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1 Sirindhorn Interntionl Institute of Technology Thmmst University Deprtment of Common nd Grdute Studies Semester: 3/2008 Instructors: Dr. Prpun Suksompong MAS 6: Lecture Notes 7 6 Applictions of the Definite Integrl Riemnn sums nd definite integrls hve pplictions tht extend fr beyond the re problem. The required clcultions cn ll be pproched by the sme procedure tht we used to find res () breking the required clcultion into smll prts, (2) mking n pproximtion for ech prt, (3) dding the pproximtions from the prts to produce Riemnn sum tht pproximtes the entire quntity to be clculted, nd then (4) tking the limit of the Riemnn sums to produce n exct result. 6. Are Between Two Curves Theorem 6.. If f nd g re continuous with f(x) g(x) throughout [, b], then the re of the region between the curves y = f(x) nd y = g(x) from to b is [f(x) g(x)] dx. The condition f(x) g(x) throughout [, b] mens tht the curve y = f(x) lies bove the curve y = g(x) nd tht the two cn touch but not cross. Therefore, the bove integrtion cn be written s [ top bottom ]dx Exmple 6.2. Find the re of the region bounded bove by y = x + 6, bounded below by y = x 2, nd bounded on the sides by the lines x = 0 nd x = 2 s shown in Figure
2 6.3. It is possible tht the upper nd lower boundries of region my intersect t one or both endpoints, in which cse the sides of the region will be points, rther thn verticl line segments. When tht occurs, you will hve to determine the points of intersection to obtin the limits of integrtion. Exmple 6.4. Find the re of the region tht is enclosed between the curves y = x + 6 nd y = x 2 s shown in Figure
3 6.2 Finding Volumes by Slicing 6.5. A right cylinder is solid tht is generted when plne region is trnslted long line or xis tht is perpendiculr to the region (Figure 6.2.3). If right cylinder is generted by trnslting region of re A through distnce h the h is clled the height (or sometimes the width) of the cylinder, nd the volume V of the cylinder is defiend to be (Figure 6.2.4). V = A h = [re of cross section] [height] Theorem 6.6. Finding Volumes by the Method of Slicing: The volume of solid cn be obtined by integrting the cross-sectionl re from one end of the solid to the other. () Consider solid bounded by two prllel plnes perpendiculr to the x-xis t x = n x = b. If for ech x in [, b], the cross-sectionl re of of the solid perpendiculr to the x-xis is A(x), then the volume of the solid is V = A(x)dx. (7) The re A(x) ws obtined by slicing through the solid with plne perpendiculr to the x-xis. Volume of the kth slb V k = A(x k ) x k. (b) Consider solid bounded by two prllel plnes perpendiculr to the y-xis t y = c n y = d. If for ech y in [c, d], the cross-sectionl re of of the solid perpendiculr to the y-xis is A(y), then the volume of the solid is V = d c A(y)dy. 99
4 Exmple 6.7. Derive the formul for the volume of right pyrmid whose ltitude is h nd whose bse is squre with slides of length Cvlieris Principle: Cvlieris principle sys tht solids with equl ltitudes nd identicl crosssectionl res t ech height hve the sme volume. This follows immeditely from the definition of volume, becuse the cross-sectionl re function A(x) nd the intervl [, b] re the sme for both solids. Definition 6.9. The solid generted by rotting plne region bout n xis in its plne is clled solid of revolution. Theorem 6.0. A Solid of Revolution Rottion bout the x-axis: Let f be continuous nd nonnegtive on [, b], nd let R be the region tht is bounded bove by y = f(x), below by the x-xis, nd on the sides by the lines x = nd x = b. The volume of the solid of revolution tht is generted by revolving the region R bout the x-xis is given by V = π [f(x)] 2 dx. (8) 00
5 Becuse the cross sections re disk shped, the ppliction of this formul is clled the method of disks or disk method. Formul (8) follows directly from (7) by setting A(x) = π [f(x)] 2. Exmple 6.. Find the volume of sphere of rdius r. As indicted in Figure 6.2., sphere of rdius r cn be generted by revolving the upper semicirculr disk enclosed between the x-xis nd bout the x-xis. x 2 + y 2 = r 2 V = r r ( ) 2dx π r2 x 2 4 = 3 πr3. Exmple 6.2. Find the volume of the solid tht is obtined when the region under the curve y = x over the intervl [, 4] is revolved bout the x-xis (Figure 6.2.0). 0
6 6.3 Volumes by Cylindricl Shells Some volume problems re difficult to hndle by the slicing method discussed erlier. In this section we will develop nother method for finding volumes A cylindricl shell is solid enclosed by two concentric right circulr cylinders (Figure 6.3.2). Theorem 6.4. Volume by Cylindricl Shells bout the y-xis: Let f be continuous nd nonnegtive on [, b] (0 < b), nd let R be the region tht is bounded bove by y = f(x), below by the x-xis, nd on the sides by the lines x = nd x = b. Then the volume V of the solid of revolution tht is generted by revolving the region R bout the y xis is given by V = 2πxf(x)dx. (9) 6.5. Observe tht we slice through the solid using circulr cylinders of incresing rdii, like cookie cutters (Figure 6.3.3b). We slice stright down through the solid perpendiculr to the x-xis, with the xis of the cylinder prllel to the y-xis. The verticl xis of ech cylinder is the sme line, but the rdii of the cylinders increse with ech slice. In this wy the solid is sliced up into thin cylindricl shells of constnt thickness tht grow outwrd from their common xis, like circulr tree rings. Unrolling cylindricl shell shows tht its volume is pproximtely tht of rectngulr slb with re A(x) nd thickness x (Figure6.3.7). 02
7 VOLUMES BY ; Use grph to estimte the x-coordintes of the points of 45. A right circulr cone with height h nd b intersection of the given curves. Then use this informtion to estimte the volume of the solid obtined by rotting bout the y-xis the region enclosed by these curves. 46. Suppose you mke npkin rings by drill 33. y 0, y x x 2 x 4 dimeters through two wooden blls (wh dimeters). You discover tht both npki 34. y x 4, y 3x x 3 height h, s shown in the figure. CAS Use computer lgebr system to find the exct volume of the solid obtined by rotting the region bounded by the given curves bout the specified line. 35. y sin 2 x, y sin 4 x, 0 x ; bout x y x 3 sin x, y 0, 0 x ; bout x Another (nd more rigorous) The wy region to see bounded tht by (9) the works given curves is to is consider rotted bout the formul of the volume of cylindricl shell directly. the specified The volume xis. Find of the cylindricl volume of the shell resulting withsolid inner by rdius ny r, outer rdius r 2, nd height h (Figure 6.2.2) is method. given by 37. y V = ( x 2 πr2 2 x 2 πr,) y 0; bout rthe + x-xis r 2 h = 2π h (r 2 r ) 38. y x 2 3x 2, y 0; bout } the {{ 2 y-xis }}{{} thickness verge rdius 39. y 5, y x 4x; bout x The corresponding Riemnn 40. x sum tht y 4, xpproximtes 0; bout x the 2 volume is given by 4. x 2 y 2 ; bout the y-xis n 42. x 2 y 2 V ; bout 2πc the k f(c x-xis k ) x k k= () Guess which ring hs more wood in (b) Check your guess: Use cylindricl sh volume of npkin ring creted by d rdius r through the center of sphe express the nswer in terms of h. 47. We rrived t Formul 2, V x b 2xf cylindricl shells, but now we cn use in prove it using the slicing method of Sec cse where f is one-to-one nd therefore tion t. Use the figure to show tht V b 2 d 2 c y d where we choose c k to be the midpoint Use cylindricl of theshells intervl to find [xthe k volume, x k ]: (x of k the + solid. x k )/2. Mke the substitution y f x nd then 43. A sphere of rdius r prts on the resulting integrl to prove th Exmple 6.6. Use cylindricl shells to find the volume of the solid torus ( donut-shped solid with rdii r nd R) shown in44. thethe figure solid below). torus ( donut-shped solid with rdii R nd r) shown in the figure y x=g(y) y=ƒ d h c R r c x= 0 b 03
8 7 Inverse Trigonometric Function Inverse trigonometric functions rise when we wnt to clculte ngles from side mesurements in tringles. They lso provide useful ntiderivtives nd pper frequently in the solutions of differentil equtions. 7.. The six bsic trigonometric functions re not one-to-one (their vlues repet periodiclly). However we cn restrict their domins to intervls on which they re one-to-one. Figure 7.7. nd Tble 7.7. summrize the bsic properties of three importnt inverse trigonometric functions. The - in the expressions for the inverse mens inverse. It does not men reciprocl. For exmple, the reciprocl of sin x is (sin x) = csc x. 04
9 Observe tht the inverse sine nd inverse tngent re odd functions: However, sin ( x) = sin (x) nd tn ( x) = tn (x). cos ( x) = π cos x. The Arc in Arc Sine nd Arc Cosine: If x = sin θ then, in ddition to being the ngle whose sine is x, θ is lso the length of rc on the unit circle tht subtends n ngle whose sine is x. So we cll θ the rc whose sine is x Inverse Function-Inverse Cofunction Identities [3][p 527]: sin x + cos x = π 2 tn x + cot x = π 2 sec x + csc x = π 2 Therefore, the derivtive of n inverse cofunction is the sme s the derivtive of corresponding inverse function with n extr negtive sign If we interpret sin x (or cos x, or tn x) s n ngle θ in rdin mesure whose sine (or cosine, or tngent, respectively) is x, nd if tht ngle is between 0 nd π/2 (or equivlently, x is positive), then we cn represent θ geometriclly s n ngle in right tringle s shown below. Such tringle revels useful identities: cos ( sin x ) = sin ( cos x ) = x 2 (20) tn ( sin x ) = tn (cos x) = x x 2 (2) sin ( tn x ) x = + x 2 (22) cos ( tn x ) = + x 2 (23) It turns out tht the identities bove re lso vlid when x is negtive. For exmple, when < x < 0, we know, from (20), tht cos ( sin ( x) ) = ( x) 2 becuse 0 < x <. The we cn simplify the LHS to be cos ( sin ( x) ) = cos ( sin (x) ) = cos ( sin (x) ). So, we get bck (20) but now for < x < 0. Note tht we lso hve to directly check tht the identity is vlid for x = 0 nd t the two endpoints. Similr resoning extends the rest of the identities. Note tht the tringle technique does not lwys produce the most generl form of n identity. For exmple, it does not give in sin ( sec x ) x2 =. x 05
10 Exmple 7.4. A derivtive formul for sin x cn be obtined vi implicit differentition (Section 3.9) or the Formul (4) for the differentition of inverse function. To use implicit differentition, we rewrite the eqution y = sin x s x = sin y nd differentite with respect to x. We then obtin d dx [x] = d [sin y] dx = cos y dy dx dy dx = cos y = cos ( sin x ), which we cn lterntively obtin directly vi Formul (4). This derivtive formul cn be simplified using (20). This yields dy dx = x 2 The method used to derive this formul cn be used to obtin derivtive formuls for the remining inverse trig. functions. Theorem 7.5. Derivtives of the inverse trigonometric functions: d dx sin x =, x 2 ( x < ) d dx cos x =, x 2 ( x < ) d dx tn x = + x 2. Theorem 7.6. Integrtion Formuls: ( 2 x dx = x 2 sin 2 + x 2 dx = ( x tn ) + C, (x 2 < 2 ) ) + C. The formuls re redily verified by differentiting the functions on the right-hnd sides. Exmple 7.7. Completing the Squre: dx = dx = 8x x 2 6 (x 4) 2 u=x 4 ( 42 u du = u ) ( ) x 4 2 sin + C = sin + C Additionl Integrtion Techniques 8. Integrtion by Prts Integrtion by prts is technique for simplifying integrls of the form f (x) g (x)dx. 8.. Integrtion by Prts: f (x) g (x)dx = f (x) g (x) f (x) g (x)dx. (24) 06
11 Sometimes it is esier to remember the formul if we write it in differentil form. Let u = f(x) nd v = g(x). Then du = f (x)dx nd dv = g (x)dx. Using the Substitution Rule, the integrtion by prts formul becomes udv = uv vdu (25) To see (24), strt with the product rule: (f(x)g(x)) = f(x)g (x)+f (x)g(x). Then, integrte both sides. The min gol in integrtion by prts is to choose u nd dv to obtin new integrl tht is esier to evlute then the originl. In other words, the gol of integrtion by prts is to go from n integrl udv tht we dont see how to evlute to n integrl vdu tht we cn evlute. In generl, there re no hrd nd fst rules for doing this; it is minly mtter of experience tht comes from lots of prctice. Note tht when we clculte v from dv, we cn use ny of the ntiderivtive. In other words, we my put in v + C insted of v in (25). Hd we included this constnt of integrtion C in (25), it would hve eventully dropped out. This is lwys the cse in integrtion by prts We hve four possible choices when pplying integrtion by prts to n integrl of the form f (x) g (x)dx : () Let u = nd dv = f(x)g(x)dx. This choice won t do becuse we hve to find v = f(x)g(x)dx which is the sme s the originl integrl. (b) Let u = f(x) nd dv = g(x)dx (c) Let u = g(x) nd dv = f(x)dx (d) Let u = f(x)g(x) nd dv = dx Therefore, we hve three useful choices. Remember tht we must be ble to redily integrte dv to get v in order to obtin the right side of the formul (25). If the new integrl on the right side is more complex thn the originl one, try different choice for u nd dv. Exmple 8.3. Use integrtion by prts to evlute x cos xdx Evluting Definite Integrls by Prts: For definite integrls, the formul corresponding to (24) is b f (x) g (x)dx = f (x) g (x)] b f (x) g (x)dx. (26) The corresponding u nd v nottion is udv = uv] b vdu (27) 07
12 It is importnt to keep in mind tht the vribles u nd v in this formul re functions of x nd tht the limits of integrtion in (27) re limits on the vrible x. Sometimes it is helpful to emphsize this by writing (27) s udv = uv] b x= vdu (28) x= Exmple 8.5. Evlute the following integrls. () xe x dx x= (b) 3 2 ln(t)dt (c) 0 tn (s)ds Exmple 8.6. x n e x dx = xn e x n x n e x dx 8.7. Tbulr Integrtion by Prts: Repeted ppliction of integrtion by prts gives n f (x) g (x)dx = f (x) G (x) + ( ) i f (i) (x) G i+ (x) + ( ) n where f (i) (x) = di dx i f (x), G (x) = g (x)dx, nd G i+ (x) = G i (x)dx. To see this, note tht i= f (x) g (x)dx = f (x) G (x) f (x) G (x) dx, f (n) (x) G n (x) dx (29) nd f (n) (x) G n (x) dx = f (n) (x) G n+ (x) f (n+) (x) G n+ (x) dx. 08
13 A convenient method for orgnizing the computtions into two columns is clled tbulr integrtion by prts. Differentite n n n n n f x g x f x G x f x G x f x G x f x G x n Integrte Exmple 8.8. Evlute To see this, xnote 2 sin(x)dx tht f x g xdx f xg x f xg xdx, nd n n n f x Gn x dx f x Gn x f xgn xdx. 2 2 x e dx x x e x 2 3x x x sin xe dx - 3 sin x e + x x 3x sin x cos xe sin xe dx 2 e + 9 Integrls like the one in the next exmple requires two integrtions sin cos x x xe by prts, followed by solving for 3x the unknown integrl. 0 - e Exmple 8.9. Evlute e x sin(x)dx x e n x e dx x e 2 n x n x n x (Integrtion by prts). 0 t dt t dt,,,, 2 3x So, the integrtion of the function is the test cse. In fct, t Exmple 8.0. If n is positive integer, () n = : ex (b) n = 2 : ex ( ) x ( x 2 2 x ) x n e x dx = ex x 2x n k=0 + e e 27 3x dt dt t t. 0 ( ) k n! k (n k)! xn k. Exmple 8.. Let f be twice differentible with f(0) = 6, f() = 5, nd f () = 2. Evlute the integrl 0 xf (x)dx sin x cos x + - e e x x 09
14 8.2 Methods for Approching Integrtion Problem 8.2. There re three bsic pproches for evluting unfmilir integrls: () Technology: Computer Algebr Systems (CAS) such s Mthemtic, Mple, nd Derive cn be used to evlute n integrl, if such system is vilble. (b) Tbles: Prior to the development of CAS, scientists relied hevily on tbles to evlute difficult integrls rising in pplictions. Such tble were compiled over mny yers, incorporting the skills nd experience of mny people. Extensive tbles pper in compiltions such s CRC Mthemticl Tbles, which contin thousnds of integrls. (c) Trnsformtion Methods: Trnsformtion methods re methods for converting unfmilir integrls into fmilir integrls. None of these methods is perfect. () CAS often encounter integrls tht they cnnot evlute nd they sometimes produce nswers tht re unnecessrily complicted. (b) Tbles re not exhustive nd hence my not include prticulr integrl of interest. (c) Trnsformtion methods rely on humn ingenuity tht my prove to be indequte in difficult problems. 8.3 Improper Integrls Our min objective in this section is to extend the concept of definite integrl to llow for () infinite intervl of integrtion nd (2) integrnds with verticl symptotes within the intervl of integrtion. Definition 8.3. Integrls over Infinite Intervls: Integrls with infinite limits of integrtion re improper integrls of Type I. () If f(x) is continuous on [, ) then (b) If f(x) is continuous on (, b] then f(x)dx = lim b f(x)dx = lim f(x)dx. f(x)dx. In ech cse, if the limit is finite we sy tht the improper integrl converges nd tht the limit is the vlue of the improper integrl. If the limit fils to exist, the improper integrl diverges, nd it is not ssigned vlue. Exmple 8.4. Evlute x dx. 2 0
15 Exmple 8.5. Evlute 0 +x dx nd 0 2 +x 2 dx. Exmple 8.6. Gmm function: Γ (n) = x n e x dx = (n )! ; n > 0. Integrtion by prts show tht Γ(n + ) = nγ(n). 0 Definition 8.7. Integrls over Infinite Intervls (con t): If f(x) is continuous on (, ) then f(x)dx = c f(x)dx + c f(x)dx where c is ny rel number. The improper integrl is sid to converge if both terms converge nd diverge if either term diverges. It cn be shown tht the choice of c is unimportnt Any of the integrls in Definitions 8.3 nd 8.7 cn be interpreted s n re if f 0 on the intervl of integrtion. If f 0 nd the improper integrl diverges, we sy the re under the curve is infinite. Exmple 8.9. Evlute +x 2 dx. Definition Integrnds with Verticl Asymptotes (Infinite Discontinuity): Integrls of functions tht become infinite t point within the intervl of integrtion re improper integrls of Type II. If f(x) is continuous on (, b] nd is discontinuous t then f(x)dx = lim c + If f(x) is continuous on [, b) nd is discontinuous t b then f(x)dx = lim c b c c f(x)dx. f(x)dx. In ech cse, if the limit is finite we sy the improper integrl converges nd tht the limit is the vlue of the improper integrl. If the limit does not exist, the integrl diverges.
16 If f(x) is continuous on the intervl [, b] except for discontinuity t point c in (, b), then f(x)dx = c f(x)dx + c f(x)dx. The integrl on the left side of the eqution converges if both integrls on the right side converge; otherwise it diverges. Exmple 8.2. Evlute the following integrls () 0 x dx 2 (b) 0 x dx Exmple The function /x is the boundry between the convergent nd divergent improper integrls with integrnds of the form x α : 0 t α dt = { α+, α >, α nd In fct, 0 t dt = t dt =. t α dt = { α+, α <, α. 2
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