Type 2: Improper Integrals with Infinite Discontinuities
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1 mth imroer integrls: tye 6 Tye : Imroer Integrls with Infinite Disontinuities A seond wy tht funtion n fil to be integrble in the ordinry sense is tht it my hve n infinite disontinuity (vertil symtote) t some oint in the intervl The simlest ses re when the funtion hs suh disontinuity t n endoint of the intervl An obvious emle is d The integrnd is not defined t = nd, in ft, the funtion hs vertil symtote there (see Figure 7) As with integrls on infinite intervls, its ome to the resue nd llow us to define seond tye of imroer integrl DEFINITION 76 (Imroer Integrls with Infinite Disontinuities) Consider the following three tyes of infinite disontinuities f () = Figure 7: The integrl f () = on the intervl [, ] is imroer beuse f () hs vertil symtote t = () (b) () If f is ontinuous on (, b] nd f () =±, then! + b b f () d = f () d! + rovided the it eists If it eists, we sy the imroer integrl onverges Otherwise it diverges If f is ontinuous on [, b) nd!b rovided the it eists b f () d =!b f () =±, then f () d If f is ontinuous on [, b] eet t some oint between nd b where it hs n infinite disontinuity, then we define b f () d = rovided both integrls onverge f () d + f () d, Three Simle Emles EXAMPLE 76 Does d eist (onverge)? Solution As we sw bove, this funtion hs vertil symtote t =, but is otherwise ontinuous on the intervl (, ] So we ly Definition 76 d = d =! +! + =! + le + =+ The integrl diverges EXAMPLE 76 Does d eist (onverge)? Solution This time the funtion hs vertil symtote t =, but is otherwise ontinuous on the intervl [, ) So we ly Definition 76 (nd use mini-substitution to do the integrl) Use u = nd du = d Imroerte Version: Mithell-5//:6:7
2 7 d =! The integrl onverges ( ) / d = ( ) /! h i = ( ) / + = + =! / EXAMPLE 76 Does tn deist (onverge)? Solution Be reful Remember tht the tngent funtion hs vertil symtote t =, but is otherwise ontinuous on the intervl [, )So / tn d=! tn d=! ln se =! [ln se() ] =+ Remember the tht sent funtion goes to + s! And sine the nturl log funtion inreses without bound s! +, the entire eression goes to + So the integrl diverges Combined with Other Tehniques Of ourse imroer integrls n our when more omlited integrls rise EXAMPLE 76 Does d eist (onverge)? Solution This time the funtion hs vertil symtote t =, but is otherwise ontinuous on the intervl (, ] So we will eventully ly Definition 76 But first it mkes sense to do the integrtion without its beuse this involves tringle substitution = se q q d = se q tn q dq = tn q Now roeed with the substitution: d = tn q se q tn q dq = se q dq = ln se q + tn q dq So lying Definition 76 = ln + d =! + d = ln! + + " = ln 8! + + ln + t # = ln( + ) ln( + ) = ln( + ) The integrl onverges Whew!
3 mth imroer integrls: tye 8 EXAMPLE 765 Evlute ln d, if it onverges Solution This time the funtion hs vertil symtote t = where the nturl log beomes undefined, but is otherwise ontinuous on the intervl (, ] So we will eventully ly Definition 76 But first it mkes sense to do the integrtion without its beuse this involves n integrtion by rts Let u = ln so du = d nd dv = d, so v = R dv = R d = Now roeed with the integrtion: ln d= ln So lying Definition 76 d = ln ln d= ln d= ln! +! + le = The integrl onverges Wht fun!! + =! + = l Ho = d = ln ln! +! + = +! + = EXAMPLE 766 Wht n you sy bout d? t ln t ln! + Solution This time the funtion hs vertil symtote t = where the beomes undefined, but is otherwise ontinuous on the intervl [, ) We will eventully ly Definition 76 It mkes sense to do the integrtion (rtil frtions) first without its beuse this involves The degree of the numertor is less thn the degree of the denomintor ( < ) nd the denomintor ftors into distint liner ftors: ( ) = ( ) = A + B = A A + B Comring the numertors of the first nd lst funtions nd solving for A nd B gives s: = A + B onstnts: = A ) A =, B = Consequently, d = d = ln ln + = ln + Imroerte Version: Mithell-5//:6:7
4 Notie tht how we hve simlified the nswer! So lying Definition 76 d =! + d = ln! + = ln! + % ln + ln 5 = The integrl diverges Sometimes you need to be etr reful with imroer integrls, s this net emle illustrtes 6 EXAMPLE 767 Wht n you sy bout d? Solution This time the funtion hs vertil symtote t = where the denomintor is, but is otherwise ontinuous on the intervls [, ) nd (, 6] (see Figure 75) We must ly rt () of Definition 76 nd slit the integrl into two iees t 6 6 d = d +!! + d We need to determine eh integrl sertely! d = ln =!! ln + ln = f () = 6 Figure 75: f () = on the intervl [, 6] is imroer beuse f () hs vertil symtote t = 6 Consequently, the entire integrl d diverges We do not need to evlute 6! + d So wht ws so hrd bout this? Well, it is esy to get this wrong beuse you might hve missed the vertil symtote t = nd done the integrtion s follows: 6 d = ln 6 = ln ln = ln If so, you would hve missed the ft the integrl ws imroer nd tully diverges webwork: Clik to try Problems 8 through Use guest login, if not in my ourse
5 mth imroer integrls: tye 77 Problems Determine these three integrls; for one use theorem to mke it quik () + d (b) d () d Evlute eh imroer integrl; determine whether it onverges or not Reuse some rt of your work in () for (f) () (d) d (b) ln d (e) 5 + d d (f ) () ( + d ) / d Evlute eh imroer integrl; determine whether it onverges or not () e d (d) d (e) (b) e + e d () d (f ) ( ) d d Use #(e) to find the verge vlue of on [, ) (Answer: /) 5 This is beutiful roblem whih sks you to use mny of the ides of the ourse, inluding l Hôitl s rule nd imroer integrls Consider the infinitely dee well formed by rotting the region bounded by the urves urve y = ln, the y-is, nd the -is in the fourth qudrnt it is filled with wter (density D = 65 lbs/ft ) How muh work W is done in emtying the well (rising wter to ground level H = ) The formul for work is b W = D (H y)( f (y)) dy where nd b the deths of the bottom nd to of the well nd = f (y) reresents the eqution of the urve s funtion of Hint: Use your work in #() [Answer: 565 ft-lbs] Some Answers () b! ln b b+ ln = ln + ln = ln (b) b! ( + b ) / =+ Diverges () b! b = b +b b! bq = b + (d)! + [(ln ) (ln ) ]= Diverges (e) 8 (f ) b! ln b b+ ln = Diverges You my need l Hôitl s rule () (b) Diverges () Diverges (d) (e) (f ) (g) () (b) b! ln( + e b )+ln = ln () b! ( b ) ( ) = (d) b! ( b)+ = (e) b! rsin b rsin = = (f ) Tringle: ln( + ) e Imroerte Version: Mithell-5//:6:7
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