5.1 A r e a s a n d D i s t a n c e s

Transcription

1 To compute n re we pproimte region b rectngles nd let the number of rectngles become lrge. The precise re is the limit of these sums of res of rectngles. Integrls

2 Now is good time to red (or rered) A Preview of Clculus (see pge ). It discusses the unifing ides of clculus nd helps put in perspective where we hve been nd where we re going. In Chpter we used the tngent nd velocit problems to introduce the derivtive, which is the centrl ide in differentil clculus. In much the sme w, this chpter strts with the re nd distnce problems nd uses them to formulte the ide of definite integrl, which is the bsic concept of integrl clculus. We will see in Chpters 6 nd 8 how to use the integrl to solve problems concerning volumes, lengths of curves, popultion predictions, crdic output, forces on dm, work, consumer surplus, nd bsebll, mong mn others. There is connection between integrl clculus nd differentil clculus. The Fundmentl Theorem of Clculus reltes the integrl to the derivtive, nd we will see in this chpter tht it gretl simplifies the solution of mn problems. 5. A r e s n d D i s t n c e s In this section we discover tht in tring to find the re under curve or the distnce trveled b cr, we end up with the sme specil tpe of limit. T h e A r e P r o b l e m We begin b ttempting to solve the re problem: Find the re of the region S tht lies under the curve f from to b. This mens tht S, illustrted in Figure, is bounded b the grph of continuous function f [where f ], the verticl lines nd b, nd the -is. =ƒ FIGURE S=s(,) b, ƒd = S =b b In tring to solve the re problem we hve to sk ourselves: Wht is the mening of the word re? This question is es to nswer for regions with stright sides. For rectngle, the re is defined s the product of the length nd the width. The re of tringle is hlf the bse times the height. The re of polgon is found b dividing it into tringles (s in Figure ) nd dding the res of the tringles. A A w h A A l b FIGURE A=lw A= bh A=A +A +A +A

3 7 CHAPTER 5 INTEGRALS However, it isn t so es to find the re of region with curved sides. We ll hve n intuitive ide of wht the re of region is. But prt of the re problem is to mke this intuitive ide precise b giving n ect definition of re. Recll tht in defining tngent we first pproimted the slope of the tngent line b slopes of secnt lines nd then we took the limit of these pproimtions. We pursue similr ide for res. We first pproimte the region S b rectngles nd then we tke the limit of the res of these rectngles s we increse the number of rectngles. The following emple illustrtes the procedure. Tr plcing rectngles to estimte the re. Resources / Module 6 / Wht Is Are? / Estimting Are under Prbol EXAMPLE Use rectngles to estimte the re under the prbol from to (the prbolic region S illustrted in Figure ). (,) = S FIGURE SOLUTION We first notice tht the re of S must be somewhere between nd becuse S is contined in squre with side length, but we cn certinl do better thn tht. Suppose we divide S into four strips S,,, nd b drwing the verticl lines S S S 4,, nd 4 4 s in Figure 4(). (,) (,) = S S S S FIGURE 4 () (b) = (,) We cn pproimte ech strip b rectngle whose bse is the sme s the strip nd whose height is the sme s the right edge of the strip [see Figure 4(b)]. In other words, the heights of these rectngles re the vlues of the function f t the right end points of the subintervls [, 4], [ 4, ], [, 4], nd [ 4, ]. Ech rectngle hs width nd the heights re ( 4), ( ), ( 4) 4, nd. If we let R 4 be the sum of the res of these pproimting rectngles, we get R 4 4 ( 4) 4 ( ) 4 ( 4) From Figure 4(b) we see tht the re A of S is less thn R 4, so A FIGURE 5 4 Insted of using the rectngles in Figure 4(b) we could use the smller rectngles in Figure 5 whose heights re the vlues of f t the left endpoints of the subintervls. (The leftmost rectngle hs collpsed becuse its height is.) The sum of the res of these

4 SECTION 5. AREAS AND DISTANCES 7 pproimting rectngles is L ( 4) 4 ( ) 4 ( 4) L 4 We see tht the re of S is lrger thn, so we hve lower nd upper estimtes for A:.875 A We cn repet this procedure with lrger number of strips. Figure 6 shows wht hppens when we divide the region S into eight strips of equl width. = (,) (,) 8 FIGURE 6 Approimting S with eight rectngles () Using left endpoints (b) Using right endpoints 8 B computing the sum of the res of the smller rectngles L 8 nd the sum of the res of the lrger rectngles R 8, we obtin better lower nd upper estimtes for A:.7475 A n L n R n So one possible nswer to the question is to s tht the true re of S lies somewhere between.7475 nd We could obtin better estimtes b incresing the number of strips. The tble t the left shows the results of similr clcultions (with computer) using n rectngles whose heights re found with left endpoints L n or right endpoints R n. In prticulr, we see b using 5 strips tht the re lies between.4 nd.44. With strips we nrrow it down even more: A lies between.85 nd.85. A good estimte is obtined b verging these numbers: A.5. (,) From the vlues in the tble in Emple, it looks s if is pproching s n increses. We confirm this in the net emple. R n = EXAMPLE For the region S in Emple, show tht the sum of the res of the upper pproimting rectngles pproches, tht is, lim nl R n n FIGURE 7 SOLUTION R n is the sum of the res of the n rectngles in Figure 7. Ech rectngle hs width n nd the heights re the vlues of the function f t the points n, n, n,..., n n; tht is, the heights re n, n, n,..., n n.

5 7 CHAPTER 5 INTEGRALS Thus R n n n n n n n n n n n n n n n The ides in Emples nd re eplored in Module 5./5./7.7 for vriet of functions. Here we need the formul for the sum of the squres of the first n positive integers: n n n n 6 Perhps ou hve seen this formul before. It is proved in Emple 5 in Appendi E. Putting Formul into our epression for, we get R n n n n n n n 6 6n R n Here we re computing the limit of the sequence R n. Sequences were discussed in A Preview of Clculus nd will be studied in detil in Chpter. Their limits re clculted in the sme w s limits t infinit (Section.6). In prticulr, we know tht lim nl n Thus, we hve n n lim R n lim nl nl 6n lim nl n n n 6 n n lim n nl 6 6 It cn be shown tht the lower pproimting sums lso pproch, tht is, lim nl L n From Figures 8 nd 9 it ppers tht, s n increses, both L n nd R n become better nd better pproimtions to the re of S. Therefore, we define the re A to be the limit of the n= R =.85 n= R Å.5 n=5 R =.44 FIGURE 8

6 SECTION 5. AREAS AND DISTANCES 7 n= L =.85 n= L Å.69 n=5 L =.4 FIGURE 9 The re is the number tht is smller thn ll upper sums nd lrger thn ll lower sums sums of the res of the pproimting rectngles, tht is, Let s ppl the ide of Emples nd to the more generl region S of Figure. We strt b subdividing S into n strips S, S,..., S n of equl width s in Figure. The width of the intervl, b is b, so the width of ech of the n strips is These strips divide the intervl [, b] into n subintervls where nd n b. The right endpoints of the subintervls re, A lim R n lim L n nl nl b n,,,,,,..., n, n, =ƒ,... S S S S i S n FIGURE... i- i... n- b Let s pproimte the ith strip S i b rectngle with width nd height f i, which is the vlue of f t the right endpoint (see Figure ). Then the re of the ith rectngle Î f( i ) FIGURE i- i b

7 74 CHAPTER 5 INTEGRALS is f i. Wht we think of intuitivel s the re of S is pproimted b the sum of the res of these rectngles, which is R n f f f n () n= b Figure shows this pproimtion for n, 4, 8, nd. Notice tht this pproimtion ppers to become better nd better s the number of strips increses, tht is, s nl. Therefore, we define the re A of the region S in the following w. The re A of the region S tht lies under the grph of the continu- f is the limit of the sum of the res of pproimting rectngles: Definition ous function A lim R n lim f f f n nl nl b (b) n=4 f It cn be proved tht the limit in Definition lws eists, since we re ssuming tht is continuous. It cn lso be shown tht we get the sme vlue if we use left endpoints: A lim nl L n lim f f f n nl (c) n=8 b In fct, insted of using left endpoints or right endpoints, we could tke the height of the ith rectngle to be the vlue of f t n number * i in the ith subintervl i, i. We cll the numbers *, *,..., n * the smple points. Figure shows pproimting rectngles when the smple points re not chosen to be endpoints. So more generl epression for the re of S is 4 A lim f * f * f n * nl (d) n= b Î FIGURE f( i *) i- i n- b * * * i * n * FIGURE This tells us to end with i=n. This tells us to dd. This tells us to strt with i=m. µ f( i )Î n i=m We often use sigm nottion to write sums with mn terms more compctl. For instnce, n f i f f f n i

8 SECTION 5. AREAS AND DISTANCES 75 So the epressions for re in Equtions,, nd 4 cn be written s follows: If ou need prctice with sigm nottion, look t the emples nd tr some of the eercises in Appendi E. A lim nl n f i i A lim nl n f i i A lim nl n f * i i We could lso rewrite Formul in the following w: n n n n i i 6 EXAMPLE Let A be the re of the region tht lies under the grph of f e between nd. () Using right endpoints, find n epression for A s limit. Do not evlute the limit. (b) Estimte the re b tking the smple points to be midpoints nd using four subintervls nd then ten subintervls. SOLUTION () Since nd b, the width of subintervl is n n So n, 4 n, 6 n, i i n, nd n n n. The sum of the res of the pproimting rectngles is According to Definition, the re is A lim R n lim nl nl n e n e 4 n e 6 n e n n Using sigm nottion we could write R n f f f n e n e e n 4 n e n e n n n e n A lim nl n n e i n i It is difficult to evlute this limit directl b hnd, but with the id of computer lgebr sstem it isn t hrd (see Eercise 4). In Section 5. we will be ble to find A more esil using different method. (b) With n 4 the subintervls of equl width.5 re,.5,.5,,,.5, nd.5,. The midpoints of these subintervls re *.5, *.75, *.5,

9 76 CHAPTER 5 INTEGRALS =e FIGURE 4 nd * 4.75, nd the sum of the res of the four pproimting rectngles (see Figure 4) is M 4 4 f * i i f.5 f.75 f.5 f.75 e.5.5 e.75.5 e.5.5 e.75.5 e.5 e.75 e.5 e So n estimte for the re is A.8557 =e With n the subintervls re,.,.,.4,...,.8, nd the midpoints re *., *., *.5,..., *.9. Thus A M f. f. f.5 f.9. e. e. e.5 e.9.86 FIGURE 5 From Figure 5 it ppers tht this estimte is better thn the estimte with n 4. T h e D i s t n c e P r o b l e m Now let s consider the distnce problem: Find the distnce trveled b n object during certin time period if the velocit of the object is known t ll times. (In sense this is the inverse problem of the velocit problem tht we discussed in Section..) If the velocit remins constnt, then the distnce problem is es to solve b mens of the formul distnce velocit time But if the velocit vries, it s not so es to find the distnce trveled. We investigte the problem in the following emple. EXAMPLE 4 Suppose the odometer on our cr is broken nd we wnt to estimte the distnce driven over -second time intervl. We tke speedometer redings ever five seconds nd record them in the following tble: Time (s) Velocit (mi h) In order to hve the time nd the velocit in consistent units, let s convert the velocit redings to feet per second ( mi h 58 6 ft s): Time (s) Velocit (ft s) During the first five seconds the velocit doesn t chnge ver much, so we cn estimte the distnce trveled during tht time b ssuming tht the velocit is constnt. If we tke the velocit during tht time intervl to be the initil velocit (5 ft s), then we

10 SECTION 5. AREAS AND DISTANCES 77 obtin the pproimte distnce trveled during the first five seconds: 5 ft s 5 s 5 ft Similrl, during the second time intervl the velocit is pproimtel constnt nd we tke it to be the velocit when t 5 s. So our estimte for the distnce trveled from t 5 s to t s is ft s 5 s 55 ft If we dd similr estimtes for the other time intervls, we obtin n estimte for the totl distnce trveled: ft We could just s well hve used the velocit t the end of ech time period insted of the velocit t the beginning s our ssumed constnt velocit. Then our estimte becomes ft If we hd wnted more ccurte estimte, we could hve tken velocit redings ever two seconds, or even ever second. 4 FIGURE 6 t Perhps the clcultions in Emple 4 remind ou of the sums we used erlier to estimte res. The similrit is eplined when we sketch grph of the velocit function of the cr in Figure 6 nd drw rectngles whose heights re the initil velocities for ech time intervl. The re of the first rectngle is 5 5 5, which is lso our estimte for the distnce trveled in the first five seconds. In fct, the re of ech rectngle cn be interpreted s distnce becuse the height represents velocit nd the width represents time. The sum of the res of the rectngles in Figure 6 is L 6 5, which is our initil estimte for the totl distnce trveled. In generl, suppose n object moves with velocit v f t, where t b nd f t (so the object lws moves in the positive direction). We tke velocit redings t times t, t, t,..., t n b so tht the velocit is pproimtel constnt on ech subintervl. If these times re equll spced, then the time between consecutive redings is t b n. During the first time intervl the velocit is pproimtel f t nd so the distnce trveled is pproimtel f t t. Similrl, the distnce trveled during the second time intervl is bout f t t nd the totl distnce trveled during the time intervl, b is pproimtel f t t f t t f t n t n i f t i t If we use the velocit t right endpoints insted of left endpoints, our estimte for the totl distnce becomes f t t f t t f t n t n i f t i t The more frequentl we mesure the velocit, the more ccurte our estimtes become, so it seems plusible tht the ect distnce d trveled is the limit of such epressions: 5 d lim nl n f t i t lim i nl n f t i t i We will see in Section 5.4 tht this is indeed true.

11 78 CHAPTER 5 INTEGRALS Becuse Eqution 5 hs the sme form s our epressions for re in Equtions nd, it follows tht the distnce trveled is equl to the re under the grph of the velocit function. In Chpters 6 nd 8 we will see tht other quntities of interest in the nturl nd socil sciences such s the work done b vrible force or the crdic output of the hert cn lso be interpreted s the re under curve. So when we compute res in this chpter, ber in mind tht the cn be interpreted in vriet of prcticl ws. 5. Eercises. () B reding vlues from the given grph of f, use five rectngles to find lower estimte nd n upper estimte for the re under the given grph of f from to. In ech cse sketch the rectngles tht ou use. (b) Find new estimtes using rectngles in ech cse. 5 =ƒ right endpoints. Sketch the grph nd the rectngles. Is our estimte n underestimte or n overestimte? (b) Repet prt () using left endpoints. 5. () Estimte the re under the grph of f from to using three rectngles nd right endpoints. Then improve our estimte b using si rectngles. Sketch the curve nd the pproimting rectngles. (b) Repet prt () using left endpoints. (c) Repet prt () using midpoints. (d) From our sketches in prts () (c), which ppers to be the best estimte?. 5 () Use si rectngles to find estimtes of ech tpe for the re under the given grph of f from to. (i) L 6 (smple points re left endpoints) (ii) R 6 (smple points re right endpoints) (iii) M 6 (smple points re midpoints) (b) Is L 6 n underestimte or overestimte of the true re? (c) Is R 6 n underestimte or overestimte of the true re? (d) Which of the numbers L 6, R 6, or M 6 gives the best estimte? Eplin. 8 =ƒ 4 ; 6. () Grph the function f e,. (b) Estimte the re under the grph of f using four pproimting rectngles nd tking the smple points to be (i) right endpoints (ii) midpoints In ech cse sketch the curve nd the rectngles. (c) Improve our estimtes in prt (b) b using eight rectngles. 7 8 With progrmmble clcultor (or computer), it is possible to evlute the epressions for the sums of res of pproimting rectngles, even for lrge vlues of n, using looping. (On TI use the Is commnd or For-EndFor loop, on Csio use Isz, on n HP or in BASIC use FOR-NEXT loop.) Compute the sum of the res of pproimting rectngles using equl subintervls nd right endpoints for n,, nd 5. Then guess the vlue of the ect re. 7. The region under sin from to 8. The region under from to 4 8. () Estimte the re under the grph of f from to 5 using four pproimting rectngles nd right endpoints. Sketch the grph nd the rectngles. Is our estimte n underestimte or n overestimte? (b) Repet prt () using left endpoints. 4. () Estimte the re under the grph of f 5 from to 5 using five pproimting rectngles nd CAS CAS 9. Some computer lgebr sstems hve commnds tht will drw pproimting rectngles nd evlute the sums of their res, t lest if * i is left or right endpoint. (For instnce, in Mple use leftbo, rightbo, leftsum, nd rightsum.) () If f s, 4, find the left nd right sums for n,, nd 5. (b) Illustrte b grphing the rectngles in prt (). (c) Show tht the ect re under f lies between 4.6 nd () If f sin sin,, use the commnds discussed in Eercise 9 to find the left nd right sums for n,, nd 5.

12 SECTION 5. AREAS AND DISTANCES 79. (b) Illustrte b grphing the rectngles in prt (). (c) Show tht the ect re under f lies between.87 nd.9. The speed of runner incresed stedil during the first three seconds of rce. Her speed t hlf-second intervls is given in the tble. Find lower nd upper estimtes for the distnce tht she trveled during these three seconds. t (s) v (ft s) Speedometer redings for motorccle t -second intervls re given in the tble. () Estimte the distnce trveled b the motorccle during this time period using the velocities t the beginning of the time intervls. (b) Give nother estimte using the velocities t the end of the time periods. (c) Are our estimtes in prts () nd (b) upper nd lower estimtes? Eplin. t (s) v (ft s) Oil leked from tnk t rte of r t liters per hour. The rte decresed s time pssed nd vlues of the rte t -hour time intervls re shown in the tble. Find lower nd upper estimtes for the totl mount of oil tht leked out. t h r t (L h) When we estimte distnces from velocit dt, it is sometimes necessr to use times t, t, t, t,... tht re not equll spced. We cn still estimte distnces using the time periods t i t i t i. For emple, on M 7, 99, the spce shuttle Endevour ws lunched on mission STS-49, the purpose of which ws to instll new perigee kick motor in n Intelst communictions stellite. The tble, provided b NASA, gives the velocit dt for the shuttle between liftoff nd the jettisoning of the solid rocket boosters. Event Time (s) Velocit (ft s) Lunch Begin roll mneuver 85 End roll mneuver 5 9 Throttle to 89% 447 Throttle to 67% 74 Throttle to 4% 59 5 Mimum dnmic pressure Solid rocket booster seprtion Use these dt to estimte the height bove Erth s surfce of the spce shuttle Endevour, 6 seconds fter liftoff. The velocit grph of brking cr is shown. Use it to estimte the distnce trveled b the cr while the brkes re pplied. 6. The velocit grph of cr ccelerting from rest to speed of km h over period of seconds is shown. Estimte the distnce trveled during this period. 7 9 Use Definition to find n epression for the re under the grph of f s limit. Do not evlute the limit. 7. f s 4, 6 8. f ln, 9. f cos, Determine region whose re is equl to the given limit. Do not evlute the limit... lim nl n 5 i i n lim nl n i (ft/s) 6 (km/h) 4n t (seconds) tn i 4n n 4 6 t (seconds). () Use Definition to find n epression for the re under the curve from to s limit. (b) The following formul for the sum of the cubes of the first n integers is proved in Appendi E. Use it to evlute the limit in prt (). n n n

13 8 CHAPTER 5 INTEGRALS CAS CAS. () Epress the re under the curve 5 from to s limit. (b) Use computer lgebr sstem to find the sum in our epression from prt (). (c) Evlute the limit in prt (). 4. Find the ect re of the region under the grph of e from to b using computer lgebr sstem to evlute the sum nd then the limit in Emple (). Compre our nswer with the estimte obtined in Emple (b). CAS 5. Find the ect re under the cosine curve cos from to b, where b. (Use computer lgebr sstem both to evlute the sum nd compute the limit.) In prticulr, wht is the re if b? 6. () Let A n be the re of polgon with n equl sides inscribed in circle with rdius r. B dividing the polgon into n congruent tringles with centrl ngle n, show tht A n nr sin n. (b) Show tht lim nl A n r. [Hint: Use Eqution.4..] 5. T h e D e f i n i t e I n t e g r l We sw in Section 5. tht limit of the form lim nl n f * i lim f * f * f n * i nl rises when we compute n re. We lso sw tht it rises when we tr to find the distnce trveled b n object. It turns out tht this sme tpe of limit occurs in wide vriet of situtions even when f is not necessril positive function. In Chpters 6 nd 8 we will see tht limits of the form () lso rise in finding lengths of curves, volumes of solids, centers of mss, force due to wter pressure, nd work, s well s other quntities. We therefore give this tpe of limit specil nme nd nottion. Definition of Definite Integrl If f is continuous function defined for b, we divide the intervl, b into n subintervls of equl width b n. We let,,,..., n ( b) be the endpoints of these subintervls nd we let *, *,..., * n be n smple points in these subintervls, so * i lies in the ith subintervl i, i. Then the definite integrl of f from to b is b f d lim nl n f * i i Becuse we hve ssumed tht f is continuous, it cn be proved tht the limit in Definition lws eists nd gives the sme vlue no mtter how we choose the smple points * i. (See Note 4 for precise definition of this tpe of limit.) If we tke the smple points to be right endpoints, then * i i nd the definition of n integrl becomes b f d lim n l n f i i

14 SECTION 5. THE DEFINITE INTEGRAL 8 If we choose the smple points to be left endpoints, then * i i nd the definition becomes Alterntivel, we could choose * i to be the midpoint of the subintervl or n other number between i nd i. Although most of the functions tht we encounter re continuous, the limit in Definition lso eists if f hs finite number of removble or jump discontinuities (but not infinite discontinuities). (See Section.5.) So we cn lso define the definite integrl for such functions. NOTE The smbol ws introduced b Leibniz nd is clled n integrl sign. It is n elongted S nd ws chosen becuse n integrl is limit of sums. In the nottion b f d, f is clled the integrnd nd nd b re clled the limits of integrtion; is the lower limit nd b is the upper limit. The smbol d hs no officil mening b itself; b f d is ll one smbol. The procedure of clculting n integrl is clled integrtion. b f d lim nl n f i i b NOTE The definite integrl f d is number; it does not depend on. In fct, we could use n letter in plce of without chnging the vlue of the integrl: b f d b f t dt b f r dr NOTE The sum n f * i i Bernhrd Riemnn received his Ph.D. under the direction of the legendr Guss t the Universit of Göttingen nd remined there to tech. Guss, who ws not in the hbit of prising other mthemticins, spoke of Riemnn s cretive, ctive, trul mthemticl mind nd gloriousl fertile originlit. The definition () of n integrl tht we use is due to Riemnn. He lso mde mjor contributions to the theor of functions of comple vrible, mthemticl phsics, number theor, nd the foundtions of geometr. Riemnn s brod concept of spce nd geometr turned out to be the right setting, 5 ers lter, for Einstein s generl reltivit theor. Riemnn s helth ws poor throughout his life, nd he died of tuberculosis t the ge of 9. tht occurs in Definition is clled Riemnn sum fter the Germn mthemticin Bernhrd Riemnn (86 866). We know tht if f hppens to be positive, then the Riemnn sum cn be interpreted s sum of res of pproimting rectngles (see Figure ). B compring Definition with the definition of re in Section 5., we see tht the definite integrl b f d cn be interpreted s the re under the curve f from to b. (See Figure.) Î =ƒ i * b b FIGURE If ƒ, the Riemnn sum µf( i *)Î is the sum of res of rectngles. FIGURE b If ƒ, the integrl j ƒd is the re under the curve =ƒ from to b.

15 8 CHAPTER 5 INTEGRALS FIGURE µf( i *)Î is n pproimtion to the net re =ƒ + + _ b + =ƒ _ FIGURE 4 b j ƒd is the net re + b If f tkes on both positive nd negtive vlues, s in Figure, then the Riemnn sum is the sum of the res of the rectngles tht lie bove the -is nd the negtives of the res of the rectngles tht lie below the -is (the res of the gold rectngles minus the res of the blue rectngles). When we tke the limit of such Riemnn sums, we get the sitution illustrted in Figure 4. A definite integrl cn be interpreted s net re, tht is, difference of res: b f d A A where A is the re of the region bove the -is nd below the grph of f, nd A is the re of the region below the -is nd bove the grph of f. NOTE 4 In the spirit of the precise definition of the limit of function in Section.4, we cn write the precise mening of the limit tht defines the integrl in Definition s follows: For ever number there is n integer N such tht b f d n f * i i for ever integer n N nd for ever choice of * i in i, i. This mens tht definite integrl cn be pproimted to within n desired degree of ccurc b Riemnn sum. b NOTE 5 Although we hve defined f d b dividing, b into subintervls of equl width, there re situtions in which it is dvntgeous to work with subintervls of unequl width. For instnce, in Eercise 4 in Section 5. NASA provided velocit dt t times tht were not equll spced, but we were still ble to estimte the distnce trveled. And there re methods for numericl integrtion tht tke dvntge of unequl subintervls. If the subintervl widths re,,..., n, we hve to ensure tht ll these widths pproch in the limiting process. This hppens if the lrgest width, m i, pproches. So in this cse the definition of definite integrl becomes b f d lim m i l n f * i i i EXAMPLE Epress lim nl n i i sin i i s n integrl on the intervl,. SOLUTION Compring the given limit with the limit in Definition, we see tht the will be identicl if we choose f sin nd i * i (So the smple points re right endpoints nd the given limit is of the form of Eqution.) We re given tht nd b. Therefore, b Definition or Eqution, we hve lim nl n i i i sin i sin d

16 SECTION 5. THE DEFINITE INTEGRAL 8 Lter, when we ppl the definite integrl to phsicl situtions, it will be importnt to recognize limits of sums s integrls, s we did in Emple. When Leibniz chose the nottion for n integrl, he chose the ingredients s reminders of the limiting process. In generl, when we write lim nl n f * i b f d i we replce lim b, * i b, nd b d. E v l u t i n g I n t e g r l s When we use the definition to evlute definite integrl, we need to know how to work with sums. The following three equtions give formuls for sums of powers of positive integers. Eqution 4 m be fmilir to ou from course in lgebr. Equtions 5 nd 6 were discussed in Section 5. nd re proved in Appendi E. 4 n n n i i 5 6 n n n n i i 6 n i i n n The remining formuls re simple rules for working with sigm nottion: Formuls 7 re proved b writing out ech side in epnded form. The left side of Eqution 8 is The right side is c c c n c n These re equl b the distributive propert. The other formuls re discussed in Appendi E n c nc i n c i c n i i i n i b i n i n b i i i i n i b i n i n b i i i i EXAMPLE () Evlute the Riemnn sum for f 6 tking the smple points to be right endpoints nd, b, nd n 6. (b) Evlute 6 d. Tr more problems like this one. Resources / Module 6 / Wht Is Are? / Problems nd Tests SOLUTION () With n 6 the intervl width is b n 6

17 84 CHAPTER 5 INTEGRALS 5 = -6 FIGURE 5 nd the right endpoints re.5,.,.5, 4., 5.5, nd 6.. So the Riemnn sum is R 6 6 i f i f.5 f. f.5 f. f.5 f Notice tht f is not positive function nd so the Riemnn sum does not represent sum of res of rectngles. But it does represent the sum of the res of the gold rectngles (bove the -is) minus the sum of the res of the blue rectngles (below the -is) in Figure 5. (b) With n subintervls we hve b n n Thus, n, 6 n, 9 n, nd, in generl, i i n. Since we re using right endpoints, we cn use Eqution : 5 = -6 A 6 d lim nl n f i lim i nl n i lim nl n i i n lim nl n i 7 n n i 8 nl lim 8 n i 54 n n 4 i n i lim nl 8 n 4 lim nl 8 n n n 4 n 6 i n i f n i n n i 54 n n n 7 n (Eqution 8 with c n) (Equtions nd 8) A FIGURE 6 j ( -6)d=A -A =_6.75 This integrl cn t be interpreted s n re becuse f tkes on both positive nd negtive vlues. But it cn be interpreted s the difference of res A A, where A nd re shown in Figure 6. Figure 7 illustrtes the clcultion b showing the positive nd negtive terms in the right Riemnn sum R n for n 4. The vlues in the tble show the Riemnn sums pproching the ect vlue of the integrl, 6.75, s nl. A

18 SECTION 5. THE DEFINITE INTEGRAL 85 n R n 5 = FIGURE 7 R Å_6.998 A much simpler method for evluting the integrl in Emple will be given in Section 5.. Becuse f e is positive, the integrl in Emple represents the re shown in Figure 8. = EXAMPLE () Set up n epression for e d s limit of sums. (b) Use computer lgebr sstem to evlute the epression. SOLUTION () Here we hve f e,, b, nd b n n FIGURE 8 So, n, 4 n, 6 n, nd From Eqution, we get i i n e d lim nl n f i i lim nl n f i i lim nl n n e i n i n n A computer lgebr sstem is ble to find n eplicit epression for this sum becuse it is geometric series. The limit could be found using l Hospitl s Rule. (b) If we sk computer lgebr sstem to evlute the sum nd simplif, we obtin n e i n e n n e n n i e n Now we sk the computer lgebr sstem to evlute the limit: e d lim nl n e n n e n n e n e e We will lern much esier method for the evlution of integrls in the net section.

19 86 CHAPTER 5 INTEGRALS EXAMPLE 4 Evlute the following integrls b interpreting ech in terms of res. () s d (b) d = œ - or + = SOLUTION () Since f s, we cn interpret this integrl s the re under the curve s from to. But, since, we get, which shows tht the grph of f is the qurter-circle with rdius in Figure 9. Therefore s d 4 4 FIGURE 9 (In Section 7. we will be ble to prove tht the re of circle of rdius r is r.) (b) The grph of is the line with slope shown in Figure. We compute the integrl s the difference of the res of the two tringles: d A A.5 (,) =- A A FIGURE _ T h e M i d p o i n t R u l e We often choose the smple point * i to be the right endpoint of the ith subintervl becuse it is convenient for computing the limit. But if the purpose is to find n pproimtion to n integrl, it is usull better to choose * i to be the midpoint of the intervl, which we denote b i. An Riemnn sum is n pproimtion to n integrl, but if we use midpoints we get the following pproimtion. Module 5./5./7.7 shows how the Midpoint Rule estimtes improve s n increses. Midpoint Rule where nd b f d n f i f f n i b n i i i midpoint of i, i EXAMPLE 5 Use the Midpoint Rule with n 5 to pproimte. d

20 SECTION 5. THE DEFINITE INTEGRAL 87 = SOLUTION The endpoints of the five subintervls re,.,.4,.6,.8, nd., so the midpoints re.,.,.5,.7, nd.9. The width of the subintervls is 5 5, so the Midpoint Rule gives d f. f. f.5 f.7 f FIGURE Since f for, the integrl represents n re, nd the pproimtion given b the Midpoint Rule is the sum of the res of the rectngles shown in Figure. At the moment we don t know how ccurte the pproimtion in Emple 5 is, but in Section 7.7 we will lern method for estimting the error involved in using the Midpoint Rule. At tht time we will discuss other methods for pproimting definite integrls. If we ppl the Midpoint Rule to the integrl in Emple, we get the picture in Figure. The pproimtion M is much closer to the true vlue 6.75 thn the right endpoint pproimtion, R , shown in Figure 7. 5 = -6 FIGURE M Å_6.756 P r o p e r t i e s o f t h e D e f i n i t e I n t e g r l b When we defined the definite integrl f d, we implicitl ssumed tht b. But the definition s limit of Riemnn sums mkes sense even if b. Notice tht if we reverse nd b, then chnges from b n to b n. Therefore f d b f d b If b, then nd so f d We now develop some bsic properties of integrls tht will help us to evlute integrls in simple mnner. We ssume tht f nd t re continuous functions.

21 88 CHAPTER 5 INTEGRALS Properties of the Integrl b. c d c b, where c is n constnt. b f t d b f d b t d c =c b. cf d c b f d, where c is n constnt re=c(b-) 4. b f t d b f d b t d b FIG URE b j cd=c(b-) f+g Propert ss tht the integrl of constnt function f c is the constnt times the length of the intervl. If c nd b, this is to be epected becuse c b is the re of the shded rectngle in Figure. Propert ss tht the integrl of sum is the sum of the integrls. For positive functions it ss tht the re under f t is the re under f plus the re under t. Figure 4 helps us understnd wh this is true: In view of how grphicl ddition works, the corresponding verticl line segments hve equl height. In generl, Propert follows from Eqution nd the fct tht the limit of sum is the sum of the limits: b FIGURE 4 b j [ƒ+ ]d= b b j ƒd+j d g Propert seems intuitivel resonble becuse we know tht multipling function b positive number c stretches or shrinks its grph verticll b fctor of c. So it stretches or shrinks ech pproimting rectngle b fctor c nd therefore it hs the effect of multipling the re b c. f b f t d lim nl n f i t i i lim nl n f i n t i i i lim nl n f i lim i nl n t i i b Propert cn be proved in similr mnner nd ss tht the integrl of constnt times function is the constnt times the integrl of the function. In other words, constnt (but onl constnt) cn be tken in front of n integrl sign. Propert 4 is proved b writing f t f t nd using Properties nd with c. EXAMPLE 6 Use the properties of integrls to evlute 4 d. f d b t d SOLUTION Using Properties nd of integrls, we hve We know from Propert tht 4 d 4 d d 4 d d 4 d 4 4

22 SECTION 5. THE DEFINITE INTEGRAL 89 =ƒ nd we found in Emple in Section 5. tht d. So The net propert tells us how to combine integrls of the sme function over djcent intervls: 5. 4 d c f d b c 4 d 4 5 f d b f d d FIGURE 5 c b This is not es to prove in generl, but for the cse where f nd c b Propert 5 cn be seen from the geometric interprettion in Figure 5: The re under f from to c plus the re from c to b is equl to the totl re from to b. EXAMPLE 7 If it is known tht f d 7 nd 8 f d, find f d. SOLUTION B Propert 5, we hve 8 8 f d 8 f d f d so 8 f d f d 8 f d 7 5 Notice tht Properties 5 re true whether b, b, or b. The following properties, in which we compre sizes of functions nd sizes of integrls, re true onl if b. Comprison Properties of the Integrl 6. If f for b, then f d. b 7. If f t for b, then f d b t d. b 8. If m f M for b, then M m b b f d M b m FIGURE 6 =ƒ b b If f, then f d represents the re under the grph of f, so the geometric interprettion of Propert 6 is simpl tht res re positive. But the propert cn be proved from the definition of n integrl (Eercise 64). Propert 7 ss tht bigger function hs bigger integrl. It follows from Properties 6 nd 4 becuse f t. Propert 8 is illustrted b Figure 6 for the cse where f. If f is continuous we could tke m nd M to be the bsolute minimum nd mimum vlues of f on the inter-

23 9 CHAPTER 5 INTEGRALS = =/e =e vl, b. In this cse Propert 8 ss tht the re under the grph of f is greter thn the re of the rectngle with height m nd less thn the re of the rectngle with height M. Proof of Propert 8 Since m f M, Propert 7 gives Using Propert to evlute the integrls on the left nd right sides, we obtin Propert 8 is useful when ll we wnt is rough estimte of the size of n integrl without going to the bother of using the Midpoint Rule. EXAMPLE 8 Use Propert 8 to estimte d. SOLUTION Becuse f e is decresing function on,, its bsolute mimum vlue is M f nd its bsolute minimum vlue is m f e. Thus, b Propert 8, or b m d b e Since e.679, we cn write m b b f d M b e e f d b e d e d M d FIGURE 7.67 e d The result of Emple 8 is illustrted in Figure 7. The integrl is greter thn the re of the lower rectngle nd less thn the re of the squre. 5. Eercises. Evlute the Riemnn sum for f,, with four subintervls, tking the smple points to be right endpoints. Eplin, with the id of digrm, wht the Riemnn sum represents.. If f ln, 4, evlute the Riemnn sum with n 6, tking the smple points to be left endpoints. (Give our nswer correct to si deciml plces.) Wht does the Riemnn sum represent? Illustrte with digrm.. If f s, 6, find the Riemnn sum with n 5 correct to si deciml plces, tking the smple points to be midpoints. Wht does the Riemnn sum represent? Illustrte with digrm. points. (Give our nswer correct to si deciml plces.) Eplin wht the Riemnn sum represents with the id of sketch. (b) Repet prt () with midpoints s the smple points. 5. The grph of function f is given. Estimte f d using four subintervls with () right endpoints, (b) left endpoints, nd (c) midpoints. f 8 4. () Find the Riemnn sum for f sin,, with si terms, tking the smple points to be right end-

24 SECTION 5. THE DEFINITE INTEGRAL 9 6. The grph of t is shown. Estimte t d with si subintervls using () right endpoints, (b) left endpoints, nd (c) midpoints. 7. A tble of vlues of n incresing function f is shown. Use the tble to find lower nd upper estimtes for f d. 5 g 5. Use clcultor or computer to mke tble of vlues of right Riemnn sums Rn for the integrl sin d with n 5,, 5, nd. Wht vlue do these numbers pper to be pproching? 6. Use clcultor or computer to mke tble of vlues of left nd right Riemnn sums L n nd R n for the integrl with n 5,, 5, nd. Between wht two e d numbers must the vlue of the integrl lie? Cn ou mke similr sttement for the integrl? Eplin. e d 7 Epress the limit s definite integrl on the given intervl lim nl n i sin i,, i lim n l n i e i i,, f lim nl n i s* i * i,, 8] CAS 8. The tble gives the vlues of function obtined from n eperiment. Use them to estimte 6 f d using three equl subintervls with () right endpoints, (b) left endpoints, nd (c) midpoints. If the function is known to be decresing function, cn ou s whether our estimtes re less thn or greter thn the ect vlue of the integrl? 9 Use the Midpoint Rule with the given vlue of n to pproimte the integrl. Round the nswer to four deciml plces. 9. s d, n f sin d, n 5.. If ou hve CAS tht evlutes midpoint pproimtions nd grphs the corresponding rectngles (use middlesum nd middlebo commnds in Mple), check the nswer to Eercise nd illustrte with grph. Then repet with n nd n. 4. With progrmmble clcultor or computer (see the instructions for Eercise 7 in Section 5.), compute the left nd right Riemnn sums for the function f sin on the intervl, with n. Eplin wh these estimtes show tht.6 5 e d, n 4 sin d.5 Deduce tht the pproimtion using the Midpoint Rule with n 5 in Eercise is ccurte to two deciml plces. sec d, n 6. lim 4 * i 6 * 5 i, 5 Use the form of the definition of the integrl given in Eqution to evlute the integrl.. d.. 5. nl n i 5 d d 6. () Find n pproimtion to the integrl d using Riemnn sum with right endpoints nd n 8. (b) Drw digrm like Figure to illustrte the pproimtion in prt (). (c) Use Eqution to evlute 4 d. (d) Interpret the integrl in prt (c) s difference of res nd illustrte with digrm like Figure Prove tht b d b Prove tht b d b. 9 Epress the integrl s limit of Riemnn sums. Do not evlute the limit d 5, 4 5 d 5 d 4 ln d 4

25 9 CHAPTER 5 INTEGRALS CAS Epress the integrl s limit of sums. Then evlute, using computer lgebr sstem to find both the sum nd the limit.. sin 5 d.. The grph of f is shown. Evlute ech integrl b interpreting it in terms of res. () (c) f d 7 f d 5 (b) (d) 6 d 5 f d 9 f d 44. Use the properties of integrls nd the result of Emple to evlute e d. 45. Use the result of Emple to evlute e d. 46. Use the result of Eercise 7 nd the fct tht cos d (from Eercise 5 in Section 5.), together with the properties of integrls, to evlute cos 5 d. Write s single integrl in the form f d: 47. f d 5 f d f d b =ƒ 48. If nd, find 5 f d 5 f d.6 4 f d If nd 9 f d 7 9 t d 6, find 9 f t d Find f d if for f for 4. The grph of t consists of two stright lines nd semicircle. Use it to evlute ech integrl. () t d (b) t d (c) t d 5 54 Use the properties of integrls to verif the inequlit without evluting the integrls sin d 4 s5 d s d sin d = 5. s d s sin d 5 4 Evlute the integrl b interpreting it in terms of res s4 d ( d 7. ( s9 ) d d Given tht s d, wht is st dt? d 5 d 4. Evlute cos d. 4. In Emple in Section 5. we showed tht d. Use this fct nd the properties of integrls to evlute 5 6 d Use Propert 8 to estimte the vlue of the integrl d 57. tn d Use properties of integrls, together with Eercises 7 nd 8, to prove the inequlit e d s 4 d 6 4 sin d 8 s d d 4 sin d 4

26 DISCOVERY PROJECT AREA FUNCTIONS 9 6. Prove Propert of integrls. 64. Prove Propert 6 of integrls. 65. If f is continuous on, b, show tht b f d b f d f f f [Hint:.] 66. Use the result of Eercise 65 to show tht f sin d f d Epress the limit s definite integrl. 67. lim [Hint: Consider f 4.] 68. i 4 nl n i n 5 lim nl n n i i n 69. Find d. Hint: Choose * to be the geometric men of i i nd i (tht is, * i s i i ) nd use the identit m m m m DISCOVERY PROJECT Are Functions. () Drw the line t nd use geometr to find the re under this line, bove the t-is, nd between the verticl lines t nd t. (b) If, let A be the re of the region tht lies under the line t between t nd t. Sketch this region nd use geometr to find n epression for A. (c) Differentite the re function A. Wht do ou notice?. () If, let A t dt A represents the re of region. Sketch tht region. (b) Use the result of Eercise 8 in Section 5. to find n epression for A. (c) Find A. Wht do ou notice? (d) If nd h is smll positive number, then A h A represents the re of region. Describe nd sketch the region. (e) Drw rectngle tht pproimtes the region in prt (d). B compring the res of these two regions, show tht A h A h (f) Use prt (e) to give n intuitive eplntion for the result of prt (c). ;. () Drw the grph of the function f cos in the viewing rectngle, b.5,.5. (b) If we define new function t b t cos t dt then t is the re under the grph of f from to [until f becomes negtive, t which point t becomes difference of res]. Use prt () to determine the vlue of t which t strts to decrese. [Unlike the integrl in Problem, it is impossible to evlute the integrl defining t to obtin n eplicit epression for t.] (c) Use the integrtion commnd on our clcultor or computer to estimte t(.), t(.4), t(.6),...,t(.8),t(). Then use these vlues to sketch grph of t. (d) Use our grph of t from prt (c) to sketch the grph of t using the interprettion of t s the slope of tngent line. How does the grph of t compre with the grph of f?

27 94 CHAPTER 5 INTEGRALS 4. Suppose f is continuous function on the intervl, b nd we define new function t b the eqution t f t dt Bsed on our results in Problems, conjecture n epression for t. 5. T h e F u n d m e n t l T h e o r e m o f C l c u l u s Investigte the re function interctivel. Resources / Module 6 / Ares nd Derivtives / Are s Function FIGURE re= =f(t) b t The Fundmentl Theorem of Clculus is ppropritel nmed becuse it estblishes connection between the two brnches of clculus: differentil clculus nd integrl clculus. Differentil clculus rose from the tngent problem, wheres integrl clculus rose from seemingl unrelted problem, the re problem. Newton s techer t Cmbridge, Isc Brrow (6 677), discovered tht these two problems re ctull closel relted. In fct, he relized tht differentition nd integrtion re inverse processes. The Fundmentl Theorem of Clculus gives the precise inverse reltionship between the derivtive nd the integrl. It ws Newton nd Leibniz who eploited this reltionship nd used it to develop clculus into sstemtic mthemticl method. In prticulr, the sw tht the Fundmentl Theorem enbled them to compute res nd integrls ver esil without hving to compute them s limits of sums s we did in Sections 5. nd 5.. The first prt of the Fundmentl Theorem dels with functions defined b n eqution of the form t f t dt where f is continuous function on, b nd vries between nd b. Observe tht t depends onl on, which ppers s the vrible upper limit in the integrl. If is fied number, then the integrl f t dt is definite number. If we then let vr, the number f t dt lso vries nd defines function of denoted b t. If f hppens to be positive function, then t cn be interpreted s the re under the grph of f from to, where cn vr from to b. (Think of t s the re so fr function; see Figure.) EXAMPLE If f is the function whose grph is shown in Figure nd t f t dt, find the vlues of t, t, t, t, t 4, nd t 5. Then sketch rough grph of. =f(t) 4 t t SOLUTION First we notice tht t f t dt. From Figure we see tht t is the re of tringle: t f t dt To find t we dd to t the re of rectngle: FIGURE t f t dt f t dt f t dt

28 SECTION 5. THE FUNDAMENTAL THEOREM OF CALCULUS 95 We estimte tht the re under f from to is bout., so t t f t dt. 4. t t t 4 t 4 t g()= g()= g()å4. FIGURE g(4)å g(5)å.7 For t, f t is negtive nd so we strt subtrcting res: 4 g t 4 t 4 f t dt 4... t 5 t 4 5 f t dt..7 4 FIGURE 4 =j f(t)dt 4 5 We use these vlues to sketch the grph of t in Figure 4. Notice tht, becuse f t is positive for t, we keep dding re for t nd so t is incresing up to, where it ttins mimum vlue. For, t decreses becuse f t is negtive. If we tke f t t nd, then, using Eercise 7 in Section 5., we hve t t dt h ƒ b t +h FIGURE 5 Notice tht t, tht is, t f. In other words, if t is defined s the integrl of f b Eqution, then t turns out to be n ntiderivtive of f, t lest in this cse. And if we sketch the derivtive of the function t shown in Figure 4 b estimting slopes of tngents, we get grph like tht of f in Figure. So we suspect tht t f in Emple too. To see wh this might be generll true we consider n continuous function f with f. Then t f t dt cn be interpreted s the re under the grph of f from to, s in Figure. In order to compute t from the definition of derivtive we first observe tht, for h, t h t is obtined b subtrcting res, so it is the re under the grph of f from to h (the gold re in Figure 5). For smll h ou cn see from the figure tht this re is pproimtel equl to the re of the rectngle with height f nd width h: so t h t hf t h t h f

29 96 CHAPTER 5 INTEGRALS Intuitivel, we therefore epect tht The fct tht this is true, even when f Fundmentl Theorem of Clculus. t h t t lim f hl h is not necessril positive, is the first prt of the We bbrevite the nme of this theorem s FTC. In words, it ss tht the derivtive of definite integrl with respect to its upper limit is the integrnd evluted t the upper limit. The Fundmentl Theorem of Clculus, Prt If f is continuous on, b, then the function t defined b t f t dt is continuous on, b nd differentible on, b, nd t f. b Proof If nd h re in, b, then t h t h f t dt f t dt f t dt h f t dt f t dt (b Propert 5) h f t dt nd so, for h, t h t h h f t dt h =ƒ m M For now let us ssume tht h. Since f is continuous on, h, the Etreme Vlue Theorem ss tht there re numbers u nd v in, h such tht f u m nd f v M, where m nd M re the bsolute minimum nd mimum vlues of f on, h. (See Figure 6.) B Propert 8 of integrls, we hve mh h f t dt Mh u =+h tht is, f u h h f t dt f v h FIGURE 6 Since h, we cn divide this inequlit b h: f u h f t dt f v h Now we use Eqution to replce the middle prt of this inequlit: f u t h t h f v Inequlit cn be proved in similr mnner for the cse h. (See Eercise 6.)

30 SECTION 5. THE FUNDAMENTAL THEOREM OF CALCULUS 97 Module 5. provides visul evidence for FTC. Now we let hl. Then ul nd vl, since u nd v lie between nd h. Therefore lim f u lim f u f hl ul nd becuse 4 f lim f v lim f v f hl vl is continuous t. We conclude, from () nd the Squeeze Theorem, tht t h t t lim f hl h If or b, then Eqution 4 cn be interpreted s one-sided limit. Then Theorem.9.4 (modified for one-sided limits) shows tht t is continuous on, b. Using Leibniz nottion for derivtives, we cn write FTC s 5 d f t dt f d when f is continuous. Roughl speking, Eqution 5 ss tht if we first integrte f nd then differentite the result, we get bck to the originl function f. EXAMPLE Find the derivtive of the function t s t dt. SOLUTION Since f t s t is continuous, Prt of the Fundmentl Theorem of Clculus gives t s EXAMPLE Although formul of the form t f t dt m seem like strnge w of defining function, books on phsics, chemistr, nd sttistics re full of such functions. For instnce, the Fresnel function S sin t dt is nmed fter the French phsicist Augustin Fresnel (788 87), who is fmous for his works in optics. This function first ppered in Fresnel s theor of the diffrction of light wves, but more recentl it hs been pplied to the design of highws. Prt of the Fundmentl Theorem tells us how to differentite the Fresnel function: S sin This mens tht we cn ppl ll the methods of differentil clculus to nlze S (see Eercise 57).

31 98 CHAPTER 5 INTEGRALS Figure 7 shows the grphs of f sin nd the Fresnel function S f t dt. A computer ws used to grph S b computing the vlue of this integrl for mn vlues of. It does indeed look s if S is the re under the grph of f from to [until.4 when S becomes difference of res]. Figure 8 shows lrger prt of the grph of S. f S.5 FIGURE 7 ƒ=sin(π / ) S()=j sin(πt@/ )dt FIGURE 8 The Fresnel function S()=j sin(πt@/)dt If we now strt with the grph of S in Figure 7 nd think bout wht its derivtive should look like, it seems resonble tht S f. [For instnce, S is incresing when f nd decresing when f.] So this gives visul confirmtion of Prt of the Fundmentl Theorem of Clculus. d EXAMPLE 4 Find 4 sec t dt. d SOLUTION Here we hve to be creful to use the Chin Rule in conjunction with FTC. Let u 4. Then d d 4 sec t dt d d u sec t dt d du u sec u du d sec 4 4 sec t dt du d (b the Chin Rule) (b FTC) In Section 5. we computed integrls from the definition s limit of Riemnn sums nd we sw tht this procedure is sometimes long nd difficult. The second prt of the Fundmentl Theorem of Clculus, which follows esil from the first prt, provides us with much simpler method for the evlution of integrls. The Fundmentl Theorem of Clculus, Prt If f is continuous on, b, then We bbrevite this theorem s FTC. b f d F b F where F is n ntiderivtive of f, tht is, function such tht F f.