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# Calculus II for Business, Life and Social Sciences MATH 2123

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This 9 page Class Notes was uploaded by Mason Larson DDS on Monday October 26, 2015. The Class Notes belongs to MATH 2123 at University of Oklahoma taught by Jonathan Lee in Fall. Since its upload, it has received 43 views. For similar materials see /class/229293/math-2123-university-of-oklahoma in Mathematics (M) at University of Oklahoma.

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Date Created: 10/26/15

55 INTEGR ATION TECHNIQUES SUBSTITUTION l f 4 34y39 X 4 X2y TL x 1 35y2x2x 4y1x8x2 4x4 36y Vlxlyl x2x 1x1 37 Consider the following functions ffx 38x5 186x3 gx 19x4 558362 x0x2 a Graph each of these functions over the interval 3 3 b Estimatethe first coordinates a b and c of the three points of intersection of the two graphs c Estimate the area between the curves over the interval a b l d Estimate the area between the curves over the interval 19 c quot In TECHNOLOGY CONNECTION Find the area of the region bounded by the given graphs 32yx6y 2xyx3 3339y39 x2 4xy V16 x2 ntegration Techniques Substitution B E c r 1 V E s Evaluate integrals using substitution The following formulas provide a basis for an integration technique called substitution 39 Solve application problems involving integration by substitution rl u ru C provided r 75 1 u d 1 equotdue C dulnuC ugtO or d ulnuC ult0 if if of l Ii I li I We will generally ctmsider u gt O Recall the Leibniz notation dydx for a derivative we gave specific definitions of the differentials dy and dx in Section 36 Recall that 42 dx fx 379 I l fol39EITIZ quotquotquot quot Kr 380 CHAPTER 5 INTEGRATION and dy f x dx We Will make extensive use of this notation in this section EXAMPLE 1 For y fx x3 find dy Solution We have i d i f x w so dy f x dx 3x2 dx i 0 EXAMPLE 2 For u gx In x find du Solution We have du I l Eg x x so i 1 d du g x dx dx or x O x x 7 EXAMPLE 3 For y f x exz find dy Solution Using the Chain Rule we have d z i f39x erquot so gt gt dy f x dx 2x6quot2 dx 9 So far the dx in f fx dx has played no role in integrating other than to indicate the variable of integration Now it will be convenient to make use of dx Consider the integral f 2xex2 dx If integration 55 INTEGRATION TECHNIQUES SUBSTITUTION 38 If we set u x2 then du 2x dx If we substitute u for x 2 and du for 2x dx the integral takes on the form fequot du feudueu C Since it follows that Jer 2 dx Iequot du equot C equot2 C In effect we have used the Chain Rule in reverse We can check the result by differ entiating The procedure is referred to as substitution or change of variable It is a trial anderror procedure that you will become more proficient with after much practice If you try a substitution that doesn t result in an integrand that can be eas ily integrated try another There are many integrations that cannot be carried out using substitution We do know that any integrations that fit rules A B or C can be done with substitution Let us consider some additional examples 2x dx EXAMPLE 4 Evaluate I 2 l x Solution 7 2x dx du Substitution l 1 f 7 In it C ln 1 x2 C 2x dx EXAMPLE 5 Evaluate 382 CHAPTER 5 INTEGRATION Solution 2x dx I du Substitution 1 x22 u2 I u 2 du u1 C 1 C Q 1 x2 1 3 d EXAMPLE 6 Evaluate f x Solution I In 3x dx f d Substitution x uZ C 2 l 3 2 nzxFC 0 EXAMPLE 7 Evaluate I xequotZ dx Solution Suppose we try u x2 then we have du 2x dx We don t have 2x dx in f xexz dx We have x dx and need to supply a 2 We do this by multiplying by 1 using 2 l 1 2 2fxex2dx32xequot2dx l 2 l 2fequot 2xdx3fe du 2 1 1 C XC ze f We do this proximate I I x2x3 l10 dx 0 55 INTEGRATION TECHNIQUES SUBSTITUTION 383 d EXAMPLE 8 Evaluate f x x 3 dx f x3 u lnuC r lnx3C 397 0 Solution Substitution With practice you will be able to make certain substitutions mentally and just write down the answer Example 8 is a good illustration of this EXAMPLE 9 Evaluate f x2x3 1 1 dx Solution 1 s b f x2x3 1 lo dx 3 f x3 1103x2 dx u Stltutlon 1 3 In du 1 Mn 2 C 3 11 1 3 11 506 1 c 1 EXAMPLE 10 Evaluate f x2x3 110 dx 0 Solution 21 First we find the indefinite integral shown in Example 9 b Then we evaluate the definite integral on 0 1 1 1 W m t 09 Du o l 33 o 1 131 1 03111 33m gt lt 1 39 l 211111 33 211 l i 39 33 3x2 dx 39 7 4 x3 3 6 dx 5 Je dx 7 J xl equot1 dx 9 Itquot dt 11 I In 4x dx x dx 39 1x dx 4 x 17 th3 17 dt ZIIJ e dx 4equot 2 23 I ln x dx x dx 39 xlnx ZIJ Vax bdx 29 Ibe quot dx 31 3x2 dx 39 1 x35 33 J7x V3 4 x2 dx 384 CHAPTER 5 INTEGRATION Evaluate Be sure to check by differentiating 239 3x2dx quot quot 1x3 4 J39e dx 6 J39e 3 dx 8 Jx4ex5 dx 10 J te Z it x dx 5x dx 1639f1 x 14 18 KB 15 dt 19 J x4 x3 x274x3 3x2 2x dx 20 x x2 x93x2 2x 1 dx e dt 2239J 3 e 2 24 I On x dx x 261 dx 39 xlnxZ 28 foax2 bdx Jpoekl x3 dx I 3239 J 2 x 7 34 f12x V5 1 I6x239dx Evaluate 1 Z 1 3 35 erquot dx 36f 3x212quot dx 0 o l 2 37 I xx2 15 dx 38 I xx1 17 dx 0 1 3 3 39 40J ezquot dx 1 1 t 1 4 2 1 3 2 3 41 J 2 x dx 42 f x 1 x x 1 1 x 3x b b 43 J equot dx 44 I Ze39zquot dx 0 0 b b 45 I me dx 46 J 1212quot dx 0 0 4 3 41f x 62dx 43f x 52dx 0 0 2 3x2 dx 7 x3 dx 49 50 0 1 x35 1 2 x 7 V7 39 I 51J 7xV 1 x1 dx SLJ 12xV51 x2 dx 0 0 APPLICATIONS 0 Business and Economics 53 Demand from marginal demand A firm has the marginaldemand function 2000p D39p m 55 INTEGRATION TECHNIQUES SUBSTITUTION 38 Find the demand function given that D 13000 0 Social Sciences when p 3 per unit 56 Divorce The divorce rate in the United States is ap39 54 Profit from marginal profit A firm has the marginal pmmmated by profit function Dt 100000e 39 25 d1 9000 3000x where Dt the number of divorces occurring at ax x2 6x 10239 time t and t the number of years measured from ix gt 1900 That is t 0 corresponds to 1900 t 39 98 dP 9000 3000x corresponds to January 9 1998 and so on D7 dx y a xz 6x 102 a Find the total number of divorces from 1900 to 1998 Note that this is 98 f Dt dt 0 1 dx b Find the total number of divorces from 1980 to r 1998 Note that this is 98 x f Dt dt 80 x 2 dx Find the totalprofit function given that P 1500 at SYNTHESIS x 3 Find the area of the shaded region x4 7 5 Value of an investment A company buys a new ma 58 chine for 250000 The marginal revenue from the x1 dx sale of products produced by the machine is projected to be I R t 4000t The salvage value of the machine decreases at the rate of has the V t 25000e 0 The total profit from the machine after T years is given by 39Evaluate Revenue Revenue Cost dx Pm from from of 59 f va 1 4x2 dx 60 sale of sale of X b product machine maChme x2 e 617dx 62 dt r r 39 e R t dt f V tdt 250000 M 99 lo 0 7 o 0 63f dt 64f anx dx t 39 X gt a Find PT f dx I t t b End POO xan x 66 e 2e dt 7 WI L x2 t3 xC 9 t23tC 2 3 5 4 7 11 e8xc 13 3c x15 c 8 4 15 2 15 10001n x C 17 x139 C 19 3x C 21 18x1 3 C 23 4e2quot C 1 2 25 3x3 x3 3x39quot3 C 27 fx x3 3x 13 7 x3 N x439 29fx 4x7 31Cx x2100 3 33 a Rx x3 3x 1 If you sell no products you 4000 make no money 35 Dp T 3 37 a 151 301 52 32 b E3 77 55 57 39 51 t3 4 41 vt 212 20 43 51 13 312 6t 10 45 51 1612 v0 50 47 mi 49 a Mt 0001t3 0112 b 6 51 t x6 2 1 51 t 8 53 x5 x4c f V31 6 5 4 2 5515 2 4t3 218lC t4 3 32 114 57 t ttCor C 4 2 4 12 73 t3 2 617x C 633 t 4tC b 59 equotquot C a 65 Exercise Set 52 358 1 8 3 8 5 41 7 i 910 11 e3 1 z 19086 13 In 3 x 10986 15 51 17 An antiderivative velocity 19 An antiderivative energy used in time t 21 An antiderivative total revenue 23 An antiderivative amount of drug in blood 25 a Cx 50x b Ax J5Ox same c As x increases the area over 0 x increases Also as x increases total cost increases y 100 C 5 so x 0x 1 2 x 27 a Cx 100x 01x2 b Rx39 100x 01x2 c Px 02362 d P0000 200000 CHAPTERS 4 5 A25 29 21 51 t3 2 b 150 33 3591 35 6 37 6 31 3 Exercise Set 53 p 368 1 4 Z 1 1 3 5e e 7193 113 95 6 15 2 2 11 3 13 i 15 4 17 95 19 12 3 34 6 5 1 1 1 21 23 7 25 173 27 3 294826 b 291390 0 k 69 so it will be on the 7th day 29 3600 31 68676000 33 90 35 a 2253169 b 218 days 37 9 39 41 13 43 8 45 12 47 6 49 4 51 5293556 53 119879 55 31416 57 5208333 59 150733 Exercise Set 54 p 377 113 5 7396111 13 15 3 17 128 19 9 21 a B b 2 4 39 1 2 2 R E l 27 96 29 WP e 25 31 6 3 8Lv 33 24961 37 a 35 167083 b a 18623 b 0 c 14594 c 645239 d 17683 3 Exercise Set 55 p 384 1 ln 7 x3 C 3 e C 5 2e C 14 2 7e C i23 C 9 e C 11 13 In 391 x C 15 ln 4 x 39 C 17 5103 18 C 19 Blocquot x3 x28 C 21 1n4 e C 23 1n x22 C or In x2 C 2 25 ln In x C 27 3 ax My2 C a 1 39 41 x34 33 4 x2 3 C 35 e 1 37 27 b 29 e C 31 l lt 1 391n4 1n21n 1n2 4111119 431 e A26 ANSWERS 1 451 47 49 51 53 Dp zoom25 p2 5000 55 a PT 200072 250000e 1 250000 or 200012 25000027 b P10 108030 5739 1 2 59 f 21 4x2 C 61 e x C 63 e1quot C 65 1nxquot3 C 67 x3 1quot2 C 69 x2 6x C 9 71 gun 4 82 73x m C 75 t In t 4 C 77 In 1 e C 1 79 1n xquot C 81 In In In x C n 1 39 97x2 9n1 83 C 14n 1 Exercise Set 56 p 393 1 1 1 1 xesquot e5quot C 3 x6 C 5 er el C 5 2 2 4 1 3 37 7 e72x e72quotC 9x 1nx x C 2 4 3 9 2 x2 x2 11x 1nx2 Corx21nx C 2 2 2 l3 x 31n x39 3 xC Letu1nx 3 dv dx and choose v x 3 for an antiderivative of v x2 x2 15 2x lnx 2xC 2 v 4 39 2 2 17 Xgt1n39x ZTX C 19 xx 2 ax 2 2 C39 x4 x4 21 1n2x EC 23 xzex erquot2equot C 25 x2elx e xe2quot C 27 equotzquot x3 x2 42x g quotC 29 e3 x4 1 3x3 gxz x 5 C 31 21112 5 33 91n9 51n5 4 351 37 Cx xx 33 2 x 35 2 39 a 10equotT T 1 1 b v 9084 e t 1 41 gx3 23 In x 2 C 43 C 45 2Vlnxi4JcC I 47 132 484t 7 141 7W5 1g4t 7 5 C 49 Let u xquot and dv ex dx Then du nxquotl dx and v equot Then use integration by parts 51 53 355986 Exercise Set 57 p 397 x 1 1 3 1 9e 3x I C 3 Ins 1 4x 5 1n C 75 x 51n5 xC 8 4 x 9 l i1 x C quot55 x 25n 5 x ln3xxlnx xC 45 435x 12 24 x e x2e5quot 5 25 125 3125 H l 13 esquot5x 1 C 15 x4 lnx C 17 lnxx27C 2 19 2 ln 5 7x39 5 5 7x m 5 x 21 1n l 439 x3 23 me244Inm Vm24C 51nx 5 2 2 C 2x 4x 27 x3equot 3x2ex 6xex 6equot C C 25 2 P 20 5p 100 20 p In 20 12 31 4lnlt x gtC 3x 2 1 1 x 33 1n C 2x 2y4 x 2 quot3 35 l x 3 ex 3 n e C Exercise Set 58 p 407 4 1 a 14914 b 08571 3 O 5 e 1 7 E 1 913 11 n1 b 96 words per minute at t 1 c 70words per minute 15 2545 million 17 a 100 after 10 m 39b 33 19 10016 21 5333 23 2159 39 13 a 90 words per minute Sun 1 5 3 f 5 5 t mil at h 9 E 12 I 15 I 17 I 19 21 23 I 24 I 25 I 26 I 27 I 29 32 33 I 34 I 35 I 37 39 I 41 42 I 43 I 44 I 45 I 46 I Test 5 Equot nle

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