1. I = 2 3. I = 4 5. I = I = 5 2

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Blaise Cameron
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1 Vrsion Homwork 2 grandi This print-out should hav 24 qustions Multipl-choic qustions may continu on th nxt column or pag find all choics bfor answring If CalC5c9c points Fx = x 5 8sin2 d, find th valu of df/dx at x = π/ df π = 5 dx 8 4 df π = 4 dx 8 4 df π = 4 dx 4 4 df π = 4 dx 5 4 df π = 5 dx 4 corrct 4 By th Fundamntal thorm of calculus, At x = π/4, thrfor, sinc F x = 5 8sin2x df π = 5 dx 4 4 π sin = 4 2 CalC5c24s 2 points Evaluat th intgral 8x x dx corrct By th Fundamntal Thorm of Calculus, [ ] Fx = F F for any anti-drivativ F of Taking w thus s that fx = 8x x Fx = 4x 2 2 x/2, 4 2 CalC7d66s points Evaluat th intgral x 2 x dx

2 Vrsion Homwork 2 grandi corrct 5 4 Thus 4 4x2 4 +C W first divid: 4x 2 x 2 = 4x x and thn intgrat trm by trm This givs 4x ] [2x x dx = 2 lnx sinc ln = 2 2 ln 2 = 2 2 CalC52s 4 points Dtrmin th intgral 8x 4x 2 dx 4x2 +C 2 4 4x2 4 +C corrct 4 4x2 4 +C 4 4x 2 4 +C 5 4x 2 4 +C St u = 4x 2 Thn in which cas du = 8xdx, u du = 4 u4 +C with C an arbitrary constant CalC56 5 points If f is a continuous function such that 9 fxdx = 6, dtrmin th valu of th intgral corrct fxdx To rduc I to an intgral of th form 9 fxdx = 6 w chang variabl St u = x Thn du = dx, whil In this cas x = = u =, x = = u = fudu = 8 kywords: IntSubst, IntSubstExam,

3 Vrsion Homwork 2 grandi CalC52s 6 points Dtrmin th intgral 5x 4 dx +x 2 +x2 4 +C 2 2 +x2 4 +C 2 +x2 4 +C 4 2 +x2 4 +C 5 +x2 4 +C corrct 6 +x2 4 +C St u = +x 2 ; thn du = 2xdx, so 5x 4 dx +x 2 x = 5 dx +x 2 4 = 5 5 u 4 du = u 4 +C +x2 4 +C kywords: Stwart5, radical function, substitution, CalC526s 7 points Dtrmin th intgral 2 cos 4 d 2 2 +cos +C sin +C 2 2 sin +C corrct sin +C cos +C cos +C St u = / Thn so du = 2 d, cosu 4u du = sinu+2u 2 +C with C an arbitrary constant 2 2 sin +C CalC7b7a 8 points Evaluat th intgral ln x x +2 dx

4 Vrsion Homwork 2 grandi corrct 6 2 St u = x +2 Thn du = x dx, whil In this cas, x = = u = x = ln2 = u = 4 4 [ u /2 du = 2 u /2] CalC7d72b 9 points Find th valu of th intgral π/6 6 2 ln ln 5 6 ln ln ln ln 7 5 cosx 5+2sinx dx corrct Sinc th intgrand has th form cosxf2sinx, fx = 5+x, th substitution u = 2sinx is suggstd For thn whil In this cas 2 du = 2cosxdx, x = = u =, x = π 6 = u = 5+u du = [ ] ln 5+u 2 2 ln 6 5 CalC769a points Dtrmin th intgral π π π corrct x 4 x dx St u 2 = x Thn 2udu = dx, whil x = = u = x = 2 = u = 2

5 Vrsion Homwork 2 grandi 5 In this cas, u 2 du [ = 2 sin u 2 π 2 4 π 6 ] 2 = 6 π CalC8aa 2 points Evaluat th intgral π/2 2 π/2 2 x 2cosx+sinxdx kywords: CalC8a5b points Evaluat th intgral lnx x 2 dx ln7+ ln7+ corrct ln7+ ln7 ln7 ln7+ Aftr intgration by parts, [ ] 7 x lnx + 7 Thus [ x lnx+ ] 7 x 2 7 ln7+ x 2 dx 2 2 π/ π/ π/ π/2 2 corrct 6 2 π/2 + 2 Aftr intgration by parts, [ ] π/2 x 2cosx+sinx π/2 x cosx 2sinxdx = π/2 2 I TovaluatI whavtointgratbyparts again For thn I = [ ] π/2 x cosx 2sinx + π/2 Consquntly x 2cosx+sinxdx = 2 π/2 +I 2 π/2 2

6 Vrsion Homwork 2 grandi 6 CalC8a4b points Evaluat th intgral /2 9 /2 2 /2 x x x 2 dx 6 /2 corrct CalC8ba 4 points Evaluat th intgral π/2 sin xcos 2 xdx 4 9 / / / corrct Sinc d dx x = x 2, Sinc sin xcos 2 x = sinxsin 2 xcos 2 x and d dx x x = x x, ths suggst applying intgration by parts to For thn, /2 [ x x x { x x } x 2 dx ] /2 = /2 /2 x x x x dx = /2 + /2 dx = sinx cos 2 xcos 2 x = sinxcos 2 x cos 4 x, th intgrand is of th form sinxfcosx, suggsting us of th substitution u = cosx For thn du = sinxdx, whil In this cas x = = u = x = π 2 = u = u 2 u 4 du 6 /2 [ u + 5 u5] = 2 5

7 Vrsion Homwork 2 grandi 7 kywords: Stwart5, indfinit intgral, powrs of sin, powrs of cos, trig substitution, CalC8b4d 5 points Dtrmin th intgral sin2 +4 sin 2 d +2cos2 4 cos2 +C 2 2cos2 4 cos2 +C corrct 2+2cos2 4 cos2 +C 4 2cos2 4 cos2 +C 5 +2cos2 4 cos2 +C 6 2 2cos2 4 cos2 +C Sinc w s that Thus sin 2 = 2 cos2, +4sin 2 = +2 cos2 2 = 5 2 cos2 sin2 4 cos2 d Sinc th intgrand is of th form sin2fcos2, fx = 4x, th substitution x = cos2 is suggstd Thus dx = 2sin2d, in which cas 4 = 4 4xdx x 2x 2 +C 2cos2 4 cos2 +C CalC8b42s 6 points Dtrmin th intgral 2sin5xsin2xdx 2 sinx 2 7 sin7x+c 2 7 sin7x sinx+c 2 7 sin7x+ 2 sinx+c 4 sinx sin7x+c corrct sin7x 2 sinx+c 6 7 sin7x+ sinx+c Sinc sin5xsin2x = 2 cosx cos7x, w s that = cosx cos7x dx sinx sin7x +C 7 sinx 7 sin7x+c

8 Vrsion Homwork 2 grandi 8 Dtrmin if CalC8ha 7 points fxdx is convrgnt or divrgnt whn x 2, x, fx = x, x >, and find its valu if convrgnt I is not convrgnt CalC8h5s 8 points Dtrmin if th impropr intgral 2 4x+ 2dx is convrgnt, and if it is, find its valu 2 corrct I not convrgnt corrct 6 2 Th intgral is impropr bcaus th intrval of intgration is infinit Thus lim t I t, I t = But for t >, I t = = t t fxdx x 2 dx+ x dx [ x] [ ] t + lnx = +lnt On th othr hand, lim t lnt = I is not convrgnt 6 Th intgral is impropr bcaus th intrval of intgration is infinit To tst for convrgnc w thus hav to dtrmin if lim I t, I t = t t 2 4x+ 2 dx, xists To valuat I t, st u = 4x+, Thn whil In this cas, I t = 4 du = 4dx, x = = u = x = t = u = 4t+ 4t+ 2 u 2 du = [ 2 4 u ] 4t+,

9 Vrsion Homwork 2 grandi 9 so that lim I t = lim t t 2 24t+ = 2 I xists and 8 th intgral is convrgnt and 2 CalC8h9a 9 points Dtrmin if th intgral 8x dx is convrgnt, and if it is, find its valu 8 corrct 2 intgral is divrgnt = Th intgral is impropr bcaus of th infinit intrval of intgration It will b convrgnt if xists Now t lim t t 8x dx = On th othr hand, 8x dx [ 8 8x] t = 8 8t + 8 lim t 8t = CalC8h2a 2 points Dtrmin if th impropr intgral tln8t dt convrgs, and if it dos, comput its valu I dos not convrg corrct Th intgral is impropr bcaus of th infinit intrval of intgration, so w st lim n I n, I n = n tln8t dt, whnvr th limit xists Now with th substitution u = ln8t w s that tln8t dt = du = lnu+c u I n = But [ lnln8t ] n = lnln8n lnln8 lim lnln8n =, n so lim n I n dos not xist Hnc I dos not convrg

10 Vrsion Homwork 2 grandi CalC6a4b 2 points Th shadd rgion in th figur y But x x 2 2xdx [ = 4 x4 x x 2] = 5 2 On th othr hand, is boundd by th graphs of fx = x x 2 x+, gx = x+ Find th ara of this rgion ara = 25 sq units 2 2 ara = 7 2 ara = 97 8 sq units corrct sq units 4 ara = 49 sq units 2 5 ara = 5 4 sq units To st up th dfinit intgrals dtrming th shadd rgion w nd to find th x- coordinats of th points of intrsction of th two graphs Now fx gx = x x 2 2x = xx 2x+ Thus th graphs intrsct whn x =,, 2, and so th rquird ara is givn by fx gxdx+ 2 x gx fxdx 2 x +x 2 +2xdx = [ 4 x4 + x +x 2] 2 = 8 th shadd rgion has ara = 7 sq units 2 CalC7d8xam 22 points Th shadd rgion in y is boundd by th x-axis and th graphs of fx = 6lnx x, x =, not drawn to scal Find th ara of this rgion ara = 5 sq units 2 2 ara = 4 sq units ara = 2 sq units 4 ara = 7 sq units 2 x

11 Vrsion Homwork 2 grandi 5 ara = sq units corrct Th graph of fx = 6lnx x crosss th x-axis whn lnx =, i, whn x = Thus th shadd rgion is th st of points { x, y : y 6lnx x, x }, and so has ara = 6lnx x dx To valuat th intgral, st u = lnx Thn whil In this cas, du = x dx, x = = u =, x = = u = 6lnx x dx = 6udu th shadd rgion has V = 6π cuunits corrct 4 V = 4π cuunits 5 V = π cuunits Th volum, V, of th solid of rvolution gnratd by rotating th graph of x = fy about th y-axis btwn y = a and y = b is givn by b V = π a fy 2 dy So whn y = and y = 2, thn 2 V = π fy 2 dy On th othr hand, y = 2 x, so fy = 2 y as a function of y Thus 2 V = π 9 [ 4 y2 dx = π 4 y] 2 V = 6π cuunits kywords: ara = [u 2] = sq units CalC7d8d 24 points Find th volum, V, of th solid of rvolution gnratd by rotating th graph of CalC6b2c 2 points Dtrmin th volum of th right circular con gnratd by rotating th lin y = 2 x about th y-axis btwn y = and y = 2 V = 5π cuunits 2 V = 2π cuunits y = 2lnx about th y-axis btwn y = and y = 4 V = 2 π cuunits 2 V = 2 π cuunits V = 4 π cuunits corrct 4 V = 4 π cuunits

12 5 V = 2 2 π cuunits 6 V = 2 4 π cuunits Th volum, V, of th solid of rvolution gnratd by rotating th graph of x = fy about th y-axis btwn y = a and y = b is givn by b V = π a Vrsion Homwork 2 grandi 2 fy 2 dy To apply this w hav first to xprss x as a function of y sinc initiallyy is givn in trms of x by y = 2lnx But aftr xponntiation, x = y 2 Thus V = π V = 4 y dy = π [ y] 4 4 π cuunits

Calculus II (MAC )

Calculus II (MAC ) Calculus II (MAC232-2) Tst 2 (25/6/25) Nam (PRINT): Plas show your work. An answr with no work rcivs no crdit. You may us th back of a pag if you nd mor spac for a problm. You may not us any calculators.