Section 6.2 The definite integrl (3/2/8) Overview: We sw in Section 6. how the chnge of continuous function over n intervl cn be clculted from its rte of chnge if the rte of chnge is step function. We lso outlined there how in other cses, chnges in the function might be determined s limits b using step function pproimtions of the rtes of chnge. In this section we use this ide in the definition of the definite integrl. Then we derive some of the bsic properties of the integrl. Topics: The definite integrl Piecewise continuous functions Integrls nd res Specil Riemnn sums Properties of definite integrls Unbounded functions Riemnn sum progrms The definite integrl The definite integrl of the function = f() from = to = b with < b is number, denoted f() d. The smbol is clled n integrl sign, the numbers nd b re the limits of integrtion, [, b] is the intervl of integrtion, nd f() is the integrnd. As we eplined t the end of the lst section, we wnt to define the integrl so tht for the function f of Figure, the integrl from = to = b equls the re of region A between the grph nd the -is where f() is positive, minus the re of region B between the grph nd the -is where f() is negtive. To ccomplish this, we pproimte = f() b step functions whose grphs form pproimtions of the two regions b rectngles, s in Figure 2. The integrl is defined to be the limit, s the number of rectngles tends to nd their widths tend to zero, of the re of the rectngles bove the -is, minus the re of the rectngles below the -is. = f() v = f() A b b c B t FIGURE FIGURE 2 8
Section 6.2, The definite integrl p. 8 (3/2/8) To construct pproimting rectngles, we strt with prtition = < < 2 < < N < N = b () of the intervl [, b]. It divides [, b] into N subintervls, [, ], [, 2 ],...,[ N, N ]. The jth subintervl is [ j, j ] for j =, 2, 3,...,N (Figure 3). We let j denote its width: [The width of the jth subintervl] = j = j j. We lso pick, for ech j, point c j in the jth subintervl tht is in the domin of f. The jth subintervl [ j, j ] FIGURE 3 j j c j j Figure 4 shows the grph of the function f of Figures nd 2, nd five rectngles tht correspond to prtition = < < 2 < 3 < 4 < 5 = b of [, b] into five subintervls nd to points c,c 2,c 3,c 4 nd c 5 in the subintervls. For j = nd 2 on the left, f(c j ) is positive, the bse of the rectngle is the jth subintervl on the -is, nd its top is t = f(c j ). For j = 3, 4, nd 5 on the right, f(c j ) is negtive, the top of the rectngle is the jth subintervl on the -is nd its bse is t = f(c j ). = f() c 3 c 4 c 3 5 4 5 = b = c c 2 2 FIGURE 4 We use the sme procedure to construct rectngles for generl prtition (). For ech j =, 2,...,N, we pick point c j in the jth subintervl such tht f(c j ) is defined. For those vlues of j for which f(c j ) is positive or zero we construct rectngle with its bottom formed b the jth subintervl nd with top t = f(c j ) (Figure 5). For those vlues of j for which f(c j ) negtive, we use the jth subintervl s the top of the rectngle nd put its bottom t = f(c j ) (Figure 6). Then the height of the rectngle is f(c j ) if fc j ) is nd is f(c j ) if f(c ) <. In ll cses the width of the jth rectngle is the width j of the subintervl. Consequentl, { f(cj ) j if f(c j ) [Are of the jth rectngle] = f(c j ) j if f(c j ) <. (2)
p. 82 (3/2/8) Section 6.2, The definite integrl = f() = f() j j j c j j f(c j ) f(c j ) j c j j [Are] = f(c j ) j [Are] = f(c j ) j FIGURE 5 FIGURE 6 Becuse of formuls (2), the sum of the res of the rectngles bove the -is minus the res of the rectngles below the -is is given b the epression, [ ] [ ] Are of the rectngles Are of the rectngles bove the -is below the -is = f(c ) + f(c 2 ) 2 + f(c 3 ) 3 + + f(c N ) N (3) Notice tht no minus re needed in (3): the res of the rectngles below the -is re utomticll subtrcted becuse the vlues f(c j ) re negtive for those rectngles. To epress the sum on the right of (3) more concisel, we use summtion nottion nd write [ ] [ ] Are of the rectngles Are of the rectngles bove the -is below the -is = j= f(c j ) j (4) The smbol f(c j ) j in (4) represents the sum of the quntities f(c j ) j for j =, 2, 3,...,N j= tht is given in (3). Summtion nottion is further illustrted in the following emple. Emple Solution () Write out the sum 6 j= j 2. (Do not perform the clcultions.) (b) Epress + 2 + 3 + 4 + 5 + 6 + 7 + 8 with summtion nottion. 6 () j 2 denotes the sum of the numbers j 2 with j =, 2, 3, 4, 5,6. Consequentl, it j= equls 2 + 2 2 + 3 2 + 4 2 + 5 2 + 6 2. (b) We write s nd let j denote the denomintors of the frctions nd obtin 8 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = j. j=
Section 6.2, The definite integrl p. 83 (3/2/8) The sum (4) is clled Riemnn sum, nd the definite (or Riemnn) integrl of f from to b, is defined to be limit of such sums: Definition (Riemnn sums nd definite integrls) () Consider function = f() defined t ll or t ll but finite number of points in [, b]. A Riemnn sum for the integrl corresponding to prtition = < < 2 < < n = b of [,b] is sum of the form, f() d j= f(c j ) j where for ech j =,2, 3,...,N, c j is point in the jth subintervl where f is defined nd j = j j is the width of the jth subintervl. (b) The definite integrl of f from to b is the limit of Riemnn sums, f() d = lim f(c j ) j (5) j= s the number N of subintervls in the prtitions tends to infinit nd their widths tend to zero, provided tht the limit eists nd is finite. Emples of this definition will be emined lter in the section. Piecewise continuous functions We consider definite integrls of functions which re piecewise continuous, ccording to the following definition. Definition 2 The function = f() is piecewise continuous on [,b] if it is defined t ll but possibl finite number of points in [, b] nd there is prtition of the intervl such tht f is continuous on the interior of ech subintervl nd hs finite limits from the right t the left endpoints of the subintervls nd finite limits from the left t the right endpoints. According to this definition, piecewise continuous function might not be defined or might be discontinuous t finite number of other points in the intervl. The function g of Figure 7 is piecewise continuous on the intervl [, 7] becuse the prtition < 3 < 5 < 7 divides it into three continuous pieces with finite one-sided limits t their endpoints: the function g is defined for ll in [, 7] ecept t = 3; it is continuous on the open intervls (,3), (3,5), nd (5,7); nd it hs finite limits s +, s 3, s 3 +, s 5, s 5 +, nd s 7. Riemnn sums nd Riemnn integrls re nmed fter the Germn mthemticin G. F. B. Riemnn (826 866), who formulted the modern definition of the integrl. This tpe of limit hs n ɛδ-formultion: the number I is the limit of the Riemnn sums if for ever positive ɛ, no mtter how smll, there is positive δ such tht the Riemnn sums differ from I b less thn ɛ for ll prtitions into subintervls of widths less thn δ.
p. 84 (3/2/8) Section 6.2, The definite integrl 8 6 4 2 = g() FIGURE 7 2 3 5 7 The eistence of definite integrls of piecewise continuous functions is estblished b the net theorem, which is proved in dvnced courses, Theorem [, b]. The Riemnn integrl f() d is defined if f is piecewise continuous on the intervl Integrls nd res Figure 8 shows region A between the grph of positive, continuous function f nd bove the -is for b. In Definition the integrl f() d is defined to be the limit of res of pproimtions of this region b collections of rectngles such tht the pproimtions become incresingl ccurte s the limit is tken. Accordingl, we define the re of A to be the integrl: = f() A [Are A] = f() d FIGURE 8 b Definition 3 (Ares) If f is piecewise continuous nd its vlues re on [, b], then the re of the region between the grph = f() nd the -is for b equls the integrl f() d. This definition is consistent with other definitions of re if the region consists of rectngles, tringles, or other figures for which there re re formuls from geometr, nd it defines the re in other cses.
Section 6.2, The definite integrl p. 85 (3/2/8) If f hs negtive, or positive nd negtive, vlues in [, b], then Riemnn sum pproimtions of f() d equl the re of rectngles tht pproimte the region between the grph nd the -is where f() is positive, minus the re of rectngles tht pproimte the region between the grph nd the -is where f() is negtive. This leds to the following generl result relting integrls to res. Theorem 2 (Integrls s res) the integrl () If f is piecewise continuous nd its vlues re on [, b], then f() d equls the negtive of the re of the region between the grph = f() nd the -is for b (Figure 9). (b) If f is piecewise continuous nd hs positive nd negtive vlues on [, b], then the integrl f() d equls the re of the region bove the -is nd below the grph where f() is positive, minus the re of the region below the -is nd bove the grph where f() is negtive. (Figure ). = f() = f() B b A c B b f() d = [Are B] f() d = [Are A] [Are B] FIGURE 9 FIGURE In the net emple the vlue of n integrl is found b using Theorem 2 nd formul from geometr.
p. 86 (3/2/8) Section 6.2, The definite integrl Emple 2 Solution Use the formul for the re of tringle to evlute ( + ) d. 3 The grph of = + is the line of slope in Figure, which intersects the -is t =. B Theorem 2, the integrl equls the re of tringle B bove the -is, minus the re of tringle A below the -is. Since tringle B is 4 units wide nd 4 units high, its re is 2 (4)(4) = 8, nd since tringle A is 2 units wide nd 2 units high, its re is 2 (2)(2) = 2. Therefore, 3 3 3 ( + ) d = [Are B] [Are A] = 8 2 = 6. = + 3 4 3 2 B FIGURE A 2 2 3 Specil Riemnn sums A Riemnn sum f(c j ) j for j= f() d is right Riemnn sum if the points c j re the right endpoints j of the subintervls of the corresponding prtition =, < 2 < < N = b. It is left Riemnn sum if the c j s re the left endpoints j, nd is midpoint Riemnn sum if the c j s re the midpoints 2 ( j + j ) of the subintervls. Emple 3 Solution Give () the right Riemnn sum nd (b) the left Riemnn sum for ( 3 + 5) d corresponding to the generl prtition =,< < 2 < < N = 5 of [, 5]. () We write, s in Definition, j = j j for the width of the jth subintervl. Since j is the right endpoint of the jth subintervl, the right Riemnn sum is [( j ) 3 5 j ] j. j= (b) Becuse j is the left endpoint of the jth subintervl, the left Riemnn sum is 3 [( j ) 3 5 j ] j. j= If the subintervls in prtition for Riemnn sum re of equl width, s in the net emple, we write insted of j for tht width.
Section 6.2, The definite integrl p. 87 (3/2/8) Emple 4 Solution Clculte () the right Riemnn sum, (b) the left Riemnn sum, nd (c) the midpoint Riemnn sum for 2 d corresponding to the prtition of [, ] into five equl subintervls. Drw the curve = 2 with the rectngles whose res give the Riemnn sums. () Becuse [, ] hs width, ech of the five subintervls in the prtition hs width = 5 =.2, nd the prtition is <.2 <.4 <.6 <.8 <. The right endpoints re.2,.4,.6,.8, nd, nd the rectngles which give the right Riemnn sum touch the curve = 2 in Figure 2 t their upper right corners. The right Riemnn sum is (.2) 2 (.2) + (.4) 2 (.2) + (.6) 2 (.2) + (.8) 2 (.2) + 2 (.2) = [(.2) 2 + (.4) 2 + (.6) 2 + (.8) 2 + 2 ](.2) = (.4 +.6 +.36 +.64 + )(.2) =.44. = 2 = 2 = 2 FIGURE 2 FIGURE 3 FIGURE 4 (b) The left endpoints of the prtition re,.2,.4,.6, nd.8 nd the left Riemnn sum equls the re of the rectngles in Figure 3. It equls ( 2 )(.2) + (.2 2 )(.2) + (.4 2 )(.2) + (.6 2 )(.2) + (.8 2 )(.2)) = [() 2 + (.2) 2 + (.4) 2 + (.6) 2 + (.8) 2 ](.2) = ( +.4 +.6 +.36 +.64)(.2) =.24. (c) The midpoint Riemnn sum is given b the res of the rectngles in Figure 4. Since the midpoints of the subintervls re.,.3,.5,.7, nd.9, the midpoint Riemnn sum is (. 2 )(.2) + (.3 2 )(.2) + (.5 2 )(.2) + (.7 2 )(.2) + (.9 2 )(.2)) = [(.) 2 + (.3) 2 + (.5) 2 + (.7) 2 + (.9) 2 ](.2) = (. +.9 +.25 +.49 +.8)(.2) =.33. A midpoint Riemnn sum with equl subintervls, s in Emple 4c, is known s Midpoint Rule pproimtion. The right Riemnn sum (Figure 2) in Emple 4 is greter thn the left Riemnn (Figure 3) bcuse = 2 is incresing for < <. The midpoint Riemnn sum (Figure 4) is the most ccurte of the three pproimtions of the integrl.
p. 88 (3/2/8) Section 6.2, The definite integrl We will derive ll of the formuls for definite integls we will need b using formuls for derivtives nd the Fundmentl Theorem of Clculus tht we discuss in the net section. Some definite integrls cn be lso clculted directl from Definition nd formuls from lgebr, s in the net emple. Emple 5 The formul 2 + 2 2 + 3 2 + + N 2 = 3 N3 + 2 N2 + 6 N (6) Solution for the sum of the squre of the first N positive integers is is derived in Eercise 5. Use it nd Definition with right Riemnn sums nd prtitions into equl subintervls to find the vlue of 2 d. For n positive integer N, the prtition of [, ] into N equl subintervls is < N < 2 N < 3 N < < N N <. The width of ech subintervl is = /N, nd the right endpoint of the jth subintervl is j = j/n. Therefore, the right Riemnn sum for prtition is ( j ) 2 = j= [ ( ) 2 ( ) 2 ( ) 2 2 3 + + + + N N N = N 3 (2 + 2 2 + N 2 ). With formul (6) we obtin j= ( ) ] 2 ( ) N N N 2 d with this ( j ) 2 = N 3 ( 3 N 3 + 2 N2 + 6 N) = 3 + N + 6N 2 (7) Becuse the widths of the subintervls tend to zero s the number N of subintervls tends to, the integrl is the limit of the Riemnn sums (7) s N : 2 d = lim N j= ( j ) 2 = lim N ( 3 + N + ) 6N 2 = 3.
Section 6.2, The definite integrl p. 89 (3/2/8) Properties of definite integrls We now give one definition nd two theorems with bsic properties of definite integrls. Becuse regions of zero width hve zero re, integrls with equl limits of integrtion re defined to be zero. Also, n integrl from b to with < b is defined to be the negtive of the integrl from to b: Definition 4 () For n function f, (b) If the integrl of f from to b is defined with < b, then b f() d =. (8) f() d = f() d. (9) Emple 6 Solution () Wht is the vlue of Wht is the vlue of 2 d? () (b) 2 d? (b) In Emple 5 we found tht 2 d = 3. 2 d = b prt () of Definition 4. 2 d = 2 d = 3 b prt (b) of the definition. Theorem 3 (Integrls over djcent intervls) the numbers, b, nd c, then If f is piecewise continuous on n intervl contining f() d + c b f() d = c f() d. () Prtil proof: Formul () holds for < b < c becuse Riemnn sum for the integrl from to b plus Riemnn sum from b to c is Riemnn sum for the integrl from to c nd the integrls re the limits of the Riemnn sums. Formul () in other cses follows b ppling Definition 4. For instnce, if < c < b, then b (9) nd (), Adding f() d = = c c f() d + f() d c c b f() d f() d. c f() d to both sides of this eqution gives () in this cse. QED b
p. 9 (3/2/8) Section 6.2, The definite integrl Theorem 3 for < b < c nd positive function f hs the geometric interprettion illustrted in Figure 5. The left side of eqution (9) is equl to the re between the grph nd the -is for c, the integrls on the right re equl to the res for b nd for b c, nd the re of the whole region is equl to the sum of the re of its two prts. = f() FIGURE 5 b c Emple 7 Solution Wht is the integrl of g from to 8 if its integrl from to 5 is 7 nd its integrl from 5 to 8 is 9? 8 g() d = g() d + 8 5 g() d = 7 + 9 = 2 Theorem 4 (Integrls of liner combintions) If f nd g re piecewise continuous on n intervl continng to b, then for n constnts A nd B, [Af() + Bg()] d = A f() d + B g() d. () Proof: An Riemnn sum for the integrl of = Af() + Bg() equls A multiplied b Riemnn sum for the integrl of f, plus B multiplied b Riemnn sum for the integrl of g: [Af(c j ) + Bg(c j )] = A f(c j ) + B g(c j ). j= j= j= Formul () follows since the integrls re the limits of their respective Riemnn sums. QED 35 35 35 Emple 8 Wht is [2p() 3q()] d if p() d = nd q() d = 2? 35 35 35 Solution With () we obtin 35 [2p() 3q()] d = 2 35 35 35 p() d 3 35 = 2() 3(2) = 4. q() d 35
Section 6.2, The definite integrl p. 9 (3/2/8) Unbounded functions A function f is bounded on n intervl [,b] if there is constnt M such tht f() M for ll in the intervl where f() is defined. Piecewise continuous functions, which re the tpe of functions we use in definite integrls, re bounded on finite intervls. The Riemnn integrl f() d is not defined if f is not bounded on [,b]. For emple, d is not defined s Riemnn integrl becuse / s + nd consequentl = / is not bounded on the intervl of integrtion [, ] (Figure 6). The Riemnn sums for this integrl cnnot hve finite limit becuse (s is illustrted in Eercise 5 below) rbitrril lrge Riemnn sums cn be constructed for n prtition b choosing c sufficientl close to. = FIGURE 6 c We will see in Section 7.7 tht integrls of certin unbounded functions cn be defined s improper integrls. These re limits of Riemnn integrls. C Riemnn sum progrms Progrms or other procedures re vilble for mn grphing clcultors nd computer clculus softwre for clculting Riemnn sums. Mn of these progrms nd procedures lso generte the grphs of functions with the rectngles whose res give the Riemnn sums. C Emple 9 () Use Riemnn sum progrm or procedure on clcultor or computer to find the midpoint Riemnn sum for 3 d with prtitions of [, ] into, 2, 5, nd equl subintervls. Use the window.25.25,.25.25 if the progrm or precedure genertes grphs. (b) Use the results of prt () to predict the ect vlue of 3 d. If f is piecewise continuous on [, b], then in the interior of ech subintervl of prtition = < < < N = b, it is equl to function tht is is continuous in the closure of tht subintervl. The ltter function is bounded on the subintervl b the Etreme Vlue Theorem, so f is bounded on [, b]. Riemnn sum progrms for Tes Instruments clcultors cn be found t the web site for the tet, www.mth.ucsd.edu/ shenk/.
p. 92 (3/2/8) Section 6.2, The definite integrl Solution () The midpoint Riemnn sum is.24875 for N =, is.2.496875 for N = 2, is.24995 for N = 5, nd is.2499875 for N =. The rectngles for N = nd 2 re shown in Figures 7 nd 8. (b) It ppers tht the Riemnn sums re pproching 4 s N increses, so we predict tht 3 d = 4. (We could confirm this with integrtion formuls tht we will derive in Section 6.5.) = 3 = 3 FIGURE 7 FIGURE 8 Interctive Emples 6.2 Interctive solutions re on the web pge http//www.mth.ucsd.edu/ shenk/.. Use re formuls from bsic geometr to find the vlue of 2. Clculte the left Riemnn sum for the unequl subintervls.) 3. Clculte the right Riemnn sum for 8 4 ( 2 2) d. 2 d for the prtition < 2 < 4 < 8 of [, 8]. (Notice Q() d reltive to the prtition of [, 2] into four equl subintervls, where = Q() is the function whose grph is shown in Figure 9. 5 = Q() 5 FIGURE 9.5.5 2 In the published tet the interctive solutions of these emples will be on n ccompning CD disk which cn be run b n computer browser without using n internet connection.
Section 6.2, The definite integrl p. 93 (3/2/8) 4. Use three rectngles of equl width with the grph = f() in Figure 2 to find the pproimte vlue of f() d. 75 5 25 = f() FIGURE 2 25 5 5 5 C 5. Use t lest three Riemnn sums, clculted with Riemnn sum procedure on clcultor or computer to predict the vlue of 6. Wht is the vlue of Eercises 6.2 A Answer provided. CONCEPTS: 3 4 F() d if 4 (5 9) d. O Outline of solution provided. F() d = 4 nd 4 3 F() d = 6? C Grphing clcultor or computer required.. () Drw the region between the curve = 2 nd the -is for nd the rectngles tht give the left, right, nd midpoint Riemnn sums for the integrl ( 2 ) reltive to the prtition <.25 <.5 <.75 <. (b) Eplin wh the left Riemnn sum is less thn the integrl, wh the right Riemnn sum is greter thn the integrl, nd wh of the three sums the midpoint Riemnn sum gives the best pproimtion of the integrl. 2. Epress the equtions () 2k d = 2 constnts k s sttements bout res of rectngles. 3. Rewrite the the eqution 4 f() d = 6 so tht it becomes sttement bout res. k d nd (b) f() d+ 4 6 k d = 3 k d for positive f() d for positive function = f() 4. Derive () in the cse of < c < b nd positive continuous f b considering res. 5. Suppose tht = 3 in Figure 6. Illustrte the fct tht Riemnn sums for d cn be rbitrril lrge b finding vlues of c such tht the rectngle in tht drwing () hs re, (b) hs re, nd (c) hs re. BASICS: 6. Use the formul for the re of tringle to find the vlue of 3 ( ) d. 7. Use the fct tht the curve = 6 2 is the upper hlf of the circle 2 + 2 = 6 of rdius 4 with its center t the origin to find the ect vlue of 6 2 d. 4
p. 94 (3/2/8) Section 6.2, The definite integrl 8. Use res to evlute 9. O Clculte 6 6 j(j )(j 2). j= P() d wherep() = { for < < 3 3 for 3 < < 4 2 for 4 < < 6.. O Epress 2 + 2 2 + 3 2 + + 99 2 with summtion nottion. 5 ( ) j 5 ( ) 2 5 ( ). Clculte ( A ), (b A ), (c) j, nd 2 j 5 j= j= j= [ 6 ( ) ] 2 ( ) j j (d) 6. 6 6 j= 2. Epress ( A ) 3 + 2 4 + 3 5 + 4 6 + + 75 77, (b) 5 + 5 4 + 5 9 + 5 6 + + 5 2, nd (c) 2 3 + ( 2 2 ) 3 + ( 23 ) 3 + ( 24 ) 3 + + ( 2 N ) 3 with summtion nottion. 3. O Clculte the right Riemnn sum for <.5 < <.5 < 2. (4 2 ) d corresponding to the prtition 4. O Follow the instructions in Eercise 3, but with the left Riemnn sum. 5. O Clculte the left Riemnn sum for the unequl subintervls.) 6. Wht is the right Riemnn sum for the unequl subintervls.) 9 7. A Clculte () the left nd (b) the right Riemnn sums for d for the prtition < 3 < 5 < 9 of [, 9]. (Notice 2 d for the prtition < 5 < 9 < of [, ].. (Notice P() d reltive to the prtition of [, 2] into four equl subintervls, where = P() is the function whose grph is shown in Figure 2. 5 = P() 5 FIGURE 2.5.5 2
Section 6.2, The definite integrl p. 95 (3/2/8) 4 8. A The grph of = R() is shown in Figure 22. Wht is the midpoint Riemnn sum for R() d corresponding to the prtition of [, 4] into four equl subintervls? 4 3 2 = R() C 9. O FIGURE 22 Predict the vlue of 2 3 4 (4 2 ) d b using Riemnn-sum procedure on clcultor or computer to clculte midpoint Riemnn sums with 2, 5, nd 2 equl subintervls. Use the window 2, 5. Use t lest three Riemnn sums, clculted with Riemnn sum procedure on clcultor or computer, to predict the vlues of the integrls in Eercises 2 through 23. C 2. O 4 2 d C 2. A π sin(π) d 24. O Wht is 25. O Wht is 7 [6f() + 3g()] d if 7 6 h() d if 7 C 22. C 23. 4 f() d = 4 nd 7 h() d = nd 6 2 2 d ( 4 3 ) d 7 h() d = 7? g() d = 5? 7 In Eercises 26 through 34 use res formul from bsic geometr to find the vlues of the integrls. Drw the region(s) whose re(s) give the integrl. 26. O ( 2) d 2 4 27. A d 28. 2 6 d.72 3. A 3. 5 32. O 7 d 3 29. d 33. d 3.96 5 3 4 34. A 2 for 2 H() d where H() = 2 9 2 for < 3 35. O Wht is Q() d if Q() d = 35 nd 2 d 25 2 d Q() d = 2?
p. 96 (3/2/8) Section 6.2, The definite integrl 36. A Wht is 37. Wht is 38. O Wht is 39. Wht is EXPLORATION: 3 3 Y () d if P() d if 4 [2f() 4g()] d if Y () d = nd P() d = nd 3 [5r() + 25s()] d if 5 3 4 f() d = nd r() d = 2 nd 5 Y () d = 25? P() d = 25? 3 4. A Use the formuls for res of rectngles nd circles to evlute 4. O Give the right Riemnn sum for equl subintervls. 42. Give the left Riemnn sum for subintervls. π 43. A Give the generl Riemnn sum for 44. Give the generl Riemnn sum for C 45. A C 46. C 47. A C 48. C 49. The integrl 4.5 (8 + 4 2 ) g() d = 2? s() d = 3? 5 (5 3 4 2 ) d. ( + 2 ) d corresponding to the prtition of [, 5] into N sin d corresponding to the prtition of [, π] into N equl 3 ( + 5 ) d. 3 π/4 π/4 sec d. d hs the vlue 7. Use Riemnn-sum procedure nd tril nd error to find positive integer N such tht the midpoint Riemnn sum with N equl subintervls for the integrl differs from the integrl b less thn. nd more thn.9. Use the Riemnn-sum progrm nd tril nd error to find positive integer N such tht the left 4 nd right Riemnn sums with N equl subintervls for + 3 d differ b more thn nd less thn.2. How much lrger is the right Riemnn sum thn the left Riemnn sum for (3 2 3 ) d () with the prtition into four equl subintervls, (b) with the prtition of [, 2] into 2 equl subintervls, nd (c) with the prtition of [, 2] into 8 equl subintervls? () Find the vlue of ( + 2) d from the formul for the re of trpezoid or from the formuls for res of rectngles nd tringles. (b) Use the Riemnn-sum progrm to clculte midpoint Riemnn sums for (+2) d with N equl subintervls for vrious positive integers N. All such Riemnn sums give the ect vlue of the integrl. Eplin.. () Eplin wh 3 d equls zero. (b) Use the Riemnn-sum progrm to clculte mid- 2 point Riemnn sums for 3 d with N equl subintervls for vrious positive integers N. 2 Wh do ll such Riemnn sums give the ect vlue of the integrl?
Section 6.2, The definite integrl p. 97 (3/2/8) 5. A Show, b writing the sum twice, once forwrd nd once bckwrds, nd dding corresponding terms, tht for n positive integer N, 5. A () Show tht for n positive integer N, + 2 + 3 + + N = 2 N2 + 2N. (2) N 3 + 3N 2 + 3N [(j + ) 3 j 3 ] = (3j 2 + 3j + ). j= j= (b) Equte the epressions on the right of the lst eqution nd use () nd (2) to show tht 52. A Derive the summtion formul, 2 + 2 2 + 3 2 + + N 2 = 3 N3 + 2 N2 + 6N. (3) 3 + 2 3 + 3 3 + + N 3 = 4 N4 + 2 N3 + 4 N2 (4) from (2) b the following rgument: The squre of width 3 +2 3 +3 3 + +N 3 cn be divided into squre of width nd L-shped gnomons of widths 2, 3,...,N s in Figure 23. Show tht the jth gnomen hs re j 3 (Figure 24), so tht the re of the lrge squre is the sum in (4). FIGURE 23 FIGURE 24 In Eercises 53 through 56 use Definition nd formuls (2) through (4) to find the ect vlues of the integrls. 53. O d 54. A d for rbitrr b > 57. Use the formul 55. 56. 2 d j 5 = 6 N6 + 2 N5 + 2 5 N4 2 N2 to evlute j= (End of Section 6.2) 3 d for rbitrr b > 6 d. This derivtion ws known to Arb mthemticins in the eleventh centur. (See Etrit du Fkhrî pr Alkrkhî b F. Woepcke, Pris: l Imprimerie Imperérile, 853, p. 6.)