REVIEW OF CHAPTER 5 MATH 114 (SECTION C1): ELEMENTARY CALCULUS

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1 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS.. Are.. Are d Estimtig with Fiite Sums Emple. Approimte the re of the shded regio R tht is lies ove the -is, elow the grph of =, d etwee the verticl lie = d =. Ufortutel, there is o simple geometric formul for clcultig the res of geerl shpes hvig curved oudries like the regio R. How, the, c we fid the re of R? We c pproimte it i simple w. Figure () shows two rectgles tht together coti the regio R. Ech rectgle hs width / d the hve heights d /4, movig from left to right. The height of ech rectgle is the mimum vlue of the fuctio f, otied evlutig f t the left edpoit of the suitervl of [, ] formig the se of the rectgle. The totl re if the two rectgles pproimtes the re A of the regio R, A + 4 = 7 8 =.875. This estimte is lrger th the true re A sice the two rectgles coti R. We s tht.875 is upper sum ecuse it is otied tkig the height of ech rectgle s the mimum (uppermost) vlue of f() for poit i the se itervl of the rectgle. I Figure (), we improve our estimte usig four thier rectgles, ech of width /4, which tke together coti the regio R. These four rectgles give the pproimtio A = 5 =.785. Figure. Suppose isted we use four rectgles cotied iside the regio R to estimte the re. Ech rectgle hs width /4 s efore, ut the rectgles re shorter d lie etirel eeth the grph of f. The fuctio f() = is decresig o [, ], so the height of ech of these rectgle is give the vlue of f t the right edpoit of the suitervl formig theses rectgles with heights equl to the miimum vlue of f() for poit i ech se suitervl gives lower sum pproimtio ti the re A = 7 =.55.

2 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS This estimte is smller th the re A sice the rectgles ll lie iside of the regio R. The true vlue of A lies somewhere etwee these lower d upper sums:.55 < A <.785. Yet other estimte c e otied usig rectgles whose heights re the vlues of f t the midpoits of their ses. This method of estimtio is clled midpoit rule for pproimtig the re. The midpoit rule gives estimte tht is etwee lower sum d upper sum, ut it is ot quite so cler whether it overestimtes or uderestimtes the re of R to e A = Figure... Averge Vlue of Noegtive Cotiuous Fuctio. m = ++ + is the verge vlue of collectio of umers,,,. Wht is the verge vlue of cotiuous fuctio f o itervl [, ]? Figure.

3 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS Emple. The verge of of si Estimte the verge of the fuctio f() = si o the itervl [, π]. (see figure.) Figure 4. So the verge vlue of si is.65 π.75 which is otied divide the estimted re π. Note. The re uder the grph of positive fuctio d the verge vlue of oegtive fuctio over itervl c ll e pproimted fiite sums. First, we sudivide the itervl ito suitervls, tretig the pproprite fuctio f s if it were costt over ech prticulr suitervl. The we multipl the width of ech suitervl the vlue of f t some poit withi it, d dd these products together. If the itervl [, ] is sudivided ito suitervls of equl widths = ( )/, d if f(c k ) is the vlue of f t the chose poit c k i the k th suitervl, this process gives fiite sum of the form f(c ) + f(c ) + f(c ) + + f(c ). The choices of the c k could mimize or miimize the vlue of f i the kth suitervl, or give some vlue i etwee.. Sigm Nottio d imits of Fiite Sums I estimtig with fiite sum i Sectio 5., we ofte ecoutered sums with m terms. I this sectio we itroduce ottio to write sums with lrge umer of terms. After descriig the ottio d sttig severl of its properties, we look t wht hppes to fiite sum pproimtio s the umer of terms pproches ifiit. Nottio 4. Sigm ottio eles us to write sum with m terms i the compct form k = The Greek letter stds for sum. The ide of summtio k tells us where the sum egis d where it eds. A letter c e used to deote the ide, ut the letters i, j, d k re customr. k is formul for the k-th term. Emple 5. Usig Sigm Nottio = i = (j ) = j= = (i ) 7 (k ) k=

4 4 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS = = Note 6. Alger Rules for Fiite Sums. Sum Rule: k + k = ( k + k ). Differece Rule: k i i k = ( k k ). Costt Multiple Rule: (A umer c) 4. Costt Vlue Rule: c = c Theorem 7. Some Summtios.. The Sum of the First Itegers: k = c k = c ( + ). The first squres: k ( + )( + ) = 6 ( ) ( + ). The first cues: k = Emple 8. Evlute.. Solutio. ( + i). ( + i) = i(i ) = + (i i) = ( + )( ). c. lim ( i ) = lim d. lim i(i ) c. lim i ( i ) = lim ( + ) i = + i = k ( i ) = + 7. ( + )( + ) 6 ( + ) d. lim = ( i ) ( + )( + ) 6 i ( + )( + ) + + = lim = lim 6 6 =. = lim j= (i ) = lim j = lim j= j ( )( + )(( ) + ) 6 ( )( ) ( + ) = lim = lim 6 6 =. Emple 9. Fid the limitig vlue of lower sum pproimtios to the re of the regio R elow the grph = d ove the itervl [, ] o the is usig equl width rectgles whose widths pproch zero d whose umer pproches ifiit. Solutio = =. We divide [, ] ito equl width suitervls [, ], [, ], [, ],, [, ]. So, lower sum is costructed with rectgles whose height over the suitervl [ k, k ] is f( k ) = ( k ), givig the sum S = f ( ) ( ) + f ( ) ( ) + + f()( ) = ( ) ( ) ( k ( ) ) ( ) k f = = =

5 k. So, REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS 5 lim S = pproimtios lso coverge to /. lim ( k ) =. A similr clcultio shows tht the upper sum. Riem Sums d Defiite Itegrls Riem Sums The theor of limits of fiite pproimtios ws mde precise the Germ mthemtici Berhrd Riem. We ow itroduce the otio of Riem sum, which uderlies the theor of the defiite itegrl studied i the et sectio. We egi with ritrr fuctio ƒ defied o closed itervl [, ]. ike the fuctio pictured i Figure 5.8, ƒ m hve egtive s well s positive vlues. We sudivide the itervl [, ] ito suitervls, ot ecessril of equl widths (or legths), d form sums i the sme w s for the fiite pproimtios i Sectio 5.. To do so, we choose - poits 5,,, Á, - 6 etwee d d stisfig To mke the ottio cosistet, we deote d, so tht The set Á 6-6. = Á 6-6 =. P =5,,, Á, -, 6 is clled prtitio of [, ]. The prtitio P divides [, ] ito closed suitervls [, ], [, ], Á, [ -, ]. The first of these suitervls is [, ], the secod is [, ], d the kth suitervl of P is [ k-, k ], for k iteger etwee d. Figure 5. kth suitervl k k The width of the first suitervl [, ] is deoted, the width of the secod [, ] is deoted, d the width of the kth suitervl is k = k - k-. If ll suitervls hve equl width, the the commo width is equl to s - d>. k k k I ech suitervl we select some poit. The poit chose i the kth suitervl [k-, k] is clled ck. The o ech suitervl we std verticl rectgle tht stretches from the -is to touch the curve t sck, ƒsckdd. These rectgles c e ove or elow the -is, depedig o whether ƒsckd is positive or egtive, or o it if ƒsckd = (Figure 5.9). O ech suitervl we form the product ƒsckd# k. This product is positive, egtive or zero, depedig o the sig of ƒsckd. Whe ƒsckd 7, the product ƒsckd# k is the re of rectgle with height ƒsckd d width k. Whe ƒsckd 6, the product ƒsckd# k is egtive umer, the egtive of the re of rectgle of width k tht drops from the -is to the egtive umer ƒsckd. Fill we sum ll these products to get SP = ƒsckd k. f() (c, f(c )) (c k, f(c k)) kth rectgle c c c k c k k (c, f(c )) (c, f(c )) FIGURE 5.9 The rectgles pproimte the regio etwee the grph of the fuctio = ƒsd d the -is. Figure 6.

6 6 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS f() The sum S P is clled Riem sum for ƒ o the itervl [, ]. There re m such sums, depedig o the prtitio P we choose, d the choices of the poits c k i the suitervls. I Emple 5, where the suitervls ll hd equl widths = >, we could mke them thier simpl icresig their umer. Whe prtitio hs suitervls of vrig widths, we c esure the re ll thi cotrollig the width of widest (logest) suitervl. We defie the orm of prtitio P, writte 7P7, to e the lrgest of ll the suitervl widths. If 7P7 is smll umer, the ll of the suitervls i the prtitio P hve smll width. et s look t emple of these ides. () EXAMPE 6 Prtitioig Closed Itervl The set P = {,.,.6,,.5, } is prtitio of [, ]. There re five suitervls of P: [,.], [.,.6], [.6, ], [,.5], d [.5, ]: f() () FIGURE 5. The curve of Figure 5.9 with rectgles from fier prtitios of [, ]. Fier prtitios crete collectios of rectgles with thier ses tht pproimte the regio etwee the grph of ƒ d the -is with icresig ccurc The legths of the suitervls re =., =.4, =.4, 4 =.5, d 5 =.5. The logest suitervl legth is.5, so the orm of the prtitio is 7P7 =.5. I this emple, there re two suitervls of this legth. A Riem sum ssocited with prtitio of closed itervl [, ] defies rectgles tht pproimte the regio etwee the grph of cotiuous fuctio ƒ d the -is. Prtitios with orm pprochig zero led to collectios of rectgles tht pproimte this regio with icresig ccurc, s suggested Figure 5.. We will see i the et sectio tht if the fuctio ƒ is cotiuous over the closed itervl [, ], the o mtter how we choose the prtitio P d the poits i its suitervls to costruct Riem sum, sigle limitig vlue is pproched s the suitervl widths, cotrolled the orm of the prtitio, pproch zero. c k Figure 7. Defiitio. et f() e fuctio defied o closed itervl [, ]. We s tht umer I is the defiite itegrl of f over [, ] if I is the limit of the Riem sums f(c k ) k s the orm P of the prtitio P goes to, tht is tht is I = lim f()d(= P f(c k ) k = I. We deote the defiite itegrl I of f over [, ] lim P f(c k ) k ). f()d, Note. Whe the coditio i the defiitio is stisfied, we s the Riem sums of f o [, ] coverge to the defiite itegrl I = f()d d tht f is itegrle over [, ].. We hve m choices for prtitio P with orm goig to zero, d m choices of poits c k for ech prtitio. The defiite itegrl eists whe we lws get the sme limit I, o mtter wht choices re mde. Whe ech prtitio hs equl suitervls, ech of width =, we will lso write lim f(c k ) c = I = f()d. The limit of Riem sum is lws tke s the orm of the prtitios pproches zero d the umer of itervls goes to ifiit.

7 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS 7 The smol for the umer I i the defiitio of the defiite itegrl is which is red s the itegrl from to of ƒ of dee or sometimes s the itegrl from to of ƒ of with respect to. The compoet prts i the itegrl smol lso hve mes: Upper limit of itegrtio Itegrl sig ower limit of itegrtio ƒsd d Itegrl of f from to The fuctio is the itegrd. f() d is the vrile of itegrtio. Whe ou fid the vlue of the itegrl, ou hve evluted the itegrl. Figure 8. Figure 9. Emple. Fid d d the re A uder = over the itervl [, ] for < <. Solutio Theorem tells us f() = is itegrle o [, ], so d eists. Sice d is idepedet o the choices of prtitio P d c i, we c divide [, ] ito suitervls with equl widths = choose c i to e the right edpoit i ech suitervl. The prtitio is P = {, +, +,, + } d C i = + i. So, ( f(c i ) = c i = + i ) ( ) + ( ) ( + ) lim d ( ) f(c i ) = lim (( ) + = ( ) + lim = ( ) + ( ) = ( ) + ) ( + ) ( ) ( ) ( + ) =. ( ) d i =

8 8 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS Thus, d =. Figure. To fid the re uder the grph = over [, ], we check Figure 6. It is es to see tht the re is equl to the re of the igger trigle (with se d height ) sutrct the re of the smller trigle (with se d height ). So, Are =. Therefore, we c see tht the defiite itegrl of over [, ] is equl to the re uder the grph = over [, ], tht is Are = d =. This leds us to itroduce the followig defiitio. Defiitio. If = f() is oegtive d itegrle over closed itervl [, ], the the re uder the curve = f() over [, ] is the itegrl of f from to, It follows esil from this defiitio tht for < d costt c. A = f()d. d = for < (Wh?). Plese check 4. Properties of Defiite Itegrls cd = c( ) d = + + +,, <. + I sectio 5., we itroduced iformll the verge vlue of oegtive cotiuous fuctio f over itervl [, ], ledig us to defie this verge s the re uder the grph of = f() divided. I itegrl ottio we write this s Averge = f()d. We c use this formul to give precise defiitio of the verge vlue of cotiuous (or itegrle) fuctio, whether positive, egtive, or oth. Altertivel, we c use the followig resoig. We strt with the ide from rithmetic tht the verge of umers is their sum divided. A cotiuous fuctio f o [, ] m hve ifiitel m vlues, ut we c still smple them i orderl w. We divide [, ] ito suitervls of equl width = d evlute f t poit c i i ech suitervls. The verge of the smpled vlues is f(c ) + f(c ) + f(c ) = f(c i ) = f(c i )(wh?) = f(c i ). The verge is otied dividig Riem sum for f o [, ] ( ). As we icrese the size of the smple d the orm of the prtitio pproch zero, the verge pproches poits of view led us to the followig defiitio. f()d (Wh?). Both

9 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS 9 Defiitio. If f is itegrle o [, ], the its verge vlue o [, ], lso clled its me, is v(f) = f()d. Emple 4. Fid the verge vlue of f() = 6 o [ 4, 4]. Solutio Sice f() = 6 is fuctio whose grph is the upper semicircle of rdius cetered t the origi. The re etwee the semicircle d the is for 4 to 4 c e computed usig the geometr formul A = πr = π4 = 8π. Sice f is cotiuous, the re is lso the itegrl of f from 4 to 4, d = 8π. Therefore, the verge vlue of f is v(f) = 4 ( 4) 6 d = 8π 8 = π. 4 4 Emple 5. Suppose tht () g() d. Solutio () g(u)du =. Figure. g()d =. Fid () g()d = g()d =. g()d, () g(u)du, (c) ( g())d, (d)

10 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS (c) (d) ( g())d = g() d = ( g())d =. g() d = g()d = =. Emple 6. Use the M-Mi Iequlit to fid upper d lower ouds for the vlue of Solutio Sice the mimum vlue of f over [, ] is d the miimum vlue of f is, ( ) )d ( ). ( + )d. ( + 5. The Fudmetl Theorem of Clculus Theorem 7. If f is cotiuous o [, ], the t some poit c [, ], f(c) = If f(t) is cotiuous fuctio o [, ], the the itegrl Emple 8. et f(t) = t, =. Fid f(t)dt for >. Theorem 9. The Fudmetl Theorem of Clculus et f(t) e cotiuous fuctio o [, ]. Prt F () = F () = d d f()d. f(t)dt defies ew fuctio F (). f(t)dt is cotiuous o [, ] d differetile o (, ) d its derivtive is f(), f(t)dt = f() for ll (, ). I other words, F () is tiderivtive of f(). Prt If F () is tiderivtive of f(), the f()d = F () F () = F () From previous theorem, we c see tht if F () is tiderivtive of f() (i.e. f()d = F () F (), d so F ()d = F () F (). Theorem. et F () e cotiuous fuctio o [, ]. We hve F () = F () + F ()d. F () = f()) the, Note:. The equtio d d f(t)dt = f() tells us tht if we first itegrte the fuctio f d the differetite the result, we get the fuctio f gi.. F ()d = F () F () tells us if we first differetite the fuctio F () d the itegrte the result, we get the fuctio F ck (djusted itegrtio costt).. Prt of the Fudmetl Theorem of Clculus tells us so to fid f()d = F () F () = F () if F () = f(), f()d we ol to () fid tiderivtive F () of f() () clculte F () F (). Emple. Fid g () (). g() = g() = si t dt (e). g() = + t dt dt (). g() = si(t)dt (c). g() = + t t si(t)dt (d).

11 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS Solutio () d () c e solved pplig Theorem 6 directl. We cosider (c) g() = et = d H() = t si(t)dt. t si(t)dt. The g() = H( ) = H. Accordig to Theorem 6, we kow tht H () = si. Chi Rule tells us g () = H ( )( ) = (si ) = si. (d) g() = si si t dt = t dt. Similr to (c), we c see tht g () = (si ) (si ) = (si ) cos. (e) c e solved comig the methods used i (c) d (d). Note: Usig the ide for solvig (c), we c prove tht ( ) d g() f(t)dt = f(g())g () f(h())h (). d Emple. Fid () π h() cos d d () 4 d. Solutio () Sice si is tiderivtive of cos, ( ) =, 4 d = 4 = 4 = 8 = 7. π cos d = si(π) si = =. () Sice Are si Are To compute the re of the regio ouded the grph of fuctio = ƒsd d the -is requires more cre whe the fuctio tkes o oth positive d egtive vlues. We must e creful to rek up the itervl [, ] ito suitervls o which the fuctio does t chge sig. Otherwise we might get ccelltio etwee positive d egtive siged res, ledig to icorrect totl. The correct totl re is otied ddig the solute vlue of the defiite itegrl over ech suitervl where ƒ() does ot chge sig. The term re will e tke to me totl re. FIGURE 5. The totl re etwee = si d the -is for p is the sum of the solute vlues of two itegrls (Emple 7). EXAMPE 7 Ccelig Ares Figure 5. shows the grph of the fuctio ƒsd = si etwee = d = p. Compute () the defiite itegrl of ƒ() over [, p]. () the re etwee the grph of ƒ() d the -is over [, p]. Figure. Solutio The defiite itegrl for ƒsd = si is give The defiite itegrl is zero ecuse the portios of the grph ove d elow the -is mke ccelig cotriutios. The re etwee the grph of ƒ() d the -is over [, p] is clculted rekig up the domi of si ito two pieces: the itervl [, p] over which it is oegtive d the itervl [p, p] over which it is opositive. p p p si d = -cos d = -[cos p - cos ] = -[ - ] =. p p si d = -cos d = -[cos p - cos ] = -[- - ] =. p p si d = -cos d = -[cos p - cos p] = -[ - s-d] = -. p The secod itegrl gives egtive vlue. The re etwee the grph d the is is otied ddig the solute vlues Are = ƒ ƒ + ƒ -ƒ = 4. Summr: To fid the re etwee the grph of = ƒsd d the -is over the itervl [, ], do the followig:. Sudivide [, ] t the zeros of ƒ.. Itegrte ƒ over ech suitervl.. Add the solute vlues of the itegrls. Figure.

12 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS Emple. (Homework) Fid the re of the regio etwee the is d the grph of f() =,. 6. Idefiite Itegrls d the Sustitutio Method Recll tht f()d deotes ll tiderivtives of f(). Sice two tiderivtives differ costt, for tiderivtive F () of f(), we hve f()d = F () + C, where C is ritrr costt. 6.. Sustitutio: ( ) Ruig the Chi Rule Bckwrds. If u = u() d, the Chi Rule tells us d d u + + = u du u+ d. Tht is + du is oe of tiderivtives of the fuctio u d. Therefore ( u du ) d = u+ d + + C, tht is u du = u+ + + C, if we thik du d is the rte of differetils du to d. The previous equtio suggests tht the simpler epressio du c e sustituted for ( ) du d d whe computig itegrl. As with differetils, whe computig itegrls we hve du = ( ) du d d. Emple 4. Fid ( + ) 5 ( + )d Solutio We set u() = +. The du = ( ) du d d = ( + )d, so tht sustitutio we hve ( + ) 5 ( + )d = u 5 du = u6 6 + C = ( + ) 6 + C. 6 Sustitutio: Ruig the Chi Rule Bckwrds The sustitutios i Emples d re istces of the followig geerl rule. THEOREM 5 The Sustitutio Rule If u = gsd is differetile fuctio whose rge is itervl I d ƒ is cotiuous o I, the ƒsgsddg sd d = ƒsud du. Proof The rule is true ecuse, the Chi Rule, F(g()) is tiderivtive of ƒsgsdd# g sd wheever F is tiderivtive of ƒ: If we mke the sustitutio u = gsd the d d Fsgsdd = F sgsdd # g sd = ƒsgsdd# g sd. ƒsgsddg sd d = d Fsgsdd d d = Fsgsdd + C = Fsud + C Chi Rule Becuse F = ƒ Fudmetl Theorem u = gsd = F sud du Fudmetl Theorem = ƒsud du F = ƒ Emple 5. Fid () si d, () Figure 4. z + z dz.

13 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS The Sustitutio Rule provides the followig method to evlute the itegrl whe ƒ d g re cotiuous fuctios: ƒsgsddg sd d,. Sustitute u = gsd d du = g sd d to oti the itegrl. Itegrte with respect to u.. Replce u g() i the result. ƒsud du. EXAMPE Usig Sustitutio cos s7u + 5d du = cos u # 7 du = 7 cos u du = 7 si u + C = si s7u + 5d + C 7 et u = 7u + 5, du = 7 du, s>7d du = du. With the ( >7) out frot, the itegrl is ow i stdrd form. Itegrte with respect to u, Tle 4.. Replce u 7u + 5. We c verif this solutio differetitig d checkig tht we oti the origil fuctio cos s7u + 5d. Figure The Itegrls of si d cos. Emple 6. Fid () si d, () cos d, (c) si cos d, (d) sec ( + )d. 7. Sustitutio i Defiite Itegrls Sustitutio Formul I the followig formul, the limits of itegrtio chge whe the vrile of itegrtio is chged sustitutio. THEOREM 6 Sustitutio i Defiite Itegrls If g is cotiuous o the itervl [, ] d ƒ is cotiuous o the rge of g, the gsd ƒsgsdd# g sd d = ƒsud du gsd Figure 6.

14 4 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS Proof et F deote tiderivtive of ƒ. The, To use the formul, mke the sme u-sustitutio u = gsd d du = g sd d ou would use to evlute the correspodig idefiite itegrl. The itegrte the trsformed itegrl with respect to u from the vlue g() (the vlue of u t = ) to the vlue g() (the vlue of u t = ). EXAMPE Solutio Evlute + d. - Sustitutio Two Methods We hve two choices. Method : Trsform the itegrl d evlute the trsformed itegrl with the trsformed limits give i Theorem 6. - ƒsgsdd# g sd d = Fsgsddd = = Fsgsdd - Fsgsdd + d = u du = u> d = c> - > d = u=gsd = Fsudd u=gsd gsd = ƒsud du. Evlute the ew defiite itegrl. cd = 4 Method : Trsform the itegrl s idefiite itegrl, itegrte, chge ck to, d use the origil -limits. gsd = et u = +, du = d. Whe = -, u = s-d + =. Whe =, u = sd + =. d d Fsgsdd = F sgsddg sd = ƒsgsddg sd Fudmetl Theorem, Prt + d = u du = u> + C et u = +, du = d. Itegrte with respect to u. = s + d > + C Replce u d = s + d > d - = cssd + d > - ss-d + d > d Use the itegrl just foud, with limits of itegrtio for. = c> - > d = cd = 4 Figure 7. Emple 7. Fid π/ Emple 8. Fid () π/4 cot csc d. t ( + t 4 ) dt, () t 4 dt.

15 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS 5 Defiite Itegrls of Smmetric Fuctios The Sustitutio Formul i Theorem 6 simplifies the clcultio of defiite itegrls of eve d odd fuctios (Sectio.4) over smmetric itervl [-, ] (Figure 5.6). () () FIGURE 5.6 () ƒ eve, - ƒsd d = ƒsd d () ƒ odd, - ƒsd d = Theorem 7 et ƒ e cotiuous o the smmetric itervl [-, ]. () If ƒ is eve, the ƒsd d = ƒsd d. - () If ƒ is odd, the ƒ() d =. - Figure 8. Upper curve f() ower curve g() FIGURE 5.7 The regio etwee the curves = ƒsd d = gsd d the lies = d =. f() g() FIGURE 5.8 We pproimte the regio with rectgles perpediculr to the -is. Figure 9.

16 6 REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS DEFINITION Are Betwee Curves If ƒ d g re cotiuous with ƒsd Ú gsd throughout [, ], the the re of the regio etwee the curves = fsd d = gsd from to is the itegrl of ( f - g) from to : A = [ƒsd - gsd] d. Whe pplig this defiitio it is helpful to grph the curves. The grph revels which curve is the upper curve ƒ d which is the lower curve g. It lso helps ou fid the limits of itegrtio if the re ot lred kow. You m eed to fid where the curves itersect to determie the limits of itegrtio, d this m ivolve solvig the equtio ƒsd = gsd for vlues of. The ou c itegrte the fuctio ƒ - g for the re etwee the itersectios. (, ) (, f()) (, g()) (, ) FIGURE 5. The regio i Emple 4 with tpicl pproimtig rectgle. EXAMPE 4 Are Betwee Itersectig Curves Fid the re of the regio eclosed the prol = - d the lie = -. Solutio First we sketch the two curves (Figure 5.). The limits of itegrtio re foud solvig = - d = - simulteousl for. - = - Equte ƒ() d g(). - - = Rewrite. s + ds - d = Fctor. = -, =. Solve. The regio rus from = - to =. The limits of itegrtio re = -, =. The re etwee the curves is A = [ƒsd - gsd] d = [s - d - s-d] d - = s + - d d = c d - = = 9 Figure.