Section 6.3 The Fundamental Theorem, Part I


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1 Section 6.3 The Funmentl Theorem, Prt I (3//8) Overview: The Funmentl Theorem of Clculus shows tht ifferentition n integrtion re, in sense, inverse opertions. It is presente in two prts. We previewe Prt I in Section 6. n prove it in this section. It els with integrls of erivtives. Prt II, which will be covere in Section 6.4, involves erivtives of integrls Topics: nother look t Section 6. The Funmentl Theorem of Clculus, Prt I nother look t Section 6. In the first section of this chpter we use the following result in pplictions involving continuous functions whose erivtives re step functions. Theorem Suppose tht function y = F() is continuous on finite close intervl [, b] n tht its erivtive r = F () is step function on [,b]. Then the region between the grph r = F () n the is for b consists of finite number of rectngles, n the chnge in the function s vlue from = to = b is given by F(b) F() = [ The re of ] [ The re of ] ll rectngles bove the is ll rectngles below the is. () For the continuous function y = F() whose erivtive is the step function in Figure, for instnce, this theorem sttes tht F(b) F() is equl to the re of the two rectngles bove the is minus the re of the three rectngles below the is. r r = F () b FIGURE The Funmentl Theorem of Clculus, Prt I We know from Theorem of Section 6. tht the ifference of res escribe in Theorem is equl to the integrl of F () from to b. Consequently, eqution () cn be rewritten in the form, F(b) F() = F (). () 98
2 Section 6.3, The Funmentl Theorem, Prt I p. 99 (3//8) Eqution () for functions with stepfunction erivtives is specil cse of the following generl result. Theorem (The Funmentl Theorem of Clculus, Prt I) Suppose tht y = F() is continuous n its erivtive r = F () is piecewise continuous on n intervl contining n b. Then F () = F(b) F() (3) or, in Leibniz nottion F = F(b) F(). (3b) Proof: We cn estblish this theorem for generl functions tht stisfy its hypotheses by using the Men Vlue Theorem from Section 4.. Recll tht the Men Vlue Theorem sttes tht if y = F() is continuous on n intervl [,b] n its erivtive eists for ll with < < b, then there is t lest one point c with < c < b such F(b) F() tht the verge rte of chnge of F for b equls its (instntneous) rte of chnge b F (c) t tht point: F(b) F() = F (c). (4) b The geometric interprettion of the Men Vlue Theorem is illustrte in Figure. Since the verge rte of chnge of F is the slope of the secnt line through the points t = n = b on the grph n F (c) is the slope of the tngent line t = c, the theorem sttes tht there is t lest one point c where the tngent line is prllel to the secnt line. y y = F() FIGURE c b To put (4) in the form we nee, we multiply both sies by the length b of the intervl. This yiels F(b) F() = F (c)(b ). (5)
3 p. (3//8) Section 6.3, The Funmentl Theorem, Prt I To estblish Theorem, we suppose first tht is less thn b n tht F () eists for ll with < < b. We consier n rbitrry prtition, = < < < < N = b of [, b]. For ech j we write, s usul, j for the chnge j j in cross the jth subintervl. The Men Vlue Theorem (5) with = j,b = j,c = c j, n j j = j gives F( j ) F( i ) = F (c j ) j for j =,,...,N. (6) To fin the chnge F(b) F() cross [, b], we the chnges (6) cross ll of the subintervls. This gives F(b) F() = N [F( j ) F( j )] = j= N F (c j ) j. (7) The sum on the right of (7) is Riemnn sum for the integrl F () corresponing to the rbitrry prtition of [,b]. Thus, we hve shown tht for ech prtition there is t lest one Riemnn sum tht equls F(b) F(). Since F is piecewise continuous on [,b], the integrl is efine n equls the limit of ll Riemnn sums s the number of subintervls in the prtitions tens to n their withs to zero. Consequently, the integrl equls F(b) F(), s is stte in (3). If F is continuous on [,b] n F () is efine on ll of (,b) ecept t one point with < < b, we cn pply the rgument bove to the intervls [, ] n [, b] to conclue tht F(b) F() = [F() F()] + [F(b) F()] = j= F () + F () = F (). This rgument cn be repete to estblish (3) in ny cse where F () oes not eist t finite number of points in (,b). Eqution (3) for b < follows from tht eqution for < b since n F(b) F() = [F() F(b)]. QED F () = b F ()
4 Section 6.3, The Funmentl Theorem, Prt I p. (3//8) Emple () Show tht the hypotheses of Theorem re stisfie for F() = { 3 + for 6 for < 3 on the intervl [, 3]. (b) Show, without using the theorem, tht the conclusion of the theorem hols in this cse by using res to evlute 3 F (). Solution () The grph of y = F() is shown in Figure 3. The function is continuous on [, 3] becuse the polynomils y = 3 + n y = 6 re continuous for ll n both hve the vlue 5 t =. The erivtive r = F (), whose grph is shown in Figure 4, equls (3 + ) = 3 for < < n equls (6 ) = for < < 3. The hypotheses of Theorem re stisfie becuse F is continuous n F is piecewise continuous on [, 3]. y y = F() r r = F () FIGURE 3 FIGURE 4 Emple (b) The integrl of F () from to 3 equls the re of the rectngle bove the is in Figure 4, minus the re of the rectngle below the is: 3 F () = (3)() ()() =. On the other hn, Figure 3 shows tht F(3) F() = 3 =. Therefore, is vli. 3 Wht is the vlue of F () n F(3) F() re equl, n the conclusion of Theorem (4 + 5)? Solution y formul (3b) with F() = 4 + 5, [ ] [ ] (4 + 5) = = = = [ 4 + 5] [ 4 + 5] = 4 = 6.
5 p. (3//8) Section 6.3, The Funmentl Theorem, Prt I Emple 3 Figure 5 shows the grph of the continuous erivtive r = G () of continuous function y = G(). Region in the rwing hs re 89 n region hs re 6. Wht is G(6) G()? 4 r r = G () 6 FIGURE 5 Solution The Funmentl Theorem (3) with G in plce of F gives G(6) G() = 6 y Theorem of Section 6., the integrl G (). 6 G () is equl to the re of region in Figure 7 between the grph of G n the is for 6 where G () is, minus the re of region between the grph n the is for 6, where G () is. Therefore, G(6) G() = [re of ] [re of ] = 89 6 = 7. It is frequently convenient to use the Funmentl Theorem in the form, F(b) = F() + F (). (8) This eqution is obtine by ing F(b) to both sies of eqution (3). Remember this formul in wors: the vlue of F t b is equl to its vlue t plus the integrl of its erivtive from to b.
6 Section 6.3, The Funmentl Theorem, Prt I p. 3 (3//8) Emple 4 tnk contins gllons of wter t t = (minutes) n wter is e t the rte of t gllons per minute t time t for t 4. How much wter is in the tnk t t = 4? Solution We let V (t) be the volume of wter t time t, mesure in gllons. Then V () = n V (t) = t. We ssume tht V is continuous on [, 4]. The Funmentl Theorem (8) with F replce by V n replce by t res V (4) = V () + 4 V (t) t = + 4 t t. (9) To fin the vlue of the integrl, we rw the grph of V (t) = t s in Figure 6. ecuse t is for t 4, the integrl is equl to the re of the tringle between the grph n the tis for t 4. Since the tringle hs with 4 n height 8, its re is (4)(8) = 6. With this vlue, eqution (9) gives V (4) = + 6 = 6. The tnk hs 6 gllons of wter in it t t = 4. r (gllons per minute) r = t 8 4 FIGURE 6 4 t (minutes) Interctive Emples 6.3 Interctive solutions re on the web pge http// shenk/. 5 ( ). Evlute the integrl Wht is W(3) if W is continuous n W is piecewise continuous on [, 3], W() = 7, n 3 W () =? In the publishe tet the interctive solutions of these emples will be on n ccompnying CD isk which cn be run by ny computer browser without using n internet connection.
7 p. 4 (3//8) Section 6.3, The Funmentl Theorem, Prt I 3. Figure 7 shows the grph of the rte of CO emissions by Chin, mesure in gigtons per yer, from 99 to 5. () ( gigton is 5 tons.) Use the grph to fin the pproimte totl CO emissions by Chin from the beginning of 99 to the beginning of 5. r (gigtons per yer) r = C (t) FIGURE t (yers) 4. The grph in Figure 8 gives the rte of chnge with respect to time of the verge price of gsoline in Sn Diego in Jnury n Februry, 8. () The price P is mesure in cents n the time t is mesure in ys with t = t the beginning of Jnury n t = 6 t the beginning of Februry 9. Region in the rwing hs re 8.6, region hs re 39.5, n region C hs re The price ws 3.9 ollrs per gllon t the beginning of Jnury. Wht ws it t the beginning of Februry 9? 3 r (cents per gllon per y) r = P (t) C FIGURE t (ys) () Dt pte from Climte Chnge the Chinese Chllenge, Science Mgzine, Februry 8, 8, p. 73. () Dt pte from Sn Diego Union Tribune, Februry 7, 8.
8 Section 6.3, The Funmentl Theorem, Prt I p. 5 (3//8) Eercises 6.3 nswer provie. CONCEPTS: O Outline of solution provie. C Grphing clcultor or computer require. se on Theorem, wht is G() if G is continuous on [, ], G( ) = 5 n G () = for < <?. Wht re the units in the clcultion, (4)(8) = 6 in the solution of Emple 4? 3. Figures 9 n show the grphs of function y = f() n its erivtive r = f (). If Theorem pplie in this cse, then f(3) f() woul equl correct n eplin why this is possible. 3 f (). Show tht the conclusion is not y y = f() r r = f () 3 3 FIGURE 9 FIGURE SICS: Evlute the integrls in Eercises 4 through 7. π/ 4. O (sin ) 5 5. ( ) π/4 [(3 + ) 4 ]. (sin ) 8. O Wht is f() if f is continuous n f is piecewise continuous on [, 6], f(6) = 8? 9. Wht is Z(4) Z(3) if y = Z(t) is continuous n Z (t) = 4t 3 for ll t? 6 f () = 5, n. Wht is f() if f is continuous n f is piecewise continuous on [, ], f() = 3, n f () = 7?. Wht is G(4) if G() is continuous n G is piecwise continuous on [ 4, 4], G( 4) = 3, n 4 G =? 4
9 p. 6 (3//8) Section 6.3, The Funmentl Theorem, Prt I. O n object moving on n sis with coorintes given in feet is t s = t time t =. The grph of its velocity v(t) = s (t) in the positive sirection is shown in Figure. The re of region in the rwing is 6.5 n the re of region is Where is the object t t = 4? v (feet per minute) v = v(t) 3 4 t (minutes) FIGURE 3. The temperture in room T = T(t) is 6 F t time t = (hours). The grph of its rte of chnge with respect to t is shown in Figure. The re of region in the rwing is.9 n the re of region is 5.4. Wht is the temperture in the room t t = 4? 5 r (egrees per hour) r = T (t) 5 FIGURE t (hours) 4. Figure 3 shows the grph of the rte r = F (t) of rel estte foreclosures in Sn Diego County s function of the time. (3) Region in the rwing hs re 5,89; region hs re 5,64; n region C hs re 3,45. se on this t how mny foreclosures where there in Sn Diego County between the beginning of 99 n the beginning of 7? r (foreclosures per yer) C r = F (t) FIGURE t (yers) (3) Dt pte from Sn Diego Union Tribune, Jnury 3, 8. Source:DtQuick Informtion Systems.
10 Section 6.3, The Funmentl Theorem, Prt I p. 7 (3//8) 5. The rte of chnge (US cents per month) of the cost of europen euro uring 7 is shown in Figure 4, where t = t the beginning of the yer. (4) One euro ws worth $.9 t the beginning of the yer. Wht ws it worth t the en of the yer? r (cents per month) r = E (t) FIGURE t (months) EXPLORTION: 6. The erivtive y = F () of function y = F() is such tht the re of region in Figure 5 is 79 n the re of region is 396. Wht re the mimum n minimum vlues of F() for 4 if F() = 5? y y = F () 4 FIGURE 5 7. Figure 5 shows the grph of the rte of chnge of the tie level in Sn Diego on Februry, 8, with t = t minight the previous night. (5) The tie level t minight ws 3 feet. Region in the rwing hs re 8, region hs re 6.5, region C hs re 6., region D hs re 6.5, n region D hs re 8. Wht were the low n high ties tht y? r (feet per hour) y = h (t) D 4 C E t (hours) FIGURE 6 8. toy rocket is shot stright up from the groun t t = (secons) with n upwr velocity of 64 feet per secon. If there were no ir resistnce, the rocket s upwr velocity woul be v(t) = 64 3t feet per secon t time t until it hits the groun. How high is the rock t t = 3, ssuming tht there is no ir resistnce? (4) Dt pte from Foreign Echnge Rtes, Feerl Reserve, 8. (5) Dt pte from Sn Diego 8 Tie Clenr, Tielines, 8.
11 p. 8 (3//8) Section 6.3, The Funmentl Theorem, Prt I 9. Figure 7 shows the grph of the rte of chnge of the number of mnufcturing employees in the Unite Sttes from 98 to. (6) Region in the rwing hs re.89, region hs re.4, n region C hs re.. There were 6.65 million mnufcturing employees in the U. S. t the beginning of. How mny were there re the beginning of 98?.4 r (million per yer) r = J (t) 98 t (yers) C FIGURE 7.5 (En of Section 6.3) (6) Dt pte from nnul Survey of Mnufctures, U.S. Census ureu, 8.
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