6.1 Integration by Parts and Present Value. Copyright Cengage Learning. All rights reserved.

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1 6.1 Intgration by Parts and Prsnt Valu Copyright Cngag Larning. All rights rsrvd.

2 Warm-Up: Find f () 1. F() = ln(+1). F() = 3 3. F() =. F() = ln ( 1) 5. F() = 6. F() = -

3 Objctivs, Day #1 Studnts will b abl to: us intgration by parts to valuat indfinit and dfinit intgrals. Homwork: p. 383: Skills warm-up: 7-10 Erciss: 5-13 ODD # u dv 5 3 d 7 ln 3 d 9 ln 1d ln 1 / 3

4 Intgration by Parts Formula A way to intgrat a product is to writ it in th form on function diffrntial of anothr function. If u and v ar diffrntiabl function of, thn u dv uv v du.

5 Intgration by Parts Following is th guidlins for intgration by parts: 5

6 Eampl 1 Intgration by Parts Find d. Solution: To apply intgration by parts, you must rwrit th original intgral in th form u dv. That is, you must brak d into two factors on part rprsnting u and th othr part rprsnting dv. Thr ar svral ways to do this. 6

7 Eampl 1 Solution cont d Th guidlins for intgration by parts suggst th first option bcaus dv = d is th most complicatd portion of th intgrand that fits a basic intgration formula and bcaus th drivativ of u = is simplr than. Nt, you can apply th intgration by parts formula as shown. 7

8 Eampl 1 Solution cont d You can chck this rsult by diffrntiating. 8

9 You try: Using Intgration by Parts Evaluat 5 3 d u dv uv d v du d Lt u 5, du 5d, d 3 dv d 3 v d 1 3 v ( ) 3 33 C C 9

10 Eampl Intgration by Parts Find ln d. Solution: In this cas, is mor asily intgratd than In. Furthrmor, th drivativ of In is simplr than In. So, you should choos dv = d. 10

11 Eampl Solution cont d Nt, apply th intgration by parts formula. 11

12 Eampl 3 Solving an Initial Valu Problm Solv th diffrntial quation dy/d = ln subjct to th initial condition y = -1 whn = 1 ln d. It is typically bttr to lt u = ln u dv ln ( ) 1 ( )ln 1 d d ( Lt u ln, 1 du d, )ln 1 dv d v v C d 1

13 13 13 C )ln ( C 1 )ln1 1 ( 1 C C 3 3 )ln ( y

14 Eampl Using Intgration by Parts Rpatdly Find d. Solution: Th factors and ar both asy to intgrat. Notic, howvr, that th drivativ of bcoms simplr, whras th drivativ of dos not. So, you should lt u = and lt dv = d. 1

15 Eampl Solution cont d Nt, apply th intgration by parts formula. This first us of intgration by parts has succdd in simplifying th original intgral, but th intgral on th right still dosn t fit a basic intgration rul. To valuat that intgral, you can apply intgration by parts again. This tim, lt u = and dv = d. 15

16 Eampl Solution cont d Nt, apply th intgration by parts formula. 16

17 Eampl Solution cont d You can confirm this rsult by diffrntiating. 17

18 Intgration by Parts Bfor starting th rciss in this sction, rmmbr that it is not nough to know how to us th various intgration tchniqus. You also must know whn to us thm. Intgration is first and formost a problm of rcognition rcognizing which formula or tchniqu to apply to obtain an antidrivativ. Oftn, a slight altration of an intgrand will ncssitat th us of a diffrnt intgration tchniqu. 18

19 Intgration by Parts Hr ar som ampls. As you gain princ in using intgration by parts, your skill in dtrmining u and dv will improv. 19

20 Intgration by Parts Th following summary lists svral common intgrals with suggstions for th choics of u and dv. 0

21 Closur Us intgration by parts. ln d Hint: u = ln and dv = 1d du = (1/)d v = (ln)() - 1 d = ln- 1d = ln + C 1

22 Objctivs, Day # Us intgration by parts to find indfinit and dfinit intgrals. Find th prsnt valu of futur incom. HW: p. 383: 6-1 EVEN, p. 38: 7, 75

23 Warm-Up Us intgration by parts rpatdly to solv th indfinit intgral. 3 d u = 3 du = 3 d dv = d v = u dv uv v du. 3-3 d u = du = d dv = d v = 3-3 d 3-3( )d 3

24 d u = du = d dv = d v = ( d) C

25 Eampl 5 Evaluat th indfinit intgral: u = ln and dv = 1d du = (1/)d v = 1 lnd (ln)() - 1 d = ln- 1d = 1 ln ; (ln ) - (1ln1 1) = 0 (-1) = 1 5

26 Prsnt Valu Th prsnt valu of a futur paymnt is th amount that would hav to b dpositd today to produc th futur paymnt. What is th prsnt valu of a futur paymnt of $1000 on yar from now? Bcaus of inflation, $1000 today buys mor than $1000 will buy a yar from now. 6

27 Prsnt Valu Th dfinition blow considrs only th ffct of inflation. Ignoring inflation, th quation for prsnt valu also applis to an intrst-baring account, whr th annual intrst rat r is compoundd continuously and c is an incom function in dollars pr yar. 7

28 Prsnt Valu You hav just won $1,000,000 in a stat lottry. You will b paid an annuity of $50,000 a yar for 0 yars. Whn th annual rat of inflation is 6%, what is th prsnt valu of this incom? Th incom function is c(t) = 50,000 Actual incom: ,000dt 0 50,000t= 50,000(0) 50,000(0) = 1,000, Howvr, you did not rciv this ntir amount right now, so th prsnt valu is: 0 50,000.06t = $58, ,000.06t dt = 8

29 Eampl 7 Finding Prsnt Valu A company pcts its incom during th nt 5 yars to b givn by c(t) = 100,000t, 0 t 5. Assuming an annual inflation rat of 5%, can th company claim that th prsnt valu of this incom is at last $1 million? Figur 6.(a) 9

30 Eampl 7 Solution Th prsnt valu is Using intgration by parts, lt dv = 0.05t dt. 30

31 Eampl 7 Solution cont d This implis that 31

32 Eampl 7 Solution cont d So, th prsnt valu is Ys, th company can claim that th prsnt valu of its pctd incom during th nt 5 yars is at last $1 million. Figur 6.(b) 3

33 Closur, chckpoint 7 A company pcts its incom during th nt 10 yars to b givn by c(t) = 0,000t, for t:[0, 10]. Assuming an annual inflation rat of 5%, what is th prsnt valu of this incom? 10 0,000t.05t dt 0 u = 0,000t dv =.05t dt du = 0,000 v = t 0,000t( t ) ,000.05t dt 10-00,000t -.05t - 8,000, t 0 = $71,

34 Objctivs, Day #3 Studnts will b abl to: us intgration by parts to valuat indfinit and dfinit intgrals. us rapid rpatd intgration or tabular mthod to valuat indfinit intgrals. Choos th bst mthod to valuat intgrals. p. 383: 18-3 EVEN, p. 38: 8 3

35 Warm-Up Solv th diffrntial quation: dy/d = (This mans you will nd to find th antidrivativ of dy/d = ) Lt u dv d., du d, v d u dv 1 d d v 35

36 36 36 d 1 dv u, Lt d v d du, v d 1 C 16 1 C 3 8 C 3 8

37 Rapid Rpatd Intgration by Parts AKA: Th Tabular Mthod Choos parts for u and dv. Diffrntiat th u s until you hav 0. Intgrat th dv s th sam numbr of tims. Multiply down diagonals. Altrnat signs along th diagonals. 37

38 Eampl 8 Rapid Rpatd Intgration by Parts Evaluat d. C u and its drivativs 0 dv and its intgrals 38

39 Eampl 9: Us th Tabular Mthod to find th ara of a rgion y = 3 y = 0 = 0 = 3 d 0 39

40 Eampl 9, continud U and its drivativs 3 3 dv and its intgrals

41 Eampl 9 continud

42 Choosing th bst mthod p. 383 #17 d #19 d

43 Choosing th bst mthod #5 Rwrit as a product 3

44 Choosing th bst mthod #7 ln Must us intgration by parts whn in numrator

45 Choosing th bst mthod #9 ln Whn or a is in dnominator, and thr is a ln in numrator, can us u substitution. 5

46 Choosing th bst mthod #31 ln d Sinc thr is an in dnominator, w must us intgration by parts. 6

u r du = ur+1 r + 1 du = ln u + C u sin u du = cos u + C cos u du = sin u + C sec u tan u du = sec u + C e u du = e u + C

u r du = ur+1 r + 1 du = ln u + C u sin u du = cos u + C cos u du = sin u + C sec u tan u du = sec u + C e u du = e u + C Tchniqus of Intgration c Donald Kridr and Dwight Lahr In this sction w ar going to introduc th first approachs to valuating an indfinit intgral whos intgrand dos not hav an immdiat antidrivativ. W bgin