Lecture for Week 14 (Secs ) Definite Integrals and the Fundamental Theorem

Transcription

1 Lecture for Week 14 (Secs ) Definite Integrls nd the Fundmentl Theorem 1

2 Integrl clculus the second hlf of the subject, fter differentil clculus hs two spects: 1. Undoing differentition. This is the problem of finding ntiderivtives, which we ve lredy discussed. 2. The study of dding things up or ccumulting something. This is very closely relted to the re topic of lst week. 2

3 The fundmentl theorem of clculus shows tht these two things re essentilly the sme. To revel the bsic ide, consider speeddistnce problem: We know tht if n object moves t constnt speed for certin period of time, then the totl distnce trveled is distnce = speed time. Suppose insted tht the speed is vrying function of time. If we consider very short time intervl, [t i 1,t i ], then the speed is pproximtely 3

4 constnt (t lest if f is continuous) nd hence we cn pproximte the distnce by distnce = speed in tht intervl (t i t i 1 ). As the speed we my choose the mximum speed in the short intervl, or the minimum, or nything in between sy f(w i ) for some w i [t i 1,t i ]. The choice won t mke ny difference in the end. (There should be picture here, but I don t hve time to drw it now.) 4

5 So the totl distnce trveled between t = nd t = b is pproximtely n f(w i ) t i, where t i t i 1. i=1 As we let n, the pproximtion should become exct. In effect, we hve concluded tht: 5

6 (A) The distnce trveled between times nd b is the re under the grph of the speed function between the verticl lines t = nd t = b (if re is defined in pproprite units, nd if the speed is lwys nonnegtive). On the other hnd, (B) The distnce function is n ntiderivtive of the speed function. 6

7 Conclusion: Ares nd ntiderivtives re very closely relted. In some sense, they re the sme thing! Before continuing we need more precise definition of the definite integrl s limit of sum, f(x)dx = lim P 0 n f(x i) x i. i=1 Tht occupies the first hlf of Sec. 6.3, nd it 7

8 involves ll the issues bout vrying the sizes x i of the strips, etc., tht I discussed lst week bout re. One importnt difference between res nd generic definite integrls: The integrnd function is llowed to be negtive in some (or ll) plces). Ares must lwys be positive (or zero), but integrls cn be negtive. In generl, f(x)dx = A + A, 8

9 where A + is the re below the grph nd bove the x xis, nd A is the re bove the grph nd below the x xis. (In Fig. 3 on p. 379, A + is yellow nd A is blue.) Note lso tht the dummy vrible in n integrl cn be ny letter tht doesn t cuse confusion: 6 2 x 3 dx = 6 2 t 3 dt. Also, this thing is not function of the vrible of integrtion, it is just number. 9

10 Here is n exmple where bd choice of letter would cuse confusion: x 2 x must not be written s x 2 2 t 3 dt ( ) x 2 x x 2 2 x 3 dx. (The integrl in ( ) is function of x, though not of t.) 10

11 Algebric properties of integrls 1. [f(x)+g(x)]dx = f(x)dx+ g(x) dx. 2. cf(x)dx = c f(x)dx for constnt c. 11

12 3. f(x)dx = c f(x)dx+ c f(x)dx. 4. dx mens 1dx nd equls b. Remrks: Formul (4) is just the re of rectngle. The others re lso rther obvious for res. Recll tht mking (3) obvious ws one of the resons for llowing strips of vrying widths x i. 12

13 5. b f(x)dx mens f(x)dx if b >. It follows tht in (3), c does not need to lie between nd b. Perhps more importnt (from the point of view of voiding mistkes) is n identity tht is not in the list: 13

14 f(x)g(x)dx = f(x)dx g(x) dx FALSE! The integrl of product is not the product of the integrls, just s (nd becuse) the derivtive of product is not the product of the derivtives. Roughly speking, every derivtive formul turns round to give n integrl formul. The integrl formul corresponding to the product rule for derivtives is integrtion by prts (Sec. 8.1). 14

15 Order properties of the integrl 1. If f(x) g(x) for ll x in [,b], then f(x)dx g(x) dx. (Here we ssume < b, of course.) In prticulr, if f is nonnegtive, then f(x)dx 0 lso. 15

16 2. If f is continuous nd f(x) 0, then 3. f(x)dx (with < b) is strictly greter thn 0 unless f(x) = 0 for ll x in [,b]. f(x)dx f((x) dx. 16

17 4. Men vlue theorem for integrls: If f is continuous, then there is z in (,b) such tht f(x)dx = f(z)(b ). 5. If m f(x) M in [,b], then m(b ) f(x)dx M (b ). 17

18 Now bck to the fundmentl theorem. Roughly speking, it sys tht differentition undoes integrtion, nd vice vers. They re inverse opertions (lmost), like squring nd tking the squre root, except tht they operte on functions insted of numbers. The two functions involved re relted s re the two redings on cr s speedometerodometer pnel. The odometer reding is the integrl of the speedometer reding. The speedo- 18

19 meter reding is the derivtive of the odometer reding. Tht is the essence of clculus! The theorem hs two prts, one for ech order of the opertions. And I stte ech prt in two versions, depending on which function (the integrl or the derivtive) tkes center stge. In stting the theorem, we ssume for simplicity tht f (the derivtive) is continuous. 19

20 Fundmentl Theorem, Prt 1: d dx x f(t)dt = f(x). Tht is, G(x) is n ntiderivtive of f. x f(x)dt 20

21 Exercise Evlute d dx d dy x 10 y 3 2udu, 1 t dt, d dv v udu. 21

22 For the first two, just pply the theorem: d dx d dy x 10 y 3 2udu = 2x, 1 t dt = 1 y. We know these fcts without necessrily knowing wht the integrls themselves re. You my know tht the first one is x 10 2udu = x 2 100, 22

23 either (the hrd wy) in nlogy to Exercise or (the esy wy) peeking hed to Prt 2 of the theorem. The second integrl requires logrithm function (see Sec. 6.6). For the third one, use the chin rule: d dv v udu = 2v 2 2v = 4v 3. (Mke sure you understnd this. The integrl is function of v 2 nd therefore of v.) 23

24 Fundmentl Theorem, Prt 2: H (x)dx = H(b) H(). Tht is, f(x)dx = H(b) H(), where H is ny ntiderivtive of f. 24

25 Now we cn fill in the remrks I mde on the previous exmple: Exercise Find x 10 y 3 2udu, 1 t dt. 25

26 Since d dx x2 = 2x, x 10 2udu = u 2 u=x u 2 u=10 u 2 x 10 = x Certinly much esier thn Use Theorem 5 on p. 387! 26

27 From Sec. 4.4 or 5.7, we know tht n ntiderivtive of 1 y is lny. (This is under the ssumption tht y > 0 in the intervl concerned. If it s negtive, we should write ln y.) So y 3 1 t dt = lnt y 3 = lny ln3. (Since 3 is positive, so is y. The formul does not pply to negtive y, becuse the logrithm is discontinuous t t = 0.) 27

28 Why did I sy tht integrtion nd differentition re lmost inverse opertions? Look bck t slide 23. It sys tht, if we think of H s function of b, then differentiting it nd then integrting it lmost gives bck H(b), but not quite: There is constnt of integrtion, H(), stuck on t the end. This compliction is inevitble, becuse given function hs mny ntiderivtives, differing by constnts. 28

29 In fct, Prt 2 is esy to prove, if you know Prt 1 nd the theorem tht two ntiderivtives (on n intervl) differ only by constnt. Becuse (Prt 1) d db f(x)dx = f(b), it must be tht f(x)dx G(b) 29

30 is n ntiderivtive of f(b); the only question is which one. If H is ny ntiderivtive (see Prt 2), then f(x)dx = H(b)+C (Theorem 2, p. 345). To find C, pply the initil condition tht f(x)dx = 0. 30

31 It tells us tht 0 = H()+C. So f(x)dx = H(b) H(). 31

32 Proving Prt 1 is hrder, nd you should red the detils in the book. The bsic ide is the sme s in the erlier rgument tht the distnce trveled is the re under the grph of the speed function, but run in reverse: we will pick the sum prt insted of building it up. 32

33 d dx By definition of derivtive, [ 1 x+h f(t)dt = lim f(t)dt h 0 h x 1 = lim h 0 h x+h x f(t)dt x f(t)dt 1 lim (re of lst strip ) h 0 h = height of lst strip = f(x). ] 33

34 Prcticl spects of the fundmentl theorem Integrls (defined by Riemnn sums) re equl to ntiderivtives ccording to the theorem. So, 1. If you re trying to integrte (in either sense) function nlyticlly (i.e., in terms of exct formuls), the ntiderivtive is lmost lwys esier to evlute thn the limit of Riemnn sums. So you use Prt 2: 34

35 f(t)dt = H(b) H() ( H (t) = f(t) ). To find the left side (which might rise in n ppliction s n re, for exmple) you clculte the right side. Any ntiderivtive H will do for this purpose, so it is permissible to leve out the + C in this context. 35

36 2. If you re integrting numericlly, sums re usully esier thn ntiderivtives. So the computer progrms for such work use some refinement of the definition x f(t)dt = lim P 0 f(t i) t i. To find the left side (which might rise in n ppliction s distnce trveled t speed f, for exmple) you compute one of the sums on the right side with smll P nd i 36

37 trust tht tht is good pproximtion to the nswer. 37