4.8 Improper Integrals


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1 4.8 Improper Inegrls Well you ve mde i hrough ll he inegrion echniques. Congrs! Unforunely for us, we sill need o cover one more inegrl. They re clled Improper Inegrls. A his poin, we ve only del wih inegrls of he form f(x) dx Before we lk bou he improper ype, le s ry o build up o i. Consider he inegrl x 2 dx Evluing he inegrl, we ge x dx = 2 x = + = Does his work for ny >?. x dx = 2 2 x = 2 + = 2 2. x dx = 2 x = + = 9 83
2 x dx = 2 x x dx = 2 x x dx = 2 x = + = 99 5 = + = 999 = 5 + = I ppers when b ges very lrge, he vlue of he inegrl pproches. Noice h + = Wh his sys is s ges very lrge, sy x 2 dx hen he vlue of h inegrl is exremely close o. And s, he vlue of he inegrl converges o. 84
3 4.8. Definiion of n Improper Inegrl, Type I. If f(x) dx exiss for every number, hen f(x) dx f(x) dx 2. If f(x) dx exiss for every number b, hen f(x) dx f(x) dx 3. If f(x) dx or we cll hem divergen. f(x) dx exis, hey re clled convergen. If hey do no exis, Exmple Find x dx Recll, we sw x 2 dx =. Le s see bou x. 85
4 x dx = x dx ln(x) ln() ln() ln() = This is ineresing. We chnged he degree of he denominor slighly, nd now he inegrl diverges. The nurl quesion his poin would be, wh does p hve o be so dx exiss? xp I m wrning you hed of ime h we do bi of lgebr hocus pocus here. Nohing oo bd hough. 86
5 dx xp x dx p x p dx [ x p+ p + p x p [ ] p p If p >, hen p >, so p. Therefore, we hve. p x dx = p p Wh hppens when p <? We lredy know wh hppens when p =. mkes If p <, hen p <. This mkes p s. Therefore, which p dx = (divergen) xp If you don see why p when, give yourself n exmple. Le p =.5..5 /2 = /2 87
6 Which mens,.5.5 = This priculr inegrl will be exremely imporn for he res of he semeser. memorize i! So Exmple xe x dx Noe, ll of hese inegrls will hve is. Mny of hem will require L Hospils Rule. xe x dx xe x dx And...his one requires Inegrion by Prs. () Le u = x, dv = e x dx (b) So du = dx, v = e x 88
7 xe x dx xex xex e x e x dx ( e e ) (e e ) e e (c) We know e = (d) We need o find e e LH = e e (e) So our finl nswer is xe x dx e e = Seems like we re going o hve fun in his secion! 2. x + x 2 + 2x dx. Le s se i up wih he i noion. x + x 2 + 2x x + dx x 2 + 2x dx 89
8 For now, we ll drop he unil fer we nidifferenie. Also, he inegrl requires bsic usubsiuion. () Le u = x 2 + 2x (b) du = 2x + 2 dx 2 du = x + dx (c) Chnge he bounds If x =, hen u = If x =, hen u = 3 (d) Subsiue! x + x 2 + 2x dx = u du = 2 ln u Le s ge h i bck in u du ln +2 u 2 3 ln ln 3 = ln 3 = 3. dx (x + 3)(x + 4) We need o use pril frcions. 9
9 (x + 3)(x + 4) = A x B x + 4 () Sr by muliply boh sides by (x + 3)(x + 4) (b) Disribue nd collec like erms = A(x + 4) + B(x + 3) (c) Mch he coefficiens = (A + B)x + (4A + 3B) A + B = 4A + 3B = (d) Solving his syem, we ge A = nd B = (e) Rewrie he inegrl x + 3 x + 4 dx x + 3 x + 4 dx [ln x + 3 ln x + 4 ] ln x + 3 x + 4 ln ln 3 4 = ln 3 4 = ln 4 3 9
10 4. + x 2 dx We don know how o do his when boh bounds re ±. Recll he following propery of inegrls where < c < b f(x) dx = c f(x) dx + c f(x) dx We use his propery now. Choose number ( nice one) beween (, ). How bou? Rewrie he inegrl + x dx = 2 + x dx x 2 dx Le s do ech inegrl seprely. () + x 2 dx dx + x2 + x dx 2 n (x) n () n () = ( π/2) = π/2 (b) + x 2 dx 92
11 dx + x2 + x dx 2 n (x) n () n () = π/2 = π/2 (c) Finl Answer dx = π/2 + π/2 = π + x2 We conclude he ype of inegrl where is bound. Now we move on o he second ype of improper inegrls Type 2 Improper Inegrls This ype of improper inegrl involves inegrls where bound is where vericl sympoe occurs, or when one exiss in he inervl.. If f is coninuous [, b) bu disconinuous b, hen, f(x) dx b f(x) dx Here s wh his ype looks like 93
12 2. If f is coninuous (, b] bu disconinuous, hen, f(x) dx + f(x) dx Here s wh his ype looks like 3. If f hs disconinuiy x = c, where < c < b hen, f(x) dx = c f(x) dx + c f(x) dx Ech inegrl is compued using () nd (2). Here s wh his ype looks like 94
13 Exmple Find dx x 2 dx This inegrl is improper becuse we hve n infinie disconinuiy (sympoe) x =. So we use () from bove. dx x 2 dx dx dx x 2 sin (x) sin () sin () = π/2 = π/2 You cn see he process is very much like he previous inegrls. Exmple Find dx x ln x 95
14 We ll sr by using usubsiuion. Le u = ln x, which gives us du = x inegrl hs bounds, we ll do he chnge of bounds now. dx. Since he u() = ln() = u(2) = ln(2) Le s ge sred wih he inegrion. dx x ln x = ln 2 + u du ln 2 u du 2 ln u ln + ln ln 2 ln + = ln 2 () = Exmple Find x 2 dx Suppose you didn check o see if here were ny infinie disconinuiies. Le s see wh we ge. 96
15 x 2 dx = x 2 = 2 + = 3/2 However, if you use he pproprie improper inegrl mehod, you ll find his is incorrec. You sr by finding he disconinuiy, which exiss x =. You hen brek i up ino wo sepre inegrls. x 2 dx = You hen inegre ech on is own. x 2 dx + x 2 dx. Find x 2 dx x 2 dx = x x + = 2. Find x 2 dx x 2 dx + x = 2 97
16 This shows which is no 3/2. x 2 dx = Now suppose you wned o inegre dx x + e 3x You ll find h is hs no niderivive. So we cn evlue his in closed form (i.e., we would hve o pproxime). The nex quesion is, does his converge o number? In oher words, does he inegrl exis? If you hink of he inegrl s re, is he re finie? Comprison Tes Suppose f nd g re coninuous funcions where f(x) g(x) for x.. If f(x) dx is convergen, hen g(x) is lso convergen. 2. If g(x) dx diverges, hen f(x) dx lso diverges. The Comprison Tes is lso vlid for improper inegrls wih infinie disconinuiies he endpoins. Le s ge bck o dx x + e 3x 98
17 The downside o he comprison es is you hve o some ide if i s going o converge or diverge. This helps figure ou if you re suppose o find nice divergen funcion or nice convergen funcion. Idenifying convergen inegrls jus kes ime. I m going o guess his inegrl converges. So I need o find nice funcion h s bigger hn, whose inegrl converges. Noe h if I cn mke he denominor smller, x + e3x i mkes he whole funcion lrger. x + e 3x e 3x when x Therefore, x + e 3x e 3x So if I cn show e dx converges, nd show for ll x (which I don 3x x + e3x hink needs explining), hen we re good. e 3x dx e 3x dx 3 e 3x 3 e e 3 = + 3 e 3 Since e nd dx converges, hen 3x x + e3x e3x dx converges x + e3x 99
e t dt e t dt = lim e t dt T (1 e T ) = 1
Improper Inegrls There re wo ypes of improper inegrls  hose wih infinie limis of inegrion, nd hose wih inegrnds h pproch some poin wihin he limis of inegrion. Firs we will consider inegrls wih infinie
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