Calculus MATH 172-Fall 2017 Lecture Notes

Clculus MATH 172-Fll 2017 Lecture Notes

These notes re concise summry of wht hs been covered so fr during the lectures. All the definitions must be memorized nd understood. Sttements of importnt theorems lbelled in the mrgin with the symbol must be memorized nd understood. Proofs tht you re expected to be ble to reproduce nd tht re fundmentl re lbelled with the symbol in the mrgin.

Contents Chpter 1. Integrtion 5 1. The definite integrl 5 2. The Fundmentl Theorem of Clculus 5 2.1. The Fundmentl Theorem of Clculus I 5 2.2. The Fundmentl Theorem of Clculus II 6 3. Chnge of vrible or the substitution rule 7 Chpter 2. Applictions of Integrtion 9 1. Are between curves 9 2. Volumes 11 3. Volumes by cylindricl shells 12 4. Work 15 5. Averge vlue of function 16 Chpter 3. Techniques of Integrtion 17 1. Integrtion by Prts 17 2. Trigonometric Integrls 20 3. Weierstrss Substitution 24 4. Trigonometric Substitution 25 5. Integrtion of rtionl functions by prtil frctions 26 5.1. Denomintors with degree exctly 2 27 5.2. Denomintors with degree exctly 3 29 Chpter 4. Improper Integrls 33 1. Integrtion on unbounded intervls 33 2. Integrbility of unbounded functions 35 3. Comprison theorems 37 Chpter 5. Sequences of Rel Numbers 39 1. Definitions nd bsic properties 39 2. Convergence Criteri 44 3. The Monotone Convergence Theorem 45 Chpter 6. Series of rel numbers 47 1. Definitions, Nottion, nd Bsic Properties 47 2. Comprison Theorems 51 3. Alternting Series 52 4. Absolutely Convergent Series 53 Chpter 7. Applictions of Integrtion II 55 1. Arc length 55 2. Are of surfce of revolution 56 Chpter 8. Prmetric equtions nd polr equtions 59 1. Prmetric curves 59 3

4 CONTENTS 1.1. Arc length of prmetric curves 60 1.2. Are enclosed by prmetric curve 60 2. Polr coordintes 61 2.1. Arc length in polr coordintes 61 2.2. Are in polr coordintes 62

CHAPTER 1 Integrtion 1. The definite integrl If (x n ) n=1 be sequence of rel numbers. We write n i=1 x i the sum of the n-th first term in the sequence, i.e. n x i := x 1 + + x n. i=1 Definition 1 (Limit of sequence). A sequence of rel numbers (x n ) n=1 is sid to converge to rel number l R, if for every ε > 0 there exists nturl number N := N(ε) N such tht for ll n N, x n l < ε. If (x n ) n=1 converges to l, we write lim x n = l. n Definition 2. [Integrble functions/definite integrl] Let f be function defined on closed intervl [, b] nd let (x i ) n i=0 be the regulr prtition of [, b], i.e. x i = + i b n for ll i {0, 1,..., n}. If lim n n i=1 f(x i ) b n exists nd gives the sme vlue for ll possible choices of points x i [x i 1, x i ], then we sy tht f is integrble on [, b]. If f is integrble on [, b] we denote the definite integrl of f on [, b] by n f(x)dx := lim f(x i ) b n n. i=1 Exmple 1. Is the function x x 2 + 1 integrble on [0, 1]? If it is wht is the vlue of 1 0 (x2 + 1)dx. 2. The Fundmentl Theorem of Clculus 2.1. The Fundmentl Theorem of Clculus I. Theorem 1 (Fundmentl Theorem of Clculus I). Let f : [, b] R be continuous on [, b]. Then, F (x) = x f(t)dt is well-defined for ll x [, b]. More over, the function F is differentible on [, b], nd F is continuous on [, b] with F (x) = f(x) for ll x [, b]. Sketch of proof. 5

6 1. INTEGRATION Exmple 2. Describe where the following function is differentible nd compute its derivtive. (1) F (x) = x 1 cos(t)dt 2.2. The Fundmentl Theorem of Clculus II. Definition 3. Let f : [, b] R. A function F : [, b] R is sid to be n nti-derivtive of f on [, b], if F is differentible on [, b] with F (x) = f(x) for ll x [, b]. Theorem 2 (Fundmentl Theorem of Clculus II). Let f : [, b] R be continuous on [, b], nd let F be n nti-derivtive of f on [, b]. Then, f(t)dt = F (b) F (). The FTC II combined with the Chin Rule hs the following useful ppliction. Theorem 3. Let I be n intervl, ϕ: I [, b] be differentible on I, nd f : [, b] R be continuous on [, b]. Then, H(x) = ϕ(x) f(t)dt is well-defined for ll x I, nd the function H is differentible on I with H (x) = f(ϕ(x))ϕ (x) for ll x I. Proof. Exmple 3. Describe where the following function is differentible nd compute its derivtive. (1) F (x) = x 2 1 tdt

3. CHANGE OF VARIABLE OR THE SUBSTITUTION RULE 7 3. Chnge of vrible or the substitution rule Theorem 4 (Chnge of vrible/substitution rule). Let φ be differentible function on [, b] such tht φ is continuous on [, b]. Let f be continuous function on the rnge of φ, i.e. on φ([, b]), then f(φ(t))φ (t)dt = φ(b) φ() f(u)du. Exercise 1. Compute the following definite integrls: (1) 3 2 3x 2 1 (x 3 x) 2 dx (2) π 3 π 6 cos(θ) sin 2 (θ) dθ (3) 1 0 x 1 + x 4 dx (4) e e 2 dx x ln(x) The substitution rule is useful to compute nti-derivtives of complicted functions. Exercise 2. Find ll the nti-derivtives for the following functions: (1) f(x) = (2x + 1)(x 2 + x + 1) 3

8 1. INTEGRATION (2) f(x) = 3 1 x (3) f(x) = x 2x 2 +3 (4) f(θ) = cos(θ) 1+sin 2 (θ)

CHAPTER 2 Applictions of Integrtion 1. Are between curves In mny cses, we cn compute the re enclosed by certin curves by setting up single integrl. The re enclosed by the curves y = f(x), y = g(x), x =, nd x = b where f, g re continuous functions on [, b] is (1) A = f(x) g(x) dx. The re enclosed by the curves x = f(y), x = g(y), y = c, nd y = d, where f, g re continuous functions on [c, d] is (2) A = d c f(y) g(y) dy. Method: To solve problem which sks to find the re of region enclosed by severl curves you need to: (1) sketch the grph of the functions representing the curves in order to understnd wht is the region under considertion, (2) determined wht will be judicious choice for the vrible of integrtion, (3) depending on how mny times the curves interlce, set up one or severl integrls tht re needed in order to compute the re, (4) find the intersection points of the curves in order to get the upper nd lower bounds in the integrl(s), (5) compute the ntiderivtive(s) needed to compute the integrl(s) using FTC II. 9

10 2. APPLICATIONS OF INTEGRATION Exercise 3. Find the re of region enclosed by the curves y = x 2 nd y = x 4. Solution. Exercise 4. Find the re of region enclosed by the curves y = 2x + 4 nd 4x + y 2 = 0. Solution.

2. VOLUMES 11 Exercise 5. Find the re of region enclosed by the curves x = 3y, x+y = 0 nd 7x + 3y = 24. Solution. 2. Volumes Computing the volume of 3-dimensionl solid cn be rther complicted nd might require the use of triple integrl. However in certin situtions, e.g. if the solid hs certin symmetries, then we cn compute the volume using simple integrl. The following formuls give the volume of solids of revolution obtined in by rotting region of the xy-plne bout one of the two xes. The volume of the solid obtined by rotting bout the x-xis the region bounded by the curves y = f(x), y = g(x), x =, nd x = b, where f g 0 re continuous functions on [, b] is (3) V = π (f(x) 2 g(x) 2 )dx. The volume of the solid obtined by rotting bout the y-xis the region bounded by the curves x = f(y), x = g(y), y = c, nd y = d, where f g 0 re continuous functions on [c, d] is (4) V = π d c (f(y) 2 g(y) 2 )dy. For rbitrry solids, we cn use the following formuls.

12 2. APPLICATIONS OF INTEGRATION Let S be solid tht lies between x = nd x = b nd A(x) be its crosssectionl re in the plne pssing through x nd perpendiculr to the x-xis. If x A(x) is continuous function on [, b], then the volume of S is (5) V = A(x)dx. Let S be solid tht lies between y = c nd y = d nd A(y) be its crosssectionl re in the plne pssing through y nd perpendiculr to the y-xis. If y A(y) is continuous function on [c, d], then the volume of S is (6) V = d c A(y)dy. Exercise 6. Compute the volume of frustum of pyrmid with height h, lower bse side, nd top side b. Solution. 3. Volumes by cylindricl shells The volume of the solid obtined by rotting bout the y-xis the region bounded by the curves y = f(x), y = g(x), x =, nd x = b, where f nd g re continuous functions on [, b] with 0 < b is (7) V = 2π x f(x) g(x) dx.

3. VOLUMES BY CYLINDRICAL SHELLS 13 The volume of the solid obtined by rotting bout the x-xis the region bounded by the curves x = f(y), x = g(y), y = c, nd y = d, where f nd g re continuous functions on [c, d] with 0 c < d is (8) V = 2π d c y f(y) g(y) dy. Exercise 7. Find the volume of the solid of revolution obtined by rotting the region enclosed by the curves y = sin(x 2 ), y = 0, x = 0, x = π, bout the y-xis. Exercise 8. Find the volume of the solid of revolution obtined by rotting the region enclosed by the curves y = x 2 6x + 10, y = x 2 + 6x 6, bout the y-xis.

14 2. APPLICATIONS OF INTEGRATION Exercise 9. Find the volume of the solid of revolution obtined by rotting the region enclosed by the curves y = x, x = 0, x + y = 2, bout the x-xis. Exercise 10. Find the volume of the solid of revolution obtined by rotting the region enclosed by the curves y = x, x = 0, x + y = 2, bout the y-xis.

4. WORK 15 Exercise 11. Find the volume of the solid of revolution obtined by rotting the region enclosed by the curves y = x 1 + x 3, y = 0, x = 0, x = 2, bout the y-xis. Exercise 12. Find the volume of the solid of revolution obtined by rotting the region enclosed by the curves x 2 + (y 1) 2 = 1, bout the y-xis. 4. Work Definition 4. Suppose tht n object moves long the x-xis in the positive from x = to x = b nd t ech point between nd b force f(x) cts on the object, where f is continuous function on [, b]. We define the work done in moving the object from to b s W b = In prticulr if f is constnt force, i.e. W b = F (b ). f(x)dx. f(x) = F for ll x b, then Remrk 1. The work done is n lgebric number nd hs to be interpreted ccordingly.

16 2. APPLICATIONS OF INTEGRATION 5. Averge vlue of function Definition 5. Let < b nd f be continuous function on [, b]. The verge vlue of f on [, b] is defined by f ve = 1 b Recll the Men Vlue Theorem. f(x)dx. Theorem 5 (Men Vlue Theorem). Let < b nd f : [, b] R. If, (i) f is continuous on [, b], (ii) f is differentible on (, b), then there exists c (, b) such tht f (c) = f(b) f() b. We will not prove the Men Vlue Theorem but we will use it to prove the Men Vlue Theorem for Integrls which sys tht if f is continuous on [, b] then there exists rel number c [, b] such tht f ve = f(c). Theorem 6 (Men Vlue Theorem for Integrls). Let f be continuous function on [, b]. Then, there exists rel number c [, b] such tht Proof. f(x)dx = f(c)(b ). Exercise 13. The temperture of metl rod, 10m long, is 5x (in o C) t distnce x from one end of the rod. Wht is the verge temperture of the rod?

CHAPTER 3 Techniques of Integrtion 1. Integrtion by Prts Theorem 7. Let f, g be rel vlued functions on [, b] such tht: (1) f nd g re differentible on [, b], (2) f nd g re integrble on [, b]. Then, (9) f (x)g(x)dx = f(b)g(b) f()g() f(x)g (x)dx. Proof. (Hint: Use the product rule nd the FTC) Remrk 2. It might be convenient in prctice to write eqution (9) s (10) f (x)g(x)dx = [f(x)g(x)] b f(x)g (x)dx. Exercise 14. Let f be rel-vlued differentible function on [0, 1] such tht f is integrble on [0, 1]. If f(0) = f(1) = 0, show tht 1 e x (f(x) + f (x))dx = 0. 0 17

18 3. TECHNIQUES OF INTEGRATION The integrtion by prt formul is extremely useful to compute ntiderivtive of functions of the form P (t)e αt, P (t) sin(αt), or P (t) cos(αt) where P (t) is polynomil. We typiclly pply the integrtion by prt with g = P in order to decrese the degree of the polynomil. If P is polynomil of degree n we will pply the formul n-times in row. Exercise 15. Compute the following integrls. (1) 1 0 x2 e x. (2) e π 1 sin(ln(x)). (3) 4 0 e x dx.

1. INTEGRATION BY PARTS 19 Exercise 16. Find ll the ntiderivtives of the function f, where: (1) f(x) = x cos(x), (2) f(x) = rcsin(x), (3) f(x) = e x cos(2x), (4) f(x) = x 3 e x2,

20 3. TECHNIQUES OF INTEGRATION (5) f(x) = cos n (x), where n is fixed nturl number. Exercise 17. Compute the volume of the solid of revolution obtined by rotting the region bounded by the curves y = e x, y = e x, x = 1, bout the y-xis. 2. Trigonometric Integrls Our gol in this section is to compute ntiderivtives of trigonometric functions. We first consider functions of the form f(x) = sin(mx) cos(nx), f(x) = sin(mx) sin(nx), f(x) = cos(mx) cos(nx) where m nd n re nturl numbers. If f(x) = sin(mx) cos(nx) use the ddition formul sin() cos(b) = 1 (sin( b) + sin( + b)). 2 Exercise 18. Compute ll the ntiderivtives of the function f(x) = cos(3x) sin(x).

2. TRIGONOMETRIC INTEGRALS 21 If f(x) = sin(mx) sin(nx) use the ddition formul sin() sin(b) = 1 (cos( b) cos( + b)). 2 Exercise 19. Compute ll the ntiderivtive of the function f(x) = sin(3x) sin(x). If f(x) = cos(mx) cos(nx) use the ddition formul cos() cos(b) = 1 (cos( b) + cos( + b)). 2 Exercise 20. Compute ll the ntiderivtive of the function f(x) = cos(3x) cos(x). We now consider functions of the form f(x) = sin m (x) cos n (x) where m nd n re nturl numbers. We will use vrious chnges of vribles depending on the prity of m or n. If f(x) = sin m (x) cos n (x) where m is odd, i.e., if f(x) = sin 2k+1 (x) cos n (x) use the chnge of vrible u(x) = cos(x).

22 3. TECHNIQUES OF INTEGRATION Exercise 21. Compute ll the ntiderivtives of the function f(x) = sin 3 (x) cos 2 (x). If f(x) = sin m (x) cos n (x) where n is odd, i.e., if f(x) = sin m (x) cos 2k+1 (x) use the chnge of vrible u(x) = sin(x). Exercise 22. Compute ll the ntiderivtive of the function f(x) = cos 5 (x) sin 2 (x). Exercise 23. Compute ll the ntiderivtive of the function f(x) = x cos 5 (x 2 ) sin 2 (x 2 ).

2. TRIGONOMETRIC INTEGRALS 23 If f(x) = sin m (x) cos n (x) where m nd n re even, i.e., if f(x) = sin 2k (x) cos 2r (x), use the formuls (1) sin 2 (x) = 1 cos(2x) 2, (2) cos 2 (x) = 1+cos(2x) 2, to progressively get rid of the exponents. In the specil cse where k = r, you cn lso use sin(x) cos(x) = sin(2x) 2. Exercise 24. Compute ll the ntiderivtives of the function f(x) = cos 2 (x) sin 2 (x). Exercise 25. Compute π 2 0 cos2 (x) sin 2 (x)dx. Exercise 26. Compute the volume of the solid of revolution obtined by rotting the region bounded by the curves y = sin(x), y = 2 sin 2 (x), x = 0, x = π 2, bout the x-xis.

24 3. TECHNIQUES OF INTEGRATION 3. Weierstrss Substitution Computing ll the ntiderivtives of trigonometric function cn lwys be reduced to computing ll the ntiderivtives of rtionl function using Weierstrss chnge of vrible t = tn( θ 2 ) for π < θ < π. Let t = tn( θ 2 ) for π < θ < π, then sin(θ) = 2t 1 + t 2, cos(θ) = 1 t2 1 + t 2, dt dθ = 1 + t2. 2 How to compute ll the ntiderivtives of rtionl function will be discussed lter on. Exercise 27. Find ll the ntiderivtives of 1 cos(θ). Exercise 28. Find ll the ntiderivtives of 1 sin(θ).

4. TRIGONOMETRIC SUBSTITUTION 25 4. Trigonometric Substitution In this section we wnt to compute the ntiderivtive of functions involving expressions of the form 2 x 2, 2 + x 2, x 2 2. In the cse where you hve n expression of the form 2 x 2 use the chnge of vrible x = sin(θ) with π 2 θ π 2 or x = cos(θ) with 0 θ π. Exercise 29. Compute 2 0 x3 4 x 2 dx. Exercise 30. Find ll the ntiderivtives of f(t) = 9 e 2t. In the cse where you hve n expression of the form 2 + x 2 use the chnge of vrible x = tn(θ) with π 2 < θ < π 2. Exercise 31. Find ll the ntiderivtives of f(x) = 1 x2 +1.

26 3. TECHNIQUES OF INTEGRATION In the cse where you hve n expression of the form x 2 2 use the chnge of vrible x = cos(θ) with 0 θ < π 2 or π θ < 3π 2. Exercise 32. Find ll the ntiderivtives of f(x) = 1 x 2 16x 2 9. 5. Integrtion of rtionl functions by prtil frctions It is possible to compute ll the ntiderivtives of rtionl functions, i.e., functions of the form f(x) = P (x) Q(x) where P nd Q re polynomils. Exercise 33. Let 0. Find ll the ntiderivtive of the function f(x) = 1 x + b Exercise 34. Let 0. Find ll the ntiderivtive of the function f(x) = 2x + b x 2 + bx + c Exercise 35. Let 0. Find ll the ntiderivtive of the function f(x) = 1 (x + b) 2

5. INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 27 Exercise 36. Let β > 0. Find ll the ntiderivtive of the function f(x) = 1 (x α) 2 + β 2 Exercise 37. Let 0. Find ll the ntiderivtive of the function f(x) = 1 x 2 + bx + c if b2 4c < 0. The generl cse cn get rther tricky nd we will restrict ourselves to functions of the form f(x) = P (x) Q(x) where Q is polynomil of degree t most 3 nd the degree of P is strictly less thn the degree of Q. 5.1. Denomintors with degree exctly 2. Two distinct roots: Let f(x) = P (x) where Q is polynomil of degree Q(x) exctly 2 nd P hs degree t most 1. To find ll the ntiderivtives of f, P (x) fctorize Q nd if f(x) =, with b, find α, β R such tht (x )(x b) f(x) = α x + β x b. Exercise 38. Compute ll the ntiderivtives of the function f(x) = 7 2x 2 + 5x 12.

28 3. TECHNIQUES OF INTEGRATION One repeted root: Let f(x) = P (x) where Q is polynomil of degree Q(x) exctly 2 nd P hs degree t most 1. To find ll the ntiderivtives of f, fctorize Q nd if f(x) = P (x) (x ), find α, β R such tht 2 f(x) = α x + β (x ) 2. Exercise 39. Compute ll the ntiderivtives of the function f(x) = 3x x 2 + 2x + 1. Irreducible denomintor: Let f(x) = P (x) where Q(x) = αx + β nd Q(x) Q(x) = x 2 + bx + c is polynomil of degree exctly 2 tht does not hve rel roots nd P hs degree t most 1. To find ll the ntiderivtives of f, write f(x) s f(x) = α 2x + b αb 2 x 2 + (β + bx + c 2 ) 1 (x + b to be ble to find 2 )2 + c b2 4 the ntiderivtives. Exercise 40. Compute ll the ntiderivtives of the function f(x) = 2x + 6 2x 2 + 2x + 7.

5. INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 29 5.2. Denomintors with degree exctly 3. Three distinct roots: Let f(x) = P (x) where Q is polynomil of Q(x) degree exctly 3 nd P hs degree t most 2. To find ll the ntiderivtives of P (x) f, fctorize Q nd if f(x) =, with, b, c ll distinct, find (x )(x b)(x c) α, β, γ R such tht f(x) = α x + β x b + γ x c. Exercise 41. Compute ll the ntiderivtives of the function f(x) = 1 x 3 + 2x 2 x + 2. One simple root nd double root: Let f(x) = P (x) Q(x) where Q is polynomil of degree exctly 3 nd P hs degree t most 2. To find ll the ntiderivtives of f, fctorize Q nd if f(x) = α, β, γ R such tht f(x) = α x + P (x) (x ) 2, with b, find (x b) β (x ) 2 + γ (x b). Exercise 42. Compute ll the ntiderivtives of the function f(x) = x 2 2x 3 x 2 45x 12.

30 3. TECHNIQUES OF INTEGRATION One triple root: Let f(x) = P (x) where Q is polynomil of degree Q(x) exctly 3 nd P hs degree t most 2. To find ll the ntiderivtives of f, fctorize Q nd if f(x) = P (x) (x ), find α, β, γ R such tht f(x) = α 3 x + β (x ) 2 + γ (x ) 3. Exercise 43. Compute ll the ntiderivtives of the function f(x) = 2x + 1 x 3 3x 2 + 3x 1. One simple root nd n irreducible fctor: Let f(x) = P (x) Q(x) where Q is polynomil of degree exctly 3. To find ll the ntiderivtives of f, fctorize P (x) Q nd if f(x) = (x )(x 2 + bx + c) where x2 +bx+c does not hve rel roots, find α, β R such tht f(x) = α x + βx + γ x 2 nd then write f(x) in the + bx + c following form f(x) = α x + β 2x + b βb 2 x 2 + (γ + bx + c 2 ) 1 (x + b to 2 )2 + c b2 4 be ble to find the ntiderivtives.

5. INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 31 Exercise 44. Compute ll the ntiderivtives of the function f(x) = 1 x 3 2x 2 + x 2.

CHAPTER 4 Improper Integrls 1. Integrtion on unbounded intervls Definition 6 (Integrble functions on unbounded intervls). Let f be function defined on the intervl [, ) such tht for every t, t f is integrble on [, t]. If lim t f(x)dx exists nd is finite we sy tht f is integrble on [, ) nd we define the improper integrl of f on [, ) s f(x)dx := lim t t f(x)dx. Let f be function defined on the intervl (, b] such tht for every t b, f is integrble on [t, b]. If lim t f(x)dx exists nd is finite we t sy tht f is integrble on (, b] nd we define the improper integrl of f on (, b] s f(x)dx := lim t t f(x)dx. Let f be function defined on the intervl (, ). If there exists c (, ) such tht f is integrble on (, c] nd on [c, ) then we sy tht f is integrble on (, ) nd we define the improper integrl of f on (, ) s f(x)dx := lim t c t f(x)dx + lim t t c f(x)dx. Remrk 3. If the improper integrl f(x)dx ( or f(x)dx or f(x)dx) exists nd is finite we sy tht the improper integrl is convergent. Otherwise we sy tht the improper integrl is divergent. Exercise 45. Is the function f : x 1 x integrble on [1, )? If it is, compute the improper integrl of f on [1, ). 33

34 4. IMPROPER INTEGRALS Exercise 46. Is the function f : x 1 x 2 the improper integrl of f on [1, ). integrble on [1, )? If it is, compute Exercise 47. For wht vlues of p is the function f : x 1 x p [1, )? When it is, compute the improper integrl of f on [1, ). integrble on Exercise 48. Is the function f : θ sin(θ)e cos(θ) integrble on [0, )? If it is, compute the improper integrl of f on [0, ). Exercise 49. For wht vlues of α is the function f : x e αx integrble on [0, )? When it is, compute the improper integrl of f on [0, ).

2. INTEGRABILITY OF UNBOUNDED FUNCTIONS 35 2. Integrbility of unbounded functions Definition 7. [Improper integrl of n unbounded functions on bounded intervl] Let < < b < nd f be function defined on the intervl [, b) t such tht for every t < b, f is integrble on [, t]. If lim t b f(x)dx exists nd is finite we sy tht f is integrble on [, b) nd we define the improper integrl of f on [, b) s f(x)dx := lim t b t f(x)dx. Let f be function defined on the intervl (, b] such tht for every < t b, f is integrble on [t, b]. If lim t f(x)dx exists nd is finite we t sy tht f is integrble on (, b] nd we define the improper integrl of f on (, b] s f(x)dx := lim f(x)dx. t + t Let c (, b) nd f be function defined on the intervl [, b] with n unbounded discontinuity t c. If f is integrble on [, c) nd on (c, b] then we sy tht f is integrble on [, b] nd we define the improper integrl of f on [, b] s f(x)dx := c f(x)dx + c f(x)dx. Remrk 4. If the improper integrl f(x)dx exists nd is finite we sy tht the improper integrl is convergent. Otherwise we sy tht the improper integrl is divergent. Remrk 5. A function tht is integrble on [, b] is in prticulr defined on [, b) nd integrble on [, t] for ll t < b nd we need to mke sure tht the two definitions for f(x)dx (Definition 2 nd Definition 7) give the sme number insuring the comptibility of the two definitions. This is indeed the cse thnks to the FTC I. Assume tht f is function tht is integrble on [, b] then F (t) = t f(x)dx is n nti-derivtive of f. On the one hnd, by the FTC I F is continuous on [, b] nd f(x)dx = F (b) F () = F (b) (ccording to Definition 2 nd the FTC t II). On the other hnd, by continuity of F F (b) = lim t b F (t) = lim t b f(x)dx = b f(x)dx (ccording to Definition 7). Therefore both definitions re comptible.

36 4. IMPROPER INTEGRALS Exercise 50. Is the function f : x 1 x the improper integrl of f on (0, 1]. integrble on (0, 1]? If it is, compute Exercise 51. Is the function f : x 1 x 2 the improper integrl of f on (0, 1]. integrble on (0, 1]? If it is, compute Exercise 52. For wht vlues of p is the function f : x 1 x p (0, 1]? When it is, compute the improper integrl of f on (0, 1]. integrble on Exercise 53. Is the function f : x ln(x) integrble on (0, 1]? compute the improper integrl of f on (0, 1]. If it is,

3. COMPARISON THEOREMS 37 3. Comprison theorems Theorem 8 (Comprison Theorem: convergence criterion). Let f, g : [, ) R such tht Then, (1) f nd g re continuous on [, ), (2) 0 f(x) g(x) for ll x, (3) g is integrble on [, ). (1) f is integrble on [, ) nd (2) f(x)dx g(x)dx. Remrk 6. The conclusion of the convergence criterion is sometimes summrized s follows: If g(x)dx is convergent, then f(x)dx is convergent. Exercise 54. Show tht 0 e x2 dx is convergent. [Hint: Split the integrl t 1 nd use the inequlity x 2 x for ll x 1] Theorem 9 (Comprison Theorem: divergence criterion). Let f, g : [, ) R such tht (1) f nd g re continuous on [, ), (2) 0 f(x) g(x) for ll x, (3) f is not integrble on [, ). Then, g is not integrble on [, ).

38 4. IMPROPER INTEGRALS Remrk 7. The conclusion of the divergence criterion is sometimes summrized s follows: If f(x)dx is divergent, then g(x)dx is divergent. Exercise 55. Is the function f : x 1 + e x x [Hint: Use Exercise 47] integrble on [1, )? Below is list of useful inequlities. (1) x x 2 for ll x 1 (2) x 2 x for ll 0 x 1 (3) x x for ll x 1 (4) x x for ll 0 x 1 (5) ln(x) x for ll x 0 (6) 1 cos(x) 1 for ll x R (7) 1 sin(x) 1 for ll x R (8) sin(x) x for ll x 0

CHAPTER 5 Sequences of Rel Numbers 1. Definitions nd bsic properties In Clculus, you re used to work with rel-vlued functions defined on the set (or subset) of rel numbers. A sequence is nothing else but rel-vlued function whose domin is the set of nturl numbers. Definition 8. A sequence of rel numbers is function from f : N R. The intuition nd the tools to study sequences re completely different from the one to study rbitrry functions nd we usully prefer to see sequence not s function defined on N but s collection of rel numbers tht is enumerted in specil order. We will thus use specil nottion insted of the functionl nottion, nd sequence whose vlues re x n := f(n) will be simply denoted by (x n ) n=1 or (x n ) n N or x 1, x 2,.... We sy tht x n is the term in the n-th position or the term of rnk n. We might lso sy tht x n is the generic term of the sequence. Every function g tht is defined on [1, ) cn give rise to sequence (x n ) n=1 simply by defining x n = g(n) for ll n 1. Definition 9. A sequence of rel numbers (x n ) n=1 is bounded bove if there exists M R such tht for ll n N, x n M. Definition 10. A sequence of rel numbers (x n ) n=1 is bounded below if there exists m R such tht for ll n N, x n m. Definition 11. A sequence of rel numbers (x n ) n=1 is bounded if there exist m, M R such tht for ll n N, m x n M. Exmple 4. (1) The sequence defined by x n = n for ll n N is not bounded bove but is bounded below by m = 1. (2) The sequence defined by x n = 2 ln(n) for ll n N is not bounded below but is bounded bove by M = 2. (3) The sequence defined by x n = ( 1) n for ll n N is bounded bove by M = 1 nd bounded below by m = 1. (4) The sequence defined by x n = 1 n for ll n N is bounded bove by M = 1 nd bounded below by m = 0. (5) The sequence defined by x n = ( 1) n 2 n for ll n N is neither bounded bove nor bounded below. (6) Definition 12. A sequence (x n ) n=1 is incresing if for ll n N x n x n+1. Definition 13. A sequence (x n ) n=1 is strictly incresing if for ll n N x n < x n+1. Exmple 5. The sequence defined by x n = n for ll n N is strictly incresing. Definition 14. A sequence (x n ) n=1 is decresing if for ll n N x n x n+1. Definition 15. A sequence (x n ) n=1 is strictly decresing if for ll n N x n > x n+1. 39

40 5. SEQUENCES OF REAL NUMBERS Exmple 6. The sequence defined by x n = 2 3n for ll n N is strictly decresing. Definition 16. A sequence (x n ) n=1 is monotonic (or monotone) if it is either incresing or decresing. Remrk 8. There re sequences tht re neither incresing nor decresing, e.g. the sequence defined by x n = ( 1) n for ll n N. To study the monotonicity of sequence we usully study the sign of x n+1 x n. If x n+1 x n > 0 for ll n 1 we cn conclude tht (x n ) n=1 is strictly incresing. If x n+1 x n 0 for ll n 1 we cn conclude tht (x n ) n=1 is incresing. If x n+1 x n < 0 for ll n 1 we cn conclude tht (x n ) n=1 is strictly decresing. If x n+1 x n 0 for ll n 1 we cn conclude tht (x n ) n=1 is decresing. For sequences whose terms re ll positive (i.e. for ll n 1, x n > 0) we cn look t the quotient xn+1 x n. If xn+1 x n > 1 for ll n 1 we cn conclude tht (x n ) n=1 is strictly incresing. If xn+1 x n 1 for ll n 1 we cn conclude tht (x n ) n=1 is incresing. If xn+1 x n < 1 for ll n 1 we cn conclude tht (x n ) n=1 is strictly decresing. If xn+1 x n 1 for ll n 1 we cn conclude tht (x n ) n=1 is decresing. Exercise 56. Show tht the sequence defined by x n = n 2 for ll n N is strictly incresing. Exercise 57. Show tht the sequence defined by x n = 2 n for ll n N is strictly decresing.

1. DEFINITIONS AND BASIC PROPERTIES 41 Exercise 58. Show tht the sequence defined by x n = 2 ln(n) for ll n N is strictly decresing. Exercise 59. Show tht the sequence defined by x n = 1 n strictly decresing. for ll n N is Exmple 7. Consider the following recursively defined sequence: { 1 if n = 1 n = 3 1 n 1 if n 2. Cn you show tht the sequence ( n ) n=1 is strictly incresing nd bounded bove by M = 3? Is the sequence bounded below? Definition 17. A sequence of rel numbers (x n ) n=1 is sid to converge to rel number l R, if for every ε > 0 there exists N := N(ε) N such tht for ll n N, x n l < ε. If sequence (x n ) n=1 converges to l R, we write lim n x n = l Remrk 9. There is ctully some freedom in the bove definition. Indeed, one cn replce x n l < ε in the definition by x n l ε or x n l < 2ε or x n l 100ε for instnce, nd yet obtin n equivlent definition. This cn be useful on occsion. Proposition 1. A sequence of rel numbers (x n ) n=1 hs t most one limit.

42 5. SEQUENCES OF REAL NUMBERS Exmple 8. (1) Let c R nd consider the sequence defined by x n = c for ll n N. We cn show tht lim n x n = c. (2) Consider the sequence defined by x n = ( 1) n for ll n N. We cn show tht (x n ) n=1 is not convergent. (3) Consider the sequence defined by x n = 1 n for ll n N. We cn show tht lim n x n = 0. Theorem 10. Every convergent sequence is bounded. Proof. Assume tht (x n ) n=1 converges to l. Then for ε = 1 there exists N N such tht for ll n N, x n l 1, nd thus l 1 < x n < l + 1. Let M := mx{x 1, x 2,..., x N 1, l + 1} nd m := min{x 1, x 2,..., x N 1, l 1}, then for ll n N, m x n M nd (x n ) n=1 is bounded. We will use repetedly the following limit theorems. Theorem 11. Let (x n ) n=1 nd (y n ) n=1 be convergent sequences. Then, the sequence (x n +y n ) n=1 is convergent nd lim n (x n +y n ) = lim n x n +lim n y n. Proof. Assume tht lim n x n = l 1 < nd lim n y n = l 2 <. Let ε > 0, then there exist N 1, N 2 N such tht for ll n N 1, x n l 1 < ε 2 nd for ll n N 2, y n l 2 < ε 2. If follows from the tringle inequlity tht x n + y n (l 1 + l 2 ) = x n l 1 + y n l 2 x n l 1 + y n l 2, nd hence for n N := mx{n 1, N 2 }, x n + y n (l 1 + l 2 ) ε 2 + ε 2 = ε. Exercise 60. Let (x n ) n=1 nd (y n ) n=1 be convergent sequences. Show tht the sequence (x n y n ) n=1 is convergent nd lim n (x n y n ) = lim n x n lim n y n. Theorem 12. Let (x n ) n=1 nd (y n ) n=1 be convergent sequences nd λ R. Then, the sequence (λx n ) n=1 is convergent nd lim n (λx n ) = λ lim n x n Proof. Assume tht lim n x n = l 1 <. If λ = 0 the equlity clerly holds. Otherwise, let ε > 0, then there exist N 1 N such tht for ll n N 1, x n l 1 < ε λ nd simply remrk tht λx n λ = λ x n l 1 λ ε λ = ε.

1. DEFINITIONS AND BASIC PROPERTIES 43 Exercise 61. Let (x n ) n=1 nd (y n ) n=1 be convergent sequences. Show tht the sequence (2x n +3y n ) n=1 is convergent nd lim n (2x n +3y n ) = 2 lim n x n + 3 lim n y n. Theorem 13. Let (x n ) n=1 nd (y n ) n=1 be convergent sequences. Then, the sequence (x n y n ) n=1 is convergent nd lim n (x n y n ) = lim n x n lim n y n. Proof. Assume tht lim n x n = l 1 < nd lim n y n = l 2 <. If follows from the tringle inequlity tht x n y n (l 1 l 2 ) = (x n l 1 )y n + l 1 (y n l 2 ) x n l 1 y n + y n l 2 l 1. Since (y n ) n=1 is convergent, nd thus bounded, there exists M > 0 such tht for ll n N, y n M. Let ε > 0. If l 1 > 0, then there exist N 1, N 2 N such tht for ll n N 1, x n l 1 < ε 2M nd for ll n N 2, y n l 2 < ε 2 l, nd hence for n mx{n 1 1, N 2 }, x n y n (l 1 l 2 ) < ε 2M M + ε 2 l l 1 1 = ε. If l 1 = 0 then for n mx{n 1, N 2 }, x n y n < ε 2M M < ε, nd the proof is complete. Theorem 14. Let (x n ) n=1 be convergent sequence. If x n 0 for ll n N nd lim n x n 0, then the sequence ( 1 x n ) n=1 is convergent nd lim n 1 x n = 1 lim x. n n Proof. Assume lim n x n = l 1 0, then for ε = l1 2 > 0 there exists N 1 N such tht for ll n N 1, x n l 1 < l1 2. It follows from the reverse tringle inequlity tht x n > l1 2 > 0 for n N 1. Also for ε > 0 there exists N 2 N such tht for ll n N 2, x n l 1 < ε l1 2 1 x n 1 l 1 = l1 xn x nl 1 < 2 l 1 xn l1 2 nd for n mx{n 1, N 2 }, l 1 < ε. Theorem 15. Let (x n ) n=1 nd (y n ) n=1 be convergent sequences. If y n 0 for ll n N nd lim n y n 0, then the sequence ( xn y n ) n=1 is convergent nd x lim x n n n lim = n y n lim y n n Proof. Assume tht lim n x n = l 1 < nd lim n y n = l 2 0. Since x n 1 yn = x n y n, the result follows by combining Theorem 13 nd Theorem 14. Exercise 62. If (x n ) n=1 is convergent show tht the sequence defined by y n = x n+1 for ll n 1 is convergent nd tht lim n y n = lim n x n, i.e., lim n x n+1 = lim n x n.

44 5. SEQUENCES OF REAL NUMBERS 2. Convergence Criteri We will now try to find wys to recognize whether or not sequence is convergent. Theorem 16. Let f : [0, ) R be function. If (x n ) n=1 is the sequence such tht x n = f(n) for ll n N if lim x f(x) = l, then, the sequence (x n ) n=1 is convergent nd lim n x n = l. Exercise 63. Let (x n ) n n=1 be the sequence defined by x n = 2 2n 2 +n+5 for ll n N. Is the sequence (x n ) n=1 convergent? If it is wht is the limit of the sequence. Exercise 64. Let (x n ) n=1 be the sequence defined by x n = n3 3n 2 +2 for ll n N. Is the sequence (x n ) n=1 convergent? If it is wht is the limit of the sequence.

3. THE MONOTONE CONVERGENCE THEOREM 45 Theorem 17 (The Squeeze Theorem). Let (x n ) n=1, (y n ) n=1, (z n ) n=1 be sequences of rel numbers. If there exists n 0 N such tht x n y n z n for ll n n 0, (x n ) n=1 nd (z n ) n=1 re convergent nd lim n x n = lim n z n = l, then, (y n ) n=1 is convergent nd lim n y n = l. Proof. Let ε > 0, then there exist N 1, N 2 N such tht for ll n mx{n 1, N 2 }, x n l < ε nd z n l < ε. Thus, if n mx{n 1, N 2, n 0 }, l ε < x n y n z n < l + ε, nd ε < y n l < ε, which shows tht (y n ) n=1 is convergent to l. Since x n x n x n the following corollry follows from the Squeeze Theorem. Corollry 1. Let (x n ) n=1 be sequences of rel numbers. If the sequence ( x n ) n=1 is convergent nd lim n x n = 0 then the sequence (x n ) n=1 is convergent nd lim n x n = 0. Exmple 9. Geometric sequences. Let r R. We cn show tht the geometric sequence (r n ) n=1 is: divergent if r 1, convergent if 1 < r < 1 with lim n r n = 0, convergent if r = 1 with lim n r n = 1, divergent if r > 1 with lim n r n = +. 3. The Monotone Convergence Theorem The Monotone Convergent Theorem reltes the convergence of sequence with its monotonicity nd boundedness. Theorem 18. Let (x n ) n=1 be sequence of rel numbers. If (x n ) n=1 is incresing nd bounded bove then (x n ) n=1 is convergent. Theorem 19. Let (x n ) n=1 be sequence of rel numbers. If (x n ) n=1 is decresing nd bounded below then (x n ) n=1 is convergent. Hint. Use the pproximtion property of suprem to show tht (x n ) n=1 converges to sup{x n : n N} in (1) or to inf{x n : n N} in (2). Combining Theorem 18 nd Theorem 19 we obtin the Monotone Convergence Theorem lso clled the Monotonic Sequence Theorem. Theorem 20. Every bounded monotone sequence is convergent. Exmple 10. Consider the following recursively defined sequence: 1 if n = 1 x n = 3 1 if n 2. x n 1 Is the sequence (x n ) n=1 convergent? If it is wht is its limit?

46 5. SEQUENCES OF REAL NUMBERS Exmple 11. Consider the following recursively defined sequence: 1 if n = 1 x n = 1 if n 2. 1 + x n 1 Assuming tht the sequence (x n ) n=1 is convergent nd tht lim n x n = l 0, compute l.

CHAPTER 6 Series of rel numbers 1. Definitions, Nottion, nd Bsic Properties Definition 18. Let ( n ) n=1 be sequence of rel numbers. For n 1 we define the n-th prtil sum, denoted s n, s follows s n = n k=1 nd we consider the sequence of prtil sums (s n ) n=1. If the sequence (s n ) n=1 is convergent we sy tht the series with generl term n is convergent nd if n lim n s n = lim n k=1 k = s, then s is clled the sum of the series with generl term n. For convenience we will use the following nottion. The series with generl term n is denoted by n. If n is convergent nd its sum is s we write n=1 n = s. In other words, n=1 n n is convenient nottion for lim n k=1 k whenever the limit exists nd is finite. Exmple 12. The series 1 n(n+2) n 1 s n = k(k + 2) = k=1 n k=1 = 1 2 ( n n=1 k is convergent nd its sum is 1 n(n + 2) = 3 4. 1 2 ( 1 k 1 k + 2 ) k=1 = 1 2 ( n k=1 1 n k 1 k + 2 ) k=1 n+2 1 k 1 i ) i=3 = 1 2 (1 1 + 1 2 + n k=3 = 1 2 (1 + 1 2 + n k=3 n 1 k ( 1 i + 1 n + 1 + 1 n + 2 )) i=3 1 n k 1 i 1 n + 1 1 n + 2 ) i=3 = 1 2 (3 2 1 n + 1 1 n + 2 ) n 1 2 (3 2 0 0) = 3 4. Therefore, 1 n(n + 2) is convergent nd 1 n(n + 2) = 3 4. n=1 47

48 6. SERIES OF REAL NUMBERS Exmple 13. The series n ln( ) is not convergent. It ctully diverges n + 1 to. n k s n = ln( k + 1 ) = k=1 = = n k=1 (ln(k) ln(k + 1)) n ln(k) k=1 n k=1 = ln(1) + = ln(1) + n ln(k + 1) k=1 n+1 ln(k) ln(i) k=2 i=2 n n ln(k) ( ln(i) + ln(n + 1)) n ln(k) k=2 i=2 n ln(i) ln(n + 1) i=2 = ln(1) ln(n + 1) = ln(n + 1) n. Therefore, n ln( ) is not convergent since it diverges to. n + 1 Theorem 21 (Divergence Test). Let n be series. If the series n is convergent then the sequence ( n ) n=1 is convergent nd lim n n = 0. The proof follows from the observtion tht n = n k=1 k n 1 k=1 k for ll n 2. 0. Exmple 14. The series 2n 3 n+5 is not convergent, since lim n 2n 3 n+5 = 2 Theorem 22 (Limit theorems for series). Let n nd b n be series. (1) If the series n nd b n re convergent then the series ( n + b n ) is convergent nd n=1 ( n + b n ) = n=1 n + n=1 b n. (2) Let λ R. If the series n is convergent then the series (λ n ) is convergent nd n=1 (λ n) = λ n=1 n. As we will see lter nlogue limit theorems for product nd quotient of sequences do not hold for series. Exmple 15 (Geometric series). For r R, the series r n 1 is clled the geometric series with rtio r. When r = 1 the n-th prtil sums re s n = n nd the geometric series with rtio 1 is not convergent. For r 1, we cn lso estimte the n-th prtil sum s n = n k=1 rk 1 explicitly. Indeed, n n n 1 rs n = r r k 1 = r k = r k + r n r 0 = k=1 k=1 k=0 n r i 1 + r n 1)( chnge of index i = k + 1) i=1 = s n + r n 1.

1. DEFINITIONS, NOTATION, AND BASIC PROPERTIES 49 And hence, s n (1 r) = 1 r n for ll n 1, nd if r 1 n s n = r k 1 = 1 rn 1 r. k=1 If r 1, (s n ) n=1 is not convergent since does not hve limit. If 1 < r < 1, (s n ) n=1 is convergent nd lim n s n = 1 1 r. If r > 1, (s n ) n=1 is divergent since lim n s n =. Therefore, If r 1, the geometric series r n 1 is not convergent. If 1 < r < 1, the geometric series r n 1 is convergent nd n=1 r n 1 = 1 1 r. If r 1, the geometric series r n 1 is divergent towrds +. Exercise 65. Is the series 1 n(n+1) convergent (nd if it is compute its sum)? Exercise 66. Is the series ln( n n+2 ) convergent (nd if it is compute its sum)? Exercise 67. Is the series 3r n 1 with r ( 1, 1) convergent (nd if it is compute its sum)?

50 6. SERIES OF REAL NUMBERS Exercise 68. Is the series 2n convergent (nd if it is compute its sum)? 3n+1 Exercise 69. Is the series 4n+1 5 n convergent (nd if it is compute its sum)? Exercise 70. Is the series 5n+1 4 n convergent (nd if it is compute its sum)?

2. COMPARISON THEOREMS 51 2. Comprison Theorems Theorem 23 (Comprison theorems for series (convergence version)). Let n nd b n be series such tht (1) 0 n b n for ll n 1, (2) the series b n is convergent. Then, the series n is convergent nd n=1 n n=1 b n. Remrk 10. If we only ssume tht 0 n b n for ll n n 0 for some n 0 N, then we cn still conclude tht the series n is convergent but it might not be true nymore tht n=1 n n=1 b n. Theorem 24 (Comprison theorems for series (divergence version)). Let n nd b n be series such tht (1) 0 n b n for ll n 1, (2) the series n is not convergent. Then, the series b n is not convergent. Theorem 25 (Limit comprison theorem). Let n nd b n be series such tht (i) n > 0 nd b n > 0 for ll n 1, (ii) there exists c > 0 such tht lim n n bn = c, then, n is convergent if nd only if b n is convergent. Then, Theorem 26 (Integrl comprison theorem). Let f : [1, ) [0, ). If (i) f is continuous on [1, ), (i) f is decresing on [1, ). i.e. for ll x y, f(x) f(y). (1) f(n) is convergent if nd only if f is integrble on [1, ) nd (2) f(n) f(x)dx f(n) whenever the limits exist. n=2 1 n=1 In order to pply the integrl comprison theorem to series with nonnegtive terms n, the generl term must be of the form n = f(n) where f is (nonnegtive) decresing function. Exmple 16. For wht vlues of s R is the series 1 n s convergent?

52 6. SERIES OF REAL NUMBERS convergent (nd if it is deter- Exercise 71. Is the series 1 (n + 1) ln(n + 1) mine its sum)? 3. Alternting Series We sy tht sequence (b n ) n=1 hs constnt sign if either b n 0 for ll n 1 or b n 0 for ll n 1. Definition 19. A series n is n lternting series if for ll n 1, n = ( 1) n b n where (b n ) n=1 is sequence of constnt sign. Exmple 17. The lternting hrmonic series ( 1) n 1 ( n 1 ) series since ( 1)n 1 n = ( 1) n, nd 1 0 for ll n 1. n n If is n lternting Theorem 27 (Alternting series theorem). Let n be n lternting series. (1) lim n n = 0, nd (2) ( n ) n=1 is decresing, then n is convergent. Exercise 72. Using the lternting series theorem, show tht the lternting hrmonic series ( 1) n 1 n is convergent.

4. ABSOLUTELY CONVERGENT SERIES 53 4. Absolutely Convergent Series Definition 20. A series n is bsolutely convergent if the series n is convergent. Remrk 11. A convergent series is not necessrily bsolutely convergent. For instnce, the lternting hrmonic series is convergent but not bsolutely convergent. Theorem 28 (Absolute convergence implies convergence). If series n is bsolutely convergent then n is convergent. Theorem 29 (The rtio test). Let n be series. n+1 (1) if lim n = L > 1 then n is not convergent. n n+1 (2) if lim n = L < 1 then n is bsolutely convergent (nd in n prticulr convergent). n+1 Remrk 12. If lim n = 1 then the rtio test is inconclusive. For n instnce, 1 n is not convergent, ( 1) n n is convergent but not bsolutely convergent, nd 1 n is bsolutely convergent. For ll these series the limit of the rtio 2 is 1. Exercise 73. Is the series ( 3) n 1 n convergent?

CHAPTER 7 Applictions of Integrtion II 1. Arc length If f : [, b] R is differentible on [, b] nd f is continuous on [, b], the length of the rc of the curve represented by y = f(x) for x b is (11) L = 1 + (f (x)) 2 dx. Exercise 74. Compute the length of the curve represented by y = x3 6 + 1 2x for 1 x 2. L = 17 12. If g : [c, d] R is differentible on [c, d] nd g is continuous on [c, d], the length of the curve represented by x = g(y) for c y d is (12) L = d c 1 + (g (y)) 2 dy. Exercise 75. Compute the length of the curve represented by x = y 3/2 for 0 y 1. 55

56 7. APPLICATIONS OF INTEGRATION II L = 13 13 8. 27 2. Are of surfce of revolution If f : [.b] [0, ) is differentible on [, b] nd f is continuous on [, b], the surfce re obtined by rotting bout the x-xis the curve y = f(x) for x b is (13) S = 2π f(x) 1 + (f (x)) 2 dx. Exercise 76. Compute the surfce re obtined by rotting bout the x-xis the curve y = cos(x) for 0 x π 3. If g : [c, d] [0, ) is differentible on [c, d] nd g is continuous on [c, d], the surfce re obtined by rotting bout the y-xis the curve x = g(y) for c y d is (14) S = 2π d c g(y) 1 + (g (y)) 2 dy. Exercise 77. Compute the surfce re obtined by rotting bout the y-xis the curve y 2 = 4x + 4 for 0 x 8.

2. AREA OF A SURFACE OF REVOLUTION 57 S = 8π 3 (10 10 2 2).

CHAPTER 8 Prmetric equtions nd polr equtions Some curves cn be described conveniently using vrious equtions. Considering circle is simple but enlightening exmple. The circle of rdius R > 0 centered t the origin of the pln cn be represented s : the curve represented by the Crtesin eqution: x 2 + y 2 = R 2 the curve tht is the combintion of the grph of two functions: y = R2 x 2 nd y = R 2 x 2 for R x R. the prmetric curve: { x(t) = R cos(t), t [0, 2π] y(t) = R sin t the prmetric curve: { x(t) = R sin(2t), t [0, π] y(t) = R cos 2t the polr curve: r(θ) = R for θ [0, 2π]. 1. Prmetric curves Let f, g : I R be functions defined on n intervl I. The collection of points M(t) = (f(t), g(t)) when t I describes curve of the 2-dimensionl spce, clled prmetric curve. A representtion of the prmetric curve is given by the system of two equtions : { x = f(t), t I y = g(t) where the vrible t is clled the prmeter. The support of the curve is the set of points M(t) = (f(t), g(t)) when t I. Prmetric curves re useful to describe complicted curves (e.g. those who re bcktrcking) nd trjectories of objects where the prmeter represents time. Every function y = f(x) (resp. x = g(y)) cn be turned into prmetric eqution by setting either x = t (resp. y = t). Exmple 18. Exercise 78. Eliminte the prmeter to find Crtesin eqution of the curve { x = 3t + 1, t R y = 5t 2. Exercise 79. Eliminte the prmeter to find Crtesin eqution of the curve { x = cos(θ), 0 θ π 3 y = 5 sin(θ). 59

60 8. PARAMETRIC EQUATIONS AND POLAR EQUATIONS 1.1. Arc length of prmetric curves. Consider prmetric curve { x = f(t), t [, b] y = g(t) such tht f nd g re differentible on [, b], nd f nd g re continuous nd not simultneously 0 on [, b]. If the curve is trversed once s t increses from to b, then the rc length of the prmetric curve is (15) L = (dx ) 2 ((f (t)) 2 + (g (t)) 2 dt = + dt ( ) 2 dy dt. dt Exercise 80. Find the length of the curve defined by { x = 1 + 3t 2, t [0, 1] y = 4 + 2t 3 1.2. Are enclosed by prmetric curve. The re of the region between prmetric curve { x = f(t), t [, b] y = g(t) tht is trced out once for t [, b] where g 0 nd the x-xis is (16) A = (17) A = g(t)f (t)dt = g(t)f (t)dt = y dx dt dt, if f 0 y dx dt dt, if f 0. Exercise 81. Find the re under the rch of the cycloid { x = r(θ sin(θ)) y = r(1 cos(θ)), for θ [0, 2π].

2. POLAR COORDINATES 61 A = 3πr 2 2. Polr coordintes In the Crtesin coordinte system the coordintes of point re given in the form (x, y). The polr coordinte system is defined by fixing n origin, the pole, nd ry emnting from the pole, the polr xis. Typiclly, the pole is tken to be the origin of the 2-dimensionl plne nd the polr xis is horizontl ry pointing towrds the right. The polr coordintes of point is given in the form (r, θ) where θ is n ngle in rdin nd r rel number. If r 0, the point with polr coordintes (r, θ) is locted t distnce r on the ry tht is rotted counterclockwise of n ngle θ strting from the polr xis. If r 0, we dopt the convention tht the point with polr coordintes (r, θ) is the point with polr coordintes ( r, θ + π). If point hs polr coordintes (r, θ) its Crtesin coordintes re (r cos(θ), r sin(θ)). If point hs Crtesin coordintes (x, y) with x > 0 nd y 0 then it polr coordintes re ( x 2 + y 2, rctn( y x )) (the other cses re treted in the book pge ). 2.1. Arc length in polr coordintes. Let r = f(θ) be polr eqution of curve between the rys θ = nd θ = b such tht f is differentible on [, b], nd f is continuous on [, b]. The rc length of the polr curve r = f(θ) between the rys θ = nd θ = b is (18) L = f(θ)2 + (f (θ)) 2 dθ = r 2 + ( ) 2 dr dθ dθ Exercise 82. Find the length of the crdioid whose polr eqution is r = 1 + sin(θ).

62 8. PARAMETRIC EQUATIONS AND POLAR EQUATIONS L = 8 2.2. Are in polr coordintes. The re of the region enclosed by the polr curve r = f(θ) nd the rys θ = nd θ = b, where f is positive continuous function nd where 0 < b 2π is (19) A = 1 2 f(θ) 2 dθ = 1 2 r 2 dθ. Exercise 83. Find the re of the region tht is enclosed by the curve r = e θ 4 nd tht lies in the sector π 3 θ π. A = e π 2 e π 3