Notes on Calculus II Integral Calculus. Miguel A. Lerma

Transcription

1 Notes on Clculus II Integrl Clculus Miguel A. Lerm

2 November 22, 22

3 Contents Introduction 5 Chpter. Integrls 6.. Ares nd Distnces. The Definite Integrl 6.2. The Evlution Theorem.3. The Fundmentl Theorem of Clculus 4.4. The Substitution Rule 6.5. Integrtion by Prts 2.6. Trigonometric Integrls nd Trigonometric Substitutions Prtil Frctions Integrtion using Tbles nd CAS Numericl Integrtion 4.. Improper Integrls 46 Chpter 2. Applictions of Integrtion More bout Ares Volumes Arc Length, Prmetric Curves Averge Vlue of Function (Men Vlue Theorem) Applictions to Physics nd Engineering Probbility 69 Chpter 3. Differentil Equtions Differentil Equtions nd Seprble Equtions Directionl Fields nd Euler s Method Exponentil Growth nd Decy 8 Chpter 4. Infinite Sequences nd Series Sequences Series The Integrl nd Comprison Tests Other Convergence Tests Power Series Representtion of Functions s Power Series 4.7. Tylor nd McLurin Series 3 3

4 CONTENTS Applictions of Tylor Polynomils 9 Appendix A. Hyperbolic Functions 3 A.. Hyperbolic Functions 3 Appendix B. Vrious Formuls 8 B.. Summtion Formuls 8 Appendix C. Tble of Integrls 9

5 Introduction These notes re intended to be summry of the min ides in course MATH 24-2: Integrl Clculus. I my keep working on this document s the course goes on, so these notes will not be completely finished until the end of the qurter. The textbook for this course is Stewrt: Clculus, Concepts nd Contexts (2th ed.), Brooks/Cole. With few exceptions I will follow the nottion in the book. If you find ny typos or errors, or you hve ny suggestions, plese, do not hesitte to let me know. You my emil me, or use the web form for feedbck on the web pges for the course: Miguel A. Lerm mlerm@mth.northwestern.edu Northwestern University Fll 22 5

6 CHAPTER Integrls.. Ares nd Distnces. The Definite Integrl... The Are Problem. The Definite Integrl. Here we try to find the re of the region S under the curve y = f(x) from to b, where f is some continuous function. Y y=f(x) S O b X In order to estimte tht re we begin by dividing the intervl [, b] into n subintervls [x, x ], [x, x 2 ], [x 2, x 3 ],..., [x n, x n ], ech of length x = (b )/n (so x i = + i x). Y O b X 6

7 .. AREAS AND DISTANCES. THE DEFINITE INTEGRAL 7 The re S i of the strip between x i nd x i cn be pproximted s the re of the rectngle of width x nd height f(x i ), where x i is smple point in the intervl [x i, x i+ ]. So the totl re under the curve is pproximtely the sum n f(x i ) x = f(x ) x + f(x 2) x + + f(x n) x. i= This expression is clled Riemnn Sum. The estimtion is better the thiner the strips re, nd we cn identify the exct re under the grph of f with the limit: n A = lim f(x i ) x n i= As long s f is continuous the vlue of the limit is independent of the smple points x i used. Tht limit is represented b f(x) dx, nd is clled definite integrl of f from to b: b n f(x) dx = lim f(x i ) x n The symbols t the left historiclly were intended to men n infinite sum, represented by long S (the integrl symbol ), of infinitely smll mounts f(x) dx. The symbol dx ws interpreted s the length of n infinitesiml intervl, sort of wht x becomes for infinite n. This interprettion ws lter bndoned due to the difficulty of resoning with infinitesimls, but we keep the nottion. Remrk: Note tht in intervls where f(x) is negtive the grph of y = f(x) lies below the x-xis nd the definite integrl tkes negtive vlue. In generl definite integrl gives the net re between the grph of y = f(x) nd the x-xis, i.e., the sum of the res of the regions where y = f(x) is bove the x-xis minus the sum of the res of the regions where y = f(x) is below the x-xis...2. Evluting Integrls. We will soon study simple nd efficient methods to evlute integrls, but here we will look t how to evlute integrls directly from the definition. Exmple: Find the vlue of the definite integrl x2 dx from its definition in terms of Riemnn sums. i=

8 .. AREAS AND DISTANCES. THE DEFINITE INTEGRAL 8 Answer: We divide the intervl [, ] into n equl prts, so x i = i/n nd x = /n. Next we must choose some point x i in ech subintervl [x i, x i ]. Here we will use the right endpoint of the intervl x i = i/n. Hence the Riemnn sum ssocited to this prtition is: n i= ( ) 2 i /n = n n 3 n i= i 2 = n 3 2n 3 + 3n 2 + n 6 = 2 + 3/n + 2/n2 6. So: x 2 dx = lim n 2 + 3/n + 2/n 2 6 = 3. In order to check tht the result does not depend on the smple points used, let s redo the computtion using now the left endpoint of ech subintervl: n ( i i= n ) 2 /n = n 3 n (i ) 2 = 2n 3 3n 2 + n = n 3 6 i= 2 3/n + 2/n2 6. So: x 2 dx = lim n 2 3/n + 2/n 2 6 = The Midpoint Rule. The Midpoint Rule consists of computing Riemnn sums using x i = (x i + x i )/2 = midpoint of ech intervl s smple point. This yields the following pproximtion for the vlue of definite integrl: b f(x) dx n f(x i ) x = x [f(x ) + f(x 2 ) + + f(x n )]. i= Exmple: Use the Midpoint Rule with n = 5 to pproximte x2 x. Answer: The subintervls re [,.2], [.2,.4], [.4,.6], [.6,.8], [.8, ], the midpoints re.,.3,.5,.7,.9, nd x = /5, so x 2 dx 5 [ ] =.65/5 =.33, which grees up to the second deciml plce with the ctul vlue /3.

9 .. AREAS AND DISTANCES. THE DEFINITE INTEGRAL The Distnce Problem. Here we show how the concept of definite integrl cn be pplied to more generl problems. In prticulr we study the problem of finding the distnce trveled by n object with vrible velocity during certin period of time. If the velocity v were constnt we could just multiply it by the time t: distnce = v t. Otherwise we cn pproximte the totl distnce trveled by dividing the totl time intervl into smll intervls so tht in ech of them the velocity vries very little nd cn cn be considered pproximtely constnt. So, ssume tht the body strts moving t time t strt nd finishes t time t end, nd the velocity is vrible, i.e., is function of time v = f(t). We divide the time intervl into n smll intervls [t i, t i ] of length t = (t end t strt )/n, choose some instnt t i between t i nd t i, nd tke v = f(t i ) s the pproximte velocity of the body between t i nd t i. Then the distnce trveled during tht time intervl is pproximtely f(t i ) t, nd the totl distnce cn be pproximted s the sum n f(t i ) t i= The result will be more ccurte the lrger the number of subintervls is, nd the exct distnce trveled will be limit of the bove expression s n goes to infinity: lim n n f(t i ) t Tht limit turns out to be the following definite integrl: tend t strt f(t) dt i=..5. Properties of the Definite Integrl. () Integrl of constnt: (2) Linerity: () (b) b b b [f(x) + g(x)] dx = cf(x) dx = c (3) Intervl Additivity b c dx = c (b ). b f(x) dx. f(x) dx + b g(x) dx.

10 .. AREAS AND DISTANCES. THE DEFINITE INTEGRAL c b b () f(x) dx + (b) (c) b f(x) dx = f(x) dx =. (4) Comprison: () f(x) c b b (b) f(x) g(x) f(x) dx = f(x) dx, f(x) dx. b f(x) dx (c) m f(x) M m (b ) f(x) dx. b b g(x) dx. f(x) dx M (b ).

11 .2. THE EVALUATION THEOREM.2. The Evlution Theorem.2.. The Evlution Theorem. If f is continuous function nd F is n ntiderivtive of f, i.e., F (x) = f(x), then b f(x) dx = F (b) F (). Exmple: Find x2 dx using the evlution theorem. Answer: An ntiderivtive of x 2 is x 3 /3, hence: [ ] x x 2 3 dx = = = Indefinite Integrls. If F is n ntiderivtive of function f, i.e., F (x) = f(x), then for ny constnt C, F (x) + C is nother ntiderivtive of f(x). The fmily of ll ntiderivtives of f is clled indefinite integrl of f nd represented: f(x) dx = F (x) + C. Exmple: x 2 dx = x3 3 + C Tble of Indefinite Integrls. We cn mke n integrl tble just by reversing tble of derivtives. () x n dx = xn+ + C (n ). n + (2) dx = ln x + C. x (3) e x dx = e x + C. (4) x dx = x ln + C. (5) sin x dx = cos x + C. (6) cos x dx = sin x + C. (7) sec 2 x dx = tn x + C.

12 (8) (9) () () (2) (3).2. THE EVALUATION THEOREM 2 csc 2 x dx = cot x + C. sec x tn x dx = sec x + C. csc x cot x dx = csc x + C. dx x 2 + = tn x + C. dx = x 2 sin x + C. dx x x 2 dx = sec x + C Totl Chnge Theorem. The integrl of rte of chnge is the totl chnge: b F (x) dx = F (b) F (). This is just resttement of the evlution theorem. As n exmple of ppliction we find the net distnce or displcement, nd the totl distnce trveled by n object tht moves long stright line with position function s(t). The velocity of the object is v(t) = s (t). The net distnce or displcement is the difference between the finl nd the initil positions of the object, nd cn be found with the following integrl t2 t v(t) dt = s(t 2 ) s(t ). In the computtion of the displcement the distnce trveled by the object when it moves to the left (while v(t) ) is subtrcted from the distnce trveled to the right (while v(t) ). If we wnt to find the totl distnce trveled we need to dd ll distnces with positive sign, nd this is ccomplished by integrting the bsolute vlue of the velocity: t2 t v(t) dt = totl distnce trveled. Exmple: Find the displcement nd the totl distnce trveled by n object tht moves with velocity v(t) = t 2 t 6 from t = to t = 4.

13 .2. THE EVALUATION THEOREM 3 Answer: The displcement is 4 [ t (t 2 3 t 6) dx = 3 t2 2 6t ] 4 ( ) ( ) 4 3 = = 32 3 ( 37 6 ) = 9 2 In order to find the totl distnce trveled we need to seprte the intervls in which the velocity tkes vlues of different signs. Those intervls re seprted by points t which v(t) =, i.e., t 2 t 6 = t = 2 nd t = 3. Since we re interested only in wht hppens in [, 4] we only need to look t the intervls [, 3] nd [3, 4]. Since v() = 6, the velocity is negtive in [, 3], nd since v(4) = 6, the velocity is positive in [3, 4]. Hence: 4 v(t) dt = = = 3 3 [ v(t)] dt (t 2 t 6) dt + [ t3 3 + t t ] 3 + = = 6 6. v(t) dt 4 3 [ t 3 (t 2 t 6) dt 3 t2 2 6t ] 4 3

14 .3. THE FUNDAMENTAL THEOREM OF CALCULUS 4.3. The Fundmentl Theorem of Clculus.3.. The Fundmentl Theorem of Clculus. The Fundmentl Theorem of Clculus (FTC) connects the two brnches of clculus: differentil clculus nd integrl clculus. It sys the following: Suppose f is continuous on [, b]. Then: () The function g(x) = x f(t) dt is n ntiderivtive of f, i.e., g (x) = f(x). (2) (Evlution Theorem) If F is n ntiderivtive of f, i.e. F (x) = f(x), then b f(x) dx = F (b) F (). The two prts of the theorem cn be rewritten like this: () (2) d dx b x f(t) dt = f(x). F (x) dx = F (b) F (). So the theorem sttes tht integrtion nd differentition re inverse opertions, i.e., the derivtive of n integrl of function yields the originl function, nd the integrl of derivtive lso yields the function originlly differentited (up to constnt). Exmple: Find d x 2 t 3 dt. dx Answer: We solve this problem in two wys. First directly: hence g(x) = x 2 [ t t 3 4 dt = 4 ] x 2 = (x2 ) 4 4 g (x) = 8x7 4 = 2x7. = x8 4,

15 .3. THE FUNDAMENTAL THEOREM OF CALCULUS 5 Second, using the FTC: h(u) = u t 3 dt h (u) = u 3. Now we hve g(x) = h(x 2 ), hence (using the chin rule): g (x) = h (x 2 ) 2x = (x 2 ) 3 2x = 2x 7.

16 .4. THE SUBSTITUTION RULE 6.4. The Substitution Rule.4.. The Substitution Rule. The substitution rule is trick for evluting integrls. It is bsed on the following identity between differentils (where u is function of x): Hence we cn write: du = u dx. f(u) u dx = f(u) du or using slightly different nottion: f(g(x)) g (x) dx = f(u) du where u = g(x). Exmple: Find + x2 2x dx. Answer: Using the substitution u = + x 2 we get u + x2 2x dx = u dx = u du = 2 3 u3/2 + C = 2 3 ( + x2 ) 3/2 + C. Most of the time the only problem in using this method of integrtion is finding the right substitution. Exmple: Find cos 2x dx. Answer: We wnt to write the integrl s cos u du, so cos u = cos 2x u = 2x, u = 2. Since we do not see ny fctor 2 inside the

17 .4. THE SUBSTITUTION RULE 7 integrl we write it, tking cre of dividing by 2 outside the integrl: cos 2x dx = 2 = 2 = 2 cos 2x 2 dx cos u u dx cos u du = 2 sin u + C (lwys remember to undo the substitution) = sin 2x + C. 2 In generl we need to look t the integrnd s function of some expression (which we will lter identify with u) multiplied by the derivtive of tht expression. Exmple: Find e x2 x dx. Answer: We see tht x is lmost, the derivtive of x 2, so we use the substitution u = x 2, u = 2x, hence in order to get u inside the integrl we do the following: e x2 x dx = 2 = 2 e x2 ( 2x) dx }{{}} {{ } e u du e u du = 2 eu + C = 2 e x2 + C. Sometimes the substitution is hrd to see until we mke some ingenious trnsformtion in the integrnd. Exmple: Find tn x dx.

18 .4. THE SUBSTITUTION RULE 8 Answer: Here the ide is to write tn x = sin x nd use tht cos x (cos x) = sin x, so we mke the substitution u = cos x, u = sin x: sin x u tn x dx = cos x dx = u dx = u du = ln u + C = ln cos x + C. In generl we need to identify inside the integrl some expression of the form f(u) u, where f is some function with known ntiderivtive. e x Exmple: Find e 2x + dx. Answer: Let s write e x e 2x + dx = k f(u) u dx (where k is some constnt to be determined lter) nd try to identify the function f, the rgument u nd its derivtive u. Since (e x ) = e x it seems nturl to chose u = e x, u = e x, so e 2x = u 2 nd e x e 2x + dx = u u 2 + dx = u 2 + du = tn u + C = tn (e x ) + C. There is no much more tht cn be sid in generl, the wy to lern more is just to prctice Other Chnges of Vrible. Sometimes rther thn mking substitution of the form u = function of x, we my try chnge of vrible of the form x = function of some other vrible such s t, nd write dx = x (t) dt, where x = derivtive of x respect to t. Exmple: Find x2 dx. Answer: Here we write x = sin t, so dx = cos t dt, x 2 = sin 2 t = cos 2 t, nd x2 dx = cos t }{{} x cos t dt = }{{} dx cos 2 t dt.

19 .4. THE SUBSTITUTION RULE 9 Since we do not know yet how to integrte cos 2 t we leve it like this nd will be bck to it lter (fter we study integrls of trigonometric functions) The Substitution Rule for Definite Integrls. When computing definite integrl using the substitution rule there re two possibilities: () Compute the definite integrl first, then use the evlution theorem: f(u) u dx = F (x) ; b f(u) u dx = F (b) F (). (2) Use the substitution rule for definite integrls: b f(u) u dx = u(b) u() f(u) du. The dvntge of the second method is tht we do not need to undo the substitution. Exmple: Find 4 2x + dx. Answer: Using the first method first we compute the indefinite integrl: 2x 2x + dx = + 2x dx (u = 2x + ) 2 = 2 u du = 3 u3/2 + C = 3 (2x + )3/2 + C. Then we use it for computing the definite integrl: 4 2x + dx = [ 3 (2x + )3/2 ] 4 = 3 93/2 3 3/2 = = 26 3.

20 .4. THE SUBSTITUTION RULE 2 In the second method we compute the definite integrl directly djusting the limits of integrtion fter the substitution: 4 4 2x + dx = 2x + 2x dx (u = 2x + ; u = 2) 2 = 9 u du 2 (note the chnge in the limits of integrtion to u() = nd u(4) = 9) [ ] 9 = 3 u3/2 = 3 93/2 3 3/2 = = 26 3.

21 .5. INTEGRATION BY PARTS 2.5. Integrtion by Prts The method of integrtion by prts is bsed on the product rule for differentition: which we cn write like this: [f(x)g(x)] = f (x)g(x) + f(x)g (x), f(x)g (x) = [f(x)g(x)] f (x)g(x). Integrting we get: f(x) g (x) dx = [f(x)g(x)] dx g(x)f (x) dx, i.e.: f(x)g (x) dx = f(x)g(x) g(x)f (x) dx. Writing u = f(x), v = g(x), we hve du = f (x) dx, dv = g (x) dx, hence: u dv = uv v du. Exmple: Integrte xe x dx by prts. Answer: In integrtion by prts the key thing is to choose u nd dv correctly. In this cse the right choice is u = x, dv = e x dx, so du = dx, v = e x. We see tht the choice is right becuse the new integrl tht we obtin fter pplying the formul of integrtion by prts is simpler thn the originl one: x e x dx = x e x e x dx = xe x e x + C. }{{}}{{}}{{}}{{}}{{}}{{} u dv u v v du Usully it is good ide to check the nswer by differentiting it: (xe x e x + C) = e x + xe x e x = xe x. A couple of dditionl typicl exmples: Exmple: x sin x dx =

22 .5. INTEGRATION BY PARTS 22 u = x, dv = sin x dx, so du = dx, v = cos x: = x sin x dx = x ( cos x) }{{}} {{ } }{{}} {{ } u dv u v = x cos x + sin x + C. ( cos x) } {{ } v dx }{{} du Exmple: ln x dx = u = ln x, dv = dx, so du = dx, v = x: x = ln x dx = ln x x }{{}}{{}}{{}}{{} u dv u v = x ln x dx x }{{} v = x ln x x + C. x dx }{{} du Sometimes we need to use the formul more thn once. Exmple: x 2 e x dx =... u = x 2, dv = e x dx, so du = 2x dx, v = e x : = x 2 e x dx = x 2 e x e x 2x dx =... }{{}}{{} u dv u = 2x, dv = e x dx, so du = 2dx, v = e x : = x 2 e x 2x e x dx = x 2 e x 2xe x + }{{}}{{} u dv 2e x dx = x 2 e x 2xe x + 2e x + C. In the following exmple the formul of integrtion by prts does not yield finl nswer, but n eqution verified by the integrl from which its vlue cn be derived. Exmple: sin x e x dx =...

23 .5. INTEGRATION BY PARTS 23 u = sin x, dv = e x dx, so du = cos x dx, v = e x : = sin x e x dx = sin x e x }{{}}{{} e x cos x dx =... u dv u = cos x, dv = e x dx, so du = sin x dx, v = e x :... = sin x e x cos x e x dx }{{}}{{} u dv = sin x e x cos x e x e x sin x dx Hence the integrl I = sin x e x dx verifies I = sin x e x cos x e x I, i.e., 2I = sin x e x cos x e x, hence I = 2 ex (sin x cos x) + C..5.. Integrtion by prts for Definite Integrls. Combining the formul of integrtion by prts with the Evlution Theorem we get: b b f(x)g (x) dx = [f(x)g(x)] b g(x)f (x) dx. Exmple: tn x dx = u = tn x, dv = dx, so du = dx, v = x: + x2 = tn x dx = [ tn x x ] }{{}}{{}}{{}}{{} u dv u v = [ tn tn ] = π 4 x + x 2 dx x + x dx }{{}}{{ 2 } v du x + x dx 2

24 .5. INTEGRATION BY PARTS 24 The lst integrl cn be computed with the substitution t = + x 2, dt = 2x dx: Hence the originl integrl is: x + x dx = t dt = 2 [ln t]2 = ln 2 2. tn x dx = π 4 ln Reduction Formuls. Assume tht we wnt to find the following integrl for given vlue of n > : x n e x dx. Using integrtion by prts with u = x n nd dv = e x dx, so v = e x nd du = nx n dx, we get: x n e x dx = x n e x n x n e x dx. On the right hnd side we get n integrl similr to the originl one but with x rised to n insted of n. This kind of expression is clled reduction formul. Using this sme formul severl times, nd tking into ccount tht for n = the integrl becomes e x dx = e x + C, we cn evlute the originl integrl for ny n. For instnce: x 3 e x dx = x 3 e x 3 x 2 e x dx = x 3 e x 3(x 2 e x 2 xe x dx) = x 3 e x 3(x 2 e x 2(xe x e x dx)) = x 3 e x 3(x 2 e x 2(xe x e x )) + C = x 3 e x 3x 2 e x + 6xe x 6e x + C.

25 Another exmple: sin n x dx =.5. INTEGRATION BY PARTS 25 sin n x sin x dx } {{ }} {{ } u dv = sin n x cos x + (n ) = sin n x cos x + (n ) (n ) cos 2 x sin n 2 dx }{{} sin 2 x sin n 2 dx sin n x dx Adding the lst term to both sides nd dividing by n we get the following reduction formul: sin n x dx = sinn x cos x + n sin n 2 x dx. n n

26 .6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS Trigonometric Integrls nd Trigonometric Substitutions.6.. Trigonometric Integrls. Here we discuss integrls of powers of trigonometric functions. To tht end the following hlf-ngle identities will be useful: sin 2 x = ( cos 2x), 2 cos 2 x = ( + cos 2x). 2 Remember lso the identities: sin 2 x + cos 2 x =, sec 2 x = + tn 2 x Integrls of Products of Sines nd Cosines. We will study now integrls of the form sin m x cos n x dx, including cses in which m = or n =, i.e.: cos n x dx ; sin m x dx. The simplest cse is when either n = or m =, in which cse the substitution u = sin x or u = cos x respectively will work. Exmple: sin 4 x cos x dx = (u = sin x, du = cos x dx) = u 4 du = u5 5 + C = sin5 x + C. 5 More generlly if t lest one exponent is odd then we cn use the identity sin 2 x+cos 2 x = to trnsform the integrnd into n expression contining only one sine or one cosine.

27 .6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 27 Exmple: sin 2 x cos 3 x dx = = sin 2 x cos 2 x cos x dx sin 2 x ( sin 2 x) cos x dx = (u = sin x, du = cos x dx) = u 2 ( u 2 ) du = (u 2 u 4 ) du = u3 3 u5 5 + C = sin3 x 3 sin5 x 5 + C. If ll the exponents re even then we use the hlf-ngle identities. Exmple: sin 2 x cos 2 x dx = ( cos 2x) ( + cos 2x) dx 2 2 = ( cos 2 2x) dx 4 = ( ( + cos 4x)) dx 2 4 = ( cos 4x) dx 8 = x 8 sin 4x 32 + C Integrls of Secnts nd Tngents. The integrl of tn x cn be computed in the following wy: sin x du tn x dx = cos x dx = = ln u + C = ln cos x + C, u where u = cos x. Anlogously cos x du cot x dx = sin x dx = u where u = sin x. = ln u + C = ln sin x + C,

28 .6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 28 The integrl of sec x is little tricky: sec x (tn x + sec x) sec x tn x + sec 2 x sec x dx = dx = dx = sec x + tn x sec x + tn x du = ln u + C = ln sec x + tn x + C, u where u = sec x + tn x, du = (sec x tn x + sec 2 x) dx. Anlogously: csc x dx = ln csc x + cot x + C. More generlly n integrl of the form tn m x sec n x dx cn be computed in the following wy: () If m is odd, use u = sec x, du = sec x tn x dx. (2) If n is even, use u = tn x, du = sec 2 x dx. Exmple: tn 3 x sec 2 x dx = Since in this cse m is odd nd n is even it does not mtter which method we use, so let s use the first one: (u = sec x, du = sec x tn x dx) = tn 2 x sec x tn x sec x dx = }{{}}{{}}{{} u 2 u du = (u 2 )u du (u 3 u) du = u4 4 u2 2 + C = 4 sec4 x 2 sec2 x + C. Next let s solve the sme problem using the second method:

29 .6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 29 (u = tn x, du = sec 2 x dx) tn 3 x sec 2 x dx = u 3 du = u4 4 + C = 4 tn4 x + C. }{{}}{{} u 3 du Although this nswer looks different from the one obtined using the first method it is in fct equivlent to it becuse they differ in constnt: 4 tn4 x = 4 (sec2 x ) 2 = 4 sec4 x 2 sec2 x +. 4 }{{} previous nswer.6.2. Trigonometric Substitutions. Here we study substitutions of the form x = some trigonometric function. Exmple: Find x2 dx. Answer: We mke x = sin t, dx = cos t dt, hence x2 = sin 2 t = cos 2 t = cos t, nd x2 dx = = = = t 2 = t 2 cos t cos t dt cos 2 t dt ( + cos 2t) dt (hlf-ngle identity) 2 sin 2t + + C 4 2 sin t cos t + 4 = t 2 + sin t sin 2 t 2 = sin x 2 + x x C (double-ngle identity) + C + C. The following substitutions re useful in integrls contining the following expressions:

30 .6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 3 expression substitution identity 2 u 2 u = sin t sin 2 t = cos 2 t 2 + u 2 u = tn t + tn 2 t = sec 2 t u 2 2 u = sec t sec 2 t = tn 2 t So for instnce, if n integrl contins the expression 2 u 2, we my try the substitution u = sin t nd use the identity sin 2 t = cos 2 t in order to trnsform the originl expression in this wy: 2 u 2 = 2 ( sin 2 t) = 2 cos 2 t. Exmple: x 3 dx = 27 9 x 2 sin 3 t cos t dt (x = 3 sin t) sin 2 t = 27 sin 3 t dx = 27 ( = 27 cos t + cos3 t 3 ( = 27 ( cos 2 t) sin t dx ) + C ) sin 2 t + 3 ( sin2 t) 3/2 + C = 9 9 x (9 x2 ) 3/2 + C. where x = 3 sin t, dx = 3 cos t dt.

31 .6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 3 Exmple: x2 dx = x2 dx (x = 3 tn t) 2 3 = 2 + tn 2 t sec2 t dt = 9 sec 3 t dt 2 = 9 4 (sec t tn t + ln sec t + tn t ) + C = 9 ( x ) x2 + ln 2x x2 + C = x 9 + 4x 2 2 where x = 3 2 tn t, dx = 3 2 sec2 t dt ln 2x x 2 + C. Exmple: x2 x sec2 t dx = sec t = tn 2 t dt sec t tn t dt (x = sec t) = tn t t + C = sec 2 t t + C = x 2 sec x + C. where x = sec t, dx = sec t tn t dt.

32 .7. PARTIAL FRACTIONS Prtil Frctions.7.. Rtionl Functions nd Prtil Frctions. A rtionl function is quotient of two polynomils: R(x) = P (x) Q(x). Here we discuss how to integrte rtionl functions. The ide consists of rewriting the rtionl function s sum of simpler frctions clled prtil frctions. This cn be done in the following wy: () Use long division of polynomils to get quotient p(x) nd reminder r(x). Then write: R(x) = P (x) Q(x) = p(x) + r(x) Q(x), where the degree of r(x) is less thn tht of Q(x). (2) Fctor the denomintor Q(x) = q (x)q 2 (x)... q n (x), where ech fctor q i (x) is either liner x+b, or irreducible qudrtic x 2 +bx+c, or power of the form (x+b) n or (x 2 +bx+c) n. (3) Decompose r(x)/q(x) into prtil frctions of the form: r(x) Q(x) = F (x) + F 2 (x) + F 3 (x) + where ech frction is of the form A F i (x) = (x + b) k or Ax + B F i (x) = (x 2 + bx + c), k where k n (n is the exponent of x + b or x 2 + bx + c in the fctoriztion of Q(x).) Exmple: Decompose the following rtionl function into prtil frctions: R(x) = x3 + x x 2 Answer:

33 () x3 + x x 2.7. PARTIAL FRACTIONS 33 = x + + x + 3 x 2 (2) x 2 = (x + )(x ). x + 3 (3) (x + )(x ) = A x + + B x. Multiplying by (x + )(x ) we get: x + 3 = A(x ) + B(x + ). Now there re two wys of finding A nd B: Method. Expnd the right hnd side of (*), collect terms with the sme power of x, nd identify coefficients of the polynomils obtined on both sides: x + 3 = (A + B) x + (B A), Hence: { = A + B (coefficient of x) 3 = A + B (constnt term) Method 2. In (*) give x two different vlues (s mny s the number of coefficients to determine), sy x = nd x =. We get: { 4 = 2B (x = ) 2 = 2A (x = ) The solution to the system of equtions obtined in either cse is A =, B = 2, so: x + 3 (x + )(x ) = x x. ( ) Finlly: R(x) = x3 + x x 2 = x + x x Fctoring Polynomil. In order to fctor polynomil Q(x) (with rel coefficients) into liner of irreducible qudrtic fctors, first solve the lgebric eqution: Q(x) =. Then for ech rel root r write fctor of the form (x r) k where k is the multiplicity of the root. For ech pir of conjugte complex roots r, r write fctor (x 2 sx + p) k, where s = r + r, p = r r, nd k is the

34 .7. PARTIAL FRACTIONS 34 multiplicity of those roots. Finlly multiply by the leding coefficient of Q(x). Note tht the eqution Q(x) = is sometimes hrd to solve, or only the rel roots cn be esily found (when they re integrl or rtionl they cn be found by Ruffini s rule, or just by tril nd error). In tht cse we get s mny roots s we cn, nd divide Q(x) by the fctors found. The quotient is nother polynomil q(x) which we must now try to fctor. So pose the lgebric eqution q(x) = nd try to solve it for this new (nd simpler) polynomil. Exmple: Fctor the polynomil Q(x) = x 6 x 5 5x 4 + 5x 3 + 7x 2 + 2x 72. Answer: The roots of Q(x) re (simple), 2 (triple) nd 3 (double), hence: Q(x) = (x )(x + 2) 3 (x 3) 2. Exmple: Fctor Q(x) = x 3 + 2x 2 + 2x +. Answer: Q(x) hs simple rel root x =. After dividing Q(x) by x + we get the polynomil x 2 + x +, which is irreducible (it hs only complex roots), so we fctor Q(x) like this: Q(x) = (x + )(x 2 + x + ) Decomposing Into Prtil Frctions. Assume tht Q(x) hs lredy been fctored nd degree of r(x) is less thn degree of Q(x). Then r(x)/q(x) is decomposed into prtil frctions in the following wy: () For ech fctor of the form (x r) k write A x r + A 2 (x r) 2 + A 3 (x r) A k (x r) k, where A... A k re coefficients to be determined. (2) For ech fctor of the form (x 2 + bx + c) k write B x + C x 2 + bx + c + B 2 x + C 2 (x 2 + bx + c) + + B kx + C k 2 (x 2 + bx + c) k where B... B k nd C... C k re coefficients to be determined.

35 .7. PARTIAL FRACTIONS 35 (3) Multiply by Q(x) nd simplify. This leds to n expression of the form r(x) = some polynomil contining the indeterminte coefficients A i, B i, C i. Finlly determine the coefficients A i, B i, C i. One wy of doing this is by identifying coefficients of the polynomils on both sides of the lst expression. Another wy is to write system of equtions with unknowns A i, B i, C i by giving x vrious vlues. Exmple: Decompose the following rtionl function into prtil frctions: R(x) = 4x5 2x 4 + 2x 3 8x 2 2x 3 (x ) 2 (x 2 + x + ) 2. Answer: The denomintor is lredy fctored, so we proceed with the next step: 4x 5 2x 4 + 2x 3 8x 2 2x 3 (x ) 2 (x 2 + x + ) 2 = A x + B (x ) + Cx + D 2 x 2 + x + + Next we multiply by the denomintor: 4x 5 2x 4 + 2x 3 8x 2 2x 3 = A(x )(x 2 + x + ) 2 + B(x 2 + x + ) 2 Ex + F (x 2 + x + ) 2. + (Cx + D)(x ) 2 (x 2 + x + ) + (Ex + F )(x ) 2 = (A + C)x 5 + (A C + D + B)x 4 + (A + 2B + E D)x 3 + ( A + 3B C 2E + F )x 2 + ( A + 2B + C D + E 2F )x + ( A + B + D + F ). Identifying coefficients on both sides we get: A + C = 4 A + B C + D = 2 A + 2B D + E = 2 A + 3B C 2E + F = 8 A + 2B + C D + E 2F = 2 A + B + D + F = 3

36 .7. PARTIAL FRACTIONS 36 The solution to this system of equtions is A = 2, B =, C = 2, D =, E =, F =, hence: 4x 5 2x 4 + 2x 3 8x 2 2x 3 (x ) 2 (x 2 + x + ) 2 = 2 x (x ) + 2x 2 x 2 + x + + x + (x 2 + x + ) Integrtion of Rtionl Functions. After decomposing the rtionl function into prtil frctions ll we need to do is to integrte expressions of the form A/(x r) k nd (Bx + C)/(x 2 + bx + c) k. For the former we get: A (x r) dx = A + C if k k (k )(x r) k A dx = A ln x r + C if k = x r The ltter re more involved, but the following re prticulrly simple specil cses: x 2 + dx = 2 rctn x + C x x 2 + dx = 2 2 ln (x2 + 2 ) + C x (x ) dx = + C (k ) k 2(k )(x ) k Exmple: Find the following integrl: x 3 x 2 7x + 8 x 2 4x + 4 dx. Answer: First we decompose the integrnd into prtil frctions: () x3 x 2 7x + 8 x 2 4x + 4 = x (2) x 2 4x + 4 = (x 2) 2. (3) x 4 (x 2) = A 2 x 2 + B (x 2) 2 x 4 = A(x 2) + B x = 2 2 = B x = 3 = A + B x 4 x 2 4x + 4

37 So A =, B = 2, nd.7. PARTIAL FRACTIONS 37 x 4 x 2 4x + 4 = x 2 2 (x 2) 2 Hence: x 3 x 2 7x + 8 x 2 4x + 4 = x x 2 2 (x 2) 2. Finlly we integrte: x 3 x 2 7x + 8 dx = (x 2) 2 = x2 2 (x + 3) dx + x 2 dx + 3x + ln x x 2 + C. 2 (x 2) 2 dx.7.5. Completing the Squre. Mny integrls contining n irreducible (no rel roots) qudrtic polynomil x 2 + bx + c cn be simplified by completing the squre, i.e., writing the polynomil s u 2 + r where u = px + q, e.g.: In generl: x 2 + bx + c = x 2 + 2x + 2 = (x + ) 2 +. If = the formul cn be simplified: x 2 + bx + c = ( x + b ) 2 2 b2 4c. 4 ( x + b 2) 2 + ) (c b2. 4 This result is of the form u 2 ± A 2, where u = x + b/2. Exmple: x 2 + 6x + dx = (x + 3) 2 + dx = du (u = x + 3) u 2 + = tn u + C = tn (x + 3) + C.

38 Exmple: x2 (x 4x + 5 dx = 2)2 + dx.7. PARTIAL FRACTIONS 38 = u2 + du (u = x 2) = tn 2 t + sec 2 t dt (u = tn t) = sec 3 t dt sec t tn t = + ln sec t + tn t + C 2 2 = u u ln u + u C = (x 2) x 2 4x (x 2 ln 2) + x2 4x C.

39 .8. INTEGRATION USING TABLES AND CAS Integrtion using Tbles nd CAS The use of tbles of integrls nd Computer Algebr Systems llow us to find integrls very quickly without hving to perform ll the steps for their computtion. However we often need to modify slightly the originl integrl nd perhps complete or simplify the nswer. Exmple: Find the following integrl using the tbles t the end of Stewrd s book: x2 dx =. x Answer: In the tbles we find the following formul No. 4: 2 u 2 du = u 2 u 2 cos u + C, hence, letting =, u = we get the nswer: x 2 dx = x x 2 cos x + C. Exmple: Find the integrl: x 2 dx = 9 + 4x 2 Answer: In the tbles the formul tht resembles this integrl most is No. 26: u 2 du 2 + u = u ( 2 + u ln u + ) 2 + u 2 + C, hence mking = 3, u = 2x: x 2 dx = u 2 du 9 + 4x u 2 = { u ( 2 + u ln u + ) } 2 + u 2 + C = x 9 + 4x2 9 (2x 8 6 ln + ) 9 + 4x 2 + C. Exmple: Find the sme integrl using Mple.

40 .8. INTEGRATION USING TABLES AND CAS 4 Answer: In Mple we enter t the prompt: > int(x^2/sqrt(9+4*x^2),x); nd it returns: x 9 + 4x2 9 ( ) rcsinh 3 x First we notice tht the nswer omits the constnt C. On the other hnd, it involves n inverse hyperbolic function: rcsinh x = ln (x + ) + x 2, hence the nswer provided by Mple is: ( x 9 + 4x2 9 ) 8 6 ln 2x x2 = 9 x 9 + 4x2 9 (2x 8 6 ln + ) 9 + 4x ln(3), so it differs from the nswer found using the tbles in constnt 9 ln(3) 32 which cn be bsorbed into the constnt of integrtion.

41 .9. NUMERICAL INTEGRATION 4.9. Numericl Integrtion Sometimes the integrl of function cnnot be expressed with elementry functions, i.e., polynomil, trigonometric, exponentil, logrithmic, or suitble combintion of these. However, in those cses we still cn find n pproximte vlue for the integrl of function on n intervl..9.. Trpezoidl Approximtion. A first ttempt to pproximte the vlue of n integrl b f(x) dx is to compute its Riemnn sum: n R = f(x i ) x. i= Where x = x i x i = (b )/n nd x i is some point in the intervl [x i, x i ]. If we choose the left endpoints of ech intervl, we get the left-endpoint pproximtion: n L n = f(x i )) x = ( x){f(x ) + f(x ) + + f(x n )}, i= Similrly, by choosing the right endpoints of ech intervl we get the right-endpoint pproximtion: n R n = f(x i ) x = ( x){f(x ) + f(x 2 ) + + f(x n )}. i= The trpezoidl pproximtion is the verge of L n nd R n : T n = 2 (L n+r n ) = x 2 {f(x )+2f(x )+2f(x 2 )+ +2f(x n )+f(x n )}. Exmple: Approximte x2 dx with trpezoidl pproximtion using 4 intervls. Solution: We hve x = /4 =.25. The vlues for x i nd f(x i ) = x 2 i cn be tbulted in the following wy: i x i f(x i )

42 .9. NUMERICAL INTEGRATION 42 Hence: L 4 =.25 ( ) =.2875, R 4 =.25 ( ) = So: T 4 = 2 (L 4 + R 4 ) = ( ) = Compre to the exct vlue of the integrl, which is /3 = Midpoint Approximtion. Alterntively, in the Riemnn sum we cn use the middle point x i = (x i + x i )/2 of ech intervl [x i, x i ]. Then the midpoint pproximtion of b f(x) dx is M n = ( x){f(x ) + f(x 2 ) + + f(x n )}. Exmple: Approximte x2 dx with midpoint pproximtion using 4 intervls. Solution: We hve: Hence: i x i f(x i ) M 4 =.25 ( ) = Simpson s Approximtion. Simpson s pproximtion is weighted verge of the trpezoidl nd midpoint pproximtions ssocited to the intervls [x, x 2 ], [x 2, x 4 ],..., [x n 2, x n ] (of length

43 .9. NUMERICAL INTEGRATION 43 2 x ech): S 2n = 3 (2M n + T n ) = [ 2(2 x){f(x ) + f(x 3 ) + + f(x 2n )} x 2 {f(x ) + 2f(x 2 ) + 2f(x 4 ) + + 2f(x n 2 ) + f(x n )} = x 3 {f(x ) + 4f(x ) + 2f(x 2 ) + 4f(x 3 ) + 2f(x 4 ) + + 2f(x 2n 2) + 4f(x 2n ) + f(x 2n )}. Exmple: Approximte x2 dx with Simpson s pproximtion using 8 intervls. Solution: We use the previous results nd get: S 8 = 3 (2M 4 + T 4 ) = ( ) = /3. 3 Note: in this prticulr cse Simpson s pproximtion gives the exct vlue in generl it just gives good pproximtion Error Bounds. Here we give wy to estimte the error or difference E between the ctul vlue of n integrl nd the vlue obtined using numericl pproximtion Error Bound for the Trpezoidl Approximtion. Suppose f (x) K for x b. Then the error E T in the trpezoidl pproximtion verifies: ] E T K(b )3 2n Error Bound for the Midpoint Approximtion. Suppose f (x) K for x b. Then the error E M in the trpezoidl pproximtion verifies: K(b )3 E M. 24n Error Bound for the Simpson s Rule. Suppose f (4) (x) K for x b. Then the error E S in the Simpson s rule verifies: E S K(b )5 8n 4.

44 .9. NUMERICAL INTEGRATION 44 Exmple: Approximte the vlue of π using the trpezoidl, midpoint nd Simpson s pproximtions of for n = 4. Estimte the error. Answer: First note tht: x 2 dx + x 2 dx = 4 [ tn x ] = 4π 4 = π, so by pproximting the given integrl we re in fct finding pproximted vlues for π. Now we find the requested pproximtions: () Trpezoidl pproximtion: T 4 = /4 {f() + 2f(/4) + 2f(/2) + 2f(3/4) + f()} 2 = For estimting the error we need the second derivtive of f(x) = 4/( + x 2 ), which is f (x) = 8(3x 2 )/( + x 2 ) 3 so we hve f (x) = 8 3x2 8(3x2 + ) + x 2 3 ( + x 2 ) 3 for x, hence 8(3 2 + ) = ( )3 E T = (2) Midpoint pproximtion: T M = {f(/8) + f(3/8) + f(5/8) + f(7/8)} 4 = The error estimte is: E M 32 ( ) =

45 (3) Simpson s rule:.9. NUMERICAL INTEGRATION 45 T S = /4 {f() + 4f(/4) + 2f(/2) + 4f(3/4) + f()} 3 = For the error estimte we now need the fourth derivtive: f (4) (x) = 96(5x 4 x 2 + )/( + x 2 ) 5, so f (4) 96(5 + + ) (x) = 536 for x. Hence the error estimte is 536 ( )5 E S =

46 .. IMPROPER INTEGRALS 46.. Improper Integrls... Improper Integrls. Up to now we hve studied integrls of the form b f(x) dx, where f is continuous function defined on the closed nd bounded intervl [, b]. Improper integrls re integrls in which one or both of these conditions re not met, i.e., () The intervl of integrtion is not bounded: [, + ), (, ], (, + ), e.g.: x dx. 2 (2) The integrnd hs n infinite discontinuity t some point c in [, b]: e.g.: lim f(x) = ±. x c x dx...2. Infinite Limits of Integrtion. Improper Integrls of Type. In cse one of the limits of integrtion is infinite, we define: or f(x) dx = lim t t f(x) dx = lim t t f(x) dx, f(x) dx. If both limits of integrtion re infinite, then we choose ny c nd define: f(x) dx = c f(x) dx + c f(x) dx. If the limits defining the integrl exist the integrl is clled convergent, otherwise it is clled divergent. Remrk: Sometimes we write [F (x)] [F (x)] = lim t [F (x)]t. s n bbrevition for

47 Anlogously: nd.. IMPROPER INTEGRALS 47 [F (x)] = lim t [F (x)] t, [F (x)] = [F (x)]c + [F (x)] c = lim t [F (x)]c t + lim t [F (x)]t c. Exmple: dx = lim x2 t t or in simplified nottion: x 2 dx = [ dx = lim ] t = lim ( t ) x2 t x + =, t [ ] = lim ( t ) x + =. t Exmple: For wht vlues of p is the following integrl convergent?: x dx. p Answer: If p = then we hve t x dx = [ln x]t = ln t, so t dx = lim dx = lim x t x ln t =, p t nd the integrl is divergent. Now suppose p : t [ ] x p+ t x dx = = { } p p + p t p If p > then p > nd { } dx = lim xp t p t =, p hence the integrl is convergent. On the other hnd if p < then p <, p > nd { dx = lim xp t p } =, t p hence the integrl is divergent. So: dx is convergent if p > nd divergent if p. xp

48 .. IMPROPER INTEGRALS Infinite Integrnds. Improper Integrls of Type 2. Assume f is defined in [, b) but Then we define b lim f(x) = ±. x b f(x) dx = lim t b t f(x) dx. Anlogously, if f is defined in (, b] but Then we define b lim f(x) = ±. x + f(x) dx = lim t + b t f(x) dx. Finlly, if f(x) hs n infinite discontinuity t c inside [, b], then the definition is b f(x) dx = c f(x) dx + b c f(x) dx. If the limits defining the integrl exist the integrl is clled convergent, otherwise it is clled divergent. Remrk: If the intervl of integrtion is [, b) sometimes we write [F (x)] b s n bbrevition for lim [F (x)]t t b nd nlogously for intervls of the form (, b]. Exmple: x dx = lim t or in simplified nottion: Exmple: Evlute t [ ] ( dx = lim 2 x = lim 2 2 ) t = 2, x t t t dx = [ 2 x ] ( = lim 2 2 ) t = 2. x t ln x dx.

49 lim x.. IMPROPER INTEGRALS 49 Answer: The function ln x hs verticl symptote t x = becuse ln x =. Hence: + ln x dx = lim t + t ln x dx = lim [x ln x x] t + t = lim {( ln ) (t ln t t)} t + = lim {t t ln t} ( t + =. lim t ln t = ) t Comprison Test for Improper Integrls. Suppose f nd g re continuous functions such tht f(x) g(x) for x. () If (2) If f(x) dx if convergent then g(x) dx if divergent then A similr sttement holds for type 2 integrls. Exmple: Prove tht Answer: We hve: e x2 dx = e x2 dx is convergent. e x2 dx + g(x) dx is convergent. f(x) dx is divergent. e x2 dx. The first integrl on the right hnd side is n ordinry definite integrl so we only need to show tht the second integrl is convergent. In fct, for x we hve x 2 x, so e x2 e x. On the other hnd: hence t e x dx = [ e x] t = e t + e, e x dx = lim t ( e t + e ) = e, so e x dx is convergent. Hence, by the comprison theorem e x2 dx is convergent, QED.

50 CHAPTER 2 Applictions of Integrtion 2.. More bout Ares 2... Are Between Two Curves. The re between the curves y = f(x) nd y = g(x) nd the lines x = nd x = b (f, g continuous nd f(x) g(x) for x in [, b]) is A = b f(x) dx b g(x) dx = Clling y T = f(x), y B = g(x), we hve: b [f(x) g(x)] dx. A = b (y T y B ) dx Exmple: Find the re between y = e x nd y = x bounded on the sides by x = nd x =. Answer: First note tht e x x for x. So: ] ) ) A = (e x x) dx = [e x x2 = (e 2 (e = e 2 = e 3 2. The re between two curves y = f(x) nd y = g(x) tht intersect t two points cn be computed in the following wy. First find the intersection points nd b by solving the eqution f(x) = g(x). Then find the difference: b f(x) dx b g(x) dx = b [f(x) g(x)] dx. If the result is negtive tht mens tht we hve subtrcted wrong. Just tke the result in bsolute vlue. 5

51 2.. MORE ABOUT AREAS 5 Exmple: Find the re between y = x 2 nd y = 2 x. Solution: First, find the intersection points by solving x 2 (2 x) = x 2 +x 2 =. We get x = 2 nd x =. Next compute: 2 Hence the re is 9/2. (x 2 (2 x)) dx = 2 (x 2 + x 2) dx = 9/2. Sometimes it is esier or more convenient to write x s function of y nd integrte respect to y. If x L (y) x R (y) for p y q, then the re between the grphs of x = x L (y) nd x = x R (y) nd the horizontl lines y = p nd y = q is: A = q p (x R x L ) dy Exmple: Find the re between the line y = x nd the prbol y 2 = 2x x 2 4 Answer: The intersection points between those curves re (, 2) nd (5, 4), but in the figure we cn see tht the region extends to the left of x =. In this cse it is esier to write x L = 2 y2 3, x R = y +, nd integrte from y = 2 to y = 4: A = 4 2 (x R x L ) dx = = = { (y + ) ( 2 y2 3) } dx ( 2 y2 + y + 4 ) dx [ y3 6 + y y = 8 ] 4 2

52 2.2. VOLUMES Volumes Volumes by Slices. First we study how to find the volume of some solids by the method of cross sections (or slices ). The ide is to divide the solid into slices perpendiculr to given reference line. The volume of the solid is the sum of the volumes of its slices Volume of Cylinders. A cylinder is solid whose cross sections re prllel trnsltions of one nother. The volume of cylinder is the product of its height nd the re of its bse: V = Ah Volume by Cross Sections. Let R be solid lying longside some intervl [, b] of the x-xis. For ech x in [, b] we denote A(x) the re of the cross section of the solid by plne perpendiculr to the x-xis t x. We divide the intervl into n subintervls [x i, x i ], of length x = (b )/n ech. The plnes tht re perpendiculr to the x-xis t the points x, x, x 2,..., x n divide the solid into n slices. If the cross section of R chnges little long subintervl [x i, x i ], the slb positioned longside tht subintervl cn be considered cylinder of height x nd whose bse equls the cross section A(x i ) t some point x i in [x i, x i ]. So the volume of the slice is V i A(x i ) x. The totl volume of the solid is n V = V i i= n A(x i ) x. Once gin we recognize Riemnn sum t the right. In the limit s n we get the so clled Cvlieri s principle: V = b i= A(x) dx. Of course, the formul cn be pplied to ny xis. For instnce if solid lies longside some intervl [, b] on the y xis, the formul becomes V = b A(y) dy. Exmple: Find the volume of cone of rdius r nd height h.

53 2.2. VOLUMES 53 Answer: Assume tht the cone is plced with its vertex in the origin of coordintes nd its xis on the x-xis. The x coordinte runs through the intervl [, h]. The cross section of the cone t ech point x is circulr disk of rdius xr/h, hence its re is A(x) = π(xr/h) 2 = πr 2 x 2 /h 2. The volume of the cone cn now be computed by Cvlieri s formul: V = h πr 2 h 2 x2 dx = πr2 h 2 [ x 3 3 ] h = πr2 h 2 h 3 3 = 3 πr2 h Solids of Revolution. Consider the plne region between the grph of the function y = f(x) nd the x-xis long the intervl [, b]. By revolving tht region round the x-xis we get solid of revolution. Now ech cross section is circulr disk of rdius y, so its re is A(x) = πy 2 = π[f(x)] 2. Hence, the volume of the solid is V = b πy 2 dx = b π[f(x)] 2 dx. Exmple: Find the volume of cone of rdius r nd height h. Answer: Assume tht the cone is plced with its vertex in the origin of coordintes nd its xis on the x-xis. This cone cn be obtined by revolving the re under the line y = rx/h between x = nd x = h round the x-xis. So its volume is V = h ( rx ) 2 h π dx = h πr 2 h 2 x2 dx = πr2 h 2 [ x 3 3 ] h = πr2 h 2 h 3 3 = 3 πr2 h. If the revolution is performed round the y-xis, the roles of x nd y re interchnged, so in tht cse the formul is V = b πx 2 dy, where x must be written s function of y. If the region being revolved is the re between two curves y = f(x) nd y = g(x), then ech cross section is n nnulr ring (or wsher)with outer rdius f(x) nd inner rdius g(x) (ssuming f(x) g(x).) The re of the nnulr ring is A(x) = π(f(x) 2 g(x) 2 ), hence the volume of the solid will be: V = b π [ (y T ) 2 (y B ) 2] dx = b π [ f(x) 2 g(x) 2] dx.

54 2.2. VOLUMES 54 If the revolution is performed round the y-xis, then: V = b π [ (x R ) 2 (x L ) 2] dy. Exmple: Find the volume of the solid obtined by revolving the re between y = x 2 nd y = x round the x-xis. Solution: First we need to find the intersection points of these curves in order to find the intervl of integrtion: { y = x 2 y = (x, y) = (, ) nd (x, y) = (, ), x hence we must integrte from x = to x = : V = π [ x 2 = π 2 x5 5 [ ( x) 2 (x 2 ) 2] ( dx = π ) x x 4 dx ] ( = π 2 ) = 3π Volumes by Shells. Next we study how to find the volume of some solids by the method of shells. Now the ide is to divide the solid into shells nd dd up their volumes Volume of Cylindricl Shell. A cylindricl shell is the region between two concentric circulr cylinders of the sme height h. If their rdii re r nd r 2 respectively, then the volume is: {}}{{}}{ V = πr2h 2 πrh 2 = πh(r2 2 r) 2 = πh (r 2 + r ) (r 2 r ) = 2πrth, where r = (r 2 + r )/2 is the verge rdius, nd t = r 2 r is the thickness of the shell. 2r t Volumes by Cylindricl Shells. Consider the solid generted by revolving round the y-xis the region under the grph of y = f(x) between x = nd x = b. We divide the intervl [, b] into n subintervls [x i, x i ] of length x = (b )/n ech. The volume V of the solid is the sum of the volumes V i of the shells determined by the prtition. Ech shell, obtined by revolving the region under y = f(x) over the subintervl [x i, x i ], is pproximtely cylindricl. Its height

55 2.2. VOLUMES 55 is f(x i ), where x i is the midpoint of [x i, x i ]. Its thickness is x. Its verge rdius is x i. Hence its volume is nd the volume of the solid is n V = V i V i 2πx i f(x i ) x, i= n 2πx i f(x i ) x. i= As n the right Riemnn sum converges to the following integrl: V = b 2πxf(x) dx = b 2πxy dx. Exmple: Find the volume of the solid obtined by revolving round the y-xis the plne re between the grph of y = x 2 nd the x-xis. Answer: The grph intersects the positive x-xis t x =, so the intervl is [, ]. Hence V = 2πxy dx = 2πx ( x 2 ) dx = 2π [ ] x 2 = 2π 2 x4 = 2π 4 (x x 3 ) dx ( 2 ) = π Revolving the Region Between Two Curves. Here we find the volume of the solid obtined by revolving round the y-xis the re between two curves y = f(x) nd y = g(x) over n intervl [, b]. The computtion is similr, but if f(x) g(x) the shells will hve height f(x i ) g(x i ), so the volume will be given by the integrl: V = b 2πx(f(x) g(x)) dx = b 2πx(y T y B ) dx. Exmple: Find the volume of the solid obtined by revolving the plne region limited by the curves y = x nd y = x 2 over the intervl [, ]. V = Answer: In [, ] we hve x x 2, so: = 2π 2πx (y T y B ) dx = 2π [ x (x 2 x 3 3 ) dx = 2π 3 x4 4 x (x x 2 ) dx ] ( = 2π 3 ) = 2π 4 2 = π 6.

56 2.2. VOLUMES 56 If the region is revolved round the x-xis then the vribles x nd y reverse their roles: V = b 2πy (x R x L ) dy Revolving Around n Arbitrry Line. If the plne region is revolved round verticl line y = c, the rdius of the shell will be x c (or c x, whichever is positive) insted of x, so the formul becomes: V = b 2π(x c)(f(x) g(x)) dx = b 2π(x c)(y T y B ) dx. Similrly, if the region is revolved round the horizontl line x = c, the formul becomes: V = b 2π(y c)(f(y) g(y)) dy = b where y c must be replced by c y if c > y. 2π(y c)(x R x L ) dy,

57 2.3. ARC LENGTH, PARAMETRIC CURVES Arc Length, Prmetric Curves Prmetric Curves. A prmetric curve cn be thought of s the trjectory of point tht moves trough the plne with coordintes (x, y) = (f(t), g(t)), where f(t) nd g(t) re functions of the prmeter t. For ech vlue of t we get point of the curve. Exmple: A prmetric eqution for circle of rdius nd center (, ) is: x = cos t, y = sin t. The equtions x = f(t), y = g(t) re clled prmetric equtions. Given prmetric curve, sometimes we cn eliminte t nd obtin n equivlent non-prmetric eqution for the sme curve. For instnce t cn be eliminted from x = cos t, y = sin t by using the trigonometric reltion cos 2 t + sin 2 t =, which yields the (non-prmetric) eqution for circle of rdius nd center (, ): x 2 + y 2 =. Exmple: Find non-prmetric eqution for the following prmetric curve: x = t 2 2t, y = t +. Answer: We eliminte t by isolting it from the second eqution: t = (y ), nd plugging it in the first eqution: i.e.: x = (y ) 2 2(y ). x = y 2 4y + 3, which is prbol with horizontl xis Arc Length. Here we describe how to find the length of smooth rc. A smooth rc is the grph of continuous function whose derivtive is lso continuous (so it does not hve corner points). If the rc is just stright line between two points of coordintes (x, y ), (x 2, y 2 ), its length cn be found by the Pythgoren theorem: L = ( x) 2 + ( y) 2, where x = x 2 x nd y = y 2 y.

58 2.3. ARC LENGTH, PARAMETRIC CURVES 58 More generlly, we pproximte the length of the rc by inscribing polygonl rc (mde up of stright line segments) nd dding up the lengths of the segments. Assume tht the rc is given by the prmetric functions x = f(x), y = g(x), t b. We divide the intervl into n subintervls of equl length. The corresponding points in the rc hve coordintes (f(t i ), g(t i )), so two consecutive points re seprted by distnce equl to L i = [f(t i ) f(t i )] 2 + [g(t i ) g(t i )] 2. We hve t = t i t i = (b )/n. On the other hnd, by the men vlue theorem f(t i ) f(t i ) = f (t i ) t for some t i in [t i, t i ]. Hence g(t i ) f(t i ) = g (t i ) t L i = [f (x i ) t]2 + [g (t i ) t]2 = [f (t i )]2 + [g (t i )]2 t. The totl length of the rc is n n L s i = [f (t i )]2 + [g (t i )]2 t, i= i= which converges to the following integrl s n : L = b [f (t)] 2 + [g (t)] 2 dt. This formul cn lso be expressed in the following (esier to remember) wy: b (dx ) 2 ( ) 2 dy L = + dt dt dt The lst formul cn be obtined by integrting the length of n infinitesiml piece of rc ds = (dx ) 2 ( ) 2 dy (dx) 2 + (dy) 2 = dt +. dt dt Exmple: Find the rc length of the curve x = t 2, y = t 3 between (, ) nd (4, 8).