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Louise Phelps
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1 Mor Tutorial at Work th problms without a calculator, but us a calculator to chck rsults. And try diffrntiating your answrs in part III as a usful chck. I. Applications of Intgration Find th volums of th solids obtaind by rotating th indicatd rgion R in th y-plan about th spcifid ais: #. W sktch th first rgion. y Rgion 5 Figur : y ln(), 2 Find th cornrs at (, ), ( 2, ), ( 2, 2) and labl thm as wll. Part (a) Rotat about th ais y =. Th cross-sctions ar washrs (annuli). Th intgral is 2 Epand as π 2 2 (ln ) 2 d + 2π π( + ln ) 2 π(2) 2 d 2 ln d 3π d and as you do th first intgral by parts, obsrv thr is cancllation with th scond intgral. Answr: π 2 + 2π. On th calculator, this is about 4.3, which is what fnint(y,x,, 2 ) will giv you if y is st to b π( + ln ) 2 4π p.

2 Mor Tutorial at (b) Rotat about th ais = 2. Using shlls w gt 2 2π( + 2)(ln ) This braks into two intgrals with th first don by parts: 2 2 2π ( + 2) ln d 2π ( + 2) d and vntually simplifis to (π/2)( ) or about 3. But it s much simplr intgratd along th y-ais using washrs: 2 π(2 + 2 ) 2 π((2 + y ) 2 dy 2 This pands to π y 2y dy which can b intgratd dirctly. #2 W sktch th scond rgion. Rgion 2 y Figur 2: y ln(), 2 (a) Rotating about th ais y = 2 w hav th intgral π π π(2 + sin ) 2 π(2) 2 d = π 4 sin + sin 2 d which coms out to 8π + π 2 /2 (about 3). (b) Rotating about th ais lin = and using shlls, w hav π 2π( + ) sin() d p. 2

3 Mor Tutorial at which can b don by parts as it stands; but it s natural to first pand ( + ) sin = sin + sin and thn brak th problm into two pics. This coms out to 4π + 2π 2 or about 32. #3 W sktch th third rgion this is a littl awkward on th graphing calculator, and th simplst approach is to us th paramtric curvs option. 5 Rgion y Figur 3: Btwn th y-ais and = y(4 y) As always, w should first idntify th cornrs of th rgion, using y(4 y) =, and labl thm in th pictur. Rotating this rgion about th y-ais, w will want to intgrat along th y-ais and work with discs: 4 π[y(4 y)] 2 dy Epanding th intgrand as 6y 2 8y 3 + y 4 ) w hav a polynomial to intgrat, and th rsult is π(6 4 3 /3 8(4 4 /4) /5) which can crtainly b simplifid furthr (to 52π/5), but in a tst situation w ar not vry intrstd in this kind of arithmtic. If you chck th answr on your calculator you can us fnint with ithr X or Y as th variabl of intgration. p. 3

4 #4 W comput th work don against gravity building a pyramid with a squar bas, 8 ft on a sid, 5 ft tall, and with dnsity 7 lbs/sq. foot. Slicing this into horizontal layrs, w can masur th hight of th layr ithr from ground lvl or from th top of th pyramid and it s a littl mor convnint to masur from th top vrt. Th usful pictur is a vrtical triangular cross sction, to work out th dimnsions of th horizontal cross sctions (squars). Sid S() Hight 5 ft Sid of bas: 8 ft Figur 4: Pyramid (Vrtical Sction) By similar triangls w s that th sid S() of th cross sction at dpth (masurd from th top) satisfis S()/8 = /5, S() = (8/5). Th mass of th cross sction at dpth with thicknss is (7)S() 2 () and th work don lifting that sction (to a hight of 5 ) is (5 ) (7)S() 2 = (7)(8/5) 2 (5 ) 2 Th total work is approimatd by th sum (7)(8/5) 2 (5 i ) 2 i i and in th limit this givs th intgral 5 (7)(8/5) 2 (5 ) 2 d W wr askd to writ this intgral down and justify it, but not to comput it. If w dcid instad to masur hight from ground lvl th prssion for th work will b similar: 5 (7)(8/5) 2 (5 ) 2 d p. 4

5 Mor Tutorial at #5 (a) How do w know that thr is a point in th intrval [5, 7] whr th function f() = ( 2 4) /2 taks on its avrag valu ovr that intrval? By th Man Valu Thorm for Intgrals, which rquirs us to chck that th function is continuous on [5, 7]. As 2 4 is continuous, and nonzro, on that intrval, its rciprocal is continuous as wll. This justifis our claim. (b) To calculat th point in qustion, w first work out th avrag, using th substitution = 2 sc θ. This brings us to 7 d = ( 2 ln + ) so that th avrag f ( is ) 2 ln Th point w ar looking for satisfis ( ) = (/2) ln which bcoms = ( ( ) ) 2 ln This turns out to b about 5.9, in othr words wll insid th pctd intrval. p. 5

6 Mor Tutorial at II. Numrical Mthods #6 How many subintrvals of [, 2] should w us to nsur an accuracy within 6 whn w approimat d using th Midpoint Rul or Simpson s Rul? W first nd stimats for th 2nd and 4th drivativs of our function f () = f (iv) () = 24 Sinc th 4th drivativ is constant w will tak K 4 = 24. For th 2nd drivativ, ithr w us th fact that it is a parabola with zros at = and = 2, or w us th mthods of calculus, to find that th maimum valu of f on [, 2] occurs at = and thus K 2 = 2. Our rror stimats bcom Error(M N ) 2(2) 3 /24N 2 Error(S N ) 24(2) 5 /8N 4 So using th Midpoint rul, w rquir 4/N 2 6 which translats into N > 4 6 = 2 3 = 2. Using Simpson s rul, w rquir 24(32)/8N 4 6 or N 4 (24)(32) 6 8 = /8 and th arithmtic is unplasant; a calculator will tll us that N = 5 will do hr. Running th numrical intgration programs on my calculator with N=25 (which for this program, givs m Simpson s rul with 5 subintrvals), I gt th prdictd accuracy and not much mor. With N = 2 my vrsion of th program crashd, but anothr vrsion gav for th Midpoint Rul, with th prdictd accuracy. p. 6

7 Mor Tutorial at #7 If th Trapzoidal Rul with 3 subin- trvals is usd to stimat 4 f() d, giving th stimat , and f () has th graph shown, giv a rang of valus guarantd to includ th tru valu of th intgral Figur 5: f From th graph, K 2 can b takn to b 4 on th intrval [, 4]. So our rror stimat is /2 9 = /. Thrfor th tru valu lis in th intrval [3.3, 3.6], or if you prfr, [3.32, 3.53] any mor prcision would b pointlss. Mayb it s π, or 22/7, or 3.5, or somthing ls... III. Tchniqus of Intgration #8 Four trigonomtric intgrals: Th first thr succumb to th normal substitutions u = cos, u = tan, or u = sc, th last on nds a rduction formula (obtaind by intgration by parts). (a) 7 cos7 5 cos5 + C (b) tan + 3 tan3 + C (c) 7 sc7 2 5 sc5 () + 3 sc3 () + C (d) 2 (sc tan + ln sc + tan ) + C #9 A mi of intgration by parts and othr things. (a): by parts. (/6) 6 (ln ) 2 (/8) 6 ln + (/8) 6 + C (b): Just a substitution! ln(ln ) + C. (c): First substitut u =, thn intgrat by parts. 2 sin( ) + 2 cos( ) + C 3 (d) Intgration by parts to rduc to + 2 d, thn us 3 (+ 2 ) = to rduc to d, and a substitution will handl that. + 2 ( 3 /3) tan () 2 /6 + (/6) ln( + 2 ) + C d () Similar, but aftr rducing to, = sin θ is ndd. 2 sin + ln( 2 ) + C (f) First u =, thn intgration by parts. + 2 p. 7

8 Mor Tutorial at C (g) Pculiar. Suspcting som wird symmtry (look at th graph) try cos() rwriting d by taking u =. + So π/2 π/2 π/2 π/2 π/2 cos() + d = π/2 cos d. cos() π/2 + d = cos() d + cos + + cos d which simplifis down to + # Intgrals that nd ithr a trigonomtric substitution (possibly complting th squar first) or th mthod of partial fractions. (a) = 5 tan θ. (/25)(tan (/5) + 5( 2 +25) ) + C (b) Partial fractions, of th form A+b C +, lading to A = /37, B = 36/37, C = /37. (/37)((/2) ln( ) ln( + ) + 6 tan (/6)) + C. 4 (c) Complting d th squar,, so substitut u =. ( ) 2 Ans: sin ( ) + C (d) Partial fractions, in th form A+B + C (+3) 2 5 A = 5/8, B = /8, C = 3/8. (5/8) ln( + 3) + (3/8) ln( 5) + 2/( + 3) + C () = 4 sin θ 6 2 sin (/4) + C (f) Complt th squar, = ( + 2) 2 + 5, u = + 2 (/ 5) tan ( +2 5 ) + C # Evaluat sin(ln ) d using two intgrations by parts. Would anothr mthod work? sin(ln ) d = sin(ln ) cos(ln ) d = sin(ln ) cos(ln ) sin(ln()) d sin(ln ) d = (/2)(sin(ln ) cos(ln )) + C Anothr mthod: th natural substitution u = ln() convrts this to u sin(u) du, a mor standard form, that still rquirs th sam tchniqu. p. 8

Solution: APPM 1360 Final (150 pts) Spring (60 pts total) The following parts are not related, justify your answers:

Solution: APPM 1360 Final (150 pts) Spring (60 pts total) The following parts are not related, justify your answers: APPM 6 Final 5 pts) Spring 4. 6 pts total) Th following parts ar not rlatd, justify your answrs: a) Considr th curv rprsntd by th paramtric quations, t and y t + for t. i) 6 pts) Writ down th corrsponding