Summer MTH142 College Calculus 2. Section J. Lecture Notes. Yin Su University at Buffalo


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1 Summer 6 MTH4 College Clculus Section J Lecture Notes Yin Su University t Bufflo
2 Contents Bsic techniques of integrtion 3. Antiderivtive nd indefinite integrls Antiderivtives Antiderivtive rules Method of substitution Nture of chin rule for differentition Method of substitution Shortcut of substitution method: liner vrition of ntiderivtive rule Integrtion by prts Shortcut of using integrtion by prts Choice of u function Definite integrls nd fundmentl theorem of Clculus Net re Fundmentl theorem of Clculus (Prt ): NewtonLeibniz formul Improper Integrls Type I: Improper integrls on infinite intervl Review of useful infinite limits Convergence of n improper integrl Type II: Improper integrl with unbounded function Review: useful limits of unbounded functions Integrtion nd Geometry. Riemnn sum nd pproximtion of definite integrls Riemnn sum Simpson s rule Efficiency of pproximtion Are between curves Riemnn sum: slicendsum strtegy Arc length of function Advnced integrtion techniques 3 3. Integrtion of rtionl functions using Prtil frctions Preprtion: prtil frctions of liner fctors nd qudrtic fctors Setup of prtil frctions Determine the unknown coefficients: Common denomintor Reduce degree in numertor: Long division of polynomils Trigonometric integrtion Integrls of form sin m x cos n xd x if m or n is n odd positive integer Integrls of form tn m x sec n xd x: if m is odd or n is even Use of Reduction formuls for other cses Additionl notes: Integrls sin m x cos n xd x when m nd n re both even nd positive Trigonometric substitution Integrls involving power of x Integrls involving x + nd x Appliction in geometry: volume nd surfce re of revolution Volume of solid of revolution Generl formul: Slicendsum Volume of revolution of one curve bout xxis: the disk method Volume of revolution of two curves bout xxis: the wsher method surfce re of revolution Slice nd sum
3 CONTENTS 5 Prmetric curves nd polr coordintes 5 5. Curves defined by prmetric equtions Prmetric curves Velocity nd tngent of prmetric curves Arc length of prmetric curve Polr coordintes Directed ngles polr coordinte system Conversion between Crtesin nd polr coordintes Polr curves Clculus in polr coordintes Slope of tngent line Arc length of polr curve Are of regions bounded by polr curves Sequence nd infinite series Sequences Definition nd representtion of sequences Limit of sequence Comptibility of convergent sequences with opertions nd continuous function Limit of sequences of positive frctionl terms Infinite Series Definition of infinite series nd its convergence A typicl series: geometric series Test for series of positive terms (I): The integrl test The integrl test nd pseries Tests for series of positive terms (II): rtio test nd root test The rtio test The root test Power series nd Tylor expnsion Power series Mclurin series: Tylor series t x = Approximtion using Tylor expnsion Convergence of power series Convergence intervl of power series Importnt Tylor series Tylor series: vrition, pproximtion, limits, derivtive nd integrl Derive Tylor series from known ones Derivtive nd integrl of Tylor series Approximting definite integrl
4 Chpter Bsic techniques of integrtion. Antiderivtive nd indefinite integrls Antiderivtives Definition.. A function F(x) is clled n ntiderivtive of f (x) if F (x) = f (x). A derivtive reltion will imply corresponding ntiderivtive reltion: For exmple, we know tht (x 3 ) = 3x. We cn sy 3x is the derivtive of x 3, in the lnguge of differentition; or x 3 is n ntiderivtive function of 3x, in the lnguge of ntidifferentition. Remrk.. () For fixed function f (x), the ntiderivtive function of f (x) is not unique. For exmple, x 3 nd x 3 + re both ntiderivtive functions of 3x. () For fixed function f (x) nd its two ntiderivtive functions F (x) nd F (x), they must differ by constnt: F (x) = F (x) + C. (3) The bove remrk mens tht if F(x) is known ntiderivtive of f (x), then we know ALL the ntiderivtive functions: F(x) + C, for ny constnt C. For exmple, since we know tht sin x is n ntiderivtive of cos x, we cn conclude tht ll the ntiderivtive functions of cos x re of the form sin x + C. This property yields the following definition. Definition.3. For given function f (x), the indefinite integrl of f (x) is the fmily of ll ntiderivtive functions of f (x). The nottion for indefinite integrl is f (x)d x = F(x) + C where F(x) is n ntiderivtive function of f (x). Exmple.4. By power rule of differentition (x 4 ) = 4x 3, we cn rewrite this reltion s n integrl reltion: 4x 3 d x = x 4 + C This reltion mens: For ny constnt C, the derivtive of x 4 + C is 4x 3, nd hence x 4 + C is lwys n ntiderivtive. The fmily of functions x 4 + C for ny constnt C gives ALL the ntiderivtive functions of 4x 3. So there is no function other thn x 4 + C whose derivtive cn be 4x 3. Antiderivtive rules Ech differentition rule gives rise to n ntidifferentition rule. Constnt rule: kd x = kx + C for fixed constnt k. 3
5 . ANTIDERIVATIVE AND INDEFINITE INTEGRALS 4 Power rule: x n d x = x n+ + C, n. n + Note: This rule pplies to power functions for most exponents, including frctions nd negtive numbers except. For exmple d x = x x + C; nd xd x = 3 x 3/ + C. logrithmic rule: d x = ln x + C x exponentil rule: trigonometric rules: other rules: e x d x = e x + C; cos xd x = sin x + C; sec xd x = tn x + C; d x = rcsin x + C; x sin xd x = cos x + C; sec x tn xd x = sec x + C; d x = rctn x + C; + x The bove rules re for specific functions. For combintion of functions, here re the two importnt linerity properties of integrtion. Sum nd difference rule: Constntmultiple rule: f (x) ± g(x)d x = k f (x)d x = k f (x)d x ± g(x)d x. f (x)d x, for constnt k. We will modify some of these rules nd develop more new ntiderivtive rules when we hve more dvnced integrtion techniques. Clss Exmple.5. Compute 3 x x + 5 d x Clss Exmple.6. () Find the following ntiderivtives: 6x d x (3) 3 + x d x () (3 5 sec x)d x Remrk.7. So fr, unlike for differentition, there is no corresponding product rule, quotient rule or chin rule for integrtion. For this reson, if the integrnd f (x) is product of two functions involving x, we cn not ply the sme splitting trick s in the sum nd difference rule. For exmple, for the integrl x cos xd x, the function is product of x nd cos x, whose ntiderivtives re x3 3 nd sin x respectively. But we cn NOT conclude: Wrong: x cos xd x = x 3 3 sin x + C We will lern the integrl version of chin rule in the next section, which is clled the method of substitution, nd lern the integrl version of product rule in section.3, which is clled the method of integrtion by prts. Additionl Exmple.8. Compute 4 x + 3 d x. x
6 . METHOD OF SUBSTITUTION 5 Solution. Decompose the function into three components 4 x, Then 4 x + 3 d x = 4 x d x x Prt Prt Prt 3 = 4 Prt x / d x x Prt nd 3, which re connected by ddition or subtrction. x d x + 3 Prt 3 d x d x + 3d x x = 4 3 x 3/ rcsin x + 3x + C (sum nd difference rule) (constntmultiple rule). Method of substitution Nture of chin rule for differentition Let us look t how we pply the chin rule in differentition gin. We sy function y = f (x) hs chin structure, or composition structure, if we cn introduce n intermedite vrible u, which is function of x, such tht y cn be decomposed into the following chin: x u y Tht is, u is function of x (clled the inside function), nd y is function of u (clled the outside function). The chin rule sys, the derivtive of y with respect to x is d y d x = d y du outside derivtive du d x inside derivtive Exmple.9. Compute the derivtive of y = sin(x 3 + ). Solution. We check tht this function hs the following chin structure: x x 3 + sin(x 3 + ) This mens, if we set the inside function u s u = x 3 +, then the function y cn be written simply s y = sin u, clled the outside function. Now using the chin rule, we hve tht d y d x = d y du du d x = cos u function of u u derivtive of u = cos(x 3 + ) 3x Thus the chin rule sys, fter introducing the vrible u to connect x nd y, the finl derivtive y cn be expressed s: y = ( function of u) (derivtive of u) So, if we wnt to derive "chin rule" for ntidifferentition, we hope tht the function we wnt to integrte cn be written s the bove product form. Now let us revisit the bove exmple by trying to compute the ntiderivtive of the nswer. Exmple.. Compute the indefinite integrl cos(x 3 + )3x d x. Solution. Step. Choose n intermedite function u. Notice tht the function we re going to integrte is cos(x 3 + )3x. We wnt to introduce new vrible u to decompose the function in the form cos(x 3 + )3x = ( function of u) (derivtive of u)
7 . METHOD OF SUBSTITUTION 6 Bsed on the lst exmple, we see tht we cn try u = x 3 +. Then its derivtive du d x = 3x, which is fctor of the integrnd cos(x 3 + )3x. And the remining fctor is cos(x 3 + ), which cn be nicely written s cos u. Step. Formlly set up substitution vrible u = x 3 +. Then the derivtive reltion du d x = 3x cn be rewritten s du = 3x d x Now we cn relly pply the substitution, nd substitute ll xterms into uterms: cos(x 3 + ) 3x d x = =cos u du cos u du This decomposition works in the sense tht fter pplying this substitution u, the integrl hs no x vribles nd it becomes pure integrl of vrible u. Step 3. Evlute the new integrl of vrible u using ntiderivtive rules: cos udu = sin u + C Notice tht the ntiderivtive bove is still function of u, becuse we were integrting n integrl with respect to vrible u. In this step we cn tret u s our new independent vrible, ignoring x. Step 4. Plce x bck into the ntiderivtive. The nswer bove is lredy correct ntiderivtive. However, since u is some vrible we introduced in the solution, which should not pper in the finlly nswer, we hve to plce x bck into the ntiderivtive. To do this, we use the substitution u = x 3 + gin. Plug in this reltion to the ntiderivtive we found: sin u + C = sin(x 3 + ) + C. Then now we hve recovered the originl function in the lst exmple. Formlly, we cn stte cos(x 3 + )3x d x = sin(x 3 + ) + C Method of substitution We cn summrize the steps needed in the method of substitution: Method.. Integrtion by substitution. Suppose we hve n indefinite integrl f (x)d x, () Observe the integrnd f (x) nd choose n inside function u s function of x. Then check if f (x) cn be decomposed into f (x) = ( function of u) constnt multiple of u (x) () Substitute u nd du = u (x)d x in the integrl. The gol is to eliminte ll xterms nd to obtin new integrl with respect to vrible u. The newe integrl should be simpler thn the originl one. (3) Evlute the new integrl with respect to the new vrible u. The ntiderivtive is function of u. (4) Use the substitution u = u(x) gin to rewrite the bove ntiderivtive s function of x. Clss Exmple.. Evlute (x + )(x 3 + 3x) d x. Clss Exmple.3. + ln x Evlute the integrl d x. x
8 . METHOD OF SUBSTITUTION 7 Additionl Exmple.4. Compute x 4x d x. Solution. Set u = 4x. Its differentil is du = 8xd x nd hence 8 du = xd x. Then substitute the integrl x d x = xd x = 4x 4x u ( 8 du) = u / du 8 seprte x from numertor = 8 u/ + C = 4x + C 4 Additionl Exmple.5. Clculte tn xd x. Solution. Write tn x = sin x. From the frction we see tht we cn set the substitution u = cos x nd then du = sin xd x cos x nd du = sin xd x. So tn xd x = sin x cos x d x = sin xd x = ( du) = ln u + C = ln cos x + C cos x u In the nswer bove, using the identity cos x = sec x nd ln x = ln x, we cn write ln cos x = ln sec x. Then we obtin very useful ntiderivtive rule for tn x, which we will use in the section of trigonometric integrtion. Formul.6. Antiderivtive rule: tn xd x = ln sec x + C Shortcut of substitution method: liner vrition of ntiderivtive rule Strt with ny known ntiderivtive rule, for exmple the power rule x 3 d x = x C Wht if the integrnd function is not the pure cubic power x 3, but the cubic power of liner function x + b? For exmple: (5x + 6) 3 d x, with x replced by 5x + 6. Let s evlute this integrl using substitution. Exmple.7. Evlute (5x + 6) 3 d x. Solution. Set u = 5x + 6 nd compute its differentil du = 5d x. Write is s 5 du = d x nd substitute the integrl (5x + 6) 3 d x = u 3 5 du = 5 u4 4 + C = (5x + 6)4 + C 5 4 Notice tht we replced the x 3 in the power rule by (5x + 6) 3, nd in the ntiderivtive function, we lso see tht x4 4 is (5x + 6)4 replced by. The only essentil difference is in the ltter cse, n dditionl multiple of 5 shows up. 4 The ide in the bove exmple cn be generlized s the trick of liner vrition: Method.8. Given n ntiderivtive rule f (x)d x = F(x) + C, if the x vrible in integrnd f (x) is replced by liner function x + b, then the indefinite integrl cn be expressed s f (x + b)d x = F(x + b) + C
9 . METHOD OF SUBSTITUTION 8 Let s try to better understnd wht hppens in the vrition. A known ntiderivtive rule f (x)d x = F(x)+C corresponds to derivtive reltion: integrtion f (x) F(x) +C Now ssume we substitute x by liner function x + b in the function f (x) on the left hnd side. To get n ntiderivtive of the new function f (x + b), we only need to lso replce x in the ntiderivtive F(x) on the right hnd side by x + b, to form F(x + b), nd then divide F(x + b) by, the coefficient of x in the liner vrition x + b. f (x) f (x + b) integrtion F(x) +C F(x + b) +C Clss Exmple.9. () Antiderivtive rule: x d x = x3 3 + C x integrtion x 3 3 +C (x + 4) +C (x + 4) d x = () Antiderivtive rule: x d x = ln x + C x 3x integrtion ln x +C +C 3x d x = Clss Exmple.. Compute the following indefinite integrls using the ide of liner vrition. () cos(3x)d x () e 3 x d x (3) (x + ) d x (4) xd x 3 Additionl Exmple.. Clculte the following integrls: () (x ) 9 d x; () d x. 5x Solution. () Antiderivtive rule: () Antiderivtive rule: x 9 d x = x + C. The liner vrition is x. Thus we hve (x ) 9 d x = (x ) x d x = rcsin x + C. The liner vrition is 5x, since 5x = (5x). Then + C 5x d x = 5 rcsin(5x) + C
10 .3 INTEGRATION BY PARTS 9 Additionl Exmple.. For ny constnt >, compute the following integrls: () d x, () + x x d x. Solution. the integrl of () Divide in both sides of the frction:, we hve +x + x d x = () Divide in both sides of the frction: integrl of, we hve x x d x = + (x/) d x = + x = / + (x/) =. Then by liner vrition of + (x/) / rctn x liner vrition: x = x+ +C = rctn x + C x = / = (x/). Then by liner vrition of the (x/) (x/) d x = / rcsin x liner vrition: x = x+ +C = rcsin x + C The bove two ntiderivtive formuls re useful in the section of trigonometric substitution. So we list the results gin: Formul.3. Antiderivtive formuls: + x d x = rctn x + C x d x = rcsin x + C.3 Integrtion by prts In this section we will lern nother importnt technique of integrtion, clled integrtion by prts. This is ctully the product formul for integrtion. Unlike the product rule in differentition, this formul is not n explicit formul, i.e., we cn not find directly the ntiderivtive of product function using integrtion by prts. This formul only gives us wy to rewrite product integrl into nother product integrl, which we hope could be simpler. Recll the product rule in differentition: (u(x)v(x)) = u (x)v(x) + u(x)v (x) Rewrite this reltion s n integrl formul: u(x)v(x) = u (x)v(x) + u(x)v (x) d x = u (x)v(x)d x + u(x)v (x)d x We isolte u(x)v (x)d x nd plce it on the left hnd side. Then we obtin the product formul in integrtion, or in other words, the method of integrtion by prts: Theorem.4. Integrtion by prts: u(x)v (x)d x = u(x)v(x) u (x)v(x)d x
11 .3 INTEGRATION BY PARTS Remrk.5. To pply integrtion by prts, we need to seprte the fctors in the integrl s product of two fctors u(x) nd v (x) nd then compute u (x) nd v(x). Then the formul cn be used to rewrite the originl integrl s u(x)v(x) minus new integrl of the form u (x)v(x)d x. This method will work when the new product integrl u (x)v(x)d x is n esier integrl so we cn work it out. Remrk.6. v(x) is ssumed to be one ntiderivtive of v (x). Since ll ntiderivtives of v (x) just differ by constnt C, we normlly choose C =, in this wy v(x) hs the simplest form, e.g. hs no constnt term. Exmple.7. Evlute the integrl x cos xd x Solution. The integrnd x cos x is clerly product of two functions x nd cos x. To use integrtion by prts, we need to identify one fctor function s u nd the other one s v. Notice tht our gol is to mke the new product integrl u (x)v(x)d x s esy s possible. For this reson, if we tke u = x nd v = cos x, then u (x) = nd v(x) = sin x nd hence the new product integrl is u vd x = sin xd x which cn be esily integrted by ntiderivtive rules. Bsed on this observtion, we write down formlly functions in the originl integrl: u = x v = cos x differentition integrtion functions in the new integrl: u = v = sin x The the product formul cn be pplied: x cos x d x = x sin x u v u v = x sin x sin x d x u v sin xd x = x sin x ( cos x) + C = x sin x + cos x + C Remrk.8. In the bove exmple, we choose to identify u = x becuse it cn be differentited to constnt. I.e., u (x) =. Thus the new product integrl u (x)v(x)d x essentilly doesn t involve product nymore, nd hence is esier. If we switch the identifiction of u nd v, by setting u = cos x nd v = x, then u = cos x v = x u = sin x v = x Then the product formul becomes x cos xd x = cos x x sin x x d x Notice tht this identity still holds, becuse we just put the components into the product formul. However, the new resulting integrl on the right hnd side still involves product of polynomil x nd sin x, which doesn t seem to be simpler thn the originl one. This is why such n identifiction of u nd d v will not eventully work in the product formul. Clss Exmple.9. Evlute xe 3x d x.
12 .3 INTEGRATION BY PARTS The following exmple shows tht some product integrls require the repeted use of integrtion by prts. Clss Exmple.3. Evlute x e 3x d x. Additionl Exmple.3. Clculte rctn xd x. Solution. Set u(x) = rctn x. Since it is the only term in the integrnd, we cn force the other fctor v (x) to be. So u(x)v (x) is still rctn x. Then u = rctn x v = u = + x v = x Then rctn xd x = x rctn x x + x d x = x rctn x ln + x + C pply substitution u=+x Shortcut of using integrtion by prts The formul of integrtion by prts indictes tht we need to differentite the fctor u nd integrte the fctor v, nd then check if the product of new pir u nd v is esier to integrte. If we hve to pply integrtion by prts more thn once, then somehow we need to differentite u twice nd integrte the other fctor v twice. So for product integrl f (x)g(x)d x, we cn list the consecutive derivtives of f (x) nd ntiderivtives of g(x) in tble like the following: differentition differentition differentition f (x) f (x) f (x) f (x) I M I M I3 M3 I4 g(x) G (x) G (x) G 3 (x) integrtion integrtion integrtion I = M I = M I3 = M3 I I3 I4 I = M M + I3 I = M M + M3 I4 differentition integrtion The bove digrm shows tht we cn express the sme product integrl s multiple wys by repetedly pplying integrtion by prts, until we stop t chosen terminl lyer. Since the expnsion fter repeted ppliction of integrtion by prts only involves one new integrl from the terminl lyer, we cn try to choose best terminl lyer such tht the product integrl on tht lyer is the esiest. Also notice the digonl entries re multiplied (without integrtion), nd the sign of the digonl multipliction is lternting. To see the sign chnge, check the following digrm
13 .3 INTEGRATION BY PARTS f (x) I g(x) differentition differentition f (x) f (x) M M G (x) G (x) integrtion integrtion I = M M + M 3 M 4 + ± I n differentition f (x) M3 G 3 (x) integrtion differentition f (n) / I n G n (x) integrtion Clss Exmple.3. Evlute x 3 sin(x)d x. Solution. x 3 sin x x 3 sin xd x differentition 3x integrtion x 3 ( cos x) cos x differentition differentition 6x integrtion 3x ( 4 sin x) sin x integrtion 6x 8 cos x cos x Thus we hve tht differentition integrtion 6 6 sin x 6 sin x d x x 3 sin xd x = x 3 ( cos x) 3x ( sin x) + 6x 4 8 cos x 6 sin x + 6 = x 3 cos x x sin x x cos x sin x + C 8 6 d x Clss Exmple.33. Compute the integrl (3x + ) (3 x) 4 d x.
14 .3 INTEGRATION BY PARTS 3 Choice of u function Depending on the type of fctor functions in product integrl, how to choose the correct u function, which is to be differentited, to mke the clcultion more efficient? Here is some trick. Method.34. The order of choice of function u: () Inverse trigonometric functions: rctn x, rcsin x, rccos x; () Logrithmic function: ln x; (3) (Lower) power/polynomil function: x n ; (4) Exponentil function: e x ; (5) Trigonometric functions: sin x, cos x. But notice the bove trick is not enough to solve ll product integrls. Sometimes we hve to nlyze the form of the functions involved in product integrl, nd figure out the best wy of using integrtion by prts. Clss Exmple.35. Compute the indefinite integrl ln xd x. Formul.36. Antiderivtive rule: ln xd x = x ln x x + C Clss Exmple.37. Compute e x sin xd x.
15 .4 DEFINITE INTEGRALS AND FUNDAMENTAL THEOREM OF CALCULUS 4.4 Definite integrls nd fundmentl theorem of Clculus Net re We firstly review the grphicl definition of definite integrl for continuous function over finite domin. It is is defined s the net re of the region bounded by the function. Definition.38. Assume f (x) is continuous function over [, b]. Consider the closed region R bounded by the curve of f (x) nd the xxis, between verticl lines x = nd x = b. Then the definite integrl b f (x)d x is defined s the net re of this region R, which is the re of prts of R tht lie bove the xxis, minus the re of the prts of R tht lie below the xxis. More concretely, for instnce, in the figure the function f (x) encloses region between x = nd x = b, which consists of three components: A, B nd C. We cn see tht region A nd C re bove xxis nd region B is below xxis. Then the definite integrl is defined s the net re b f (x)d x = Are(A) + Are(C) Are(B) Clss Exmple.39. seprte regions: In this figure, suppose we know the re of the Are(A) =, Are(B) = 3, Are(C) = 7, Are(C) = 4 Then the depending on the bounds of the integrls, we hve 6 f (x)d x = Are(A) Are(B) = 3 = ; 6 f (x)d x = f (x)d x = Remrk.4. A prticulr cse. If the function f (x) is positive (or nonnegtive) over the intervl [, b], then the enclosed region by f (x) must lie bove xxis. In this cse, the vlue of definite integrl b f (x)d x is exctly the (geometric) re of the enclosed region. We will see tht the fundmentl theorem of clculus will enble us to compute the vlue of definite integrl by using ntiderivtives. With the fundmentl theorem, the clcultion is purely nlyticl. A mjor ppliction of integrtion is to derive the re formul for some shpes on plne (such s the re of disk). The ide is to use the definition bckwrds nd express the re s definite integrl. Then pply the fundmentl theorem to solve it. We re going to see such ppliction fter few sections. Fundmentl theorem of Clculus (Prt ): NewtonLeibniz formul We hve defined two types of integrls: indefinite integrls nd definite integrls. Even though they hve similr nottion, t this point, they re essentilly irrelevnt. Indefinite integrl is defined for finding the ntiderivtive function, which is purely nlyticl, while definite integrl is defined geometriclly using re. The reson tht these two types of opertions now use very similr nottion, is due to the powerful fundmentl theorem of Clculus, which reltes definite integrl to its indefinite integrl. Theorem.4. Fundmentl theorem of Clculus (prt ): NewtonLeibniz formul b f (x)d x = F(x) x=b x= = F(b) F() where F(x) is one nderivtive function of f (x).
16 .4 DEFINITE INTEGRALS AND FUNDAMENTAL THEOREM OF CALCULUS 5 The left hnd side is the definite integrl of f (x) over [, b]. The right hnd tells us how to compute it: evlute n ntiderivtive of f (x) over bounds [, b]. More formlly, two steps re needed: () Find one ntiderivtive of f (x), nd cll it F(x). We know there re infinitely mny ntiderivtives nd ll ntiderivtives differ by constnt C. Since we only need one ntiderivtive, in prctice usully we just tke C = so tht the ntiderivtive F(x) hs the simplest form. () Evlute F(x) t x = b nd x = nd then compute their difference F(b) F(). Clss Exmple.4. Evlute 4 x + 3x d x. Clss Exmple.43. Assume f (x) = x. Compute the definite integrls (x )d x, (x )d x nd (x )d x. Solution. hve An ntiderivtive function F(x) cn be chosen s F(x) = 3 x 3 x. By the fundmentl theorem of Clculus, we x= (x )d x = 3 x 3 x x= x= (x )d x = 3 x 3 x (x )d x = 3 x 3 x x= x= x= = 3 3 = 3 3 = = 3 ; 3 3 = 4 3 ; 3 3 = 3 ; Remrk.44. The grph of f (x) = x is prbol. The figure shows tht between x = nd x =, the region A bounded by the function is below the xxis. This is why the integrl (x )d x hs negtive vlue. The region B between x = nd x = is bove xxis. Thus (x )d x hs positive vlue. In other words, we know tht Are(A) = 3 nd Are(B) = 4 3. The lst integrl (x )d x, by the grphicl definition, equls the net re bounded by the curve between x = nd x =, which is Are(B) Are(A) = = 3 This mtches the nswer from the clcultion using fundmentl theorem. Remrk.45. When we pply the method of substitution to definite integrl, we cn del with the bounds in two wys. We use n exmple of integrl (x + ) d x to illustrte it. Notice tht the substitution is u = x + with du = d x. () (Recommended) Clculte the ntiderivtive function in terms of x. Then plug in the bounds for xvlues. We ignore the bounds x = nd x = for the step of computing ntiderivtive. Then F(x) = (x + ) d x = u du = 6 u3 = (x + )3 6 Then plug in the xvlue of the bounds: (x + ) d x = F(x) x= x= = 6 ( + )3 6 ( + )3 = = 98 6 = 49 3
17 .5 IMPROPER INTEGRALS 6 () When we set up substitution u = u(x), we cn trnsform the bounds using u s well. The upper bound for u is u = u() nd the lower bound is u(b). Then we cn directly use them when the ntiderivtive is clculted s ufunction. Set u = x +. Since the xbounds re x = to x =. Then the corresponding ubounds re from lower u = x + = + = 3 to upper u = x + = + = 5. Now pply the substitution to get the ntiderivtive interms of u: F(u) = (x + ) d x = u du = 6 u3 Then plug in the uvlue of the bounds: F(u) u=5 u=3 = = = 98 6 = 49 3 Both wys of deling with the bounds must led to the sme finl vlue..5 Improper Integrls The definite integrls we hve encountered so fr only involve finitevlued functions nd finite intervls of integrtion. In this section, we will see tht we cn generlize the definite integrls to the functions with infinite vlue or infinite domin intervl. Those generlized integrls re clled improper integrls. We will discuss the following two types of improper integrls: Type I: improper integrls such tht the intervl of integrtion is infinite; Type II: improper integrls such tht the integrnd f (x) is unbounded on finite intervl [, b]. Type I: Improper integrls on infinite intervl Let s strt with simple exmple. Exmple.46. Look t the function f (x) =. This function is continuous everywhere except x =. The positive brnch x of the grph is decresing curve pproching the xxis becuse lim =. x + x Consider the re of the region under this function from x =. The following figures show the regions bounded on the right by x = 3, x = 5 nd x =. 3 x d x = x d x =.3 x d x =.4 In the bove three integrls, we keep the lower bound x = unchnged but push the right bound from x = 3 to x = 5 nd to x =. As we increse the upper bound, the bounded region gets extended to the right. This is why the vlue of definite integrls shown bove lso increses. Now nturl question rises: how fr cn we push the right bound? And wht will hppen if we push the right bound to infinity? I.e. the upper bound of this integrl becomes x = +. In this cse the integrl will be written s + d x. x Geometriclly, this integrl will lose the right bound, nd represents the re of n infinitely long hlfopen strip: the whole region under the function f (x) = x to the right of x = :
18 .5 IMPROPER INTEGRALS 7 Improper integrl + x d x Such strip hs infinite length. But does this men its totl re is lso infinite? In fct, the re in this exmple is finite, which equls.5. We re going to tret the improper integrl (re of infinite strip) s limit of definite integrls (re of finite regions). In fct, We ssume tht we keep the left bound x =, nd set the right bound to be prmeter x = b for constnt b. Since b is ssumed to be finite constnt, the integrl b d x is still definite x integrl, nd it equls the re of the region between x = nd x = b (see the figure). We cn use the fundmentl theorem to compute it: b x d x = x=b x x= = b }{{} F(b) = b }{{} F() This reltion sys tht the re from x = to x = b is b, for ny choice of right bound b. Now we cn push the right bound x = b to + by tking the limit b + : Trnslting the re to integrls: nd the computtion follows by tking limit: + x d x = lim b + Are from x = to x = + = lim b + + b x d x = lim b + x d x = lim b + Therefore the re of such infinite strip is, which is finite number. Are from x = to x = b b x d x = b lim b + b This limit equls = = Remrk.47. In the bove exmple, we cn lso choose to keep the right bound fixed nd push the left bound to. In this cse the region will be extended ll the wy to the left to form hlfopen infinite strip. More extremely, we cn even push both bounds to the two directions to infinities. Tht is, push the left bound to nd right bound to +. Then now this integrl will represent the re of the region between the function nd xxis, which is n infinite strip with open ends on both directions. Now we give the forml definition of n improper integrl of type I. Definition.48. We will cll n integrl b f (x)d x of f (x) improper over infinite intervl if = nd/or b = +. To mke this definition work, we need to ssume f (x) is continuous over [, b].
19 .5 IMPROPER INTEGRALS 8 + f (x)d x = lim b + b f (x)d x b f (x)d x = lim b f (x)d x + f (x)d x = lim b + b f (x)d x Method.49. Computtion of improper integrls of type I: Step. Find the ntiderivtive of f (x), sy F(x); Step. Evlute the ntiderivtive F(x) long the bounds s for definite integrl. If the upper bound is +, we use b s the upper bound insted. If the lower bound is, we use s the lower bound insted. Then by fundmentl theorem, we cn obtin n expression in terms of nd/or b; Step 3. Tke the limit s nd/or b +. Review of useful infinite limits From the bove discussion, we see tht n improper integrl of type I is simply the combintion of definite integrl nd n infinite limit. Therefore here we review some useful infinite limits. Grph Function x e x ln x polynomil rctn x x Notdefined ± π x ± π Convergence of n improper integrl Clss Exmple.5. Evlute the improper integrl 3 x d x.
20 .5 IMPROPER INTEGRALS 9 Clss Exmple.5. + Evlute the improper integrl e x d x. In ll the bove exmples, the finl vlue of the improper integrls re ll finite rel numbers. This mens the extended infinite regions in bove exmples ll hve finite re. However, not ll improper integrls hve finite vlue. In this cse when n improper integrl is not finite, we sy it is divergent. Exmple.5. Evlute the improper integrl e x d x. Solution. Step. Using the substitution u = x, we cn compute the ntiderivtive function F(x) = e x d x = e x + C Step. Replce the lower bound by. Evlute the ntiderivtive F(x) from x = nd x = : x= F(x) = F() F() = e e = + e. x= Step 3. Push the left bound to. Tht is, we tke the limit s goes to : e x d x = lim + e = + lim e Note tht when, the exponent +. Then e x will pproch +. Tht is, it doesn t hve finite limit. Therefore e x d x = + lim e = + (+ ) = + From the grph of e x, we cn see tht s x decreses to, this function will get higher nd higher. If we extend the region to the left, we will obtin n infinite wide strip with incresing width on the left side. Therefore the re of this strip cn not be finite number. This exmple shows tht the vlue of n improper integrl cn be finite or infinite. Like the vlue of limit, we define the convergence of n improper integrl: Definition.53. An improper integrl is sid to be convergent is its vlue is finite number. An improper integrl is clled divergent if its vlue is infinity or does not exist. Clss Exmple.54. Evlute + x d x nd determine its convergence. (x 3 + ) 3
21 .5 IMPROPER INTEGRALS Additionl Exmple.55. Evlute x e x3 d x nd determine its convergence. Solution. Using substitution u = x 3 nd differentil reltion du = 3x d x nd 3 du = x d x, we cn compute the ntiderivtive: F(x) = x e x3 d x = e u ( 3 du) = 3 eu = 3 e x3 The integrl is improper t the lower bound. Thus we firstly replce it by x = : x e x3 d x = x= 3 e x3 = 3 e 3 3 e Then tke the limit s. Notice when, we hve tht 3 nd the exponent 3 +. Hence the exponentil function e 3 will pproch +. Therefore this integrl is divergent. x e x3 d x = lim x= x e x3 d x = lim 3 e 3 3 e = + Type II: Improper integrl with unbounded function Improper integrls lso occur when the integrnd becomes infinite somewhere in the intervl of integrtion. Exmple.56. Look t the function f (x) = x. The grph is shown in the figure. From its grph we see tht f (x) is defined only for x > nd s x pproches from the right, the function x will pproch +. Now consider the integrl x d x. It looks like regulr definite integrl. But ctully f (x) = x is not well defined t the left bound x =. Using the re definition of integrls, we see tht x d x represents the re of n infinitely tll strip (such strip hs finite width but hs infinite height). Therefore the corresponding re my be finite or infinite. In this cse we lso cll it improper. Notice tht over [, ], the function f (x) = x is not continuous only t x =. Thus we cn firstly consider the re from positive number close to, sy c, to the right bound x =. Over [c, ], the integrl c x is regulr definite integrl. So we cn evlute it by pplying the fundmentl theorem: c d x = x x x= x=c = c = c The bove reltion sys tht the re of the region bounded between ny smll positive number x = c nd x = is c. Wht we wnted is the re of the strip between x = x =. Thus limit process c + will work. I.e., Trnslting the re to integrls: Are from x = to x = = lim c + Are from x = c to x = x d x = lim c + This implies tht such strip with infinite height hs totl re. Now we give the forml definition of the improper integrl of type II. c x d x = lim c +( c) = ( lim c + c) = = Definition.57. We will cll n integrl b f (x)d x of f (x) improper with unbounded integrnd, if f (x) is unbounded t point c inside [, b].
22 .5 IMPROPER INTEGRALS b f (x)d x = lim c + b c f (x)d x b f (x)d x = lim c b c f (x)d x b f (x)d x = p = lim c p f (x)d x + c b p f (x)d x f (x)d x + lim d p + b d f (x)d x Clss Exmple.58. Compute the improper integrl x d x. x Solution. Notice tht this integrl is improper becuse s x, the function pproches +. And the upper bound x = is only improper point of the function over [, ]. Thus we replce the upper bound by x = c nd compute the definite integrl c x d x = ln x x=c x= = ln ln c = ( ln c ) ( ln ) Now push the upper bound x = c to x = by tking the limit c : c x d x = lim c Thus this improper integrl is divergent. x d x = lim c ln ln c = ln lim c ln c = + pproches Review: useful limits of unbounded functions Here re some limits which re quite useful for the improper integrls of type II. () lim x + x = +, lim x x = ; () For r >, lim x + x r = nd lim x x r = whenever the function x r is welldefined in the limiting process. (3) lim ln x =. x + Clss Exmple.59. Compute the improper integrl 4 x d x.
23 Chpter Integrtion nd Geometry. Riemnn sum nd pproximtion of definite integrls Riemnn sum Recll.. The grphicl definition of definite integrl: If f (x) is continuous function on [, b], then f (x) will enclose some regions with xxis, between x = nd x = b. Then the definite integrl b f (x)d x is defined to be the net re: b f (x)d x = Are(Regions bove xxis) Are(Regions below xxis) In prticulr, if f (x) is positive on [, b], then the region enclosed by f (x) must be bove xxis, nd in this cse, the vlue of definite integrl is exctly the re: If f (x) > on [, b] b f (x)d x = Are(Region under f (x) between x = nd x = b) Now we consider numericl wy of pproximting the re using limits. () Divide the closed intervl [, b] into n equl subintervl. The width of ech piece is b n, denoted by x. All prtition points re denoted by =,,, 3,, n = b List ll the n subintervls: [, ], [, ], [, 3 ],, [ n, n ] () Drw verticl lines through ll the prtition points,,,, n, which will divide the region into n mny columns.
24 . RIEMANN SUM AND APPROXIMATION OF DEFINITE INTEGRALS 3 (3) For ech column, repet the following process. Tke the third column for exmple. Pick n rbitrry number, clled x 3, inside the third subintervl [, 3 ] on the x xis. Then drw verticl line through x 3. Remember tht we wnt to find the re of this green column. It is not rectngle becuse the roof of the column is not flt. (4) Now sketch rectngle in the third strip. The rectngle uses the subintervl [, 3 ] on xxis s the bottom edge, nd uses the height of f (x) t the chosen point x 3 s the height. Then we obtin such rectngle, which looks close the ctully column, but might lose some smll portion of re or gin some extr re. We cll it n pproximting rectngle, nd its re cn be esily computed: Are = height width = f (x 3 ) x (5) Repet this process for ll the columns in the region. Since we divided the whole region into nmny columns over nmny subintervls, we pick one representtive point from ech subintervl, nd sketch pproximting rectngle using the height t the representtive point. # of columns subintervl Rep. point height of rectngle Are of rectngle st column [, ] x f (x ) f (x ) x nd column [, ] x f (x ) f (x ) x 3rd column [, 3 ] x 3 f (x 3 ) f (x 3 ) x nth column [ n, n ] x n f (x n ) f (x n ) x Then the totl re of such n rectngles is f (x ) x + f (x ) x + f (x 3 ) x + + f (x n ) x =(f (x ) + f (x ) + f (x 3 ) + f (x n )) x n = f (x i ) x i= which is clled the Riemnn sum. Totl Are Definition.. Suppose f is defined on closed intervl [, b]. The intervl [, b] is divided into n subintervls of equl length x = b n. If x i is ny point in the ith subintervl [ i, i ], for i =,,, n, then the sum n f (x i ) x = f (x ) x + f (x ) x + + f (x n ) x i= is clled Riemnn sum for f on [, b].
25 . RIEMANN SUM AND APPROXIMATION OF DEFINITE INTEGRALS 4 Let s try n exmple. Exmple.3. Suppose f (x) = x nd we wnt to compute Riemnn sum for 4 xd x. Geometriclly, since x is positive function, this definite integrl represents the re of the region shown in the figure, which is bounded by the function x between x = nd x = 4. Step. Suppose now we wnt to divide the intervl [, 4] into 5 equl pieces. Then the width of ech subintervl is denoted by x, which cn be computed: x = 4 =.6 5 Then we cn explicitly list ll 5 subintervls in [, 4]: [,.6], [.6,.], [.,.8], [.8, 3.4], [3.4, 4]. x=.6 x=.6 x=.6 x=.6 x=.6 The region for re now cn be divided into 5 columns, on the 5 subintervls we just listed. Step. In ech subintervl on the xxis, we pick representtive number. The choice is rbitrry, s long s it is contined in the corresponding subintervl. We hve 5 subintervls. So we cll the corresponding representtive points x, x, x 3, x 4 nd x 5. For instnce, we cn pick representtive points s follows: [,.6], [.6,.], [.,.8], [.8, 3.4], [3.4, 4]. x =. x =.6 x 3 =.4 Step 3. Then we cn sketch n pproximting rectngles on ech subintervl, whose height is chosen to be the function vlue f (x) = x t the representtive point. We cn list the results in the following tble: # of columns subintervl Rep. point x i height of rectngle f (x i ) Are of rectngle f (x i ) x st column [,.6]....6 nd column [.6,.] rd column [.,.8] th column [.8, 3.4] th column [3.4, 4] x 4 =3.3 Then we dd the re from ll 5 rectngles, to obtin the Riemnn sum: Riemnn sum = f (x ) x f (x ) x f (x 3 ) x f (x 4 ) x f (x 5 ) x x 5 =3.74 Remrk.4. The choice is representtive points is rbitrry. We only need the kth representtive point to belong to the kth subintervl. For instnce, in the bove exmple, we my choose different set of representtive points: [,.6], [.6,.], [.,.8], [.8, 3.4], [3.4, 4]. x =.3 x =.89 x 3 =.55 nd the corresponding Riemnm sum is x 4 =3 x 5 =3.5 Riemnn sum = f (x ) x f (x ) x Thus we see tht the Riemnn sum definitely depends on the choice of representtive points. f (x 3 ) x f (x 4 ) x f (x 5 ) x Definition.5. If we constntly choose the left endpoint (right endpoint, midpoint) for ll subintervls, we cll the corresponding Riemnn sum the left (right, midpoint) Riemnn sum. More precisely, Riemnn sum is clled
26 . RIEMANN SUM AND APPROXIMATION OF DEFINITE INTEGRALS 5 the left Riemnn sum, if ech representtive point x i is chosen to be the left endpoint of the ith subintervl. the right Riemnn sum, if ech representtive point x i is chosen to be the right endpoint of the ith subintervl. the midpoint Riemnn sum, if ech representtive point x i is chosen to be the midpoint of the ith subintervl. Left Riemnn sum Right Riemnn sum Midpoint Riemnn sum In the bove exmple of 4 xd x, for 5 subintervls, the left, right, midpoint Riemnn sum re: Subintervl [,.6] [.6,.] [.,.8] [.8, 3.4] [3.4, 4] Left endpoint Right endpoint Midpoint L = = R = = M = = The grph shows the difference when we choose different set of representtive points. Remrk.6. In the bove exmple, we t first divide the intervl [, 4] into 5 subintervls. In generl, we cn do tht for rbitrrily mny pieces. The number of pieces is denoted by n. Then in this cse, we will hve n subintervls, n column rectngles, n representtive points, nd n summnds in the Riemnn sum. The following grphs show the midpoint Riemnn sum for n =5, nd subintervls. We cn see esily tht s the number of subintervls increses, the sum of the pproximting rectngles will get closer to the ctul re under the function, nd so we know tht the limit of Riemnn sum will become the vlue of definite integrl. The ctul vlue of the definite integrl 4 xd x in the exmple cn be computed by the fundmentl theorem: 4 x=4 xd x = 3 x 3/ = 4 3 = And the Riemnn pproximtions for vrious n lso shows the tendency of convergence s n +. The bove property tht Riemnn sum converges to the definite integrl is quite criticl. We list it here: x= Theorem.7. For continuous function f (x) on [, b], we hve b f (x)d x = lim (Riemnn sum for n subintervls) n + n = lim f (x i ) x n + Here x i is point in the ith subintervl of prtition of [, b] into n prts. i=
27 . RIEMANN SUM AND APPROXIMATION OF DEFINITE INTEGRALS 6 Thus we cn use Riemnn sum to pproximte definite integrls. This method is extremely useful becuse mny functions don t hve explicit ntiderivtive functions. Thus we cn t pply properly the fundmentl theorem to clculte the vlue of definite integrl involving such functions. Therefore n pproximtion is needed. Clss Exmple.8. Compute the left, right, nd midpoint Riemnn sum of the integrl x 3 + d x using 4 subintervls. Simpson s rule Wee see tht to pproximte definite integrl, we tret it s the net re bounded by the function. The ide of Riemnn sum is to divide the bounded region into columns nd tret ech one s rectngle. Now nother wy to pproximte the re is similr but treting columns s prbols. The corresponding formul is clled Simpson s rule. The key feture of Simpon s rule is tht it converges to the ctul vlue much fster thn the Riemnn sum. So it is more efficient. As before, for definite integrl b number. Fct.9. Here re some fcts bout prbol: () Three points in plne determine prbol. f (x)d x, we divide the intervl into n pieces. Here we need to ssume tht n is n even () If prbol is determined by points (x, y ), (x, y ), (x 3, y 3 ) for x 3 x = x x = h >, the net re bounded by the prbol between x nd x 3 is Are = h 3 (y + 4 y + y ) You cn prove this formul by using polynomil interpoltion nd integrtion. The endpoints of the n subintervls of [, b] re denoted by,,,, n. The points on the grph of f (x) t these endpoints re denoted by P = (, f ( )), P = (, f ( )),, P n = ( n, f ( n )). Since n is even, we hve even mny columns. This time we consider two columns t once. Since three points determine prbol, we use points P, P nd P to drw prbol, nd pproximte the column re over [, ] by this prbol. Then repet this process for points P, P 3 nd P 4, nd then points P 4, P 5 nd P 6, until ll columns re finished. The pproximting prbolic re by P, P, P in first two columns is x 3 (f ( ) + 4 f ( ) + f ( )). The pproximting prbolic re by P, P 3, P 4 in the second two columns is x 3 (f ( ) + 4 f ( 3 ) + f ( 4 )). The pproximting prbolic re by P n, P n, P n in the lst two columns is x 3 (f ( n ) + 4 f ( n ) + f ( n )). Then the totl pproximting re is the sum, which is clled the Simpon s rule: Method.. b Simpon s rule for pproximting b f (x)d x: f (x)d x x 3 where n is n even number nd x = b n. f ( ) + 4 f ( ) + f ( ) + 4 f ( 3 ) + f ( 4 ) + + f ( n ) + 4 f ( n ) + f ( n ) Notice tht in the bove summtion, the first nd lst summnd hs coefficient, nd the coefficients for the middle terms re 4,, 4,, 4,,, 4, with 4 nd lternting. Exmple.. Let s pproximte 4 xd x using 6 subintervls in Simpon s rule. Solution. We divide [, 4] into 6 equl subintervls. So the width of ech one is 4 6 =.5. So ll the endpoints re x=.5.5 x=.5 x=.5.5 x=.5 3 x= x=
28 . AREA BETWEEN CURVES 7 Then the pproximtion using Simpon s rule is 4 xd x.5 3 /3 ( ) f ( ) 4 f ( ) f ( ) 4 f ( 3 ) f ( 4 ) 4 f ( 5 ) f ( 6 ) Efficiency of pproximtion We wnt to compre how ccurte n pproximtion is. So we firstly define the notion of error: Definition.. The error of n pproximtion is the defined to be the bsolute vlue of the difference: Error = Actul vlue Approximtion Compre the results we hd in the exmple 4 xd x using midpoint Riemnn sum nd Simpson s rule for vrious vlues of n. Remember the ctul vlue of this definite integrl is 4 3 = : Midpoint Riemnn sum Error Simpson s Rule Error n = n = n = n = On one hnd, we see tht s n increses, both of the pproximtion methods re pproching the ctul vlue, with error decresing. This shows why the pproximtion works. We cn improve ccurcy bsiclly by incresing the vlue of n. More importntly, on the other hnd, we cn lso see tht the pproximtion by Simpson s with n = 6 lredy behves better thn the midpoint Riemnn sum with n = since the error by Simpson s rule is smller. Also the pproximtion by Simpson s with n = is better thn the midpoint Riemnn sum with n = 5. In other words, we cn sy tht the pproximtion by Simpsons s rule converges to the ctul vlue lot fster thn the Riemnn sum. I.e., to gurntee the required ccurcy, Simpson s rule requires smller n. The efficiency of lgorithm becomes very criticl when we require high ccurcy, becuse s n gets quite lrge, it might tke long time for computers to finish the itertion of n steps. At lest in our knowledge, now we know Simpson s pproximtion is more efficient thn Riemnn sum. In generl it is big brnch in mthemtics nd engineering for scientists to design other lgorithms with higher efficiency.. Are between curves We hve relted the re of the region bounded by one function to its definite integrl. Now we consider the region bounded by two functions. A typicl cse is shown in the figure. The region is bounded bove by curve f (x) nd bounded below by g(x), between verticl lines x = nd x = b. To derive the integrl formul, we firstly pply the ide of Riemnn sum, to slice the shded region verticlly into smll columns.
29 . AREA BETWEEN CURVES 8 Riemnn sum: slicendsum strtegy. () Divide the closed intervl [, b] into n equl subintervls. The width of ech piece is lso x = b n. All prtition points re denoted by List ll the n subintervls: =,,, 3,, n = b [, ], [, ], [, 3 ],, [ n, n ] Drw verticl lines through ll the prtition points,,,, n, which will divide the region into nmny columns. () For ech column, repet the following process. Tke the third column for exmple. Pick n rbitrry number, clled x 3, inside the third subintervl [, 3 ] on the xxis. Then drw verticl line through x 3. Sketch rectngle in the third strip. The rectngle tkes the subintervl [, 3 ] on xxis s the bottom width, nd uses the distnce between function f (x) nd g(x) t the chosen point x 3 s the height. Then we obtin such rectngle, which looks close the ctully re column. We cll it n pproximting rectngle, nd its re is: Are = height width = (f (x 3 ) g(x 3 )) x (3) Repet this process for ll the columns in the region. Cll x, x,, x n the representtive points in the subintervls. Then the totl re of such n rectngles is (f (x ) g(x )) x + + (f (x n ) g(x n )) x n = (f (x i ) g(x i )) x i= As n gets lrger, the totl re of the pproximting rectngles will be closer to the exct re between curves f (x) nd g(x). Thus, by the ide of Riemnn sum, s n pproches +, we obtin definite integrl: Theorem.3. Suppose f (x) nd g(x) re continuous functions on [, b], with f (x) g(x). The re of the region bounded by the grph f (x) nd g(x) on [, b] is b Are = (f (x) g(x))d x Remrk.4. region. To use the re formul, we need to identify f with the top curve nd g with the bottom curve of the bounded Exmple.5. Find the re of the region shown in the figure, which is bounded by y = x + nd y = x [, ]. on the intervl
30 . AREA BETWEEN CURVES 9 For some region, the integrtion intervl [, b] is not directly provided. We hve to solve for intersection points to get the bounds. Exmple.6. Find the re of the region bounded by f (x) = 3 + x x, nd g(x) = x +, shown in the figure. Exmple.7. Find the re of the region bounded by f (x) = 3 + x x, nd g(x) = x +, shown in the figure. Exmple.8. Find the re of the region shown in the figure, which is bounded by y = x nd y = x 3. Additionl Exmple.9. Find the re of the region bounded by the curve y = sin x, y = cos x x = nd x = π. Solution. The region consists of two components, denoted by A nd B. We need to consider the re seprtely becuse for region A, cos x is the top curve, nd for B, sin x is the top curve. The two regions re seprted t the intersection point. It cn be solved: Then sin x = cos x sin x cos x = tn x = x = rctn = π 4 Are(A) = = π/4 (cos x sin x)d x + Are(A) x=π/4 sin x + cos x x= π/ π/4 (sin x cos x)d x Are(B) x=π/ + cos x sin x = x=π/4
31 .3 ARC LENGTH OF A FUNCTION 3.3 Arc length of function The grph of continuous function is curve. For finite curve segment, how to mesure the length of rc? For exmple, in the figure, the rc of the grph of f (x), between points A nd B, defined over [, b]. As usul, we wnt to use the ide of Riemnn sum nd express the rc length s definite integrl. To do tht, we gin divide [, b] into n equl subintervls, with endpoints =,,,, n, n = b. Then we cll P, P,, P n the points on the curve defined t these endpoints. Now link the two consecutive rc points P i nd P i+ with line segment. Then we obtin polyline grph. We cn see tht s n pproches +, the polyline will definitely converge to the ctul rc. Thus the polyline length will lso converge to the ctul rc length. Now let s mesure the polyline length. Ech subintervl hs the sme width x. For exmple, in the first columnn, sketch right tringle using P nd P, s in the figure. Then the length of segment P P is, P P = ( x) + ( y) We wnt to express y s function of x. To do tht, we pply the Men vlue theorem for differentition: y x = f ( ) f ( ) = f (x ) for some x in the intervl [, ]. Then y = f (x ) x, nd hence P P = ( x) + ( y) = ( x) + (f (x ) x) = ( + (f (x )) ) ( x) = ( + (f (x )) ) x Repet this process for the line segments in ll columns. In the ith column, we will obtin point x i in [ i, i ], such tht the totl length of the polyline is: ( + (f (x )) ) x + ( + (f (x )) ) x + + ( + (f (x n )) ) x = n ( + (f (x i )) ) x Therefore, by the covergence theroem of Riemnn sum, we know tht the polyline rc length will converge to the definite integrl b + (f (x)) d x, which is the following theorem: i= Theorem.. nd x = b is If f (x) is continuously differentible on [, b], then the length of the curve of y = f (x) on between x = L = b + (f (x)) d x Clss Exmple.. Find the rc length of y = x 3 on x 4. Additionl Exmple.. Find the rc length of the curve y = x ln x 8 on x 3.
32 .3 ARC LENGTH OF A FUNCTION 3 Solution. Notice tht the derivtive f (x) = x. Then the rc length is 8x 3 3 L = + f (x) d x = + (x 3 8x ) d x = + 4x + 64x = 4 = (x + 8x )d x = x + x=3 8 ln x = 8 + ln 3 8 x= 3 x + d x 8x Additionl Exmple.3. The upper hlf circle centered t the origin cn be written s f (x) = x for x. The derivtive f (x) is f x (x) =. The rc length of the hlf circle is x x + (f (x)) d x = + d x = + x x x d x = x d x = x d x = rcsin x x= x= = rcsin rcsin( ) = π ( π ) = π (This is convergent improer integrl) This shows tht the rc length of hlf circle of rdius is π. Then the perimeter of circle of rdius is π. This confirms our common geometry knowledge. Remrk.4. Since the integrl formul contins squre root function, its ntiderivtive mostly is hrd to find. So the method of pproximtion is more populr, such s Riemnn sum nd Simpson s rule we lerned in the first section. Additionl Exmple.5. Use Simpson s rule with subintervls to estimte the rc length of the curve y = x between x = nd x =. Solution. The derivtive of y is y =. The rc length cn be set up s x L = + f (x) d x = + x d x 4 To use Simpon s rule, we divide the intervl of integrtion [, ] into subintervls. Ech subintervl hs length x =.. Then the endpoints re =, =., =., 3 =.3, 4 =.4, 5 =.5, 6 =.6, 7 =.7, 8 =.8, 9 =.9, = Then by the formul of Simpson s rule (Method.), we hve the pproximtion of the rc length L = + x d x 4. 3 = Remrk.6. It is well known tht, given two points on plne, the shortest curve connecting such two points must be the line segment. We cn prove this property when the curve is ssumed to be differentil function f (x). More formlly, we know if f (x) connects two given points (, A) nd (b, B), then its rc length is L = b + (f (x)) d x Then mong ll such curves f (x), the shortest one will occur exctly when f (x) is constnt. I.e., when f (x) is stright line. We cn t prove it in the context of single vrible clculus. After lerning multivrible Clulus, you will be ble to understnd the proof. The proof is direct consequence of n importnt theorem: fundmentl theorem of Clculus of vrition. You cn the introductory informtion on the following Wikipedi pge:
33 Chpter 3 Advnced integrtion techniques 3. Integrtion of rtionl functions using Prtil frctions In this section we re going to see how to evlute n indefinite integrl of rtionl function, i.e., n integrl of form N(x) d x for two polynomils N(x) nd D(x) with no common fctor, such s the following integrls: D(x) x d x, 5x 4 x 3 + x d x, x x 4 + x + d x Preprtion: prtil frctions of liner fctors nd qudrtic fctors For n integrl N(x) D(x) d x of rtionl function, here we consider the following two typicl types, depending on the form the the denomintor function D(x): A Type I (Liner fctors): (x + b) d x. k Denomintor D(x) : liner fctor, defined s power of liner function (x + b), i.e., of the form (x + b) k. Exmples: x +, x, x 3, (3x + 4), (x ) 4. Numertor N(x): constnt A. Strtegy for integrtion : Trick of liner vrition. Clss Exmple 3.. () 3x 5 d x () 3 (x 3) d x Type II (Qudrtic fctors): Ax + B d x with >. x + Denomintor D(x) : A qudrtic fctor is of form x +. Exmples: x +, x + 9. Numertor N(x): Ax + B with two constnts A nd B. Strtegy for integrtion : For qudrtic fctor of the form x +, we cn brek the frction into two: nd then pply the following ntiderivtive rules: Ax + B x + = A x x + + B x + 3
34 3. INTEGRATION OF RATIONAL FUNCTIONS USING PARTIAL FRACTIONS 33 x x + d x = ln x + + C x + d x = rctn x + C Clss Exmple 3.. 4x + 3 x + 9 Therefore, we cn integrte the sum nd difference of the prtil frctions of the bove two types. Clss Exmple 3.3. x + 3 x + 6 ( 3x) + 5x + d x 4 x + 4 Setup of prtil frctions Now we re going to del with the integrtion of generl rtionl function degree of N(x) is less thn the degree of D(x). N(x) d x, ssuming: D(x) The cse when degree N(x) is higher will be discussed t the end of this section. To begin with, we need to fctor the denomintor function D(x) into product of liner fctors of form (x + b) k (Type I) nd qudrtic fctors x + (Type II). We need to fctor D(x) completely until ll fctors re irreducible. Exmple 3.4. Fctor the following polynomils into irreducible fctors nd determine the type of the fctors: () D(x) = x x + () D(x) = x 3 9x (3) D(x) = x 3 4x + 4x (4) D(x) = x 4 + 4x (5) D(x) = x 4 Method 3.5. Set up of prtil frctions. For the integrl N(x) D(x) d x, dd the prtil frctions for ll fctors of D(x). The wy to set up prtil frctions for ech fctor is s follows: To ech liner fctor (x + b) k of degree k, we ssocite set of prtil frctions of the form A x + b + A (x + b) + A 3 (x + b) + + A k 3 (x + b) k with undetermined constnts A, A,, A k To ech qudrtic fctor of the form x + for >, we ssocite frction: where A nd B re constnts to be determined. Ax + B x +
35 3. INTEGRATION OF RATIONAL FUNCTIONS USING PARTIAL FRACTIONS 34 Exmple 3.6. () x + (x ) () x 3 9 (3) x x 4 + 4x (4) 4x + 3 x 4 Set up prtil frctions for the following rtionl functions. Remrk 3.7. The setup of form of prtil frctions only depends on the denomintor function D(x). The numertor will be used lter to determine the unknown constnts A, B,etc. Determine the unknown coefficients: Common denomintor Exmple 3.8. Express x + x + s prtil frctions. Then compute its integrl (x ) (x ) d x. Solution. () Use the common denomintor to expnd the prtil frctions: 6x 4 (x ) = A x + B A(x ) = (x ) (x ) + B A(x ) + B = = Ax A + B (x ) (x ) (x ) Compre the two sides of the bove equlity. They hve the sme denomintor, then the numertors hve to be the sme: 6x 4 = Ax + ( A + B) Since the two sides re equl s functions, we cn compre the coefficients in front of the sme power of x, nd obtin two equtions: A = 6, A + B = 4 Solve this system for A nd B. We get tht Put them bck to the form of prtil frctions: A = 3, B = 6x 4 (x ) = 3 x + (x ) () Then its integrl cn be clculted esily from the trick of liner vrition. 6x 4 (x ) d x = 3 x d x + (x ) d x = 3 ln x + x + C Clss Exmple 3.9. Evlute the integrl x 4 d x Additionl Exmple 3.. Evlute the integrl d x for >. x Solution. Fctor x = (x + )(x ) nd set up prtil frctions s x = A x + + B A(x ) + B(x + ) (A + B)x + (B A) = = x (x + )(x ) (x + )(x )
36 3. INTEGRATION OF RATIONAL FUNCTIONS USING PARTIAL FRACTIONS 35 Now compre the numertor of the two sides: The solution is. Then the integrl cn be computed: x d x = x + + (A + B)x + (B A) = A + B =, (B A) = A =, B = x = ln x + C x + d x = ln x + + ln x + C The bove formul is sometimes useful in the section of trigonometric substitution. So we list the result gin, together with similr one: Formul 3.. Antiderivtive formuls: x d x = ln x + C x + x d x = ln x + + C x Clss Exmple 3.. Evlute the integrl 4 5x 3x 3 x d x Additionl Exmple 3.3. Evlute the integrl 3x 8 x 3 + 4x d x Solution. The denomintor cn be fctored s D(x) = x(x + 4), with liner fctor x nd n irreducible qudrtic fctor x + 4. So we set up the form of prtil frctions s Combine the frctions by unifying the denomintors: 3x 8 x 3 + 4x = A x + Bx + C x + 4 A x + Bx + C x + 4 = A(x + 4) + (Bx + C)x = Ax + 4A + Bx + C x = (A + B)x + C x + 4A x(x + 4) x(x + 4) x(x + 4) Compring the coefficients in numertors, we obtin system of equtions: A + B =, C = 3, 4A = 8 which leds to the solution set A =, B =, C = 3 Then the originl integrl cn be evluted: 3x 8 x + 3 x 3 + 4x d x = x d x + x + 4 d x = x d x + x x x + 4 d x = ln x + ln x rctn x + C
37 3. INTEGRATION OF RATIONAL FUNCTIONS USING PARTIAL FRACTIONS 36 Additionl Exmple 3.4. Evlute the integrl x 3 x 4 + 9x. Solution. Fctor the denomintor D(x) = x 4 + 9x = x (x + 9). It contins liner fctor x = (x ), nd qudrtic fctor x + 9. To the liner fctor x, we ssocite prtil frctions A x + B x To the qudrtic fctor x + 9, we ssocite prtil frction C x + D x + 9. Thus we set up the prtil frctions: x 3 + x + 3 x 4 + 9x = A x + B x + C x + D x + 9 = A x(x + 9) x (x + 9) + B (x + 9) (C x + D) x + x x + 9 = Ax(x + 9) + B(x + 9) + (C x + D)x x (x + 9) Compre the numertor on both sides. We obtin equions: Then we hve the vlue of the undetermined constnts: A + C =, B + D =, 9A =, 9B = 3 = (A + C)x 3 + (B + D)x + 9Ax + 9B x (x + 9) Then the originl integrl is x 3 x 4 + 9x d x = 3 A =, B = 3, C =, D = 3 x + x + 3 x + 9 d x = 3x + ln x rctn x 3 + C Remrk 3.5. More generl, n irreducible qudrtic fctor cn be of the form x + bx + c with b 4c <. For such fctor, we cn set the prtil frctions s Ax + B x + bx + c. To integrte it, we need to complete the squre for x + bx + c into (px + q) + r. Then pply the substitution u = px + q for integrtion. For exmple, for the integrl 4x 5 d x, the denomintor x x + 5 cn be written s (x ) + 4. Then we cn pply x x+5 substitution u = x with x = u + nd du = d x to rewrite the integrl 4x 5 4(u + ) 5 4u x x + 5 d x = du = u + 4 u + 4 du = 4 ln u + 4 rctn u + C = ln (x ) + 4 rctn x + C Reduce degree in numertor: Long division of polynomils So fr in rtionl integrl N(x) D(x), we ssumed tht the degree of N(x) is less thn tht of D(x). If the degree of N(x) is lrger thn or equl to the degree of D(x), we need to pply the long division to reduce the degree of the numertor. More precisely, we wnt to decompose the frction N(x) R(x) = Q(x) + D(x) D(x) such tht the degree of x in the numertor R(x) is smller thn the degree of x in the denomintor Q(x). To do tht, we re going to divide N(x) by D(x) using long division to express N(x) = Q(x) D(x) + R(x) quotient divisor reminder Exmple 3.6. Reduce the frction x 4 + 3x 3 + x + 3x + nd integrte it.
38 3. TRIGONOMETRIC INTEGRATION 37 Solution. Apply the long division by divide x 4 + 3x 3 + by x + 3x + : The long division implies tht x 3x +5 quotient divisor x + 3x + x 4 +3x 3 +x +x + Thus the originl rtionl function cn be reduced to x 4 +6x 3 4x 3x 3 4x +x 3x 3 9x 6x 5x +6x + 5x +5x + 9x 8 reminder (x 4 + 3x 3 + ) = (x 3x + 5) (x + 3x + ) + ( 9x 8) quotient divisor x 4 + 3x 3 + x + 3x + = (x 9x 8 3x + 5) + x + 3x + Q(x) R(x) D(x) reminder So now the remining frction function on the right hnd side stisfies the degree condition nd we cn set up prtil frctions: 9x 8 x + 3x + = A x + + B (A + B)x + (A + B) = x + (x + )(x + ) This implies two equtions: A + B = 9, The solution is A =, nd hence the originl integrl cn be computed: A + B = 8 B = x 4 + 3x 3 + x + 3x + d x = (x 3x + 5) + 9x 8 x + 3x + = (x 3x + 5) + x + + x + d x d x = 3 x 3 3 x + 5x + ln x + ln x + + C 3. Trigonometric integrtion In this section we study how to integrte functions defined by combining powers of trigonometric functions in certin form, such s sin xd x, tn 3 xd x, tn 3 x sec 4 xd x Knowing how to integrte such combined trigonometric functions is key fctor to the next section, clled trigonometric substitution, in which section we trnsform some hrd integrls into trigonometric integrls.
39 3. TRIGONOMETRIC INTEGRATION 38 Integrls of form sin m x cos n xd x if m or n is n odd positive integer We consider the integrls the form sin m x cos n xd x for integers m nd n. Here m, n cn be zero. The constnt m represents the power on sin x nd n represents the power on cos x. An esy cse to integrte is when m or n is n odd positive integer. The following integrls re of this type: sin 5 x cos 7 xd x, sin 3 x cos xd x, cos 5 xd x, cos 3 x sin 4 x d x The key to these integrls is to convert sin x to cos x or convert cos x to sin x by the following identities: Fct 3.7. Here is fmous trigonometric identity sin x + cos = The bove identity implies the following two reltions for conversion between sin x nd cos x: sin x = cos x cos x = sin x (3.) (3.b) We lso need the derivtive identities (sin x) = cos x (cos x) = sin x Let s use n exmple to see how to use the bove identities. Exmple 3.8. Consider the integrl sin 4 x cos 3 d x. Here the power of cos x is 3, which is n odd positive integer. Now brek cos 3 x into two fctors: cos x nd cos x. Then the first fctor cos x cn be converted to sine function sin x by 3.b, nd the second fctor cos x is exctly the derivtive of sin x: sin 4 x cos 3 x =cos x cos x d x = sin 4 x cos x cos x = sin x derivtive of sin x Thus if we now set substitution function u = sin x with du = cos xd x, then the originl integrl cn be converted to sin 4 x cos 3 xd x = sin 4 x( sin x) cos xd x = = 5 u5 7 u7 + C = 5 sin5 x 7 sin7 x + C du d x u 4 ( u )du = (u 4 u 6 )du The bove exmple implies the following trick. If the power of cos x is odd. We cn sve one fctor of cos x nd combine it with d x to obtin d(sin x). The other evenpower of cos x cn be converted by 3.b. Then substitution u = sin x cn be pplied. sin m x cos odd xd x = sin m x cos even convert to sine by (3.b) cos xd x d(sin x) Similrly, if the power of sin x is odd, then we cn lso sve fctor of sin x. Then the remining evenpower of sin x cn be converted to cosine functions by 3. nd the lst fctor sin x cn be bsorbed into the differentil of u = cos x: sin odd x cos n xd x = sin even x cos n x sin xd x convert to d(cos x) cosine by (3.) We cn summrize the method of integrtion by setting substitution s follows. Method 3.9. Strtegy for integrls of type sin m x cos n xd x for m or n odd nd positive.
40 3. TRIGONOMETRIC INTEGRATION 39 () If the power of sin x (i.e., m) is odd: Set u = cos x with du = sin xd x. Use identity sin = cos x (3.) to convert sine to cosine. () If the power of cos x (i.e., n) is odd: Set u = sin x with du = cos xd x. Use identity cos = sin x(3.b) to convert cosine to sine. Clss Exmple 3.. Evlute sin 5 xd x. Clss Exmple 3.. Evlute cos 3 x sin 4 x d x. Integrls of form tn m x sec n xd x: if m is odd or n is even Now we look t integrls which re formed by powers of tn x nd sec x. The reson we put tn x nd sec x together in the integrl nd hope to develop some strtegies, is tht, like sin x nd cos x, there re mny reltions between tn x nd sec x. Fct 3.. Trigonometric identities: sec x = tn x + tn x = sec x (3.) (3.b) Derivtive rules: (tn x) = sec x, The following two ntiderivtive rules re lso quite useful: (sec x) = tn x sec x tn xd x = ln sec x + C (3.3) sec xd x = ln sec x + tn x + C (3.4) Like the cse of sine nd cosine functions, we hve two typicl cses when we cn use substitution method: when m is odd nd positive, nd when n is even nd positive. Method 3.3. () Strtegy for integrls of type tn m x sec n xd x if m is odd nd positive. Set u = sec x with du = tn x sec xd x. Sve one copy of tn x nd sec x for d(sec x). Use identity tn x = sec x (3.b) to convert the remining tngent power to function of sec x. () Strtegy for integrls of type tn m x sec n xd x if n is even nd positive. Set u = tn x with du = sec xd x. Sve two copies of sec x for d(tn x). Use identity sec x = tn x + (3.) to convert the remining secnt power to function of tn x. Clss Exmple 3.4. π/4 Evlute integrls: tn 3 x sec 4 xd x.
41 3. TRIGONOMETRIC INTEGRATION 4 Additionl Exmple 3.5. Evlute the integrl tn 3 xd x. Solution. Since in this integrl tngent hs n odd positive power, we should sve copy of tn x sec x for substitution u = sec x. The originl integrl doesn t hve ny secnt copy. So we cn mke up one by writing = sec x sec x : tn tn 3 x tn x sec x (sec x ) xd x = d x = tn x sec xd x sec x sec x Now set u = sec x with differentil du = tn x sec x. Then (sec tn 3 x ) u xd x = tn x sec xd x = du sec x u = (u u )du = u ln u + C = (sec x) ln sec x + C Use of Reduction formuls for other cses Here we consider the following two cses: sin m x cos n xd x when m or n is n even nd positive integer. tn m x sec n xd x when m is n even nd positive integer. For those integrls of power of only one trigonometric function, there re recursive formuls, clled reduction formuls: Formul 3.6. Reduction formuls: sin n d x = n sinn x cos x + n sin n xd x n cos n d x = n cosn x sin x + n cos n xd x n tn n d x = n tnn x tn n xd x sec n d x = n tn x secn x + n sec n xd x n cot n d x = n cotn x cot n xd x csc n d x = n cot x cscn x + n csc n xd x n (3.5) (3.5b) (3.5c) (3.5d) (3.5e) (3.5f) They re clled reduction formuls becuse the power involved in the new integrl on the right hnd side hs been reduced by, from n to n. These reduction formuls cn be proved by integrtion by prts. As n exmple, we prove the first one. All other reduction formuls cn be proved in similr wy. Proof. For integrl sin n xd x, write sin n x = sin n x sin x nd set up components in integrtion by prts u = sin n x u = (n ) sin n x cos x v = sin x v = cos x
42 3. TRIGONOMETRIC INTEGRATION 4 Then sin n xd x = sin n x ( cos x) (n ) sin n x cos x ( cos x)d x = sin n x cos x + (n ) sin n x cos x d x = sin x = sin n x cos x + (n ) sin n xd x (n ) sin n xd x Combining the terms of sin n xd x on both sides, we obtin n sin n xd x = sin n x cos x + (n ) sin n xd x nd hence sin n xd x = n sinn x cos x + n sin n xd x n Remrk 3.7. () By repetedly using reduction formuls, we cn bsiclly integrte ll powers of pure sin x, cos x, tn x nd sec x. () For the integrl of tngent powers nd secnt powers, the ntiderivtive rule of tn x nd sec x re useful. These rules re lredy listed in equtions (3.3) nd (3.4). (3) If the integrnd is negtive power of trig function, we cn use the reciprocl reltion to convert negtive power to positive power: sin x reciprocl csc x, cos x reciprocl sec x, tn x reciprocl cot x Clss Exmple 3.8. Use reduction formuls to evlute the following integrls: () cos xd x () cos 4 xd x (3) cos 3 x d x Now we develop the method for the cses of integrls we mentioned before. Method 3.9. Use of reduction formuls in trigonometric integrls. For sin m x cos n xd x with m or n positive nd even: () Apply conversion of squre of sine or cosine by (3.b) nd (3.) () Express the integrl s sum of powers of trig functions. (3) Then pply the reduction formuls. For tn m x sec n xd x with m is positive nd even: () Apply conversion of squre of tngent by (3.b). () Express the integrl s sum of powers of trig functions. (3) Then pply the reduction formuls. Clss Exmple 3.3. Evlute the integrl sin x cos xd x. Exmple 3.3. Compute the integrl tn x sec xd x.
43 3. TRIGONOMETRIC INTEGRATION 4 Solution. The power of tn x is even nd the power of sec x is odd. So the regulr substitution method cn t work. So we use the reltion 3.b to write tn x into sec x : tn x sec xd x = (sec x ) sec xd x = The ltter integrl is mentioned in the ntiderivtive rule (3.4): The former one cn be reduced by reduction formul (3.5d): Therefore the originl integrl equls sec 3 xd x = tn x sec x + sec xd x = ln sec x + tn x + C sec 3 xd x sec xd x sec xd x (3.4) = tn x sec x + ln sec x + tn x + C tn x sec xd x = tn x sec x + ln sec x + tn x ln sec x + tn x + C = tn x sec x ln sec x + tn x + C Additionl Exmple 3.3. π/ Evlute the integrl sin 4 x cos xd x. Solution. Use conversion (3.b) cos = sin x to trnsform the integrl: π/ sin 4 x cos xd x = π/ sin 4 x( sin x)d x = π/ sin 4 xd x (I) π/ which involves two integrls of sine powers. So we cn pply the reduction formul (3.5) or sine: n = : n = 4 : sin xd x = sin x cos x + sin 4 xd x = 4 sin3 x cos x d x = sin x cos x + x + C sin xd x = 4 sin3 x cos x sin 6 xd x (II) sin x cos x + x + C n = 6 : = 4 sin3 x cos x 3 8 sin x cos x x + C sin 6 xd x = 6 sin5 x cos x sin 5 xd x = 6 sin4 x cos x sin3 x cos x 38 sin x cos x + 38 x + C Plug in the bounds: = 6 sin5 x cos x 5 4 sin3 x cos x 5 5 sin x cos x x + C (I) = Then the originl definite integrl is π/ π/ sin 4 x cos xd x = sin 4 xd x = 3π π/ 6, (II) = sin 6 xd x = 5π 3 π/ sin 4 xd x π/ sin 6 xd x = 3π 6 5π 3 = π 3 Additionl notes: Integrls sin m x cos n xd x when m nd n re both even nd positive Now we ssume both of the powers of sin x nd cos x re positive even number. The following integrls re of this type: sin x cos 4 xd x, sin 4 xd x We hve seen tht we cn use reduction formuls to compute these integrls. Now we develop nother method, using double ngle identities, which help us reduce the power of sin x nd cos x.
44 3. TRIGONOMETRIC INTEGRATION 43 Fct Double ngle identities: sin x = ( cos x) (3.6) cos x = ( + cos x) (3.6b) After using the double ngle identities, ech squre of sin x or cos x cn be reduced to expressions involving cos(x). So nthpower of sin x or cos x cn be reduced to functions involving nthpower of cos x. So the highest power hs been reduced to hlf. We cn keep doing this until the cosine power becomes odd, in which cse we cn pply the method in the previous section. Additionl Exmple Evlute the integrl sin xd x. Solution. By double ngle identity for sin x, we hve cos x sin xd x = d x = ( cos x)d x = x sin x + C Additionl Exmple π/ (Compre with Exmple. 3.3) Evlute the integrl sin 4 x cos xd x. Solution. Apply 3.6 nd 3.6b, we hve sin 4 x cos xd x = ( cos x) ( + cos x) = 8 ( cos x + cos (x))( + cos x)d x = ( cos(x) cos (x) + cos 3 (x))d x 8 = ( cos(x) 8 ( + cos 4x) + cos3 (x)d x = ( 8 cos(x) cos 4x)d x + cos (x) cos(x)d x = 8 x sin(x) 8 sin(4x) + sin(x) 6 sin3 (x) + C = 8 x 8 sin(4x) 6 sin3 (x) + C Then the definite integrl π/ sin 4 x cos xd x = 8 x 8 sin(4x) x=π/ 6 sin3 (x) = π 3 x=
45 3.3 TRIGONOMETRIC SUBSTITUTION Trigonometric substitution How to confirm the fct tht the re of circle of rdius is π? By the grphicl definition of definite integrls, we cn formulte the upper hlf circle function s y = x, nd to compute the definite integrl Are of hlf disk of rdius = x d x However, bsed on the integrl techniques we hve lerned so fr, we cn t compute the ntiderivtive of x esily. Thus in this section, we re going to see how to find integrls involving functions of power of x, x + nd x, with positive constnt. Actully we hve known few integrl rules involving these functions: x + d x = rctn x + C x x + d x = ln x + + C (by substitution u = x + ) x d x = rcsin x + C x d x = ln x + C ( see Exmple (3.) nd Formul (3.)) x + But for more integrls, such s 4 x d x, (x + ) d x, x x 9 d x new integrtion method is needed. The method is clled trigonometric substitution. We cn going to set x s n pproprite trigonometric function, nd express the integrl s trigonometric integrl, which hs been discussed in gret detils in previous section. Integrls involving power of x Given n integrl contins x, for positive constnt, if we ssume x is of the form x = sin θ with differentil du = cos θ dθ, then by the conversion reltion (3.b), we hve x = ( sin θ) = ( sin x) = cos x If we ssume θ is in [ π, π ], then cos θ, nd then the squre root of bove reltion: x = cos θ = cos θ So we cn successfully get rid of the squre root. Clss Exmple Compute the integrl x 3 9 x d x.
46 3.3 TRIGONOMETRIC SUBSTITUTION 45 Remrk When we wnt to return to the originl independent vrible x in the ntiderivtive, we cn use the elementry definition of trigonometric functions. We cn trck ll six trigonometric functions in right tringle. sin θ = Opposite Hypotenuse cos θ = Ajcent Hypotenuse tn θ = Opposite Ajcent = sin θ cos θ reciprocl reciprocl reciprocl csc θ = Hypotenuse Opposite sec θ = Hypotenuse Ajcent cot θ = Ajcent Opposite = cos θ sin θ Exmple Compute the integrl 4 x d x, which is the re of the upper hlf disk of rdius. Integrls involving x + nd x For other types, involving functions of form x + nd x, we use similr ide. The generl strtegies re listed in the following tble. Expression Substitution differentil Trnsformtion squre root x x = sin θ du = cos θ dθ x = cos θ x + x = tn θ du = sec θ dθ x + = sec θ x x = sec θ du = tn θ sec θ dθ x = tn θ x = cos θ x + = sec θ x = tn θ Clss Exmple Clculte x + 4d x. The bove integrl cn be generlized to the following ntiderivtive formul Formul 3.4. Additionl Exmple 3.4. x + d x = x x + + ln x + x + + C (3.7) Compute the integrl (x + 9) d x.
47 3.3 TRIGONOMETRIC SUBSTITUTION 46 Solution. This integrl involves functions of form x + with = 3. So we cn set up substitution x = 3 tn θ, nd then x = 3 tn θ, d x = 3 sec θ dθ, x + 9 = 9 sec θ The originl integrl cn be substituted: (x + 9) d x = (9 sec θ) 3 sec θ dθ = 7 sec θ dθ = cos θ dθ ( by definition cos θ = 7 sec θ ) = 7 sin θ cos θ + θ + C ( by cosine reduction formul (3.5b)) By ssumption x = 3 tn θ, we hve tht sin θ = the originl integrl equls (x + 9) d x = 7 = 54 x x + 9 3x x rctn x 3 x x + 9 nd cos θ = 3 x + 9. Then 3 x rctn x + C 3 + C Clss Exmple 3.4. Compute the integrl x 9 d x. Additionl Exmple Compute the integrl x x 4 d x. Solution. This integrl involves x 4 with = 4 nd =. So we cn set x = sec θ.then x = sec θ, d x = tn θ sec θ, x 4 = tn θ Now the originl integrl cn be expressed s x x 4 d x = ( sec θ) tn θ tn θ sec θ dθ = 4 sec 3 θ dθ By reduction formul by reduction formul (3.5d): sec 3 θ d x = tn θ sec θ + Notice tht sec θ = x nd tn θ = x 4. Then we hve x x 4 d x = 4 sec 3 θ dθ = 4 tn θ sec θ + x 4 = x + ln x x C x 4 + ln x + = x sec θ dθ (3.4) = tn θ sec θ + ln sec θ + tn θ + C ln sec θ + tn θ + C x 4 + C ( by log property ln(/b) = ln ln b)
48 Chpter 4 Appliction in geometry: volume nd surfce re of revolution 4. Volume of solid of revolution Generl formul: Slicendsum Consider solid object tht extends in the xdirection from x = nd x = b. Imgine cutting through the solid, perpendiculr to the xxis t point x, nd suppose the re of the cross section creted by the cut is A(x). We wnt to mesure the volume of this solid. To do this, we first divide [, b] into n equl subintervls of width x. We now mke cuts throught the solid t ech endpoint, which produces n slices of thickness x. In the ith subintervl, pick representtive point x i, nd mesure the crosssectionl re A(x i ) t x i. Treting ech slice s cylinder, then the volume of the ith slice is pproximtely equl to A(x i ) x. Then by the ide of Riemnn sum, the totl volume is pproximtely V A(x ) x + A(x ) x + + A(x n ) x = n A(x i ) x As the number of slices increses (n ), the thickness of the ech slice will pproch, nd the pproximte volume by n slices will pproch the exct volume: V = lim n + i= n A(x i ) x By the theorem of convergence of Riemnn sum to definite integrl, we hve the following conclusion: i= Theorem 4.. Suppose solid object extends from x = nd x = b nd the cross section t ny x is given by function A(x). Then the volume of the solid is b V = A(x)d x Volume of revolution of one curve bout xxis: the disk method We now consider specific type of solid known s solid of revolution. Suppose f (x) is continuous function with f (x) on n intervl [, b]. Let R be the region bounded by f (x) nd xxis, between x = nd x = b. Now revolve the region R round the xxis. As R revolves once round xxis, it sweeps out solid of revolution in the threedimensionl spce. Our gol is to find the volume of this solid. We use the slicendsum method introduced bove. Since the volume is formed from revolution, t ny point x in [, b] on xxis, the perpendiculr cross section is disk, with rdius exctly f (x). Thus the crosssectionl re is A(x) = π(f (x)), nd by the bove theorem, the volume of the solid of revolution is 47
49 4. VOLUME OF SOLID OF REVOLUTION 48 Formul 4.. b V = π(f (x)) d x Exmple 4.3. Let R be the region bounded by the curve f (x) = x, the xxis, yxis, nd verticl line x = 3. Find the volume of the solid of revolution obtined by revolving R bout the xxis. Exmple 4.4. The function y = x for x represents the upper hlf circle centered t the origin with rdius. The rc is rotted bout the xxis to form sphere of rdius. Find the volume of this sphere. Solution. By the volume formul, we hve V = π( x ) d x = π ( x )d x = π x x 3 x= 3 x= = 4 3 π3 This exmple gives the formul for the volume of sphere of rdius, which is 4 3 π3. Volume of revolution of two curves bout xxis: the wsher method Now let s ssume tht the region R is bounded by two positive functions f (x) nd g(x) between x = nd x = b, where f (x) is the top curve nd g(x) is the bottom curve. Agin R is revolved bout xxis to generte solid of revolution. We pply gin the slicendsum method s in the previous cse. By slicing the solid into slices, we see tht ech slice is pproximtely n nnulus, with outer rdius f (x) nd inner rdius g(x). Thus the crosssectionl re is A(x) = π(f (x)) π(g(x)) = π(f (x) g(x) ) nd then we obtin the formul of the volume of revolution: Formul 4.5. b V = π(f (x) g(x) )d x Exmple 4.6. Let R be the region bounded by the grphs of f (x) = x + nd g(x) = x + 3 in the first qudrnt between x = nd x =. Compute the volume of the solid when R is revolved bout xxis.
50 4. SURFACE AREA OF REVOLUTION 49 Additionl Exmple 4.7. Find the volume of the solid obtined by rotting the region shown in the figure bout the xxis. Solution. The perpendiculr cross section of the solid t ny x is n nnulus, whose outer rdius is determined by the height of function y = x nd inner rdius is determined by the function y = x. Thus we cn set up n integrl to compute the volume: π[x (x ) ]d x = (x x 4 )d x = π 5 Remrk 4.8. When we use the formul for the volume of solid of revolution, f (x) nd g(x) represent the outer rdius nd inner rdius from the the curves to the rottion xis. Therefore, if the region revolves round n rbitrry horizontl line (not necessrily the xxis), we need to mesure the distnce of the curves f (x), g(x) nd the rottion xis. Exmple 4.9. (Compre with Exmple (4.7).) Find the volume of the solid obtined by rotting the region shown in the figure bout the line y =. Solution. The cross section t ny x is n nnulus, centered t the line y =. Thus the outer rdius of such nnulus is the distnce from the curve y = x to rottion xis y =, which is ( x ), nd the inner rdius is the distnce from y = x to line y =, which is ( x). Thus by the volume formul, we hve V = = π = π π[( x ) ( x ) ]d x outer rdius inner rdius (4 x + x 4 ) (4 x + x )d x (x 4 3x + x)d x = 8π 5 4. surfce re of revolution Slice nd sum A surfce of revolution is formed when curve f (x) is rotted bout the xxis. We wnt discuss the re of the surfce of revolution. We gin pply the slicendsum method by cutting the surfce into pieces by cross sections t endpoints of subintervls. Then we wnt to mesure the re of the surfce in ech slice. Over the ith subintervl [ i, i ], the corresponding slice of surfce is bnd, with rdius on left side f ( i ) nd rdius on the right side f ( i ). Such bnd is piece of truncted cone. Here is some bsic fct in geometry bout the surfce re of truncted cone:
51 4. SURFACE AREA OF REVOLUTION 5 Fct 4.. For truncted cone with verge rdius r = r + r height s, the lterl re of the surfce re is nd slnt A = πrs Now tret ech slice of surfce s truncted cone. When n is lrge, the width x is very smll, in ech subintervl [ i, i ], we cn pick ny point x i nd use f (x i ) s the representtive verge rdius of the cone, nd use the length of line segment connecting P i nd P i s s. I.e., by Pthegoren theorem nd discuss in section (.3), we hve P i P i = ( x) + ( y) + f ((x i )) x Thus the re of the pproximting truncted cone on the ith subintevrl is A i = π f (x i ) + f ((x i )) x nd the totl pproximting surfce re is i=n A = π f (x i ) + f ((x i )) x i= As n goes to +, the bove summtion will converge to the exct surfce re of revolution by f (x). By the convergence of Riemnn sum, we hve the following result: Formul 4.. If f is positive nd hs continuous derivtivem. The surfce re of the surfce obtined by rotting the curve y = f (x) between x b, bout xxis, is S = b π f (x) + (f (x)) d x Clss Exmple 4.. the resulting surfce. Let the rc of the function y = x from (, ) nd (4, 4) revolve bout the xxis. Find the re of Exmple 4.3. The function y = x for x represents the upper hlf circle centered t the origin with rdius. The rc is rotted bout the xxis to form sphere of rdius. Find the surfce re of this sphere. Solution. By the chin rule, the derivtive is d y d x = x. Then by the x formul of surfce re, we hve S = = π x π x + d x = π x d x = π = 4π x x d x This exmple shows tht the surfce re of sphere of rdius is 4π.
52 4. SURFACE AREA OF REVOLUTION 5 Additionl Exmple 4.4. Rotte the rc of curve f (x) = e x between x = nd x = bout the xxis. Find the surfce re of the resulting solid. Solution. The derivtive of f (x) is f (x) = e x. Then by the formul of surfce re, we hve S = πe x + (e x ) d x Set up substitution u = e x with du = e x d x nd pply formul (3.4). We hve πe x + (e x ) d x = + u du = π e x = π u u + + ln u + + C e x + + ln ex + + C Evlute the ntiderivtive over [, ]. We hve the surfce re: S = e = π πe x + (e x ) d x = π e x + e 4 + ln(e4 + ) e e x + + x= ln ex + + e ln(e + ) x=
53 Chpter 5 Prmetric curves nd polr coordintes 5. Curves defined by prmetric equtions Prmetric curves So fr we hve used functions of the form y = f (x) to describe curves in the x yplne. However, to be function, the curve hs to pss the verticl line test, which mens tht for ech xvlue, there cn be only t most one yvlue. For exmple, circle in the x yplne is such curve tht cn t be represented by function. Insted, we hve to express the circle s two functions: the upper hlf circle nd lower hlf circle. Now we re going to describe curves described s prmetric equtions, which will enble us to describe common nd exotic curves. Given curve in the x yplne, we re going to be tret the curve s the trjectory of moving prticle with respect to the time vrible t. We know tht every point in the x yplne is represented by pir (x, y). Thus t ny given time t, the xvlue nd yvlue of the loction of the prticle is function of time t. Thus we cn express the curve s the following form x = x(t), y = y(t), for t in the domin This is clled prmetric curve. The two functions x(t) nd y(t) re clled prmetric equtions with prmeter t. In other words, we decompose the motion of the prticle on the plne s the composition motion long xdirection nd motion long ydirection. We cn put the coordinte functions in n ordered pir nd write it s vector function r (t) = (x(t), y(t)) Exmple 5.. Consider the following prmetric curve x = t t, y = t +. Ech vlue of t gives point (x, y) on the curve, defined by the bove prmetric equtions. For instnce, when t =, we hve x = nd y = nd so the loction of the prticle t time t = is (, ). To see the curve, we first compute some smple points by choosing some representtive t: t x y So we cn plot the bove points (x, y) on the x yplne. Then join these points to produce curve. As we plot more smple points, the curve joining them will be more precise. As t increses, the moving point (x, y) moves long the curve from the lower right corner to the upper right corner. Thus the curve is oriented. 5
54 5. CURVES DEFINED BY PARAMETRIC EQUATIONS 53 Exmple 5.. Sketch the prmetric curve x = cos t, y = sin t, for t π Pick some representtive tvlues to locte the (x, y) position of the prticle t t. t x y π 4 π 3π 4 π 5π 4 3π 7π 4 π The curve is circle of rdius. We cn ctully see it from the reltion x + y = ( cos t) + ( sin t) = 4(sin t + cos t) = 4 Notice tht s t increses from to π, the point (x, y) moves round the circle counterclockwise from the point (, ) bck to (, ). Exmple 5.3. Consider the following prmetric curves: () x = cos t, y = sin t, t 6π () x = cos t, y = sin t, t π (3) x = sin t, y = cos t, t π These three prmetric curves ll stisfy the squre reltion x + y = 4. So ll the curves on the x yplne re the circle of rdius. Cn you figure out the difference of these prmetric curves? Remrk 5.4. () The curve consists of ll points of coordintes (x, y) s t vries. We cn t directly see the tvlue of points on the curve in the x yplne. () Prmetric curve is oriented. The orienttion is determined by the incresing direction of t. (3) A prmetric curve cn intersect itself, nd even repet prt of its pth. The curve on the x yplne only shows the trce of the motion. Fct 5.5. The stndrd prmetriztion of circle of rdius is x = cos t, y = sin t Over the tintervl [, π], the curve trvels long the circle for one round from the point (, ) to itself, in the counterclockwise direction. Exmple 5.6. A lot of nice curves cn be prmetrized. Here re few exmples. x =.5 cos t cos 3t y =.5 sin t sin 3t x = 6. sin t sin 3.t y = 6. cos t cos 3.t x = sin t +. sin 5t. sin 6.3t y = cos t +. cos 5t +. cos 6.3t x = t cos t y = t sin t
55 5. CURVES DEFINED BY PARAMETRIC EQUATIONS 54 Remrk 5.7. () Every function f (x) cn be treted s nturl prmetric curve, by declring x(t) = t nd y = f (t). () For some prmetric equtions, we cn try to eliminte t nd obtin direct Crtesin reltion between vribles x nd y. For exmple, for the prmetric curve x = t t, y = t we cn use the y expression y(t) = t to express t s t = y + nd then put this reltion in the xexpression to eliminte t: x = (y + ) (y + ) = y + y This is now reltion between x nd y. But not ll prmetric curves cn be written s Crtesin reltions since isolting t from either equtions my be impossible, or t hs to be written s mny brnches. For exmple, we cn t eliminte t in the prmetric equtions x = t + sin t, y = t + cos t to get the Crtesin reltion. Exmple 5.8. Eliminte the prmeter t to find Crtesin eqution of the curve: x = t, y = t Exmple 5.9. Eliminte the prmeter t to find Crtesin eqution of the curves: Wht is the difference between these curves? () x = t, y = t, () x = sin t, y = sin t Velocity nd tngent of prmetric curves For prmetric curve defined by x = x(t) nd y = y(t), the motion of the prticle is decomposed into the motion long xdirection nd motion long yxis. So the derivtives d x nd d y will represent the velocity long the xdirection nd d t d t ydirection. Then the velocity of the prticle is lso n ordered pir, denoted by d r d t : which is lso clled the vector derivtive. velocity: v (t) = d r d t d x = d t, d y d t Exmple 5.. Suppose we re given the prmetric curve for r (t) = (t, t 3 ) Find the point nd velocity of the curve t time t =. Solution. The coordinte functions re x(t) = t nd y(t) = t 3. At time t =, the position of the prticle is r () = (x(), y()) = (, ) The vector derivtive is At time t =, the velocity is d r d t d x d t = t, t= d y d t = 6t = (, 6 ) = (, 6) Remrk 5.. Recll tht the instntneous velocity is pproximtely the verge velocity over smll time chnge. In the bove exmple, if the prticle moves within time intervl t = nd t =.. Then the initil nd terminl loction is r () = (x(), y()) = (, ), r (.) = (x(.), y(.)) = (.,.66) Then the displcement long xdirection is x = x(.) x() =. nd displcement long yxis is y = y(.) y() =.66
56 5. CURVES DEFINED BY PARAMETRIC EQUATIONS 55 Then the verge coordinte velocity from t = nd t =. is velocity long xdirection = x t =.. =. velocity long ydirection = y t =.66. = 6.6 The verge velocity nd instntneous velocity d r d t () = (, 6) re quite close. Definition 5.. By Pythgoren thoerem, the speed of the motion is defined s speed = v (t) = d r d t = d x d t + d y d t The speed is sclr function of t. The other concept relted to prmetric curve, is the direction of the velocity, which is exctly the slope of the tngent line of the curve in the x yplne. The slope of the tngent line is the limit of rtio of y nd x. Thus we hve slope of tngent line: d y d x = d y/d t d x/d t Exmple 5.3. (Exmple. 5. continued.) Compute the speed, slope nd eqution of the tngent line of the curve t t =. Solution. We hve found tht the vector velocity t t = d r d x =, d y = (, 6) d t d t d t t= t= t= Then we hve by definition speed t t = : v() = (, 6) = + 6 = 4 = d y slope t t = : = d y/d t = 6 d x d x/d t = 3 t= Then by the pointslope eqution of stright line, we hve the eqution of the tngent line t point (, ) with slope 3: t= tngent line t t = : y = 3(x ) Exmple 5.4. Show tht the following circulr prmetric curves hve constnt speed. () r (t) = (cos t, sin t), t π, () r (t) = (cos t, sin t), t π Then observe tht the second curve trvels t twice the speed s the first curve. Exmple 5.5. A prmetric curve is defined s r (t) = (t, t 3 3t) () Find the point, velocity, speed nd the eqution of tngent line t this point when t =. () Find when the prticle psses through the point (3, ), nd compute the equtions of tngent line. (3) Find the points on the curve where the tngent line is horizontl.
57 5. CURVES DEFINED BY PARAMETRIC EQUATIONS 56 Arc length of prmetric curve Recll tht in section., we use the trick of slicendsum to derive the rc length formul. The ide of to brek the rc into pieces, nd pproximte the rc length over ech subintervl by line segment. If we divide the tintervl of the prmetric curve [, b] into n equl subintervls of equl width t. Cll the endpoints of the subintervls t =, t, t,, t n = b, nd corresponding points on the curve P i = (x(t i ), y(t i )) for i =,,, n. Then by Pythgoren theorem, the length of the segment joining consecutive points is P i P i = xi yi ( x i ) + ( y i ) = + ( t) = t t x i t yi + t t The totl length of the polyline is the sum, nd s n + it will converge to the ctul rc length between t = nd t = b: n x i b yi d x d y Arc length = lim + t = + d t n + t t d t d t i= Theorem 5.6. t = to t = b is Exmple 5.7. If prmetric curve is described s r (t) = (x(t), y(t)) for t [, b]. Then the rc length of the curve from b d x d y L = + d t d t d t () The circle of rdius cn be described by prmetric curve r (t) = ( cos t, sin t), t π Then its rc length (perimeter) is L = π ( sin t) + ( cos t) d t = π (sin t + cos t)d t = = π d t = t t=π t= = π This proves the wellknow perimeter formul of circle. () The grph of the prmetric curve r (t) = ( cos t, sin t) for t 4π on the x yplne is lso circle of rdius. But the rc length is 4π L = ( sin t) + ( cos t) d t = t t=4π = 4π t= This integrl gives twice the rc length of circle, becuse s t increses from to 4π, the point ( cos t, sin t) trverses the circle TWICE. This is why the rc length is twice the perimeter. Note 5.8. Summry of concepts of prmetric curve. Ctegory Formul Exmple Prmetric curve r (t) = (x(t), y(t)) Curve: r (t) = (t, t 3 ) Point/position t time t (x(t), y(t)) At t = : position r () = (, ) Derivtive/vector velocity t time t d r v (t) = = ( d x d t d t, d y d t ) At t = : velocity v () = (, ) Speed t time t d x d y v (t) = + At t = : speed v () = 5 d t d t d y Slope of tngent/direction of motion d x = d y/d t At t = : slope m = d y d x/d t d x = t= b d x d y Arc length L = + d t Between t = nd t = : L = 4 + 9t d t d t d t
58 5. POLAR COORDINATES Polr coordintes Wht we hve encoutered is the Crtesin (rectngulr) coordinte system. Any point in plne cn be uniquely represented by pir of numbers (x, y), where x nd y re the projection of the point (x, y) on the xxis nd yxis. Directed ngles We cll the xxis the polr xis nd tke it s the initil side of n ngle. Then we rotte the initil side round the origin clockwise or couterclockwise. When the side stops moving, we cll it the terminl side, nd we obtin n ngle from the initil side nd terminl side. The orienttion of the resulting ngle is defined s direction of rottion. We sy n ngle is positive, if the initil side is rotted counterclockwise, nd n ngle is negtive, if the initil side is rotted clockwise. Positive ngle Negtive ngle How to mesure the size of the resulting ngle? We drw the unit circle centered t the origin. The genertion of the ngle corresponds to the motion of point from (, ) round the unit circle. Then the size of the ngle is defined s the trveling circulr length of the moving point. The unit for the ngle is clled rdins. We know tht the circulr length of whole unit circle is π. Thus we define the rdin mesure by π rdins = 8 From this reltion, the degree mesure of some typicl ngles cn be converted to rdin mesure. Rdins π 6 π 4 π 3 π π 3 3π 4 5π 6 Degrees π 7π 6 5π 4 4π 3 3π 5π 3 7π 4 π 6 π Fct 5.9. An obvious nd importnt fct is tht ll ngles with the sme terminl side differ by multiple of π. For exmple, ngles, π, 6π, 4π ll hve the sme terminl side, nd ll ngles π 4, 9π 4, 7π 4 hve the sme terminl side. The following figures show some directed ngles.
59 5. POLAR COORDINATES 58 polr coordinte system Like Crtesin coordintes, polr coordintes re used to locte points in the plne. Ech point in the plne will be represented by pir of numbers (r, θ). Let s now introduce the polr coordintes. Given ny two rel numbers r nd θ, we first sketch ry representing the directed ngle θ from the origin, where the initil side is still chosen to be the positive xxis. Then long the ry of terminl side of θ, we move for r units. The destintion is the point represented by the pir (r, θ). For the ordered pir (r, θ), θ is clled the ngulr coordinte, nd r is clled the rdil coordinte, becuse r mesures the distnce of the point from the origin. Exmple 5.. Here re exmples of points in polr coordintes. Remrk 5.. A mjor difference from the rectngulr Crtesin coordintes, is tht every point in the plne cn be uniquely represented by pir (x, y) of xvlue nd yvlue. But in polr coordintes, ech point ctully cn be represented by lot of combintions of r nd θ. For fixed r, polr coordintes (r, θ) nd (r, θ + integer multiple of π) represent the sme point. This cn be esily seen becuse ngles differing by n integer multiple of π hve the sme terminl side. For n ngle θ, polr coordintes (r, θ) nd ( r, θ + π) represent the sme point. Notice tht we llow the rdil coordinte to be negtive. When it is negtive, we ctully extend the ry of the terminl side of θ, nd wlk long the opposite direction of the terminl side by r units. So we my sy tht the rdil coordinte is lso directed. If r is positive, the point lies in the positive direction of the terminl side of ngle θ, nd if r is negtive, the point lies in the negtive direction of the terminl side of the ngle θ. Thus the wlking long the opposite of direction of θ by r units is the sme s wlking long the direction of θ + π by ( r) units. Exmple 5.. Here re exmples of polr coordintes representing the sme point in the plne. Clss Exmple 5.3. Plot the following points in polr coordintes: A(, 5π 4 ), B(, π 3 ), C(, 3π )
60 5. POLAR COORDINATES 59 Conversion between Crtesin nd polr coordintes We often need to convert between Crtesin nd polr coordintes. The conversion equtions emerge when we look t the generl definition of the trigonometric functions cos θ = x r, sin θ = y r, tn θ = y x for directed ngles in circle. Bsed on these reltions, we my derive the conversion formuls. Formul 5.4. Conversion between Crtesin nd polr coordintes Polr coordintes to Crtesin coordintes: If point hs polr coordintes (r, θ), then its Crtesin coordintes re x = r cos θ, y = r sin θ Crtesin coordintes to Polr coordintes: If point hs Crtesin coordintes (x, y), then polr coordintes stisfy r = x + y, tn θ = y x Then the sign of r nd the vlue of θ cn be determined by the qudrnt where the point (x, y) lies. Exmple 5.5. () Express the point (, π ) with polr coordintes in Crtesin coordintes. 3 () Express the point with Crtesin coordintes (, ) in polr coordintes. Solution. () We identify r = nd θ = π 3. Then the Crtisin coordintes re x = r cos θ = cos( π 3 ) = = y = r sin θ = sin( π 3 ) = ( 3 ) = 3 Therefore in Crtesin coordintes, this point is represented by (, 3). () From the Crtesin coordintes x = nd y =, we cn compute the reltions r = x + y = + ( ) =, tn θ = y x = = Notice tht the point lies in the fourth qudrnt. So we cn tke combintion of positive rdius r nd n ngle θ in the fourth qudrnt. Thus we my tke r =, θ = π 4 which is one polr representtion of point (, ). Also, by Remrk 5., there re infintely mny polr representtions of this point. representtions lso describe the sme point (, ): For instnce, the following polr r =, θ = 7π 4, or r =, θ = 3π 4 Polr curves In Crtesin coordintes, function is dependent reltion y = f (x), which cn be represented by curve in the x yplne. In polr coordintes, the function r = f (θ) between coordintes r nd θ lso gives curve. We cll this polr curve. More generlly, ny eqution between the coordintes r nd θ is lso clled polr curve. To see the polr curve defined by the function r = f (θ), we notice tht for every directed ngle θ, there is point defined on the ry of the terminl side θ. The rdil distnce r of this point from the origin is exctly defined by the function vlue of f (θ). In other words, the point in polr coordintes t ngle θ is (f (θ), θ). Let s first look t two extreme exmples. Recll tht in Crtesin coordintes, constnt function y = c represents horizontl line with height c, nd constnt eqution x = c represents verticl line with xvlue c. How bout the constnt functions in polr coordintes?
61 5. POLAR COORDINATES 6 Clss Exmple 5.6. Sketch the following curves in polr coordintes: () r = ; () θ = π 3. From the bove exmple, we hve the following fct bout the constnt functions in polr coordintes. Fct 5.7. In polr coordintes: constnt function r = c represents the circle of rdius c centered t the origin. constnt function θ = c represents stright line through the origin with slope ngle θ. Exmple 5.8. Sketch the polr curve =r = + sin θ. Solution. To plot the curve, like the prmetric curves, we cn first plot few smple points on the curve by picking some typicl ngles θ: θ sin θ π 6 π 4 π π 3 3 π 3 3 3π 4 5π 6 π 7π 6 5π 4 4π 3 3π 3 5π 3 3 7π 4 π 6 π r = + sin θ Now we plot these polr points in the x yplne. Then connect these points with ngle θ incresing. Then we obtin the grph of the polr curve. Notice tht before nd fter the rngle θ π, sin θ will repet. So the points curve will repet s well. So we sy the curve hs period π. Remrk 5.9. Insted of plotting discrete points nd link them to the curve, if we only need rough sketch of curve, we cn first sketch the grph of the function y = + sin x in the Crtesin coordintes, treting θ s the horizontl xis x nd r s the verticl xis y. This will enble us to red t glnce the chnge of r tht correspond to incresing vlues of ngle θ. For instnce, s θ vries from to π, the rdius r increses from to 4. Then we cn drw the piece of curve in the first qudrnt with rdius chnging from to 4 s ngle θ increses.
62 5. POLAR COORDINATES 6 Exmple 5.3. Here re some interesting polr curves. Let s see more exmples to get fmilir with sketching polr curves. Exmple 5.3. Sketch the polr curve r = cos θ nd determine its miniml period. The function y = cos θ shown below. Exmple 5.3. Sketch the polr curve r = cos θ. The function y = cos x in Crtesin coordintes is shown s follows:
63 5.3 CALCULUS IN POLAR COORDINATES 6 Additionl Exmple Sketch the polr curve r = + sin(3x). Solution. The function y = + sin(3x) is shown here. Then we cn sketch the polr curve. 5.3 Clculus in polr coordintes Now we hve the groundwork needed to explore clculus in polr coordintes. Fmilir topics, such s slopes of tngent lines, rc length, nd re bounded by curves, cn be revisited in different setting in polr coordintes. Slope of tngent line We re going to use the knowledge of clculus of prmetric curves tht re estblished in section 5., becuse polr curve cn be treted s prmetric curve under the conversion from polr coordintes to Crtesin coordintes. Indeed, if r = f (θ) is polr curve defined for θ in n intervl [, b], then we cn use θ s the prmeter nd express the curve s x(θ) = r cos θ = f (θ) cos θ, y(θ) = r sin θ = f (θ) sin θ, for θ b From the definition 5. of slope of tngent line in prmetric curves. We know tht t given point θ, the slope cn be expressed s d y d x = d y/dθ d x/dθ To compute the derivtives of x nd y with respect to θ, we need to pply the product rule. Exmple For the polr curve r = + sin θ, find the slope of the tngent line t ngle θ = π 3. Solution. Convert polr curve to Crtesin prmetric curve using prmeter θ: x(θ) = ( + sin θ) cos θ, y(θ) = ( + sin θ) sin θ At ngle θ = π 3, the corresponding point in Crtesin coordintes is x = ( + sin π 3 ) cos π 3 = 3 ( + ) = y = ( + sin π 3 ) sin π = ( + ) = By product rule, their derivtives with respect to θ is d x d t = cos θ cos θ + ( + sin θ) ( sin θ) = cos θ sin θ sin θ d y = cos θ sin θ + ( + sin θ) cos θ = sin θ cos θ + cos θ d t
64 5.3 CALCULUS IN POLAR COORDINATES 63 At the ngle θ = π 3, the slope of tngent line is d y d x θ= π 3 = d y/dθ d x/dθ Thus the eqution of tngent line is θ= π 3 sin θ cos θ + cos θ = cos θ sin θ sin θ y θ= π 3 = 3 ( ) = x ( 3 ) = 3 4 ( 3 + ) ( 3 + ) = Arc length of polr curve Agin, think of the polr curve s prmetric curve: x(θ) = r(θ) cos θ, y(θ) = r(θ) sin θ Then by product rule, the derivtives of x(θ) nd y(θ) with respect to θ re b d x dθ = r (θ) cos θ + r(θ)( sin θ) d y dθ = r (θ) sin θ + r(θ) cos θ By Theorem 5.6 of rc length formul of prmetric curve for θ in [, b], we hve b d x d y b L = + dθ = dθ dθ = (r(θ)) + (r (θ)) dθ This gives the rc length formul for polr curve: (r (θ) sin θ + r(θ) cos θ) + (r (θ) cos θ r(θ) sin θ) dθ Theorem The length of polr curve r = r(θ) for θ b is b d r L = [r(θ)] + dθ dθ Clss Exmple The polr curve r = represents circle of rdius. Compute its perimeter. Exmple in the figure. The polr curve r = θ is spirl curve, s shown in the figure. Compute the length of the piece of rc shown
65 5.3 CALCULUS IN POLAR COORDINATES 64 Additionl Exmple Find the length of the polr curve r = + sin θ. Solution. The polr curve hs been sketched in Exmple 5.8. The miniml period for the curve is π. Thus by the rc length formul, we hve π L = [r(θ)] + [ d r π dθ ] dθ = ( + sin θ) + ( cos θ) dθ = π = π sin θ + 4 sin θ + 4 cos θ dθ = + sin θ dθ π sin θ dθ The remining integrl is difficult nd nonstndrd. So the clcultion is omitted. The integrtion cn be done by π method using hlfngle substitution. The nswer to the integrl + sin θ dθ is 8. Thus the rc length is 6. Are of regions bounded by polr curves In chpter, we hve seen how to compute re of the region under function f (x) over the xintervl [, b]. Here the bounds x = nd x = b represent the two verticl boundries of the region. Now consider wht hppens if we use polr function r = r(θ) for θ in the intervl [, b]. The function r = r(θ) is polr curve, nd the boundry vlues θ = nd θ = b re two rys strting from the origin. Then the polr curve nd the boundry rys will enclose some region. So we wnt to discuss how to compute its re. We divide the ngle rnge [, b] into n equl ngles of width θ, with terminls sides =,,, n, n = b. Then ll these terminl sides will divide the region into n mny slices. We re going to pproximte ech slice s sector of circle. Recll tht the re of section of circle of rdius r nd centrl ngle θ is θ r. In the ith slice, we tke representtive point θ i in the ith ngle rnge [ i, i ], nd use the rdil vlue of the polr curve r(θ i ) to be the representtive rdius of the sector. Then the pproximte re of the ith slice is Approximte re of ith slice = (r(θ i)) θ Then the totl re of ll pproximting slices is the form of Riemnn sum n totl pproximting re = (r(θ i)) θ i= As n +, the sum bove will converge the ctul re of the shded region. And by the convergence theorem of Riemnn sum, we hve the following theorem: Theorem nd θ = b is Given polr curve r = r(θ), the re of the region bounded by the polr curve between ngle bounds θ = A = b [r(θ)] dθ As we cn think of the re under function f (x) between x = nd x = b s being swept out by verticl segment of height f (x) from the verticl edge t x = to the verticl edge x = b, here we cn lso think of the re formul s the ccumultion of rotting ry through the origin inside the polr curve r = r(θ) tht strts with ngle θ = nd ends with ngle θ = b. When we use the re formul, we should crefully determine the integrtion bounds θ = nd θ = b if they re not given. The wy to determine them is to focus on the piece rc r = r(θ) on the boundry of the region, then check the ngle rnge [, b] on which the rc is continuously defined.
66 5.3 CALCULUS IN POLAR COORDINATES 65 Exmple 5.4. Sketch the polr curve r = + cos θ using the Crtesin function y = + cos θ, nd compute the re of the region bounded by the polr curve. Exmple 5.4. Compute the re of the shded region bounded by the polr curve r = cos θ. Similr to the discuss of re between two curves f (x) nd g(x), here we cn lso tlk bout the re of the region bounded by two polr curves r = r (θ) nd r = r (θ). Then it is very esy to derive the following generl result: Theorem 5.4. Let R be the region bounded by the grph of two polr curves r = r (θ) nd r = r (θ), between two ngle bounds θ = nd θ = b with r (θ) r (θ). Then the re of R is b A = [ r (θ) ] [ r (θ) ] dθ outer rdius inner rdius Exmple r =. Compute the re of the shded regions, which re bounded by the curve r = + cos θ nd the unit circle Exmple Compute the re of the shded regions, which re bounded by the curve r = sin θ nd the unit circle r =.
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