5.4. The Fundamental Theorem of Calculus. 356 Chapter 5: Integration. Mean Value Theorem for Definite Integrals

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It connects integrtion nd differentition, enling us to comute integrls using n ntiderivtive of the integrnd function rther thn tking limits of Riemnn sums s we did in Section 5.

1 56 Chter 5: Integrtion 5.4 The Fundmentl Theorem of Clculus HISTORICA BIOGRAPHY Sir Isc Newton (64 77) In this section we resent the Fundmentl Theorem of Clculus, which is the centrl theorem of integrl clculus. It connects integrtion nd differentition, enling us to comute integrls using n ntiderivtive of the integrnd function rther thn tking limits of Riemnn sums s we did in Section 5.. einiz nd Newton eloited this reltionshi nd strted mthemticl develoments tht fueled the scientific revolution for the net ers. Along the w, we resent the integrl version of the Men Vlue Theorem, which is nother imortnt theorem of integrl clculus nd used to rove the Fundmentl Theorem. c f() f(c), verge height FIGURE 5.6 The vlue ƒ(c) in the Men Vlue Theorem is, in sense, the verge (or men) height of ƒ on [, ]. When ƒ Ú, the re of the rectngle is the re under the grh of ƒ from to, ƒscds - d = ƒsd d. Men Vlue Theorem for Definite Integrls In the revious section, we defined the verge vlue of continuous function over closed intervl [, ] s the definite integrl ƒsd d divided the length or width - of the intervl. The Men Vlue Theorem for Definite Integrls sserts tht this verge vlue is lws tken on t lest once the function ƒ in the intervl. The grh in Figure 5.6 shows ositive continuous function = ƒsd defined over the intervl [, ]. Geometricll, the Men Vlue Theorem ss tht there is numer c in [, ] such tht the rectngle with height equl to the verge vlue ƒ(c) of the function nd se width - hs ectl the sme re s the region eneth the grh of ƒ from to. THEOREM The Men Vlue Theorem for Definite Integrls If ƒ is continuous on [, ], then t some oint c in [, ], ƒscd = ƒsd d. -

2 5.4 The Fundmentl Theorem of Clculus 57 f() Averge vlue / not ssumed Proof If we divide oth sides of the M-Min Inequlit (Tle 5., Rule 6) s - d, we otin min ƒ ƒsd d m ƒ. - Since ƒ is continuous, the Intermedite Vlue Theorem for Continuous Functions (Section.6) ss tht ƒ must ssume ever vlue etween min ƒ nd m ƒ. It must therefore ssume the vlue s>s - dd ƒsd d t some oint c in [, ]. The continuit of ƒ is imortnt here. It is ossile tht discontinuous function never equls its verge vlue (Figure 5.7). FIGURE 5.7 A discontinuous function need not ssume its verge vlue. EXAMPE Aling the Men Vlue Theorem for Integrls Find the verge vlue of ƒsd = 4 - on [, ] nd where ƒ ctull tkes on this vlue t some oint in the given domin vsƒd = = ƒsd d - s4 - d d = - 4 d - d 4 FIGURE 5.8 The re of the rectngle with se [, ] nd height 5> (the verge vlue of the function ƒsd = 4 - ) is equl to the re etween the grh of ƒ nd the -is from to (Emle ). = 4s - d - - = 4 - = 5. Section 5., Eqs. () nd () The verge vlue of ƒsd = 4 - over [, ] is 5>. The function ssumes this vlue when 4 - = 5> or = >. (Figure 5.8) In Emle, we ctull found oint c where ƒ ssumed its verge vlue setting ƒ() equl to the clculted verge vlue nd solving for. It s not lws ossile to solve esil for the vlue c. Wht else cn we lern from the Men Vlue Theorem for integrls? Here s n emle. EXAMPE Show tht if ƒ is continuous on [, ], Z, nd if then ƒsd = t lest once in [, ]. The verge vlue of ƒ on [, ] is vsƒd = ƒsd d =, ƒsd d = - B the Men Vlue Theorem, ƒ ssumes this vlue t some oint c H [, ]. - # =.

3 58 Chter 5: Integrtion Fundmentl Theorem, Prt If ƒ(t) is n integrle function over finite intervl I, then the integrl from n fied numer H I to nother numer H I defines new function F whose vlue t is Fsd = ƒstd dt. () re F() f(t) FIGURE 5.9 The function F() defined Eqution () gives the re under the grh of ƒ from to when ƒ is nonnegtive nd 7. t For emle, if ƒ is nonnegtive nd lies to the right of, then F() is the re under the grh from to (Figure 5.9). The vrile is the uer limit of integrtion of n integrl, ut F is just like n other rel-vlued function of rel vrile. For ech vlue of the inut, there is well-defined numericl outut, in this cse the definite integrl of ƒ from to. Eqution () gives w to define new functions, ut its imortnce now is the connection it mkes etween integrls nd derivtives. If ƒ is n continuous function, then the Fundmentl Theorem sserts tht F is differentile function of whose derivtive is ƒ itself. At ever vlue of, d d Fsd = d ƒstd dt = ƒsd. d To gin some insight into wh this result holds, we look t the geometr ehind it. If ƒ Ú on [, ], then the comuttion of F sd from the definition of the derivtive mens tking the limit s h : of the difference quotient Fs + hd - Fsd. h f() f(t) h FIGURE 5. In Eqution (), F() is the re to the left of. Also, Fs + hd is the re to the left of + h. The difference quotient [Fs + hd - Fsd]>h is then roimtel equl to ƒ(), the height of the rectngle shown here. t For h 7, the numertor is otined sutrcting two res, so it is the re under the grh of ƒ from to + h (Figure 5.). If h is smll, this re is roimtel equl to the re of the rectngle of height ƒ() nd width h, which cn e seen from Figure 5.. Tht is, Fs + hd - Fsd hƒsd. Dividing oth sides of this roimtion h nd letting h :, it is resonle to eect tht Fs + hd - Fsd F sd = lim h: h = ƒsd. This result is true even if the function ƒ is not ositive, nd it forms the first rt of the Fundmentl Theorem of Clculus. THEOREM 4 The Fundmentl Theorem of Clculus Prt If ƒ is continuous on [, ] then Fsd = ƒstd dt is continuous on [, ] nd differentile on (, ) nd its derivtive is ƒsd; F sd = d ƒstd dt = ƒsd. d ()

4 5.4 The Fundmentl Theorem of Clculus 59 Before roving Theorem 4, we look t severl emles to gin etter understnding of wht it ss. EXAMPE Aling the Fundmentl Theorem Use the Fundmentl Theorem to find () d cos t dt d () d d + t dt (c) (d) d d d d 5 if = t sin t dt if = cos t dt () () d cos t dt = cos d d d + t dt = + Eq. with ƒ(t) = cos t (c) Rule for integrls in Tle 5. of Section 5. sets this u for the Fundmentl Theorem. d d = d 5 t sin t dt = d d d - t sin t dt 5 =- d t sin t dt d Rule (d) The uer limit of integrtion is not ut. This mkes comosite of the two functions, u = cos t dt nd u =. Eq. with ƒstd = + t We must therefore l the Chin Rule when finding d>d. d d = d # du du d = d u du = cos u # du d = coss d # = cos 5 = - sin cos t dt # du d

5 6 Chter 5: Integrtion EXAMPE 4 Constructing Function with Given Derivtive nd Vlue Find function = ƒsd on the domin s ->, >d with derivtive tht stisfies the condition ƒsd = 5. The Fundmentl Theorem mkes it es to construct function with derivtive tn tht equls t = : Since sd = tn t dt =, we hve onl to dd 5 to this function to construct one with derivtive tn whose vlue t = is 5: Although the solution to the rolem in Emle 4 stisfies the two required conditions, ou might sk whether it is in useful form. The nswer is es, since tod we hve comuters nd clcultors tht re cle of roimting integrls. In Chter 7 we will lern to write the solution in Emle 4 ectl s We now give roof of the Fundmentl Theorem for n ritrr continuous function. Proof of Theorem 4 We rove the Fundmentl Theorem ling the definition of the derivtive directl to the function F(), when nd + h re in (, ). This mens writing out the difference quotient nd showing tht its limit s h : is the numer ƒ() for ech in (, ). When we relce Fs + hd nd F() their defining integrls, the numertor in Eqution () ecomes + h Fs + hd - Fsd = ƒstd dt - ƒstd dt. The Additivit Rule for integrls (Tle 5., Rule 5) simlifies the right side to so tht Eqution () ecomes ƒsd = tn t dt + 5. Fs + hd - Fsd h d d = tn = tn t dt. cos = ln ` cos ` + 5. Fs + hd - Fsd h + h ƒstd dt, = [Fs + hd - Fsd] h = + h ƒstd dt. h () (4)

6 5.4 The Fundmentl Theorem of Clculus 6 According to the Men Vlue Theorem for Definite Integrls, the vlue of the lst eression in Eqution (4) is one of the vlues tken on ƒ in the intervl etween nd + h. Tht is, for some numer c in this intervl, + h ƒstd dt = ƒscd. h (5) As h :, + h roches, forcing c to roch lso (ecuse c is tred etween nd + h). Since ƒ is continuous t, ƒ(c) roches ƒ(): Going ck to the eginning, then, we hve lim ƒscd = ƒsd. h: (6) df d = lim Fs + hd - Fsd h: h + h = lim ƒstd dt h: h = lim ƒscd h: = ƒsd. Definition of derivtive Eq. (4) Eq. (5) Eq. (6) If = or, then the limit of Eqution () is interreted s one-sided limit with h : + or h : -, resectivel. Then Theorem in Section. shows tht F is continuous for ever oint of [, ]. This concludes the roof. Fundmentl Theorem, Prt (The Evlution Theorem) We now come to the second rt of the Fundmentl Theorem of Clculus. This rt descries how to evlute definite integrls without hving to clculte limits of Riemnn sums. Insted we find nd evlute n ntiderivtive t the uer nd lower limits of integrtion. THEOREM 4 (Continued) The Fundmentl Theorem of Clculus Prt If ƒ is continuous t ever oint of [, ] nd F is n ntiderivtive of ƒ on [, ], then ƒsd d = Fsd - Fsd. Proof Prt of the Fundmentl Theorem tells us tht n ntiderivtive of ƒ eists, nmel Gsd = ƒstd dt. Thus, if F is n ntiderivtive of ƒ, then Fsd = Gsd + C for some constnt C for 6 6 ( Corollr of the Men Vlue Theorem for Derivtives, Section 4.). Since oth F nd G re continuous on [, ], we see tht F() = G() + C lso holds when = nd = tking one-sided limits (s : + nd : - d.

7 6 Chter 5: Integrtion Evluting Fsd - Fsd, we hve Fsd - Fsd = [Gsd + C] - [Gsd + C] = Gsd - Gsd The theorem ss tht to clculte the definite integrl of ƒ over [, ] ll we need to do is:. Find n ntiderivtive F of ƒ, nd. Clculte the numer ƒsd d = Fsd - Fsd. The usul nottion for Fsd - Fsd is deending on whether F hs one or more terms. = ƒstd dt - ƒstd dt = ƒstd dt - = ƒstd dt. Fsd d or cfsd d, EXAMPE 5 Evluting Integrls () () (c) ->4 4 cos d = sin d sec tn d = sec d = sec - sec - -/4 4 = d = c/ + 4 d = sin - sin = - = 4 The rocess used in Emle 5 ws much esier thn Riemnn sum comuttion. The conclusions of the Fundmentl Theorem tell us severl things. Eqution () cn e rewritten s d ƒstd dt = df d d = ƒsd, which ss tht if ou first integrte the function ƒ nd then differentite the result, ou get the function ƒ ck gin. ikewise, the eqution = cs4d / d - csd/ + 4 d = [8 + ] - [5] = 4. df dt dt = ƒstd dt = Fsd - Fsd ss tht if ou first differentite the function F nd then integrte the result, ou get the function F ck (djusted n integrtion constnt). In sense, the rocesses of integr-

5.4 The Fundmentl Theorem of Clculus 6 tion nd differentition re inverses of ech other. The Fundmentl Theorem lso ss tht ever continuous function ƒ hs n ntiderivtive F.

8 5.4 The Fundmentl Theorem of Clculus 6 tion nd differentition re inverses of ech other. The Fundmentl Theorem lso ss tht ever continuous function ƒ hs n ntiderivtive F. And it ss tht the differentil eqution d>d = ƒsd hs solution (nmel, the function = F() ) for ever continuous function ƒ. Totl Are The Riemnn sum contins terms such s ƒsc k d k which give the re of rectngle when ƒsc k d is ositive. When ƒsc k d is negtive, then the roduct ƒsc k d k is the negtive of the rectngle s re. When we dd u such terms for negtive function we get the negtive of the re etween the curve nd the -is. If we then tke the solute vlue, we otin the correct ositive re. EXAMPE 6 Finding Are Using Antiderivtives Clculte the re ounded the -is nd the rol = FIGURE 5. The re of this rolic rch is clculted with definite integrl (Emle 6). which gives We find where the curve crosses the -is setting The curve is sketched in Figure 5., nd is nonnegtive on The re is - = = = s + ds - d, = - or =. s6 - - d d = c6 - - d - = = 5 6. The curve in Figure 5. is n rch of rol, nd it is interesting to note tht the re under such n rch is ectl equl to two-thirds the se times the ltitude: s5d 5 4 = 5 6 = 5 6. [-, ]. Are sin Are To comute the re of the region ounded the grh of function = ƒsd nd the -is requires more cre when the function tkes on oth ositive nd negtive vlues. We must e creful to rek u the intervl [, ] into suintervls on which the function doesn t chnge sign. Otherwise we might get cncelltion etween ositive nd negtive signed res, leding to n incorrect totl. The correct totl re is otined dding the solute vlue of the definite integrl over ech suintervl where ƒ() does not chnge sign. The term re will e tken to men totl re. FIGURE 5. The totl re etween = sin nd the -is for is the sum of the solute vlues of two integrls (Emle 7). EXAMPE 7 Cnceling Ares Figure 5. shows the grh of the function ƒsd = sin etween = nd Comute () the definite integrl of ƒ() over [, ]. () the re etween the grh of ƒ() nd the -is over [, ]. =.

9 64 Chter 5: Integrtion The definite integrl for ƒsd = sin is given sin d = -cos d = -[cos - cos ] = -[ - ] =. The definite integrl is zero ecuse the ortions of the grh ove nd elow the -is mke cnceling contriutions. The re etween the grh of ƒ() nd the -is over [, ] is clculted reking u the domin of sin into two ieces: the intervl [, ] over which it is nonnegtive nd the intervl [, ] over which it is nonositive. sin d = -cos d sin d = -cos d = -[cos - cos ] = -[- - ] =. = -[cos - cos ] = -[ - s -d] = -. The second integrl gives negtive vlue. The re etween the grh nd the is is otined dding the solute vlues Are = ƒ ƒ + ƒ - ƒ = 4. Summr: To find the re etween the grh of = ƒsd nd the -is over the intervl [, ], do the following:. Sudivide [, ] t the zeros of ƒ. Are 5. Integrte ƒ over ech suintervl.. Add the solute vlues of the integrls. Are 8 8 EXAMPE 8 Finding Are Using Antiderivtives Find the re of the region etween the -is nd the grh of ƒ() = - -, -. FIGURE 5. The region etween the curve = - - nd the -is (Emle 8). First find the zeros of ƒ. Since the zeros re =, -, nd (Figure 5.). The zeros sudivide [-, ] into two suintervls: [-, ], on which ƒ Ú, nd [, ], on which ƒ. We integrte ƒ over ech suintervl nd dd the solute vlues of the clculted integrls. - ƒsd = - - = s - - d = s + ds - d, s - - d d = c d - s - - d d = c d The totl enclosed re is otined dding the solute vlues of the clculted integrls, = - c d = 5 = c d - =-8 Totl enclosed re = 5 + ` - 8 ` = 7.

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp. MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.