqution of the form dy (36) = f(y, t). dt The function f describes the evolutionry lw of our system (synonyms: evolutionry lw, dynmicl lw, evolution eqution for y )... Exmple: crbon dting. Suppose we hve fossil, nd we wnt to know how old it is. All living things contin crbon, which nturlly occurs in two isotopes, C 4 (unstble) nd C (stble). A long s the living thing is live it ets & breths, nd its rtio of C to C 4 is kept constnt. Once the thing dies the isotope C 4 decys into C t stedy rte tht is proportionl to the mount of C 4 it contins. Let y(t) be the rtio of C 4 to C t time t. The lw of rdioctive decy sys tht there is constnt k > such tht dy(t) = ky(t). dt

68 68 3. FIRST ORDER DIFFERENTIAL EQUATIONS Solve this differentil eqution (it is both seprble nd first order liner: either method works) to find the generl solution y(t; C) = Ce kt. After some lb work it is found tht the current C 4 /C rtio of our fossil is y now. Thus we hve y now = Ce ktnow = C = y now e ktnow. Therefore our fossil s C 4 /C rtio t ny other time t is/ws y(t) = y now e k(tnow t). This llows us to compute the time t which the fossil died. At this time the C 4 /C rtio must hve been the common vlue in ll living things, which cn be mesured let s cll it y life. Then t the time t demise when our fossil becme fossil we would hve hd y(t demise ) = y life. Hence the ge of the fossil would be given by y life = y(t demise ) = y now e k(tnow t demise) = t now t demise = k ln x life x now.. Exmple: dting leky bucket. A bucket is filled with wter. There is hole in the bottom of the bucket so the wter strems out t certin rte. h(t) the height of wter in the bucket A re of cross section of bucket re of hole in the bucket v velocity with which wter goes through the hole. re = A v h(t) We hve the following fcts to work with: The mount (volume) of wter in the bucket is A h(t); The rte t which wter is leving the bucket is v(t); Hence dah(t) = v(t). dt In fluid mechnics it is shown tht the velocity of the wter s it psses through the hole only depends on the height h(t) of the wter, nd tht, for some constnt K, v(t) = Kh(t). The lst two equtions together give differentil eqution for h(t), nmely, dh(t) dt = A Kh(t). To mke things bit esier we ssume tht the constnts re such tht A K =. Then h(t) stisfies (37) h (t) = h(t). This eqution is seprble, nd when we solve it we get dh h = = h(t) = t + C.

69 . APPLICATIONS OF DIFFERENTIAL EQUATIONS 69 h(t) t Figure 3. Severl solutions h(t; C) of the Leking Bucket Eqution (37). Note how they ll hve the sme vlues when t. This formul cn t be vlid for ll vlues of t, for if we tke t > C, the RHS becomes negtive nd cn t be equl to the squre root in the LHS. But when t C we do get solution, h(t; C) = (C t). This solution describes bucket tht is losing wter until t time C it is empty. Motivted by the physicl interprettion of our solution it is nturl to ssume tht the bucket stys empty when t > C, so tht the solution with integrtion constnt C is given by { (C t) when t C (38) h(t) = for t > C. The sng ppers gin. (See 3. nd 3.4.) Note tht we hd to divide by h to find the solution. This is not llowed when h =. It turns out tht h(t) = is solution to the differentil eqution. The solution h(t) = stisfies h() =, nd our experience with differentil equtions so fr would led us to believe tht this should therefore be the only solution tht stisfies h() =. However, every solution from (38) with C lso stisfies h() =. This problem is therefore different from the differentil equtions we hve delt with up to now. Nmely, prescribing the vlue y() = does not single out one solution of the differentil eqution (37)..3. Het trnsfer. We ll know tht het flows from wrm to cold: if we put cold spoon in hot cup of coffee, then het will flow from the coffee to the spoon. How fst does the spoon wrm up? According to physics the rte of chnge of the spoon s temperture is proportionl to the difference in temperture between the coffee nd the spoon. So, if T c nd T s re the temperture of the coffee nd the spoon, respectively, then dt s (39) = K ( ) T s T c. dt Here K is constnt tht depends on the shpe nd mteril of the spoon, how much of the spoon is in the coffee, etc., but not on the tempertures T s nd T c. If we ssume tht the spoon is smll, then whtever smll mount of het it extrcts from the coffee will not chnge the coffee temperture. Under tht ssumption we my ssume tht T c is

70 7 3. FIRST ORDER DIFFERENTIAL EQUATIONS in out constnt nd the differentil eqution (39) is both seprble nd liner so tht we hve two methods for solving it. If the coffee itself is lso cooling or wrming up then T c will depend on time nd the eqution (39) becomes dt s (4) dt + KT s = KT c (t). If we know T c (t) then this is still liner differentil eqution for T s nd we cn solve it..4. Mixing problems. Consider continer contining wter nd vinegr. If wter nd vinegr re flowing in nd out of the continer, then the concentrtion of vinegr in the continer will chnge with time ccording to some differentil eqution. Which differentil eqution describes the vinegr content of the continer depends on the precise detils of the set-up. As n exmple, let us ssume tht the continer hs fixed volume V = liters. This mens tht the mount of liquid tht flows in during ny time intervl must equl the mount of liquid flowing out of the continer (the liquid is incompressible, i.e. its density is fixed.) Suppose furthermore tht mixture of 5% vinegr nd 95% wter flows into the continer t 4 liters per minute. And suppose lso tht the liquid in the continer is thoroughly mixed, so tht the liquid tht flows out of the continer hs the sme vinegr/wter mixture s the entire continer. Problem: Let D(t) be the frction of the liquid in the continer tht is vinegr. How does D(t) chnge with time? Solution: Insted of trcking the concentrtion D(t) of vinegr we will look t the totl mount of vinegr in the continer. This mount is D(t)V. To find the differentil eqution we consider how much vinegr flows in nd out of the continer during short time intervl of length t (i.e. between time t nd t + t): in: The liquid volume flowing into the tnk during time t is 4 t liters. Since the vinegr concentrtion of the in-flowing liquid is 5%, this mens tht 5% 4 t = t liters of vinegr enter the continer. out: Since 4 t liters of liquid enter the continer, the sme mount of liquid must lso leve the continer. The liquid tht leves is the well-stirred mixture, nd thus it contins D(t) 4 t liters vinegr. In totl we see tht the chnge in vinegr content during time t is (4) ( vinegr content ) = t 4D(t) t. To find the chnge in concentrtion we divide by the volume of the continer t 4D(t) t D = = t 5 D(t) 5 t. We find the rte of chnge by dividing by t nd letting t : dd (4) dt = 5 5 D. This eqution is gin both liner nd seprble so we hve two methods for solving it.. Problems

71 . PROBLEMS 7 K =. Abbrevite C = A K. (b) If in n experiment one found tht. Red Exmple. on Leky bucket dting gin. In tht exmple we ssumed tht A K =. () Solve diffeq for h(t) without ssuming A the bucket empties in seconds fter being filled to height cm, then how much is the constnt C?. A popultion of bcteri grows t rte proportionl to its size. Write nd solve differentil eqution tht expresses this. If there re bcteri fter one hour nd bcteri fter two hours, how mny bcteri re there fter three hours? 3. Rbbits in Mdison hve birth rte of 5% per yer nd deth rte (from old ge) of % per yer. Ech yer rbbits get run over nd 7 rbbits move in from Sun Pririe. () Write differentil eqution tht describes Mdison s rbbit popultion t time t. (b) If there were, rbbits in Mdison in 99, how mny re there in 994? 4. [Group Problem] Consider cup of soup tht is cooling by exchnging het with the room. If the temperture of the soup is T s(t), nd if the temperture of the room is T r, then Newton s cooling lw clims tht dt s = K(T s T r) dt for some constnt K >. () Wht units does K hve, if we mesure temperture in degrees Fhrenheit, nd time in seconds? (b) Find the generl solution to Newton s cooling lw. Wht units does the rbitrry constnt in your solution hve? Wht is the limit of the temperture s t? (c) The soup strts t 8 o F, nd sits in room whose temperture is 75 F. In five minutes its temperture hs dropped to 5 F. Find the cooling constnt K. When will its temperture be 9 F? 5. [Group Problem] A lke gets heted by dy when the sun is out, nd loses het t night. It lso exchnges het with the Erth. The Erth s temperture is T e, nd the lke s temperture is T l (t). These effects together cn be modeled by differentil eqution for T l (t) dt l dt = K(T l T e) + S sin(πt). Here the first term on the right represents Newton s cooling lw, while the second term ccounts for the heting nd cooling by rdition: if sin(πt) > then it s dy nd the sun is heting the lke, if sin(πt) < then it s night nd the lke is rditing het into the cold sky. Time t is mesured in dys. Tempertures re in degrees Celsius. () Assuming K = S =, nd T e = find the generl solution to the differentil eqution. (b) Drw the direction field for the differentil eqution when K = S = nd T e =. (First try to do this by hnd. Then use computer.) (c) Does lim t T l (t) exist? Consider the seprte terms in the generl solution you found in prt (), nd see if ny of these terms hve limit s t. (d) Find the solution for rbitrry K nd S (you my ssume K nd S re positive.) 6. Retw is mysterious living liquid; it grows t rte of 5% of its volume per hour. A scientist hs tnk initilly holding y gllons of retw nd removes retw from the tnk continuously t the rte of 3 gllons per hour. () Find differentil eqution for the number y(t) of gllons of retw in the tnk t time t. (b) Solve this eqution for y s function of t. (The initil volume y will pper in our nswer.) (c) Wht is lim t y(t) if y =? (d) Wht should the vlue of y be so tht y(t) remins constnt?