3.6 Rolle s Theorem; Mean Value Theorem

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1 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 7.6 Rolle s Theorem; Men Vlue Theorem In this section we will discuss result clled the Men Vlue Theorem. The theorem hs so mn importnt consequences tht it is regrded s one of the mjor principle is clculus Theorem. (Rolle s Theorem) Let f be function tht stisfies the following three hpotheses:. f is continuous on the closed intervl [,b].. f is differentible on the open intervl (, b).. f() = f(b) Then there is number c (,b) such tht f (c) =. The following Figure shows the grphs of three such functions. In ech cse it ppers tht there is t lest one point ( c,f(c) ) on the grph where the tngent is horizontl nd therefore f (c) =. c b c b c c b Emple. Verif tht the function f() = +, [,] stisfies the three hpotheses of Rolle s Theorem on the given intervl. Then find ll numbers c tht stisf the conclusion of Rolle s Theorem. Theorem. (Men Vlue Theorem) Let f be function tht stisfies the following hpotheses:. f is continuous on the closed intervl [,b].. f is differentible on the open intervl (, b). Then there is number c (,b) such tht or, equivlentl, Emple. Let f (c) = f(b) f() b f(b) f() = f (c)(b ) f() = + 4, [,] Find ll numbers c tht stisf the conclusion of the Men Vlue Theorem.

2 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 8 Eercise.6. Verif tht the function stisfies the three hpotheses of Rolle s Theorem on the given intervl. Then find ll numbers c tht stisf the conclusion of Rolle s Theorem. () f() = 4+, [,4] (b) f() = ++5, [,] (c) f() = sinπ, [,] (d) f() = +6, [ 6,]. Use the grph of f to estimte the vlues of c tht stisf the conclusion of the Men Vlue Theorem for the intervl [, 8]. = f(). Verif tht the function stisfies the hpotheses of the Men Vlue Theorem on the given intervl. Then find ll numbers c tht stisf the conclusion of the Men Vlue Theorem. () f() = ++5, [,] (b) f() =, [,] (c) f() = e, [,] (d) f() = +, [,4] Answer to Eercise.6. () (b) ± (c) ± 4,± 4 (d) 4..8,., 4.4, 6.. () (b) (c) ln[( e 6 )/6] (d).7 Indeterminte Forms nd L Hospitl s Rule In generl, if we hve limit of the form f() lim g() where both f() nd g() s, then this limit m or m not eist nd is clled n indeterminte form of tpe. In this section we introduce sstemtic method, known s l Hospitl s Rule, for the evlution of indeterminte forms. Moreover, if we hve limit of the form lim f() g()

3 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 9 where both f() (or ) nd g() (or ) s, then this limit m or m not eist nd is clled n indeterminte form of tpe. L Hospitl s Rule lso pplies to this tpe of indeterminte form. Theorem. (L Hospitl s Rule) Suppose f nd g re differentible nd g () ner (ecept possible t ). Suppose tht or tht limf() = nd lim g() = limf() = ± nd limg() = ± ( In other words, we hve n indeterminte form of tpe or.) Then f() lim g() = lim f () g () if the limit on the right side eists ( or is or ). Note:. L Hospitl s Rule ss tht the limit of quotient of functions is equl to the limit of the quotient of their derivtives, provided tht the given conditions re stisfied. It is especill importnt to verif the conditions regrding the limits of f nd g before using l Hospitl s Rule.. L Hospitl s Rule is lso vlid for one-sided limits nd for limits t infinit or negtive infinit; tht is, cn be replced b n of the following smbols:, +,,. Emple. Find lim ln. Emple.4 Clculte lim tn. Emple.5 Find lim tn ln(+). Emple.6 Clculte lim + ln cot.

4 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 4 Indeterminte Products If limf() = nd limg() = (or ), thenitisn tclerwhtthevlueof limf()g(), if n, will be. This kind of limit is clled n indeterminte form of tpe. We cn del with it b writing the product fg s quotient: fg = f /g or fg = g /f This converts the given limit into n indeterminte form of tpe or l Hospitl s Rule. so tht we cn use Emple.7 Evlute lim (tn lnsin). π/ Emple.8 Evlute lim ( sin csc ). Indeterminte Differences If lim f() = nd lim g() =, then the limit [ ] lim f() g() is clled n indeterminte form of tpe. To find the limit of this form, we tr to convert the difference into quotient (for instnce, b using common denomintor or rtionliztion, or fctoring out common fctor) so tht we hve n indeterminte form of tpe or. Emple.9 Compute lim + Emple. Compute lim + Indeterminte Powers [ ]. sin ( ). ln Severl indeterminte forms rise from the limit [ ] g() lim f(). lim f() = nd lim g() = tpe. lim f() = nd lim g() = tpe

5 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 4. lim f() = nd lim g() = ± tpe Ech of these three cses cn be treted either b tking the nturl logrithm: let = [ f() ] g(), then ln = g()lnf() or b writing the function s n eponentil: [ f() ] g() = e g()lnf(). In either method we re led to the indeterminte product g()lnf(), which is of tpe. Emple. Find lim (tn) sin. Emple. Clculte lim (π/) (tn)cos. Emple. Clculte lim ( ) /. Find the following limit. +. lim 4 +. lim 4 e 5. lim sin sin 7. lim 9. lim e. lim sinπ. lim e 5. lim +ln ( ) 7. lim ( 9. lim ln+ ) +. lim +sin Eercise.7. lim lim +4+ sin 6. lim 8. lim e. lim ln. lim 4. lim sin cos 6. lim + ln 8. lim cot ( +. lim +(/) ). lim ( ) /

6 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 4 Answer to Eercise π e e

7 Chpter 4 Integrtion 4. Antiderivtives; The Indefinite Integrl Antiderivtives Definition 4. A function F is clled n ntiderivtive of f on n intervl I if F () = f() for ll in I. For instnce, let f() =. It isn t difficult to discover n ntiderivtive F() = becuse F () = = f(). But the function G() = + lso stisfies G () =. Therefore, both F nd G re ntiderivtives of f. Indeed, n function of the form H() = +C, where C is constnt, is n ntiderivtive of f. Question! Are there n others? Answer. No. Thus, if F nd G re n two ntiderivtives of f, then F () = f() = G () so G() F() = C, where C is constnt. We cn write this s G() = F()+C, so we hve the following result. Theorem 4. If F is n ntiderivtive of f on n intervl I, then the most generl ntiderivtive of f on I is F()+C where C is n rbitrr constnt. The Indefinite Integrl The process of finding ntiderivtives is clled ntidifferentition or integrtion. Thus, if d [F()] = f() (4.) d 4

8 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 44 then integrting (or ntidifferentiting) the function f() produces n ntiderivtive of the form F() + C. To emphsize this process, Eqution (4.) is recst using integrl nottion. f()d = F()+C (4.) where C is n rbitrr constnt. For emple, d = +C is equivlent to d d [ ] = Note tht if we differentite n ntiderivtive of f(), we obtin f() bck gin. Thus, d d [ ] f()d = f() (4.) The epression f()d is clled n indefinite integrl. The elongted s tht ppers on the left side of (4.) is clled n integrl sign, the function f() is clled the integrnd, nd the constnt C is clled the constnt of integrtion. The differentil smbol, d, in the differentition nd ntidifferentition opertions d d [ ] nd [ ]d serves to identif the independent vrible. If n independent vrible other thn is used, s t, then the nottion must be djusted ppropritel. Thus, d [F(t)] = f(t) nd [f(t)]d = F(t)+C d re equivlent sttements. Here re some emples of derivtive formuls nd their equivlent integrtion formuls: Derivtive Formul d d [ ] = d d [ ] = d dt [tnt] = sec t Equivlent Integrtion Formul d = +C d = +C sec tdt = tnt+c Integrtion Formuls Some of the most importnt integrtion formuls re given in the following Tble.

9 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 45 n d = n+ n+ +C e d = e +C sind = cos+c sec d = tn+c sectnd = sec+c d = sin +C d = sec +C d = ln +C d = ln +C, > cosd = sin+c csc d = cot+c csccotd = csc+c + d = tn +C Properties of the Indefinite Integrl Our first properties of ntiderivtives follow directl from the simple constnt fctor, sum, nd difference rules for derivtive. Theorem 4. Let f nd g hve ntiderivtives (indefinite integrls) nd let c be constnt. Then (i) cf()d = c f()d (ii) (iii) [f()+g() ] d = [f() g() ] d = f()d+ f()d Emple 4. Evlute (e +5 / )d. g() d g() d Emple 4. Evlute (cos+sec )d. Emple 4. Evlute sin () cos d (b) t 4 t t dt t 4

10 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 46 Eercise 4.. Find the most generl ntiderivtive of the function. () f() = 5 (b) f() = +π (c) f() = 5/4 (d) f() = / (e) f() = (f) f() = 4 5 (g) f() = (h) f() = (i) f() = Evlute the integrl nd check our nswer b differentiting. () 4 d (b) ( +)d (c) ( 4 )d (d) (+) d (e) d (f) ( ) d 4 / ( +) (g) d (h) d / (i) (sin cos)d (j) sectnd (k) 5sec d (l) (e )d (m) (cos /)d (n) (5 e ) d e + (o) d (p) /4 ( 5/4 4)d e Answer to Eercise 4.. () 5+C (b) +π+c (c) 4 9 9/4 +C (d) +C (e) +C (f) C (g) C (h) + +C (i) 4 + +C. () 5 5 +C (b) + +C (c) 5 5 +C (d) (+) +C (e) / +C (f) + +C (g) / 9 / +C (h) 9 9/ / + / +C (i) cos sin+c (j) sec+c (k) 5tn+C (l) e +C (m) sin ln +C (n) 5 +e +C (o) e +C (p) 5 5/ 6 5 5/4 4. Integrtion b Substitution In this section we shll discuss technique, clled substitution, which cn often be used to trnsform complicted integrtion problems into simpler ones.

11 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 47 u-substitution The method of substitution hinges on the following formul in which u stnds for differentible function of. [ f(u) du ] d = f(u) du (4.4) d To justif this formul, let F be n ntiderivtive of f, so tht or, equivlentl, d [F(u)] = f(u) du f(u)du = F(u)+C (4.5) If u is differentible function of, the chin rule implied tht d d [F(u)] = d du [F(u)] du d = f(u)du d or, equivlentl, [ f(u) du ] d = F(u)+C (4.6) d Formul (4.4) follows from (4.5) nd (4.6). The following emple illustrtes how Formul (4.4) is used. Emple 4.4 Evlute ( +) 5 d. In generl, suppose tht we re interested in evluting h() d It follows from (4.4) tht if we cn epress this integrl in the form h()d = f(g())g ()d then the substitution u = g() nd du/d = g () will ield [ h()d = f(u) du ] d = f(u)du d With good choice of u = g(), the integrl on the right will be esier to evlute thn the originl. In prctice, this substitution process is crrier out s follows:

12 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 48 Guideline for u-substitution Step. Mke choice for u, s u = g(). Step. Compute du/d Step. Mke the substitution u = g(), du = g ()d At this stge, the entire integrl must be in terms of u; no s should remin. If this is not the cse, tr different choice of u. Step 4. Evlute the resulting integrl. Step 5. Replce u b g(), so the finl nswer is in terms of. Emple 4.5 Evlute 4 sin( 5 )d. Emple 4.6 Evlute Emple 4.7 Evlute cos4 sin4d. 4 5 d.... Emple 4.8 Evlute sec d. Emple 4.9 Evlute d. Eercise 4.. Evlute the integrl b mking the given substitution. () cosd, u = (b) ( +)d, u = + 4 (c) d, u = + (+) ( +) (d) d, u = +

13 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 49. Evlute the indefinite integrl () ( +) 4 d d (c) (e) (g) d 5 +4 d ++ (b) (+)( +) d (d) d (f) (h) + + d d ( +) (i) cosd (j) cos sin+d (k) sin( sin )d (l) d cos (m) cos 4 sind (n) sin(cos+) /4 d (o) sectn +secd (p) cose sin d (q) e +e d (r) e + d d 4 (s) (t) ln (ln+) d cot (u) csc d (v) sin(cos ) d e (w) sec e tnd () d e +e + + () d (z) + +7 d Answer to Eercise 4.. () sin+c (b) ( +) / +C (c) /(+) +C (d) ( +) 4 +C. () 5 ( +) 5 +C (b) 4 ( +) 4 +C (c) ( )/ +C (d) +C (e) ln 5 +C (f) ln + +C (g) ++ +C (h) ( +) +C (i) sin+c (j) (sin+)/ +C (k) cos( )+C (l) cos+c (m) 5 cos5 +C (n) 4 7 (cos+)7/4 +C (o) (+sec)/ +C (p) e sin +C (q) (+e ) / +C (r) e + +C (s) ln ln +C (t) 4(ln+) +C (u) (cot)/ +C (v) 4 (cos )4 +C (w) sec +C () ln(e +e )+C () tn + ln(+ )+C (z) (+7) ln +7 +C

14 S S MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 5 4. The Definite Integrl Definition of Are The first gol in this section is to give mthemticl definition of re. We begin b ttempting to find the re of the region S tht lies under the curve = f() from to b. In generl, we strt b subdividing S into n strips S,S,...,S n of equl width s in Figure. = f() S S i S n... i i... n b The width of the intervl [,b] is b, so the width of ech of the n strips is = b n These strips divide the intervl [, b] into n subintervls [, ], [, ], [, ],..., [ n, n ] where = nd n = b. The right-hnd endpoints of the subintervls re = +, = +, = +,... 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-hnd endpoint. Then the re of the ith rectngle is f( i ). Wht we thing 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 f( i ) Notice tht this pproimtion ppers to become better nd better s the number of strips increses, thtis, sn. Therefore, we define thereaoftheregions inthefollowing w. Definition 4. The re A of the region S tht lies under the grph of the continuous function f is the limit of the sum of the res of pproimting rectngles: A = lim n R n = lim n n f( i ) i= i=

15 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 5 Emple 4. Find the re under the curve over the intervl [, ]. We now hve tht limit of the form n lim n i= = 9 f( i ) = lim n [ f( ) +f( ) + +f( n ) ] rises when we compute n re. However this tpe of limit occurs in wide vriet of situtions even when f is not necessril positive function. In the net Chpter we will see tht limit of this form lso rise in finding volumes of solids. We therefore give tpe of limit specil nme nd nottion. Definition 4. 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 choose simple points,,..., n in these subintervls, so i lies in the ith subintervl [ i, i ]. Then the definite integrl of f from to b is b n f()d = lim f( i ) (4.7) n Note:. In the nottion b the smbol f()d, f() is clled integrnd, nd i= ws introduced b Leibniz nd is clled n integrl sign, nd b re clled the limit of integrtion; is the lower limit nd b is the upper limit, the smbol d hs no officil mening b itself. The procedure of clculting n integrl is clled integrtion.. The definite integrl b 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. Becuse we hve ssumed tht f is continuous, it cn be proved tht the limit in Definition 4. lws eists nd gives the sme vlue no mtter how we choose the smple points i.

16 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 5 4. The sum n f( i) i= tht occurs in Definition 4. 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 4. () ). B compring Definition 4. with the definition of re, we see tht the definite integrl b = f() from to b ( see Figure 4. (b) ). f()d cn be interpreted s the re under the curve = f() i b () Figure 4.: b (b) If f tke on both positive n negtive vlues, then the Riemnn sum is the sum of res of the rectngles tht lie bove the -is nd the negtive of the res of the rectngles tht lie below the -is. When we tke the limit of such Riemnn sums, we get the sitution illustrted in the following Figure. + = f() b + 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. 5. Although we hve defined b f()dbdividing[,b]intosubintervlsofequlwidth, there re situtions in which it is dvntgeous to work with subintervls of unequl width. When we defined the definite integrl, we implicitl ssumed tht < b. But the definition s limit of Riemnn sums mkes sense even if > b. Notice tht if we reverse

17 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 5 nd b, thn chnges from (b )/n to ( b)/n. Therefore If = b, then = nd so b b f()d = f() d (4.8) f()d = We now develop some bsic properties of integrls. We ssume tht f nd g re continuous on n intervl [,b]. Theorem 4. (Properties of the Integrl) b b b b cd = c(b ), where c is n constnt b cf()d = c f()d, where c is n constnt [ ] b f()+g() d = f()d+ [ ] b f() g() d = f()d b b g() d g() d Theorem 4.4 If f nd g re continuous on n intervl [,b] nd c is n constnt, then b f()d = c f()d+ b c f()d. For the cse where f() nd < c < b, this theorem cn be seen from the geometric interprettion in the following Figure. = f() c b The re under = f() from to c plus the re from c to b is equl to the totl re from to b. Theorem 4.5 Let f nd g be continuous on n intervl [,b] nd g() f() for ll in [,b]. Then b g()d b f()d

18 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 54 Emple 4. If it is known tht 8 6 g()d =, find ( f()+g() ) d. f()d =, Eercise 4. 5 f()d = 4,. Epress the limit s definite integrl on the given intervl. () lim n (b) lim n (c) lim n (d) lim n n i sin i, [,π] i= n i= e i + i, [,5] n [( i ) 5 i ], [,] i= n i, [,4] i= 6 5 g()d =, nd. Use the form of the definition of the integrl given in (4.7) to evlute the integrl. () (c) (e) 5 5. Prove tht 4. Prove tht d (+ )d (+ )d b b d = b. d = b. (b) (d) (f) 5 (+)d ( )d d 5. Write the given sum or difference s single integrl in the form b f()d. 6. If () (c) (e) 8 f()d+ f()d+ f()d+ 6 f()d =.8 nd f()d f()d f()d (b) (d) f()d f()d =., find f()d f()d+ 5 f()d. f()d f()d

19 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn If. () π f()d =,, 4 sind (b) f()d = 7, nd 4 f()d =, find Answer to Eercise 4. 5 e d (c) +. () (b) 4 (c) (d) 4 (e) 7.5 (f) () f()d (b) f()d (c) ( 5)d (d) f()d (d) 4 f()d. d f()d (e) 4.4 The Fundmentl Theorem of Clculus f()d In the previous section we defined the concept of the definite integrl but did not give n generl methods for evluting them. In this section we shll give method for using ntiderivtives to evlute definite integrls. Theorem 4.6 (The First Fundmentl Theorem of Clculus). If f is continuous on [,b], then the function F defined b F() = f(t)dt, b is continuous on [,b] nd differentible on (,b), nd F () = f(). Using Leibniz nottion for derivtive, the result in Theorem 4.6 cn be epressed b the formul d f(t)dt = f() d Emple 4. Find the derivtive of the function g() = Emple 4. Find the derivtive of the function g() = Emple 4.4 Let F() = ( ) e t + dt. Find F (). 4 +t dt. sectdt.

20 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 56 Theorem 4.7 (The Second Fundmentl Theorem of Clculus). If f is continuous on [,b], then b ] b f()d = F(b) F() = F() where F is n ntiderivtive of f, tht is, function such tht F = f. Emple 4.5 Evlute Emple 4.6 Evlute 8 π/4 d. +cos θ cos θ dθ. Eercise 4.4. Use the First Fundmentl Theorem of Clculus to find the derivtive. () f() = (t t+)dt (b) g() = tdt (c) g() = (e) f() = (g) h() = (i) = / t sintdt (d) F() = (e t +)dt (f) f() = rctntdt (h) = u du (j) = +u cos(t )dt cost t ln(t +)dt dt tsintdt. Evlute the definite integrls using the Second Fundmentl Theorem of Clculus. () (c) (e) (g) (i) (k) 4 π π/ 9 5 d (b) ( )d 4 d (d) ( +)d d 4 5 +d sectnd d (f) (h) (j) (l) π/ π π/ ( + / )d sind (sin cos)d (e e )d

21 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 57 (m) (o) (q) (s) (t) 9 8 π/ π/6 d (n) ( + ) sin d (p) d (r) π/4 (e )d 6 + d cos d {, f()d, where f() =, < { f()d, where f() = 4, < 5, Answer to Eercise 4.4. () f () = + (b) g () = + (c) g () = sin (d) F () = cos( ) (e) f () = (e 4 +) (f) f () = ln( +) (g) h () = rctn(/)/ (h) = cos (j) = 7/ sin( ) (sin )/( 4 ) (i) = ( ) +( ). () 64 (b) (c) 6 (d) 88 (e) 7 8 (f) 5 (g) (h) (i) Does not eist (j) (k) ln (l) e+e (m) 8 ln (q) 5 (r) (s) 6 (t).7 (n) e (o) π 9 + (p) π 4.5 Evluting Definite Integrls b Substitution There is onl one slight difference in using substitution for evluting definite integrl: If ou chnge vribles, ou must lso chnge the limits of integrtion to correspond to the new vrible. Tht is, when ou introduce the new vrible u = g(), ou must lso chnge the limits of integrtion from = nd = b to the corresponding limits for u : u = g() nd u = g(b). We hve b f(g())g ()d = g(b) g() f(u)du. Emple 4.7 Evlute Emple 4.8 Evlute π/8 e sin 5 cosd. +ln d.

22 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 58 Evlute the definite integrl, if it eists Eercise 4.5 ( ) 5 d. (+ ) 5 d 4. π +d ( +) d 4cos (sin+) d cosπd 6. π/ + π/ d 8. cotd d + π/. sin d. cos d 4. + d ( > ) π/4 4 e 4 e d d (+) d ln Answer to Eercise ln 9. ln ( )

23 Chpter 5 Applictions of Definite Integrl 5. Are Between Two Curves In this section we use integrls to find res of regions tht lie between the grphs of two functions. Consider the region tht lies between two curves = f() nd = g() nd between the verticl lines = nd = b, where f nd g re continuous functions nd f() g() for ll in [,b]. = f() b The re A of this region is A = b = g() [ ] f() g() d (5.) Emple 5. Find the re of the region enclosed b =, = 4, nd =. Emple 5. Find the re of the region bounded b = nd = for 4. 4 Emple 5. Find the re bounded b the grphs of = + nd = + for. 59

24 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 6 d = g() c = f() Some regions re best treted b regrding s function of. If region is bounded b curves with equtions = f(), = g(), = c, nd = d, where f nd g re continuous nd f() g() for c d, then its re is A = d c [ f() g() ] d (5.) Emple 5.4 Find the re of the region bounded b the grphs of = nd = +. Emple 5.5 Find the re of the region bounded b the curves =, = +5, =, nd =.. Find the re of the shded region. () Eercise 5. = + (b) = + =

25 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 6 (c) = =. Find the re of the region bounded b the given curves. () = +, =, (b) =, =, (c) = +, = 9, (d) = e, =, (e) = +, = ( ), (f) =, =, (g) = cos, = sin, π/ (h) =, =,. Find the re of the region enclosed b the given curves. () =, = 7 (b) = +, = (c) =, = + (d) =, = (e) =, =, = (f) =, = + (g) =, =, = (h) =, =, = 6, = (i) =, = (j) = cos, = /π (k) =, = /( +) (l) =, = 4. () 6 (b) 4 (c) 9 Answer to Eercise 5.. () (b) 4 (c) 9.5 (d) e (e) (f) (g) (h) 9 6. () 64 (b) 6 (c) 7 4 (d) (e) (f) 6 (g) (h) 8 (i) 8 (j) π (k) π (l) 8 5

26 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 6 5. Volumes b Slicing: Disks nd Wshers In this section we will use definite integrl to find volumes of solid of revolution. Method of Disks Suppose tht f() nd f is continuous on [,b]. Tke the region bounded b the curve = f() nd the -is, for b nd revolve it bout the -is, generting solid. = f() = f() b b We cn find the volume of this solid b slicing it perpendiculr to the -is nd recognizing tht ech cross section is circulr disk of rdius r = f(). We then hve tht the volume of the solid is V = b π[f()] }{{} cross-sectionl re = πr d Emple 5.6 Find the volume of the solid obtined b rotting the region bounded b = from = to = 4 bout the -is. In similr w, suppose tht g() nd g is continuous on the intervl [c,d]. Then, revolving the region bounded b the curve = g() nd the -is, for c d, bout the -is genertes solid. b b = g() = g() c c Once gin, notice from Figure tht the cross sections of the resulting solid of revolution re circulr disks of rdius r = g(). The volume of the solid is then given b V = d c π[g()] }{{} cross-sectionl re = πr d

27 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 6 Emple 5.7 Find the volume of the solid obtined b rotting the region bounded b = from = to = bout the -is. Method of Wshers There re two complictions tht cn be found in the tpes of volume clcultions we hve been studing. The first of these is tht ou m need to compute the volume of solid tht hve cvit or hole in it. The second of these occurs when region is revolved bout line other thn the -is or the -is. Suppose tht f nd g re nonnegtive continuous function such tht g() f() for b nd let R be the region enclosed between the grphs of these functions nd the lines = nd = b. = f() = g() b When this region is revolved bout the -is, it genertes solid hving nnulr or wsher-shped cross sections. Since the cross section t hs inner rdius g() nd outer rdius f(), its volume of the solid is V = b } π {[f()] [g()] d Suppose tht u nd v re nonnegtive continuous function such tht v() u() for c d. If R the region enclosed between the grphs of = u() nd = v() nd the lines = c nd = d. d = v() c = u()

28 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 64 When this region is revolved bout the -is, it lso genertes solid hving nnulr or wsher-shped cross sections. Since the cross section t hs inner rdius v() nd outer rdius u(), its volume of the solid is V = d c } π {[u()] [v()] d Emple 5.8 The region R enclosed b the curves = 4 nd =. Find the volume of the solid obtined b rotting the region R () bout the -is (b) bout the line = (c) bout the line = 7, nd (d) bout the line =. Emple 5.9 Find the volume of the solid obtined b rotting the region bounded b = + nd the line = bout the line =. Eercise 5.. Findthevolumeofthesolidobtinedbrottingtheregionboundedb =, =, nd = () bout the -is (b) bout the line =.. Find the volume of the solid obtined b rotting the region bounded b =, =, nd = () bout the -is (b) bout the line = 4.. Find the volume of the solid obtined b rotting the region bounded b = e, =, =, nd = () bout the -is (b) bout the line =. 4. Findthevolumeofthesolidobtinedbrottingtheregionboundedb =, =, nd = () bout the -is (b) bout the -is. 5. The region R enclosed b the curves =, the -is, nd the -is. Find the volume of the solid obtined b rotting the region R () bout the -is, (b) bout the -is (c) bout the line =, (d) bout the line =, (e) bout the line =, nd (f) bout the line =. 6. The region R enclosed b the curves =, =, nd =. Find the volume of the solid obtined b rotting the region R () bout the -is, (b) bout the -is, (c) bout the line =, (d) bout the line =, (e) bout the line =, nd (d) bout the line =.. () 8π 4. () π 5 (b) 8π (b) π 7 Answer to Eercise 5.. () π 5 (b) 4π 5 ( e 4. () πe +π (b) π +4e 9 ) 5. () 9π (b) 9π (c) 8π (d) 6π (e) 8π (f) 6π 6. () π (b) π 5 (c) π 6 (d) 7π 5 (e) 7π 6 (f) π 5

29 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn Volumes b Clindricl Shells A clindricl shells is solid enclosed b two concentric right-circulr clinders. h r r The volume V of clindricl shll hving inner rdius r, outer rdius r, nd height h cn be written s V = [re of cross section] [height] = (πr πr )h = π(r +r )(r r )h ( ) r +r = π h (r r ) But r +r is the verge rdius of the shell nd r r is its thickness, so V = π [verge rdius] [height] [thickness] This formul cn be used to find the volume of solid of revolution. Let R be plne region bounded bove b continuous curve = f(), bounded below b the -is, nd bounded on the left nd right, respectivel, b the line = nd = b. = f() = f() b The volume of the solid generted b revolving R bout the -is is given b b V = π }{{} rdius f() }{{} height }{{} d thickness Let R be plne region bounded bove b continuous curve = g(), the -is, nd the line = c nd = d.

30 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 66 d = g() c The volume of the solid generted b revolving R bout the -is is given b V = π d c g() }{{}}{{} rdiusheight d }{{} thickness Emple 5. Find the volume of the solid obtined b rotting the region bounded b = nd = in the first qudrnt bout the -is. Emple 5. The region bounded b the line = ( r h), the -is, nd = h is revolved bout the -is, thereb generting cone (ssume r >, h > ). Find it volume b the shell method. Eercise 5.. Use the method of clindricl shells to find the volume generted b rotting the region bounded b the given curve bout the -is. () =, =, =, = (b) = e, =, =, = (c) =, =. Usethemethodofclindriclshellstofindthevolumeofthesolidobtinedbrotting the region bounded b the given curve bout the -is. () = +, =, =, = (b) =, = 9 (c) =, =, + =. Use the pproprite method to find the volume generted b rotting the region bounded b the given curve bout the specified is. () =, =, = ; bout -is (b) =, = 5, = ; bout = 5 (c) = 4 +, =, = ; bout -is (d) =, =, = ; bout -is (e) =, =, = ; bout =

31 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 67 Answer to Eercise 5.. () π (b) π( /e) (c) 64π/5. () π/ (b) 944π/5 (c) 5π/6. () 6 5 π (b) 4 5 π (c) π (d) π (e) 8π 5.4 Length of Plne Curve Arc Length Problem. Suppose f is continuous on [,b] nd differentible on (,b). Find the rc length L of the curve = f() over the intervl [,b]. In order to solve this problem, we begin b prtitioning the intervl [,b] into n equl pieces: = < < < < n = b, where i i = = b n, for ech i =,,...,n. Between ech pir if djcent points on the curve, ( i,f( i )) nd ( i,f( i )) we pproimte the re length l i b the stright-line distnce between the two points. f( i )+ l i f( i )+ i i From the usul distnce formul, we hve l i ( i i ) +[f( i ) f( i )]. Since f is continuous on ll of [,b] nd differentible on (,b), f is lso continuous on the subintervl [ i, i ] nd is differentible on ( i, i ). Recll tht b the Men Vlue Theorem, we hve f( i ) f( i ) = f (c i )( i i ), for some number c i ( i, i ). This give us the pproimtion l i ( i i ) +[f( i ) f( i )] = ( i i ) +[f (c i )( i i )] = +[f (c i )] ( i i ) }{{} = +[f (c i )].

32 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 68 Adding together the lengths of these n line segments, we get n pproimtion of the totl rc length, n L +[f (c i )]. i= Notice tht s n gets lrger, this pproimtion should pproch the ect rc length, tht is, n L = lim +[f (c i )]. n i= You should recognize this s the limit of Riemnn sum for +[f ()], so tht the rc length is given b the definite integrl: whenever the limit eists. L = b +[f ()] d, Emple 5. Find the rc length of the curve = / from (,) to (4,8). Find the length of the curve.. = ( +) /,. = ,. = ln(sec), π/4 4. = ln( ), 5. = cosh, Eercise 5.4 Answer to Eercise ln( +) 4. ln 5. sinh

33 Chpter 6 Techniques of Integrtion 6. Integrtion b Prts Ever differentition rule hs corresponding integrtion rule. For instnce, the Substitution Rule for integrtion corresponds to the Chin Rule for differentition. The rule tht corresponds to the Product Rule for differentition is clled the rule for integrtion b prts. The Product Rule stte tht if f nd g re differentible functions, then d [ ] f()g() = f()g ()+g()f () d In the nottion for indefinite integrls this eqution becomes [f()g ()+g()f () ] d = f()g() or f()g ()d+ g()f ()d = f()g() We cn rerrnge this eqution s f()g ()d = f()g() g()f ()d Thisformul isclled theformul for integrtion b prts. Letu = f() ndv = g(). Then the differentil re du = f ()d nd dv = g ()d, so, b the Substitution Rule, the formul for integrtion b prts becomes udv = uv vdu Emple 6. Find cosd. Note. Our im in using integrtion b prts is to obtin simpler integrl thn the one we strt with. Thus, in Emple 6. we strt with cosd nd epressed it in terms of the 69

34 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn 7 simpler integrl sind. If we hd chosen u = cos nd dv = d, then du = sind nd v = /, so integrtion b prts gives cosd = (cos) + sind Although this is true, sind is more difficult integrl thn the one we strted with. In generl, when deciding on choice for u nd dv, we usull tr to choose u = f() to be function tht becomes simpler when differentited (or t lest not more complicted) s long s dv = g ()d cn be redil integrted to give v. Emple 6. Find cosln(sin)d. Emple 6. Find Emple 6.4 Find sind. e sin(+)d. If we combine the formul for integrtion b prts with Prt of the Fundmentl Theorem of Clculus, we cn evlute definite integrls b prts. Assuming f nd g re continuous, nd using the Fundmentl Theorem, we obtin b Tht is, if u = f() nd v = g(), then ] b b f()g ()d = f()g() g()f ()d b ] b b udv = uv vdu. Emple 6.5 Find / tn d. Eercise 6. Evlute the integrl.. e d.. sin4d 4. lnd e d

35 MA (Section 75, 75): Prepred b Dr. Archr Pcheenburwn cosd 6. e sin4d 7. (ln) d 8. coscosd 9. sec d. cosln(sin)d. cos(ln ) d. cos d. sin d 4. sind e d 6. ln d 8. ln d 4 (ln) d Answer to Eercise 6. e 4 e +C. ln 9 +C. 4 cos4+ 6 sin4+c 4. e 9 e 7 e +C 5. cos+ 9 cos 7 sin+c 6. 7 e sin4 4 7 e cos4+c 7. (ln) ln++c 8. sincos cossin+c 9. tn+ln cos +C. sinln(sin) sin+c. [sin(ln)+cos(ln)]+c. cos +C. cos +sin +C 4. 4 sin cos ln 7. ln4 8. e (ln) 64 6 ln+ 5 5

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