CALCULUS II LECTURE NOTES. Contents. 1. Notation 3

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1 CALCULUS II LECTURE NOTES MATTHEW BATES Contents. Nottion 3. Section 5. : The Definite Integrl 4.. Wht is Are? 4.. Riemnn Integrls 5 3. Section 5.3 : Fundmentl Theorem of Clculus 9 4. Section 5.4 : Indefinite Integrls nd Net Chnge Theorem Indefinite Integrls Net chnge 5. Section 5.5 : Substitution Rules Indefinite Integrls Using Substitution Definite integrls using substitution Symmetry 9 6. Section 6. : Are between curves 3 7. Section 6. : Volume Revolution Section 7. : Integrtion by Prts 4 9. Section 7. : Trigonometric Integrls 45. Section 7.3 : Trigonometric Substitution 5. Section 7.4 : Prtil Frctions 55. Section 7.8 : Improper Integrls 6 3. Section. : Sequences 66

2 MATTHEW BATES 4. Section. : Series 7 5. Section.3 : Integrl Test nd Estimtes of Sums Estimting the Sum of Series Section.4 : The Comprison Test : Alternting Series Section.6 : Absolute Convergence, Rtio nd Root Test Rtio Test Root Test Rerrngements Section.8 : Power Series 89. Section.9 : Representing Function by Power Series 9. Section. : Tylor nd Mclurin Series 94. Section. : Curves Defined by Prmetric Equtions Section. : Clculus With Prmetric Curves 4. Section.3 - Polr Coordintes 6 5. Section.4 : Are nd Lengths of Polr Curves

3 CALCULUS II LECTURE NOTES 3. Nottion During lectures I will use some nottion/shorthnd which you my not of seen before. I do this for two resons; firstly, it gretly increses the speed t which I cn write mthemtics, secondly, t some point you will hve to lern it if you ever wnt to red more dvnced mths texts. - R, Q, Z represent the rel numbers, rtionl numbers nd integers respectively. - [, b] mens the set of rel numbers, x, such tht x b. - x R is short hnd for x is in R, or equivlently, x is rel number. - cts is short hnd for, continuous. - s.t. is short hnd for such tht. - f : [, b] R mens f is rel vlued function defined on the intervl, [, b]. - x is short hnd for for every x. - x is short hnd for there exists n x. - f n is short hnd for function. - thm is short hnd for theorem. - def is short hnd for definition. Exmple.. f : R R, s.t. f(x) > x This is short hnd for, f is function which is defined over the rel numbers, such tht f(x) > for every rel number, x. Exmple.. x [, ] s.t. g(x) This is short hnd for, there is rel number, x between nd such tht g(x). If you find this stuff bit confusing, don t worry. This wont be on ny exm or homework, I included it since it is life skill which everyone should hve.

4 4 MATTHEW BATES. Section 5. : The Definite Integrl In this chpter I will give quick review of Riemnn sums. All of this stuff should be fmilir to you. Mke sure you understnd it!.. Wht is Are? Are is obvious for rectngles, tringles, polygons, etc. But we define the re of other shpes by the limit of pproximtion. Definition.. The re of shpe is the limit of the pproximting res. Definition. (Riemnn Sum). Given function f : [, b] R, we define its n th left (right, mid) Riemnn sum on [, b] s follows. Let x b n, n L n f(x i ) x, where x i + R n i n f(x i ) x, where x i + i n M n i i(b ). n i(b ). n f(x i ) x, where x i + b n i(b ) +. n Definition.3 (Riemnn Integrl). Given function f : [, b] mthbbr, we define its Riemnn Integrl on [, b] s the limit of ny Riemnn sum, if such limit exists. i.e. Some nottion: b f(x)dx def lim n L n lim n R n lim n M n. - f is clled the integrnd - is the lower bound - b is the upper bound - b f is sometimes used s shorthnd for b f(x)dx. Definition.4 (Are Under Curve). We define the re under curve to be b f(x) dx.

5 CALCULUS II LECTURE NOTES 5 Exmple.5. x dx /. Exmple.6. x dx π/4. Exmple.7. π cos(x) dx. Exmple.8. Find the re under the curve y e x on the intervl [, ]. Are lim n i lim n n lim n n n ( e i/n) ( ) n ( + e /n + (e /n) (e /n) ) n (( e /n ) n ) e /n e lim n n ( e /n ) e. This is long nd technicl process... Riemnn Integrls Remrk.9. Note, the Riemnn sum mesures signed re. Consider the following picture:

6 6 MATTHEW BATES Theorem. (Properties of the Integrl). ) ) 3) 4) Proof. b b b b c dx c(b ) (f + g) cf c b (f g) b f b f + f b b g g ) c dx lim n n i c b n lim c(b ) n c(b ). ) 3) Exercise. f + g lim (f + g) x n ( ) lim f x + g x n lim f x + lim g x. n n 4) Exercise.

7 CALCULUS II LECTURE NOTES 7 Exmple (3 + 4x x) dx dx x dx 9 7 x dx. Theorem. (Additive property of the integrl). b f(x) dx + c b f(x) dx c f(x) dx. Proof. Proof by picture: The following theorem gives us some comprison properties of the integrl. Theorem.3 (Bounding Integrls). ) If f(x), then ) If f g, then b b f(x) dx. f(x) dx b 3) if m f M, then m(b ) g(x) dx. b f(x) dx M(b ). Exmple.4. Suppose tht we wnted to prove tht do this by n ppliction of the previous theorem. e x dx exists nd is finite. We cn

8 8 MATTHEW BATES Notice tht e x is decresing, so mx{e x } e. Therefore, e x dx ( ). Question: Which functions re integrble? By definition, the limit must exist. For exmple, f(x) /x is not integrble on [, ] since dx is not finite. x Theorem.5. If f : [, b] R is continuous, then f is integrble. Proof.??

9 CALCULUS II LECTURE NOTES 9 3. Section 5.3 : Fundmentl Theorem of Clculus In this section we will tlk bout wht is (in my opinion) the most importnt (nd under pprecited) theorems in this course. I m tlking bout the fundmentl theorem of clculus. One wy in which the fundmentl theorem of clculus (henceforth FTC) is mzing is tht it estblishes connection between differentil (differentition) clculus nd integrl (integrtion) clculus. Tke moment to think bout this; differentition is concerned with tngent lines, while integrtion dels with re. Why, priori, should these two things be relted? Definition 3.. Let f : [, b] R be continuous. Define new function, g : [, b] R by, g(x) x f(t)dt. At first, this my look like strnge wy to define function, but such functions re ctully more common thn you might expect. You cn think of g s mesuring the re under f from to x. Exmple 3.. Let f(t) 7, for t. Then g(x) x 7 dx 7(x ). Exmple 3.3. Let f : [, 4] R be the function with the following grph... If g(x) is defined s bove, we see tht, g() g() g() g(3) g(4) Exmple 3.4. Let f(t) t. Is there nice wy to express g(x) x f(t)dt? If we consider g(x) s the re under f from to x we see tht this g(x) is ctully the re of right ngle tringle with side length x. Thus, g(x) x.

10 MATTHEW BATES Notice, in the lst exmple, g (x) f(x). Why would this be true? Lets think bout this for moment. By definition g g(x + h) g(x) (x) lim. h h Notice tht, g(x + h) g(x) is the re under the grph of f between x nd x + h. As we let h go to zero, this re strts to look like rectngle, with height f(x) nd width h. Thus, g(x + h) g(x) hf(x) f(x). h h (NOTE... this is obviously not proof! It is only included to give you intuition bout the proof). We re now redy for the big theorem, The Fundmentl Theorem of Clculus (FTC). Theorem 3.5 (Fundmentl Theorem of Clculus I). Let f : [, b] R be continuous function. If, x g(x) f(t)dt, then g is continuous nd g f. Proof. Suppose h > nd x, x + h (, b), then g(x + h) g(x) x+h x x+h x f(t)dt f(t)dt + f(t)dt. x x+h x f(t)dt f(t)dt x f(t)dt Thus, g(x + h) g(x) x+h f(t)dt. h h x Since f is cts on [x, x + h] it must ttin mximum nd minimum somewhere in [x, x + h]. i.e. f(x m ) f(x) f(x M ) for some x m, x M [x, x + h]. Thus, so, hf(x m ) f(x m ) h x+h x x+h x f(t)dt hf(x M ) f(t)dt f(x M ).

11 We now consider wht hppens s we let h go to zero. Note, CALCULUS II LECTURE NOTES lim f(x m) f(x), h since f(x) is continuous. Similrly for f(x M ). Thus, when we tke limits of the bove inequlity, we see thus, g f f(x) g (x) f(x) One lst thing to check, wht hppens if x or b? In such cse, we consider the one sided limit nd follow the sme procedure. Another wy of writing FTC is s d dx x f(t)dt f(x). Lets get tste of the gretness of the FTC by doing some simple exmples. Exmple 3.6. Find the derivtive, with respect to x, of x + sin (t)dt. If we let f(x) then, by FTC, we see the nswer is f(x). + sin (x) Exmple 3.7. Find g (x) where g(x) x + t dt. Let f(t) + t, then by the FTC, we see tht g (x) + x. Exmple 3.8. Find, d x 3 cos(u)du. dx

12 MATTHEW BATES Let y x 3. Then, using the chin rule we see, d dx x 3 dy dx cos(u)du ( d y cos(u)du dy dy dx cos(y) 3x cos(x 3 ). ) Exmple 3.9. Find, Let y x 4. Using the chin rule we see, d x 4 sec(s) ds. dx x 4 d sec sds dx dy d y sec(s)ds dx dy dy dx sec(y) 4x 3 sec(x 4 ). We cn derive generl formul for, Note, d dx b(x) (x) f(t)dt. d dx b(x) d dx d dx (x) f(t)dt ( b(x) f(t)dt b(x) f(t)dt d dx (x) (x) b (x)f(b(x)) (x)f((x)). f(t)dt ) f(t)dt

13 Exmple 3.. Clculte, d 3x te t dt. dx sin(x) Using the bove formul we see tht the nswer is, CALCULUS II LECTURE NOTES 3 6x(3x )e 3x cos(x) sin(x)e sin(x). Exmple 3.. Find the derivtive of the following indefinite integrl, 3+x 7x sin(t)dt By the bove remrk, we see the nswer is (3 + x ) sin(3 + x ) (7x) sin(7x) x sin(3 + x ) 7 sin(7x). Notice tht if f is n integrble function, then g(x) x f(t)dt, is perfectly well defined function. Thus, we should tret it like one. For exmple, Exmple 3.. Clculte, If we write f(t) : t by FTC I we see tht the nswer is i.e. d dx x ( t ) h(u)du dt. h(u)du, then we cn rewrite the question s, d dx x x f(x), f(t)dt, h(u)du. The FTC lso helps/llows us to clculte integrls.

14 4 MATTHEW BATES Theorem 3.3 (Fundmentl Theorem of Clculus II). If f is continuous nd F f then, b f(x)dx F (b) F (). A function F such tht F f is clled n ntiderivtive of f. Proof. Let F f, define g(x) Note, by FTC I, g f. Thus, nd thus for some constnt C. Moreover, Thus, g() x f(x)dx. (g F ), F g + C, f(x)dx. F (b) F () (g(b) + C) (g() + C) g(b) g() b f(x)dx. Why is this gret? It lets us clculte/evlute integrls! If we wnt to know ll we need to do is find some F such tht Lets introduce some useful nottion, F (x) b b f F f. def F (b) F (). Exmple 3.4. Find, Note, 4 xdx. d x dx x.

15 CALCULUS II LECTURE NOTES 5 Thus, by FTC II, the integrl is x 4 8. Exmple 3.5. Find, Note, Thus, by FTC II, the integrl is 7 e x dx. d dx ex e x. e x e e 7. 7 Exmple 3.6. Find, Note, Thus, by FTC II, the integrl is π π/ cos(x)dx. d sin(x) cos(x). dx sin(x) π π/. Exmple 3.7. Find, Note, Thus, by FTC II, the integrl is log(x) 5 3 x dx. d dx log(x) x. 5 3 log(5) log(3). These re very stndrd exmples, they should be second nture to you. In wy, ll other exmples re built up from these ones. Lets do some hrder exmples.

16 6 MATTHEW BATES Exmple 3.8. Find, First we expnd the brckets, (t )(3t + )dt. (3t t )dt ( t 3 t t ) ( ) 3. Exmple 3.9. Find, Note, Thus 5 x dx. ( ) d dx 3 x 3 x. 5 x dx 5 3 x 3 3 ()3/ 3 ()3/ 3 ( ). Exmple 3.. Find, Note, Thus π 4 π 4 sec (x)dx. d dx tn(x) sec (x). sec (x)dx tn(x). π 4

17 Exmple 3.. Find, Note, this is the sme s, CALCULUS II LECTURE NOTES 7 x / x / dx, ( ) 3 x3/ 4x / ( 3 ()3/ 4() 3/ x x dx. ) ( ) 3 ()3/ 4() 3/ It is not lwys obvious tht certin question requires integrtion to solve, for exmple Exmple 3.. Find, Note, this is the sme s lim n lim n n i nd this is Riemnn sum, i.e. it is equl to which we cn esily solve, x 4 4 n i i 3 n 4. ( ) i 3, n n x 3 dx, 4.

18 8 MATTHEW BATES 4. Section 5.4 : Indefinite Integrls nd Net Chnge Theorem 4.. Indefinite Integrls We hve seen tht evluting n integrl boils down to finding n nti-derivtive. Lets give some nottion for nti-derivtive. Definition 4.. If F f, then we sy tht F is n nti-derivtive of f. We use the following nottion to denote this, f(x)dx F (x), or, f F. Lets list some esy nti derivtives, k kx x n xn+ n+ e x e x sin(x) cos(x) cos(x) sin(x) You hve to know other, more complicted trig nti derivtives s well. sec (x) tn(x) csc (x) cot(x) sec(x) tn(x) sec(x) csc(x) cot(x) csc(x) x sin (x) x + tn (x) Exmple 4.. (7x 5 sec (x))dx 7x6 6 tn(x) + C. Exmple 4.3. cos(x) sin (x) dx cos(x) sin(x) sin(x) dx csc(x) cot(x)dx csc(x) + C.

19 CALCULUS II LECTURE NOTES 9 Exmple 4.4. (e 3x sin(t))dx 3 e3x x sin(t) + C. Exmple 4.5. ( 3x + ) + x dx x 3 x + tn (x) + C. Exmple 4.6. (x 3)(4x + 5)dx (8x x 5)dx 8 3 x3 x 5x + C. Exmple 4.7. π/4 π/4 + cos (θ) cos dθ sec (θ) + dθ (θ) π/4 (tn(θ) + θ) + π 4. Exmple 4.8. Wht hppens to the nti-derivtive if I trnslte function? e.g. cos(θ + 3)dθ. Notice tht, d dθ sin(θ + 3) (θ + 3) cos(θ + 3) cos(θ + 3). We cn generlize this exmple, i.e. cos(θ + K) sin(θ + K) + C, similrly,... we could go on. sin(θ + K) cos(θ + K) + C, sec (θ + K) tn(θ + K) + C,

20 MATTHEW BATES Lets formlise the bove observtion. Theorem 4.9. Suppose f : [, b] R is integrble, nd F (x) f(x)dx, then, F (x + K) f(x + K)dx. Proof. Differentite F (x + K) nd see wht you get, s required. d dx F (x + K) (x + K) F (x + K) f(x + K) 4.. Net chnge Note, there is nother wy of writing FTC, i.e. b F (x)dx F (b) F (). This formliztion is useful in mny prcticl pplictions. The ide is, if you give me the rte t which something chnges, then I cn tell you how much it hs chnged during given time period. These sorts of problems rise nturlly in mny different res; popultion growth, het diffusion, interest on svings ccount, nd mechnics. Exmple 4.. Suppose the function s(t) gives the displcement of prticle from fixed point t time t. It is cler tht d s(t) v(t), dt where v(t) is the velocity function. Thus, if you re in sitution where you re only given the velocity of given prticle, using FTC, you cn sy something bout the distnce which the prticle trvels in given time period. i.e. t t v(t)dt s(b) s(). This is the NET chnge in displcement during the time t nd t. Note, this is not the totl distnce trvelled by the prticle! This is becuse the integrl mesures signed distnce. In order to find the totl distnce trvelled, we need to integrte the bsolute vlue of the velocity.

21 CALCULUS II LECTURE NOTES Exmple 4.. Thus, if we hve the velocity function, v(t) sin(t), then during the time period to π, the net displcement is given by, The totl distnce trvelled is, π π sin(t) dt sin(t)dt cos(t) π. π π π sin(t) dt + sin(t)dt + cos(t) 4. π π π π + cos(t) sin(t) dt sin(t)dt π π Exmple 4.. Suppose the velocity of prticle is given by, v(t) t 4t, for t 8. ) Find the net displcement ) Find the totl distnced trvelled. ) Net Displcement s(8) s() 8 s (t)dt 8 v(t)dt. ( ) 8 3 t3 t t... ) Totl distnce trvelled. We first find when the velocity is positive nd when it is negtive. t 4t (t 6)(t + ). Which is positive when t < or t > 6. So the totl distnce trvelled is,... 8 t 4t dt 6 ( t + 4t + )dt (t 4t )dt

22 MATTHEW BATES Exmple 4.3 (Bcteri Popultion). Suppose the popultion of some bcteri t time t is, nd tht the growth rte t time t is t. Wht is the popultion of bcteri t time: t, t, t? Let P (t) be the popultion of the bcteri t time t. We re given tht P (t) t, nd P (). Thus, P () P () P (t)dt t dt e ln()t dt ( ) ln() eln()t ln() ( ) ln().

23 CALCULUS II LECTURE NOTES 3 5. Section 5.5 : Substitution Rules 5.. Indefinite Integrls Using Substitution By now we understnd the importnce of finding nti-derivtives. The trouble is tht this is esier sid thn done. For exmple, find n nti-derivtive for ech of the following functions. x x x 5 3x ye +y z 3 sin(3z 4 ) 4 x4 By the end of this chpter you will be ble to find nti derivtive for some of these functions. Lets exmine the first function, x x. We know how to integrte x, but we don t know how to del with x. Let u x. So du xdx. Substituting this into the integrl, gives x x dx u du 3 u 3 3 (x ) 3. In generl we hve the following result, Theorem 5. (Substitution). Suppose F f, then f(g(x))g (x)dx F (g(x)). Another wy of sying this is, if u g(x), then f(g(x))g (x)dx f(u)du. Proof. Chin rule. Exmple 5.. e 6x dx.

24 4 MATTHEW BATES Use the substitution u 6x. This gives du 6xdx. Thus the integrl becomes, e u 6 du eu 6 + C e6x 6 + C. Exmple 5.3. x e x3 dx. Use the substitution u x 3. This gives, du 3x dx. Thus the integrl becomes, e u 3 du eu 3 + C ex3 3 + C. Exmple 5.4. cos(x)e sin(x) dx. Use the substitution u sin(x). This gives, du cos(x)dx. Thus the integrl becomes, e u du e u + C e sin(x) + C. Exmple x cos(x 3 + 7)dx. Use the substitution u x This gives du 3x dx. Thus the integrl becomes, cos(u)du sin(u) + C sin(x 3 + 7) + C. Exmple 5.6. cos 4 (x) sin(x)dx.

25 CALCULUS II LECTURE NOTES 5 Use the substitution u cos(x). This gives, du sin(x)dx. Thus the integrl becomes, u 4 du u5 5 + C cos5 (x) 5 + C. Exmple 5.7. x 4x dx. Use the substitution u 4x. This gives du 8xdx. Thus the integrl becomes, du 8 u u + C 8 4 4x + C. Exmple 5.8. x 5 x + dx. Use the substitution u x +. This gives du xdx. Note, x u, thus x 4 x x + dx (u ) x udx (u ) udu (u u + ) udu (u 5/ u 3/ + u / )du Exmple 5.9. tn(x)dx. Use the substitution u cos(x). This gives, du sin(x)dx. Thus the integrl becomes, sin(x) cos(x) dx u du ln(u) + C ln(cos(x)) + C ln(sec(x)) + C.

26 6 MATTHEW BATES Exmple 5.. (3x 7) 3 dx. Use the substitution u 3x 7. This gives, du 3dx. Thus the integrl becomes, 3 u3 du u C (3x 7) C. Exmple 5.. Let u x 8. x 7 dx. + x6 5.. Definite integrls using substitution We will now exmine methods for clculting definite integrl using substitution. Theorem 5. (Substitution). If f, g re continuous functions, then, b f(g(x))g (x)dx g(b) g() f(u)du. Proof. Suppose F (x) f(x). Then, (F (g(x))) g (x)f (g(x)) g (x)f(g(x)). By FTC, Similrly, b g(b) g() f (g)g dx F (x) f(u)du F (x) b g(b) g() F (g()) F (g(b)). F (g()) F (g(b)). Lets do some exmples of definite integrls using substitution.

27 CALCULUS II LECTURE NOTES 7 Two wys to do definite integrls using substitution. Either first find the indefinite integrl (in terms of the originl vrible) then substitute in the limits, or do substitution nd plug in the new limits. Exmple 5.3. x( + x ) dx. Let u + x, then du xdx. The integrl becomes, u u du. Exmple x + dx. Let u 3x +, so du 3dx. The integrl becomes, 5 u 3 du 9 u3/... 5 Exmple 5.5. dx (x + ). Let u x +, so du dx. The integrl becomes, 3 u 3. du u 3

28 8 MATTHEW BATES Exmple 5.6. e x + e x dx. Let u e x +, so du e x dx. The integrl becomes, +e u du +e ln(x) ln( + e) ln(). Exmple 5.7. e ln(x) x dx. Let u ln(x), so du xdx. The integrl becomes, u. udu Exmple 5.8. Let u 5 + x. x 5 + x dx. Exercises (ln(x)) dx, u ln(x) x tn (3θ) + 9θ dθ, u tn (3θ) sin(ln(3x 6x )) 3x dx, u ln(3x )

29 CALCULUS II LECTURE NOTES Symmetry Sometimes we cn exploit the symmetry of given function gin informtion bout n integrl without ctully clculting it. Exmple 5.9. π sin(θ)dθ. Exmple x 5. Exmple 5.. x dx xdx. Definition 5.. Let f be function. We sy tht f is even if f(x) f( x). We sy tht f is odd if f(x) f( x). Theorem 5.3. If f is n even function, then If f is n odd function then, f f. f. Proof. Do it yourself This simple theorem cn sve us lot of work sometimes. Exmple 5.4. x is even. x 3 is odd. sin(x) is odd. cos(x) is even.

30 3 MATTHEW BATES Exmple x tn(x) + x + 4x 4 dx Exmple 5.6. Show, π xf(sin(x))dx π π f(sin(x))dx Consider the substitution u π x, so du dx. The integrl becomes, Thus, π π (π u)f(sin(π u))du π (π u)f(sin(π u))du πf(sin(u))du π uf(sin(u))du. π π xf(sin(x))dx πf(sin(u))du Exmple 5.7. Do the substitution, u x. x ( x) b dx x b ( x) dx. Exmple 5.8. Show, π/ Do the substitution, u π/ x. f(cos(x))dx π/ f(sin(x))dx.

31 CALCULUS II LECTURE NOTES 3 6. Section 6. : Are between curves We use the sme resoning s we did to clculte the re under curve. i.e., the signed re between two functions, f > g, for x in [, b], is Note, this is the sme s n lim n i n Are lim (f(x n i ) g(x i )) x. i (f(x i )) x lim n n (g(x i )) x i b f g Thus, if f > g then the re between f nd g is (f g). Note, in the specil cse where g, we hve the stndrd integrl. Exmple 6.. Find re between the curves, f(x) x nd g(x) x. First find intersection points. We solve for x, x x so x, so x ±. So, the re is given by, ( x x )dx (x 3 ) x ( ) 3 ( 3 ) Exmple 6.. Fine the re between e x nd x, on [, ]. First note tht e x > x on [, ], thus the re is given by, (e x x)dx ) (e x x ( e ) ( ) e 3.

32 3 MATTHEW BATES Exmple 6.3. Fine the re enclosed by, x 3 nd 3 x on the intervl [, ]. If f(x) is not lwys bigger thn g(x), then to work out the re enclosed between f nd g we need to split the integrl. i.e. Definition 6.4. The re between f(x) nd g(x) on [, b] is, b f(x) g(x) dx. Exmple 6.5. Find the re bounded between cos(θ) nd sin(θ) on the intervl [, π/]. Note tht between [, π/4] cos(x) is lrger thn sin(x), nd on [π/4, π/], sin(x) is lrger thn cos(x). Thus the re is π/4 (cos(θ) sin(θ))dθ + (sin(θ) + cos(θ)) π/4 π/ π/4 (sin(θ) cos(θ))dθ + ( cos(θ) sin(θ)) π/ π/4 ( / + / ) + ( + / + /). Exmple 6.6. Find the re enclosed between x 3 x nd x. First find the intersection points, x 3 x, this gives x or x ±. Thus the re is given by, (x 3 x) xdx + x (x 3 x)dx Sometimes it is more convenient to think of curve s function of y insted of s function of x, i.e. x g(y).

33 CALCULUS II LECTURE NOTES 33 Exmple 6.7. x y. x sin(y) x e y x y. Now insted of integrting with respect to x, we cn integrte with respect to y. d c (x R x L )dy Exmple 6.8. Find the re enclosed between, y x nd y x + 6. First find the points of intersection, (x ) x + 4, so x 4x 5, so x 5,. So the points of intersection re, (5, 4) nd (, ). We integrte with respect to y. 4 ( y ) 6 (y + ) dy 4 ( ) y + y + 4 dy ( ) y y 4 + 4y Exmple 6.9. Fine the re enclosed by 4x + y, nd x y. Exmple 6.. Fine the re enclosed by x y 4, nd x e y, between y nd y.

34 34 MATTHEW BATES 7. Section 6. : Volume How do we define volume? We sked this sme question for re erlier. We know the volume of some solids, e.g. cubes, prllelepipeds, cylinders etc... Given generl solid, S, how do we define its re? Consider cross section of the S, i.e. the intersection of S nd plne. This is closed curve, nd we know how to find the re of this. We cn slice S into finitely mny slices. It is cler the re of S is the sum of the res of the slices. If we slice thin enough, then ech slice is pproximtely prism, with bse equl to cross section nd height the width of the slice. If x is the width of the slice nd A(x) is the cross sectionl re of S t given x, then ech slice hs volume pproximtely equl to ha(x). Thus the volume is pproximtely, A(xi ) x. We define the volume to be the limit of the sum s x, if it exists. i.e. the volume Vol(S) lim A(xi ) x x A(x)dx. Lets think bout cross sections of some common solids. Sphere. Torus. Cylinder. Cube. Pyrmid. Exmple 7.. Let S be cylinder, rdius r, nd length h. Then the volume is equl to h A(x)dx h hπr. As expected. πr dx Exmple 7.. Find th volume of the unit sphere. Plce the sphere t the origin. Wht is A(x)? The cross sections re ll circles, so we need only to find the rdii. Pythgors gives us tht x + r r x.

35 CALCULUS II LECTURE NOTES 35 So, A(x) πr π πx. So the volume is, (π πx )dx (πx π 3 x3) π π 3 4π 3. Exmple 7.3. Clculte the volume of squre bsed pyrmid with bse length l nd height h. We need to clculte the cross sectionl re, A(x). All of the cross sections re squres, so we need only clculte their side length. Let s be the side length of the squre t x units from the pex of the pyrmid. Then, s x l h, so s lx h. So the volume is h l x h dx l x 3 3h h l h Revolution Note, we cn think of the sphere re circle rotted round the x-xis. This is generl construction. Consider the grph of function y f(x). We cn rotte this round the x-xis to form solid. Wht is the volume of this solid? We need to find the cross sectionl re. Note tht t ech x, the cross section is circle, with rdius f(x). Thus, the volume is given by, π b (f(x)) dx. Exmple 7.4. Find the volume of solid formed by rotting the curve y x round the x-xis, from to. Using the bove formul, we see the volume is given by, π( x) dx π. πxdx

36 36 MATTHEW BATES Exmple 7.5. Find the volume of revolution (round the x-xis) of the grph of y 3x between x nd x. Answer is ( π 9x ) dx π(3x 3 ) 3π. Exmple 7.6. Find the volume of revolution (round the x-xis) of the grph of y sin(3x) between x nd x π/3. Answer is π π/3 sin (3x)dx. To clculte this integrl we first do substitution u 3x, so the integrl becomes, Now use the trig identity, π 3 π π 3 π sin (u)du. cos(θ) sin (θ). π sin (u)du π cos(u) du 3 π ( x 3 sin(u) ) π 4 π ( π ) 3 π ( ) 3 π /6. Exmple 7.7. Clculte the volume of x tn(y) rotted round the y-xis for y in [, π/4]. π/4 π tn (y)dy π π/4 ( sec (y) ) dy π (tn(y) y) π( π/4). π/4

37 CALCULUS II LECTURE NOTES 37 Exmple 7.8. Find the volume of the solid formed by rotting the region bounded by y x 3, y, x nd x, round the line x. Consider the cross sectionl re, in terms of y. Note tht we cn write the eqution of the curve s, x y /3. Thus Thus, the volume is, A(y) π( y 3 ) π() π(3 4y 3 + y 3 ). π(3 4y 3 + y 3 )dy π (3y 3y 4/3 + 3 ) 5 y 5 3 π ( ) 3. 5 We now consider the problem of finding the volume between two solids. For exmple, consider two cylinder, one inside the other, wht is the volume enclosed between them? It is simply the volume of the lrger minus the volume of the smller. Exmple 7.9. Two cones, ech with pex t the origin. Lrger hs slope, the smller hs slope. Wht is the volume enclosed between x nd x? We hve two methods for solving this problem. Method I : Find the cross sectionl re of the enclosed re nd integrte it. Cross section is concentric circles, with rdii x nd x. Thus, A(x) 3πx. Integrting this gives, 3πx dx π. Method II : Find volume of smller nd subtrct from volume of lrger. 4πx dx πx dx π. Exmple 7.. Cylinder rdius, enclosing squre prism of side length. Find re enclosed between nd. Find expression for cross sectionl re, A(x) π. Thus, the volume is given by, (π )dx π.

38 38 MATTHEW BATES Exmple 7.. Find the volume of revolution enclosed by y x nd y x, rotted round the x-xis. First we must find the points of intersection, x, nd x. Thus, the volume is given by, π( x) dx ( ) x πx 4 ( ) x 5 dx π π 5 π π 5 3π. Generl formul for the volume enclosed by two surfces of revolution. Volume π (f (x) g (x))dx. Proof. Do it lter. NOTE!! This is not the sme s Volume π (f(x) g(x)) dx. This is very common mistke, don t mke it. Exmple 7.. Fine the volume enclosed between y x nd y x, rotted round the x-xis. The points of intersection re obviously x nd x, thus the volume is, πx dx (x ) dx π ( x 3 3 ( π 3 5 ) ( ) x 5 π 5 ). Exmple 7.3 (Sphere Inside A Cylinder). Consider the unit sphere, sitting inside cylinder of rdius nd height. Find the volume enclosed. Consider the grphs, y nd y x. Note, the revolution is exctly wht we wnt.

39 Thus, the desired volume is CALCULUS II LECTURE NOTES 39 Volume π π π 3. ( ( x ) ) dx (x x + x3 3 ) Exmple 7.4 (Torus). Fine the volume of torus, with rdius, nd mjor rdius 3. Consider the grphs, y 3 + x nd y 3 x. These intersect t x ±. The desired volume is the volume enclosed by these two functions. i.e. Volume π (3 + ) ( x 3 ) x dx π x dx This is mess, but you cn solve it with the current techniques.

40 4 MATTHEW BATES 8. Section 7. : Integrtion by Prts Integrtion by prts is the integrtion rule which corresponds to the product rule for differentition. i.e. d dx (f(x)g(x)) f (x)g(x) f(x)g (x), thus, d (f dx (f(x)g(x))dx (x)g(x) + f(x)g (x) ) dx so, f(x)g(x) (f (x)g(x) ) dx + (f(x)g (x) ) dx. Rerrnging this gives, (f(x)g (x) ) (f dx f(x)g(x) (x)g(x) ) dx. This is wht we cll integrtion by prts. It is sometimes esier to remember when written in shorthnd, udv uv vdu. Exmple 8.. Clculte x sin(x)dx. Let u x, dv sin(x)dx, then du dx nd v cos(x). So by integrtion by prts the integrl is equl to, x cos(x) cos(x)dx x cos(x) + sin(x) + C. Note tht the hrdest prt is knowing wht to mke u nd dv. The generl ide is to mke u something which becomes simpler fter differentiting. For exmple, if we chose u sin(x) nd dv xdx in the previous exmple, then we would get x sin(x)dx x sin(x) which is perfectly true, but just not helpful. x cos(x)dx, Exmple 8.. te t dt

41 CALCULUS II LECTURE NOTES 4 Exmple 8.3. t e t dt Exmple 8.4. ln(x)dx Let u ln(x), dv dx, then du xdx, nd v x. Thus the integrl becomes, x ln(x) x dx x ln(x) x + C. x Exmple 8.5. e x sin(x)dx. Let u e x, dv sin(x)dx, then du e x dx, nd v cos(x). Thus the integrl becomes, e x sin(x)dx e x cos(x) + e x cos(x)dx, Now do integrtion by prts on e x cos(x)dx. u e x, dv cos(x)dx, nd du e x dx, v sin(x). Thus, ( e x sin(x)dx e x cos(x) + e x sin(x) ) e x sin(x)dx. It my look like we re going in circles but now we cn solve for e x sin(x)dx. i.e. e x sin(x)dx (ex sin(x) e x cos(x)). This is stndrd trick nd it is importnt tht you understnd it. Exmple 8.6. ln(5x + 3)dx. Let u ln(5x + 3), dv dx, then du 5 5x+3dx nd v x. The integrl becomes, 5x x ln(5x + 3) 5x + 3 dx 5x x ln(5x + 3) dx x ln(5x + 3) 5x x + 3 dx x ln(5x + 3) x + 3 ln(5x + 3). 5

42 4 MATTHEW BATES Exmple 8.7. (x + )e x dx. Theorem 8.8. Integrtion by prts for definite integrls, b f(x)g (x)dx f(x)g(x) b b f (x)g(x)dx. Exmple 8.9. dx tn (x)dx. Let u tn (x), dv dx, then du, nd v x. Thus, +x tn (x)dx x tn (x) To evlute Thus, x x + x dx dx we mke the substitution, u + x, then du xdx. Thus, +x x + x dx u du ln(u) ln(). tn (x)dx x tn (x) ln() π 4 ln(). Exmple 8.. x cos(πx)dx Let u x, dv cos(πx)dx, then du dx nd v π sin(πx). So the integrl becomes, x π sin(πx) π sin(πx)dx π sin(π/) + π cos(πx) π π.

43 Exmple 8.. CALCULUS II LECTURE NOTES 43 (ln(x)) dx Let u (ln(x)), dv dx, then du ln(x) x 3 x, nd v. So the integrl becomes, x x (ln(x)) ln(x) + x 3 dx, now do integrtion by prts gin to solve nd v. So we hve tht, x (ln(x)) x 3 dx x (ln(x)) x (ln(x)) x 3 ln(x) x 3 dx. Let u ln(x), dv x 3 dx, then du x, + + ( ln(x) x 3 dx x ln(x) 8 (ln()) + ln() ) + x 3 dx 4 x ln() (ln() + ) Exmple 8.. e x ln(x)dx. Let u ln(x), dv xdx, thus du (/x)dx nd v x /. So the integrl is, (ln(x)x e e /) x x dx e e x / dx e / (x e /4) e / e /4 + /4 e /4 /4. Exmple 8.3. y dy ey

44 44 MATTHEW BATES Let u y, dv e y dy, then du dy, nd v e y. So the integrl becomes, ye y ( e 4 e y e 4 e ( e ). e y dy )

45 CALCULUS II LECTURE NOTES Section 7. : Trigonometric Integrls This section is ll bout tricks for integrting trigonometric functions. You should know the following identities, sin (x) + cos (x) cos(x) cos (x) sin (x) cos (x) ( + cos(x)) tn (x) + sec (x) sin(x) sin(x) cos(x) sin (x) ( cos(x)) + cot (x) csc (x) Exmple 9.. Clculte, cos 3 (x)dx. We first use the identity cos (x) sin (x), cos 3 (x)dx cos(x)( sin (x))dx cos(x)dx sin (x) cos(x)dx sin(x) sin (x) cos(x)dx Use the substitution, u sin(x), then du cos(x)dx. sin(x) u du sin(x) u3 3 sin(x) sin3 (x) + C. 3 Exmple 9.. Clculte, sin 5 (x) cos (x)dx. sin(x)( cos (x)) cos (x)dx sin(x)( cos (x) + cos 4 (x)) cos (x)dx sin(x)(cos (x) cos 4 (x) + cos 6 (x))dx,

46 46 MATTHEW BATES Use the substitution, u cos(x), then du sin(x)dx. u + u 4 u 6 du 3 u3 + 5 u5 7 u7 + C 3 cos3 (x) + 5 cos5 (x) 7 cos7 (x) + C. Exmple 9.3. Clculte, sin (x) cos 5 (x)dx. sin (x)( sin (x)) cos(x)dx sin (x)( sin (x) + sin 4 (x)) cos(x)dx (sin (x) sin 4 (x) + sin 6 (x) ) cos(x)dx Use the substitution, u sin(x), then du cos(x)dx. (u u 4 + u 6 )du ( 3 u3 5 u5 + 7 u7 ) 6 sin3 (x) sin5 (x) + 4 sin7 (x) + C. Exmple 9.4. Clculte, cos (x)dx. Use the identity, cos (x) ( + cos(x)), cos (x)dx + cos(x)dx x + sin(x) + C. 4

47 CALCULUS II LECTURE NOTES 47 Exmple 9.5. Clculte, sin 4 (x)dx. Use the identity sin (x) ( cos(x)), sin 4 (x)dx (sin (x)) dx ( ) cos(x) dx 4 cos(x) + 4 cos (x)dx 4 cos(x) + ( 4 + ) cos(4x) dx 4 cos(x) cos(4x)dx 3 8 x 4 sin(x) + sin(4x) + C. 3 Exmple sin (θ) cos (θ)dθ. Exmple 9.7. Clculte, tn 6 (θ) sec 4 (θ)dθ. Use the identity sec (x) tn (x) +, tn 6 (θ) sec 4 (θ)dθ Use the substitution u tn(θ), then du sec (θ)dθ. u 8 + u 6 du tn 6 (θ)(tn (θ) + ) sec (θ)dθ (tn 8 (θ) + tn 6 (θ)) sec (θ)dθ 9 u9 + 7 u7 + C 9 tn9 (θ) + 7 tn7 (θ) + C.

48 48 MATTHEW BATES Exmple 9.8. Clculte, tn 5 (θ) sec 7 (θ)dθ. Note, we cn t write this s some function of tn times sec, since 7 is odd. But we cn write this s some function of sec times tn sec. Use the identity tn (x) sec (x), tn 5 (θ) sec 7 (θ)dθ tn(θ)(sec (θ) ) sec 7 (θ)dθ tn(θ) sec(θ)(sec (θ) ) sec 6 (θ)dθ Let u sec(θ), then du tn(θ) sec(θ)dθ. (u ) u 6 du (u 4 u + )u 6 du u u 8 + u 6 du u 9 u9 + 7 u7 + C sec (θ) 9 sec9 (θ) + 7 sec7 (θ) + C Exmple 9.9. Clculte, π/4 tn(θ) sec 3 (θ)dθ. Use the substitution, u sec(θ), so du tn(θ) sec(θ), π/4 tn(θ) sec 3 (θ)dθ u du u Exmple 9.. Clculte, π/4 tn(θ) sec 4 (θ)dθ.

49 Use the identity sec (x) tn (x) +, π/4 tn(θ) sec 4 (θ)dθ Let u tn(θ), so du sec (θ). CALCULUS II LECTURE NOTES 49 π/4 π/4 tn(θ)(tn (θ) + ) sec (θ)dθ (tn 3 (θ) + tn(θ)) sec (θ)dθ u 3 + udu ( u4 4 + u ) Recll the ngle sum formul, From these eqution we cn derive, sin(a + B) sin(a) cos(b) + sin(b) cos(a) cos(a + B) cos(a) cos(b) sin(a) sin(b). sin(a) cos(b) (sin(a B) + sin(a + B)) sin(a) sin(b) (cos(a B) cos(a + B)) cos(a) cos(b) (cos(a B) + cos(a + B)). Exmple 9.. Clculte, sin(4x) cos(5x)dx. Using the bove formuls we see tht the integrl is equl to, (sin( x) + sin(9x))dx (cos( x) cos(9x)) + C. 9 Exmple 9.. Clculte, sin(3x) sin(5x)dx.

50 5 MATTHEW BATES (cos( x) cos(8x))dx (cos(x) cos(8x))dx ( sin(x) sin(8x)) + C. 8 Exmple 9.3. tn 3 (x)dx. tn(x) + tn(x) sec (x)dx ln(sec(x)) + tn (x) + C. Exmple 9.4. sec(x) + tn(x) sec(x)dx sec(x) sec(x) + tn(x) dx sec (x) + tn(x) sec(x) dx sec(x) + tn(x) ln sec(x) + tn(x) + C. Exmple 9.5. x sec(x) tn(x)dx. Use integrtion by prts, with u x, dv sec(x) tn(x)dx, so du dx, nd v sec(x). x sec(x) sec(x)dx x sec(x) ln(sec(x) + tn(x)) + C.

51 CALCULUS II LECTURE NOTES 5. Section 7.3 : Trigonometric Substitution Sometimes it is use full to do substitution bckwrds, i.e. to clculte f(x)dx mke substitution x g(t), so the integrl becomes f(g(t))g (t)dt. Exmple.. Clculte, x dx. Let x sin(t), then dx cos(t)dt. Thus the integrl becomes, sin (t) ( cos(t) ) dt sin (t) ( cos(t) ) dt cos (t)dt ( + cos(t))dt t + sin(t) + C. 4 We now need to trnslte bck to the originl coordinte, x. sin (x/) + sin(t) cos(t) + C 4 sin (x/) + x cos(t) + C sin (x/) + x sin (t) + C sin (x/) + x (x/) + C. Exmple.. Clculte, 9 x dx. x

52 5 MATTHEW BATES Use the substitution, x 3 sin(t), then dx 3 cos(3t)dt. The integrl becomes, 9 9 sin (t) 9 sin 3 cos(t)dt (t) cos (t) sin (t) dt cot (t)dt ( csc (t) ) dt cot(t) t + C. Exmple.3. Clculte, x x + 4 dx. Let x tn(t), then dx sec (t)dt. The integrl becomes, 4 tn (t) tn (t) + sec (t)dt 4 tn (t) sec(t) sec (t)dt 4 tn (t) sec(t)dt cos(t) 4 sin (t) dt Let u sin(t), then du cos(t)dt. 4u du 4u + C 4 sin(t) + C x C. 4 x Exmple.4. Clculte, x dx.

53 CALCULUS II LECTURE NOTES 53 Let x sec(t), then dx sec(t) tn(t)dt, nd the integrl becomes, sec(t) tn(t)dt sec (t) sec(t)dt ln sec(t) + tn(t) + C ln x x + x + C. Exmple.5. Clculte, u u du Let u cos(t), then du sin(t)dt, nd the integrl becomes, cos(t) ( ) sin(t) dt cos (t) cos(t) dt sec(t)dt ln sec(t) + tn(t) + C u ln u + u. Exmple.6. Clculte, 3 x x dx. Complete the squre to rewrite this s, dx. 4 (x + ) Now use the substitution, x sin(t), then dx cos(t)dt, nd the integrl becomes, 4 4 sin (t) cos(t)dt dt t + C sin ( x + ) + C.

54 54 MATTHEW BATES Exmple.7 (Complete the Squre). 4 6x x dx. Complete the squre to get, 7 (x 3) dx, now do the substitution x 3 7 sin(θ). Exmple.8 (Trick Question). Clculte, t dt. 6 t Use stndrd substitution. i.e. u 6 t, so du tdt, nd the integrl is, du u u / + C.

55 CALCULUS II LECTURE NOTES 55. Section 7.4 : Prtil Frctions Definition. (Rtionl Function). A rtionl function is rtio of polynomils. Exmple.. For exmple the following re rtionl functions, 3x x 3, + x + x + x 3 3x x Notice tht, thus, x + x x + x (x )(x ) 3x 4 x 3x +, 3x 4 x 3x + dx x + x dx ln(x ) + ln(x ) + C. Given rtionl function, we try to write it s sum of simple rtionl functions, ones which we cn integrte. Definition.3 (Proper Frction). A rtionl function, f(x) P (x) Q(x) is clled proper if deg(p ) deg(q), nd clled improper otherwise. Where the degree of polynomil, P (x), is the highest power of x ppering in P (x). If you hve n improper frction you cn do long division to write it s polynomil plus proper frction. i.e. f(x) P (x) R(x) S(x) + Q(x) Q(x). Exmple.4 (Long Division). We consider the improper rtionl function, Do long division, So, x 3 + x x. (x ) x 3 + x + x +... x 3 + x x x + x + x.

56 56 MATTHEW BATES So, x 3 + x x dx x + x + x dx 3 x3 + x + x ln(x ) + C. Exmple.5 (Distinct Liner). Clculte, x + x x 3 + 3x x dx. First we fctorise the denomintor, x 3 + 3x x x(x + 3x ) x(x )(x + ), We wnt to write x + x x 3 + 3x x A x + A x + A 3 x +, for some A, A, A 3 R. Solve for the A i, A (x )(x + ) + A (x)(x + ) + A 3 (x)(x ) x + x, quick wy to find the A i is to cunningly choose x so tht most of the terms vnish. x then the eqution becomes, A ( )(), so A. x then the eqution becomes, A ( )( 5 ) 4, so A 5. x then the eqution becomes, A 3 ( )(5), so A. Thus the integrl becomes, x + x x 3 + 3x x dx x + 5 x x + dx ln(x) + ln(x ) ln(x + ) + C. Exmple.6 (Distinct Liner). Find x 4 3x 3 + x + x x dx. 3x + Note, this is n improper frction, so the first thing we do is some long division, So, We now exmine (x )(x ). (x 3x + ) x 4 3x 3 + x + x... x 4 3x 3 + x + x x 3x + x + x x 3x +. x x 3x+, nd try to write it s sum of simpler frctions. Note, x 3x+ x x 3x + A x + A x,

57 CALCULUS II LECTURE NOTES 57 then, A (x ) + A (x ) x. x : then A ( ), so A. x : then A () 3, so A 3. Thus, the integrl becomes, x 4 3x 3 + x + x x dx x + x 3x + x 3x + dx x x + 3 x 3 x3 ln(x ) + 3 ln(x ) + C. Depending on how the denomintor fctors chnges how we write rtionl function s sum of prtil frctions. Theorem.7. Any polynomil cn be written s product of liner terms nd irreducible qudrtic terms. Theorem.8 (Prtil Frctions). Let f(x) P (x) Q(x) be proper rtionl function. There re four possible cses to consider. Cse : Q(x) fctors into distinct liner terms, i.e. Then write, Q(x) ( x b )( x b )... ( n x b n ). P (x) Q(x) A A n + +. x b n x b n Cse : Q(x) fctors into liner fctors with some repeted, i.e. Then write, Q(x) ( x b ) r ( x b ) r... ( n x b n ) rn. P (x) Q(x) A, A, + x b ( x b ) + + A,n ( x b ) r + A, A, + x b ( x b ) + + A,n ( x b ) r. + A n, A n, + n x b n ( n x b n ) + + A n,n ( n x b n ). rn

58 58 MATTHEW BATES Cse 3 : Q(x) hs non-repeted qudrtic term, i.e. Then use, for the qudrtic term. Q(x) Q(x)(x + bx + c). Ax + B x + bx + c, Cse 4 : Q(x) hs repeted qudrtic term. Then do the nlogous thing to wht hppened in cse. Exmple.9 (Repeted Liner). Write the form of the prtil frction decomposition of the function x 3 x + x (x ) 3 A x + B x + C x + D (x ) + E (x ) 3. Exmple. (Repeted Liner). Write the form of the prtil frction decomposition of the function x 3 x + 5x + 4. First do some long division to get tht, x 3 x + x x x + 4 x + 5x + 4. Now fctorise the denomintor, x + 5x + 4 (x + )(x + 4). Thus the nswer is, x 5 + A x + + B x + 4. Exmple. (Repeted Liner). Write the form of the prtil frction decomposition of the function 8x + (x + ) 3 (x + 3). The nswer is, A x + + B (x + ) + C (x + ) 3 + Dx + E x F x + G (x + 3).

59 Exmple. (Repeted Liner). Clculte, x 4 x + 4x + x 3 x x + dx. First do long division, CALCULUS II LECTURE NOTES 59 (x 3 x x + ) x 4 + x 3 x + 4x +... So the integrl becomes, 4x x + + x 3 x x + dx. We now concentrte on the frction. Fctorise the denomintor, x 3 x x + (x )(x ) (x ) (x + ). Thus we consider prtil frctions of the form, 4x x 3 x x + A x + Putting everything over common denomintor gives tht B (x ) + C x +. 4x A(x )(x + ) + B(x + ) + C(x ), Letting x we see tht 4 B, so B. Letting x we see tht 4 4C, so C Thus, 4x A(x )(x + ) + 4(x + ) (x ), so A. So the integrl becomes, x 4 x + 4x + x 3 x x + dx x + + x + (x ) x + dx x / + x + ln(x ) ln(x + ) + C. x Exmple.3 (Distinct Qudrtic). Clculte, x x + 4 x 3 + 4x dx. Fctorise the denomintor, x 3 + 4x x(x + 4), so we need to consider prtil frctions of the form, x x + 4 x 3 + 4x A x + Bx + C x + 4. So, x x + 4 A(x + 4) + Bx + Cx (A + B)x + Cx + 4A. So A, C, nd B. The integrl becomes, x x + 4 x 3 + 4x dx x + x x + 4 dx x ln(x) + x + 4 dx x + 4 dx ) + C. ln(x) + ln(x + 4) rctn ( x

60 6 MATTHEW BATES Exercise.4. Clculte the following, (x 3)(x + ) dx x (x + 4)(x ) dx x x 7 dx Exmple.5. Clculte, 6 x 9 x 4 dx. First we mke the substitution, u x, so x u nd dx udu. So the integrl becomes, 6 9 x x 4 dx u u 4 udu ( u u 4 ) du 4 u 4 du u u + du 4 + (ln(u ) + ln(u + )) 3 ( ) ( ) + ln ln. 6 5

61 CALCULUS II LECTURE NOTES 6. Section 7.8 : Improper Integrls Does it mke sense to tlk bout the re of n unbounded shpe? Exmple.. Clculte the re under the grph of f(x) x form x to x. Infinity isn t rel number, so we cn t just clculte the ntiderivte then plug in infinity. Wht we cn do is estimte the re by considering for lrger nd lrger vlues of t. t t x dx, dx x x t t. So for lrger nd lrger vlues of t, this estimte gets closer nd closer to. So it is resonbly to sy, dx. x Definition. (Type Improper Integrls). t f(x)dx lim t f(x)dx, if it exists. b f(x)dx lim b t f(x)dx, if it exists. t f(x)dx f(x)dx + b f(x)dx, if it exists. Exmple.3. Clculte, x dx. Exmple.4. Clculte, x 3 dx. Exmple.5. Clculte, for generl p R. x p dx,

62 6 MATTHEW BATES Theorem.6. converges if p >, nd diverges otherwise. x p dx, Exmple.7. Clculte, e x dx. Exmple.8. Clculte, xe x dx. Exmple.9. Clculte, xe x dx. Exmple.. Clculte, x dx π. + Exmple.. Clculte, sin(x)dx. There re other wys in which the grph of function cn be unbounded, i.e. infinite discontinuities. Definition. (Type Improper Integrls). If f(x) hs verticl symptote t x c where < c < b, then Sinilrly, And, When such limits exist. b c b c f(x)dx def lim t c f(x)dx def lim t c+ f(x)dx def c t b t f(x)dx + f(x)dx. f(x)dx. b c f(x)dx.

63 CALCULUS II LECTURE NOTES 63 Exmple.3. Clculte, x dx. Exmple.4. Clculte, 7 3 x 3 dx. Exmple.5. Clculte, 3 x dx. This function hs symptote t x, so we hve to split this into two integrls, 3 x dx Lets concentrte on the first term, Thus the integrl doesn t exist. dx lim x t lim t x dx + t 3 x dx t ln x x dx. lim ln t ln t lim ln( t) t. Note tht in the previous exmple it is importnt to split the integrl, if you didn t relise tht it ws n improper integrl nd just tried to integrte it nively you would get, 3 3 dx ln x x ln() ln() ln(). Which is incorrect. Exmple.6. Clculte, ln(x)dx.

64 64 MATTHEW BATES This hs n verticl symptote t x, so the integrl equls, ln(x)dx lim ln(x)dx t + t ( lim x ln(x) t + t lim ( t ln(t) + t) t + lim t + ln(t) /t /t lim t + /t lim t t +. t dx ) Exmple.7. Clculte, e x x dx First find n ntiderivtive for e x. Use the substitution u /x. x e x x dx e u du e u e /x. So, e x t e x dx lim x t x dx t lim t e/x lim t e/t e lim s es e e. Theorem.8 (Comprison Theorem for Improper Integrls). If f(x) g(x) for ll x, then if f(x)dx exists then so does g(x)dx, nd f(x)dx g(x)dx.

65 Exmple.9 (Comprison Theorem). The integrl diverges. CALCULUS II LECTURE NOTES 65 4 ln(x) x dx,

66 66 MATTHEW BATES 3. Section. : Sequences Definition 3.. A Sequence is ny order list of numbers, {,,... }. Sometimes denoted, { n } n, { n }. One could think of sequence s function f : N R. It is importnt to note tht sequence doesn t need to follow ny sort of pttern. Exmple 3. (Sequences s lists). n n th digit of π. {3,, 4,, 5,... }. Exmple 3.3 (Sequences s functions). {n} n n n {,, 3, 4,... } {/n} n n /n {, /, /3, /4,... } {sin(nπ/3)} n n sin(nπ/3) { 3/, 3/,, 3/, 3/,,... }. Exmple 3.4 (Sequences recursively defined). The sequence defined by, nd n n + n is clled the Fiboncci sequence. It hs lots of strnge properties. For exmple, not tht {,,, 3, 5, 8, 3,, 34,... } 3, 3 5, 5 3 8, Exmple 3.5 (Pictures of sequences). One cn think of sequences pictorilly, this cn be helpful to understnd convergence.

67 CALCULUS II LECTURE NOTES 67 Definition 3.6 (Limit of sequence). We define the limit of the sequence, { n } s n to be, if lim n l n ɛ >, N > such tht, n > N n l < ɛ. We lso sometimes write this s, n l. If sequence hs limit, we sy tht it converges, else we sy tht it diverges. If lim n n, then we sy tht the sequence diverges to infinity. You should think of the sequence n converging to l to men tht no mtter how close you wnt the sequence to be to l, there is some number, N >, such tht if n > N then the sequence is lwys t lest tht close. Theorem 3.7. If f(x) is some function, nd n f(n) for ll n, then lim n lim f(x). n x Exmple 3.8. If n n, then lim n n lim x x. Exmple 3.9. Agin, let n n. We will show tht n using the ɛ N definition. Fix some ɛ >. We wnt to find some N >, such tht n > N < ɛ. n This is true when n > ɛ. So let N [ ɛ ]. Theorem 3.. If f(x) is continuous function then, lim f( n) f ( lim n n n ).

68 68 MATTHEW BATES Exmple 3.. Clculte the limit of the following sequence, Divide the numertor nd denomintor by n, n 4 + 5n n 3n n 4 n 3n + 5 n n 3 5/3. Exmple 3.. Clculte the limit of the following sequence, Use the function f(x) /x. b n /n. Exmple 3.3. Clculte the limit of the following sequence, c n ( ) n. The limit doesn t exist. To see this not tht if the limit did exist, ( ) n l for some l, then for ɛ /, there must exist some N > such tht So in prticulr, nd n > N ( ) n l < /. ( ) N l < /, ( ) N+ l < /. So, / < l < 3/ nd 3/ < l < /, which is impossible. Exmple 3.4. Clculte the limit of the following sequence, d n sin(nπ/). Note, tht d n {,,,,,,,,... } so the limit doesn t exist for the sme reson s the previous exmple. Definition 3.5. We sy tht sequence diverges to infinity, n, if M >, N >, such tht, n > N n > M.

69 Exmple 3.6. lim ( + n n3 ). To show this let M >, we wnt to find N >, such tht, i.e. n 3 > M, so n > 3 M. CALCULUS II LECTURE NOTES 69 n > N + n 3 > M Theorem 3.7 (Limit Lws). If lim n < nd lim b n <, then, lim( n + b n ) lim( n ) + lim(b n ), lim(c n ) c lim( n ), lim( n b n ) lim( n ) lim(b n ), lim( n /b n ) lim( n )/ lim(b n ). Note, it is very importnt tht both lim( n ) <, nd lim(b n ) <, other wise these equtions re not true. For exmple, lim() lim(n ) lim(n) lim(/n). n Theorem 3.8 (Squeeze Rule). If n b n c n for ll n > n for some n >, then, lim( n ) lim(b n ) lim(c n ). Exmple 3.9. Clculte the limit of the following sequence, Note, so, So by the Squeeze rule, Thus, n sin(nπ/). n sin(nπ/), /n sin(nπ/)/n /n. lim( /n) lim(sin(nπ/)/n) lim(/n). lim(sin(nπ/)/n).

70 7 MATTHEW BATES Exmple 3.. Clculte the limit of, Note tht, n! n n. n! n n n. Theorem 3. (Absolute Convergence Theorem for Sequences). If lim n n, then lim n n. Definition 3.. A sequence n is clled... Incresing : if n+ n for ll n. Decresing : if n+ n for ll n. Monotonic : if it is incresing or decresing. Bounded : if there exists some K > such tht K < n < K for ll n. Exmple 3.3. Determine whether the sequence is bounded, incresing, decresing, or not monotonic. n 4n + Exmple 3.4. Determine whether the sequence is bounded, incresing, decresing, or not monotonic. ( ) n n Theorem 3.5 (Monotone Convergence Theorem). Every bounded monotonic sequence converges. Theorem 3.6 (Geometric Sequence). if r > lim n rn if < r DNE if r. Exmple 3.7. Does the following sequence converge? n ( )3n+ 3 n

71 CALCULUS II LECTURE NOTES 7 4. Section. : Series Wht does it men when we write deciml number? So wht does it men when we write n infinite deciml? This is n infinite sum. How do we mke sense of infinite sums in generl? Definition 4. (Prtil Sums). Given sequence n, we define s k k i i def k. We denote this sum by s k, nd cll it the k th prtil sum of the sequence. Exmple 4.. n i i n n(n + ). Definition 4.3 (Series). Given sequence, the ssocited series is defined to be, n def i lim i lim s n. n n i You cn think of this s n infinite sum. i Exmple 4.4. Clculte, Drw squre. i ( ) n. Exmple 4.5. Clculte, Look t prtil sums. n. i

72 7 MATTHEW BATES Exmple 4.6. Clculte, ( ) n. Look t prtil sums. They don t converge. i Definition 4.7 (Geometric Series). A series of the form is clled geometric series. + r + r + r 3 + r 4 + r i, i Theorem 4.8 (Geometric Series). r i i { Diverges if r r if r < Proof. Note, So, nd thus, s n + r + r + + r n + r n, rs n r + r + r r n + r n, s n rs n r n, s n ( rn ). r Exmple 4.9. Show tht i ( ) i Exmple 4.. Clculte ( ) i 3 i

73 CALCULUS II LECTURE NOTES 73 Exmple 4.. Clculte i 3 i i Exmple 4.. Clculte sin i (x). i Exmple 4.3. For wht vlues of x does the following converge? And when it does converge, wht is it s limit? x n x if x <, if x, n DNE if x. Exmple 4.4. For wht vlues of x does the following converge? And when it does converge, wht is it s limit? (x ) n 3 n. This is geometric series, with common rtion x 3. Thus (x ) n if x x 3 <, 3 3 n if n x 3, DNE if x 3. Simplifying gives, 3 (x ) n 5 x if x (, 5), 3 n if x 5, n DNE if x. n Theorem 4.5 (Series Lws). If n nd b n re both convergent series, then, (n + b n ) ( n ) + (b n ) (cn ) c ( n ) Exmple 4.6. Clculte n + 3 n 5 n.

74 74 MATTHEW BATES Exmple 4.7 (Prtil Frctions / Telescoping Series). Clculte n(n + ). Use prtil frctions to write this s, Now clculte the prtil sums, ( s k Thus, n ) + n n ( n ) n + ( ) ( + 3 k ) k + k + ( n ) ( lim(s k ) lim ). n + k + Exmple 4.8 (Hrmonic Series). The Hrmonic series is defined s, n n. We will show tht the ssocited series is divergent. s s + s s s k + k Thus, n n lim(s k). This is very importnt exmple since it is n exmple of sequence, n, where n, but n does not converge. Theorem 4.9 (Divergence Test). If n, then n diverges. Equivlently, if n converges then n. Proof. If n l <, then lim s k l. Now note tht, n s n s n l l.

75 CALCULUS II LECTURE NOTES 75 Note the direction of the impliction! The bove theorem sys tht if n < then n. It does not sy n then n <!!! Exmple 4.. Clculte the following limits, n n 3n + ( ) 3 n π n + n ( ) n ln n + n n n n n Exmple 4.. Write.37 s frction ( ) ( ) 99

76 76 MATTHEW BATES 5. Section.3 : Integrl Test nd Estimtes of Sums It is very hrd to clculte the exct vlue of generl series. So on settles for estimtes. Moreover, it is often useful to know tht series converges, even if you don t know it s exct vlue. Exmple 5.. We will show tht n n, converges. There is no simple formul for the prtil sums of this series, but we cn still prove tht it converges by thinking of it geometriclly. Consider the function, n f(x) x. Thinking of the series s the re of bunch of rectngles, we see tht, n + dx. x Thus the series converges. (The exct vlue is ctully π 6, but this is very hrd to show.) Exmple 5.. We will show tht diverges. Consider the function, n n, f(x) x. Thinking of the series s the re of bunch of rectngles, we see tht, n dx. x Thus the series diverges. n Theorem 5.3 (Integrl Test). If f(x) is positive, continuous, decresing function, nd n f(n), then the series converges iff the integrl f(x)dx, converges. n n