# MAT 168: Calculus II with Analytic Geometry. James V. Lambers

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1 MAT 68: Clculus II with Anlytic Geometry Jmes V. Lmbers Februry 7,

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3 Contents Integrls 5. Introduction Differentil Clculus nd Quotient Formuls Integrl Clculus nd Product Formuls Ares nd Distnces The Are Problem Sigm Nottion Computing the Exct Are The Distnce Problem The Definite Integrl Evluting Integrls The Midpoint Rule Properties of the Definite Integrl The Fundmentl Theorem of Clculus Prt : Differentition Undoes Integrtion Prt : Integrtion Undoes Differentition Sttement of the Fundmentl Theorem of Clculus Differentition nd Integrtion s Inverse Processes Computing Are Using the Fundmentl Theorem of Clculus Net Are vs. Totl Are Displcement vs. Distnce Averge Vlues Indefinite Integrls The Net Chnge Theorem The Substitution Rule Definite Integrls Symmetry

4 4 CONTENTS Techniques of Integrtion 67. Integrtion by Prts Trigonometric Integrls Trigonometric Substitution Integrtion of Rtionl Functions Simple Proper Rtionl Functions The Method of Prtil Frctions Approximte Integrtion Riemnn Sums Simpson s Rule nd Other Approximtion Methods Improper Integrls Infinite Intervls Discontinuous Integrnds A Comprison Test for Improper Integrls Applictions of Integrtion 9 3. Ares Between Curves Volume by Slices Volume by Shells Arc Length

5 Chpter Integrls. Introduction In your previous clculus course, you lerned bout differentil clculus, which is the study of the rte of chnge of one quntity with respect to nother. We briefly review the min ides of differentil clculus before introducing the closely relted brnch of mthemtics known s integrl clculus, which is the subject of this course... Differentil Clculus nd Quotient Formuls Suppose tht n object moves in stright line during some intervl in time. How cn we compute the velocity of the object t ny prticulr point in time within this intervl? If we know tht the object s velocity does not chnge over time, then we cn use the simple formul velocity = distnce trveled elpsed time = finl position initil position. (.) finl time initil time Since the velocity is known to be constnt during the given time intervl, we cn conclude tht this formul yields the velocity of the object t ny point in time between the initil time nd the finl time. Wht if the object is not necessrily moving t constnt velocity? In this cse, eqution (.) is only useful for computing the verge velocity of the object between the initil time nd the finl time. To compute the velocity t prticulr instnt t, we must be more resourceful. We cn pproximte this quntity, which is clled the instntneous velocity, by using eqution (.) over some smll intervl in time tht contins t, such s the intervl [t, t + h], where h is some smll positive number. 5

6 6 CHAPTER. INTEGRALS For concreteness, we let s(t) be the function tht describes the reltionship between the position of the object nd time. Specificlly, for ech rel number t between the initil time nd the finl time, s(t) is the position of the object, t time t, on the stright line long which it is moving. Then, the velocity of the object t time t cn be pproximted by s(t + h) s(t ) = s(t + h) s(t ). (.) (t + h) t h From eqution (.), we cn see tht this quntity is ctully the verge velocity of the object on the intervl [t, t + h], but since h is smll, it is resonble to ssume tht the object s velocity cnnot vry much during such short period of time, nd we cn conclude tht the instntneous velocity t time t cn be pproximted by this verge velocity. The smller the vlue of h, the less the object s velocity cn vry over the intervl [t, t + h]. Therefore, it is resonble to conclude tht s we choose smller nd smller vlues of h, the verge velocity over the intervl [t, t + h] becomes better pproximtion of the instntneous velocity of the object t time t. We cn therefore define the instntneous velocity t time t to be the limit of the verge velocity s the width of the intervl [t, t +h], which is h, pproches zero. Specificlly, if we denote this velocity by v(t ), then s(t + h) s(t ) v(t ) = lim. (.3) h h Informlly, we re defining the instntneous velocity t time t to be the verge velocity over n infinitely smll intervl in time contining the instnt t. In generl, if two quntities x nd y re relted by the eqution y = f(x), where f is given function, then the instntneous rte of chnge of y with respect to x when x = x is given by the derivtive of f(x) t x = x. The derivtive of f(x) t x = x, which we denote by f (x ), is defined by f f(x + h) f(x ) (x ) = lim, (.4) h h provided this limit exists. The derivtive of function f(x) is itself function, denoted by f (x), tht is defined t every point x t which the limit in eqution (.4) exists nd whose vlue t x is the quntity f (x ) defined in eqution (.4). This notion of the derivtive s function llows us to compute derivtives of mny functions using differentition rules, which is much more efficient thn using the definition of the derivtive directly.

7 .. INTRODUCTION 7 Exmple Suppose tht bll is thrown stright up with n initil velocity of ft/s, nd tht the initil height of the bll is 6 ft. Then, the height, in feet, of the bll t seconds fter it hs been thrown cn be described by the function s(t), where s(t) = 6t + t + 6, t. (.5) Our gol is to compute the velocity of the bll 5 seconds fter it hs been thrown. After 5 seconds, the height of the bll is s(5) = 6 ft. Since the height of the bll hs chnged by 6 6 = ft in 5 seconds, it is tempting to use the simple formul for velocity in eqution (.) to conclude tht the velocity of the bll during this time is /5 = ft/s, but since the initil velocity of the bll is ft/s, we see tht the velocity is clerly not constnt, so this formul cnnot be used directly to compute the velocity t ny prticulr point in time. Insted, we cn pproximte the velocity of the bll fter 5 seconds by using the formul in eqution (.) to compute the verge velocity between t = 5 nd t = 6, which is s(6) s(5) 6 5 = 3 6 = 76 ft/s. (.6) However, we cn obtin more ccurte pproximtion by computing the verge velocity between t = 5 nd t = 5.5, which is s(5.5) s(5) = = 68 ft/s. (.7) We cn use this pproch to obtin the exct velocity t t = 5 using little bstrction. If we pproximte this velocity by computing the verge velocity between t = 5 nd t = 5 + h, where h is ssumed to be some smll, positive number, then we hve s(5 + h) s(5) (5 + h) 5 = [ 6(5 + h) + (5 + h) + 6] 6 h = 6(5 + h + h ) h h = 4 6h 6h h h = 6h 6h h = 6h 6. (.8)

8 8 CHAPTER. INTEGRALS Tking the limit s h pproches zero, we find tht the velocity of the bll fter 5 seconds is 6 ft/s. In this cse, however, the function s(t) tht describes the height of the bll is function whose derivtive cn be computed using differentition rules. In prticulr, we cn use the Power Rule, the Sum Rule, nd the Constnt Multiple Rule to obtin v(t) = s (t) = 3t +. (.9) Since velocity is the rte of chnge of position, nd s(t) describes the position, or height, of the bll s function of time, its derivtive v(t) = s (t) describes the velocity of the bll s function of time. Tht is, v(t), for t, represents the velocity of the bll t seconds fter it hs been thrown. Substituting t = 5 into v(t) yields v(5) = 6, so we conclude, s before, tht the velocity of the bll fter 5 seconds is 6 ft/s. Bsed on this exmple, we cn summrize the dvntge tht the derivtive provides us with regrd to the use of quotient formuls such s the formul in eqution (.). The derivtive, being the limit of quotient of differences, llows quotient formuls to be pplied to more generl problems. For exmple, the cost to produce single unit of product, known s mrginl cost, cn be obtined by the simple quotient formul mrginl cost = cost to produce ll units number of units produced, (.) but this formul is vlid only if the cost to produce ech unit is the sme. This is rrely the cse, due to the distinction between fixed costs, such s the cost of buying mnufcturing equipment, tht re independent of the number of units, nd vrible costs, such s workers wges or costs to mintin equipment, which increse s the number of units increses. Therefore, in order to compute the mrginl cost, it is importnt to recognize tht mrginl cost is the rte of chnge of totl production cost with respect to the number of units produced. Therefore, given function C(x) tht describes the totl production cost in terms of the number of units, which is denoted by x, the mrginl cost fter producing x units is given by the derivtive C (x). This derivtive cn be interpreted s the cost of producing one dditionl unit, given tht x units hve lredy been produced.

9 .. INTRODUCTION 9.. Integrl Clculus nd Product Formuls It is nturl to sk whether other types of formuls cn be generlized to more difficult problems using limits. For exmple, consider the formul for the re A of rectngle of width w nd height h, A = wh. (.) How cn we use this formul to compute the re of shpe tht is not rectngle? We cn proceed by pproximting the given shpe by number of rectngles. Then, we cn pproximte the re of the shpe by using eqution (.) to compute the re of ech rectngle, nd then dding ll of these res together. The ccurcy of this pproch depends on how well we hve pproximted the shpe using rectngles. If we use more, smller rectngles insted of fewer lrge ones, then the pproximte re obtined by dding the res of these rectngles is likely to be more ccurte. Therefore, we cn define the re of the shpe to be the limit of this pproximte re s the number of rectngles becomes infinite, with ech rectngle becoming infinitely smll. Exmple Figure. shows how rectngles cn be used to pproximte given shpe, so tht the re of the shpe cn esily be pproximted by computing the res of ll of the rectngles using eqution (.). Using four rectngles, we obtin n pproximte re of 5.75 squre units, wheres with eight rectngles, we obtin n pproximte re of squre units. Lter in this course, we will lern how to compute the exct re of this shpe, which is 4.6 squre units. Note tht the re obtined using eight rectngles is much more ccurte thn tht obtined using only four rectngles. As nother exmple, suppose we rerrnge the formul for velocity described in eqution (.) to obtin the formul distnce trveled = velocity elpsed time. (.) As with eqution (.), this formul is vlid when the velocity is constnt, but wht if it is not? Agin, we need to be more resourceful. We cn pproximte the distnce trveled by dividing the intervl of time into smll subintervls, nd then using eqution (.) to pproximte the distnce trveled during ech of these subintervls. Then, we cn dd ll of these distnces together to pproximte the distnce trveled during the entire intervl of time.

10 CHAPTER. INTEGRALS Figure.: Approximtion the shpe bounded by the curve y = x + nd the lines y =, x = nd x =, using four rectngles (left plot) nd eight rectngles (right plot). This pproch yields resonble pproximtion to the distnce trveled becuse, s previously mentioned, the velocity cnnot vry much during very short periods of time, nd therefore eqution (.) yields n ccurte pproximtion of the distnce trveled during ech subintervl. As we divide the originl time intervl into more nd more subintervls, our pproximtion of the distnce becomes more nd more ccurte. Therefore, we cn define the exct distnce trveled to be the limit of this pproximte distnce s the number of subintervls becomes infinite, provided tht ech subintervl becomes infinitely smll. In both cses, we wished to compute some quntity z using product formul of the form z = xy. In the first exmple, z is re, x is width nd y is height; in the second, z is distnce, x is time nd y is velocity. However, in both cses, the formul z = xy is only vlid if x nd y re constnt

11 .. AREAS AND DISTANCES vlues tht re independent of one nother. If this is not the cse, we cn still use this formul on smller instnces of the sme problem, nd dd the results together. The process of pproximting the quntity z in this mnner, nd tking the limit s the number of smller problems becomes infinite, is clled integrtion, nd the exct vlue of z is known s definite integrl. The study of definite integrls nd their computtion is clled integrl clculus. We now relte the definite integrl to the derivtive by the following sttement tht summrizes the usefulness of the definite integrl in the sme wy tht the usefulness of the derivtive ws described erlier. The definite integrl, being the limit of sum of products, llows product formuls to be pplied to more generl problems. In this course, we will use the definite integrl to expnd the usefulness of number of product formuls, including not only the formuls for re nd distnce tht we hve discussed, but lso formuls for computing quntities such s the volume of solid object, or the work tht is required to move n object given distnce. As we will see, the concept of the definite integrl closely prllels the concept of the derivtive. Both concepts re defined using limit, which is the essentil ingredient tht llows us to obtin n exct vlue for the quntity we wish to compute, insted of mere pproximtion. Furthermore, both derivtives nd definite integrls cn be viewed s functions, which leds to the development of rules tht cn be used to compute these exct vlues with much greter efficiency thn by computing the pproprite limit directly. The significnce of this one dvntge cnnot be overstted, for it hs served s one of the most fundmentl ctlysts for scientific dvncement over the pst few centuries. This benefit is direct result of the resourcefulness nd dediction of the mthemticins who developed differentil nd integrl clculus in order to solve the problems of their dy s efficiently nd ccurtely s possible. We would do well to emulte their virtues in our efforts to solve the problems of tody.. Ares nd Distnces There re mny cses in which some quntity is defined to be the product of two other quntities. Some exmples re:

12 CHAPTER. INTEGRALS If n object is trveling t constnt velocity v, then the distnce d tht it trvels within time spn of length t is given by the reltion d = vt. By definition, rectngle of width w hs constnt height h. The re A of the rectngle is given by the formul A = wh. The mss m of n object with volume v nd density d is given by the formul m = dv. Unfortuntely, in mny pplictions, we cnnot necessrily ssume tht certin quntities such s velocity or density re constnt, nd therefore we cnnot use formuls such s d = vt directly. However, we cn use them indirectly in these more difficult cses by employing the most fundmentl concept of clculus, the limit... The Are Problem Suppose we wish to compute the re A of shpe tht is not rectngle. To simplify the discussion, we ssume tht the shpe is bounded by the verticl lines x = nd x = b, the x-xis y =, nd the curve defined by some continuous function y = f(x), where f(x) for x b. If the function f(x) is not constnt on the intervl [, b], then the shpe is not rectngle, so we cnnot compute its re by using the formul A = wh directly. However, we cn use this formul indirectly. We begin by ttempting to pproximte the shpe by rectngle, since three sides of the boundry of the shpe do form portion of rectngle. How should we choose this rectngle? We consider this question in the following exmple. Exmple 3 Suppose tht we wish to pproximte the shpe shown in Figure. by rectngle. Certinly, the bse of the rectngle should coincide with the bse of the shpe, since it is line segment connecting the points (, ) nd (, ). It follows tht the width of the rectngle is units. Choosing the height, however, is more difficult. Figure.3 illustrtes the consequences of two choices for the height. If the height of the rectngle is determined by the height of the shpe on the left side, then the resulting rectngle hs height, so the re of the rectngle is. As cn esily be seen from the figure, this vlue is too smll, since significnt portion of the shpe lies outside the rectngle. On the other hnd, if the height of the rectngle is determined by the height of the shpe on the right side, then the resulting rectngle hs height 5, so the re of the rectngle is. Clerly, this vlue is too lrge, becuse the rectngle contins not only the entire

13 .. AREAS AND DISTANCES 3 Figure.: Region bounded by the verticl lines x = nd x =, the horizontl line y =, nd the curve y = x +. In this section we will lern how to compute the exct re of this region. shpe, but lso significnt re outside of the shpe. In summry, ll we know so fr is tht the re of the shpe is number between nd. The reson why neither of these estimtes re close to the exct re is tht the height of the shpe vries significntly between x = nd x =, wheres the height of rectngle is constnt. However, if we exmine ny smll subintervl [c, d] of [, ], we find tht the function f(x) = x + tends not to vry much within the subintervl. Therefore, single rectngle could yield n ccurte pproximtion to the re bounded by y = f(x), x = c, x = d nd y =. The discussion in the preceding exmple suggests tht we cn obtin better pproximtion to the re of non-rectngulr shpe by pproximting the shpe itself with number of rectngles, insted of just one

14 4 CHAPTER. INTEGRALS Figure.3: Attempts to pproximte the shpe from Figure. by rectngle. In the left plot, the height of the rectngle is the height of the left side of the shpe, wheres in the right plot, the height of the rectngle is equl to the height of the right side of the shpe. rectngle. Then, we cn use the formul A = wh to compute the re of ech rectngle, nd dd these res together. We begin by dividing the intervl [, b] into n subintervls of equl width x = (b )/n. These subintervls hve endpoints [x, x ], [x, x ],..., [x n, x n ], where x i = + i x, for i =,,,..., n. Then, we define n rectngles s follows: for ech i =,,..., n, the ith rectngle hs bse defined by the line segment connecting the points (x i, ) nd (x i, ). It follows tht ech rectngle hs width of x, since, for i =,,..., n, x i x i = ( + i x) ( + (i ) x) = [i + (i )] x = x. (.3) We cn choose the height of ech rectngle using similr pproch s with single rectngle: setting the height of the rectngle equl to the height of

15 .. AREAS AND DISTANCES 5 the shpe t some point. Which point should we choose? We consider the first rectngle, whose bse is the line segment connecting (, ) to ( + x, ). The purpose of this rectngle is to pproximte the re of portion of the shpe. This portion is the region bounded by the verticl lines x = nd x = + x, the horizontl line y =, nd the curve y = f(x). To pproximte the re of this portion with the re of the given rectngle, we cn proceed s before, nd set the height of the rectngle equl to the height of either the left side or the right side of this portion of the shpe. For concreteness, we choose the right side, in which cse the height of the rectngle is f( + x), or f(x ). Therefore, the re of this first rectngle is f(x ) x. Proceeding in this fshion with the other rectngles, we choose the height of the ith rectngle to be f(x i ), for i =,,..., n. Then, the re of the ith rectngle is f(x i ) x. From the res of these n rectngles, we obtin n pproximtion A n for the re of the shpe: A A n = f(x ) x + f(x ) x + + f(x n ) x. (.4) Exmple 4 Figure. illustrtes this pproch to pproximting the re of the shpe shown in Figure., using n = 4 nd n = 8 rectngles. In the cse where n = 4, the width of ech rectngle is ( )/4 = /, nd the heights of the rectngles re given by f(/), f(), f(3/), nd f(), where f(x) = x +. Computing the res of these four rectngles nd dding them together, we obtin the vlue In the cse where n = 8, the width of ech rectngle is ( )/8 = /4, nd the heights of the eight rectngles re equl to f(i/4), for i =,,..., 8. Computing the res of these eight rectngles nd dding them together, we obtin the vlue As cn be seen from Figure., in both cses, the pproximte re tht we hve computed is too lrge, becuse the union of the rectngles contins the entire shpe s well s some dditionl re. However, these estimtes of the re re much more ccurte thn our previous estimtes of nd tht were obtined by pproximting the shpe with single rectngle... Sigm Nottion For convenience, we will write summtions such s the expression for A n in eqution (.4) using sigm nottion: A n = f(x ) x + + f(x n ) x = n f(x i ) x. (.5) i=

16 6 CHAPTER. INTEGRALS The Σ symbol indictes tht the expression to the right, in this cse f(x i ) x, is to be evluted for certin vlues of some index vrible, nd the resulting vlues re to be dded. The index vrible is indicted below the Σ; in this cse, the index vrible is i. The strting vlue of the index vrible is lso indicted below the Σ; in this cse, the strting vlue is. The index vrible counts from the strting vlue to the finl vlue, which is indicted bove the Σ; in this cse, the finl vlue is n. In summry, the index vrible i ssumes the vlues,,..., n. For ech of these vlues of i, the vlue of the expression f(x i ) x is computed, nd ll of these vlues re dded together. In generl, sigm nottion is useful for concisely describing the sum of consecutive elements of sequence m, m+, m+,..., n, where m nd n re integers. Using this nottion, the sum of ll of these elements cn be written s n i = m + m+ + m+ + + n. (.6) i=m The index vrible i specified below the Σ effectively counts from the strting vlue m to the finl vlue n. For ech vlue of i between m nd n, including m nd n, the vlue of i is included in the sum. It should be noted tht the letter i is not lwys used s the index vrible, though this is the most common choice. It should lso be noted tht n need not be greter thn m. If m = n, then the summtion includes only one term, which is m. If m > n, then the summtion does not include ny terms, nd therefore the vlue of the sum is zero...3 Computing the Exct Are We now return to the problem of computing the re A of non-rectngulr shpe. If we pproximte the shpe by n rectngles using the pproch described previously, we obtin the pproximte re A n given in eqution (.5). As cn be seen from Figure., we cn obtin more ccurte pproximtion by choosing lrger vlue for n. Why is this the cse? The nswer lies in the fct tht ech rectngle is used to pproximte the re of portion of the originl shpe. Since ech rectngle hs the sme width x = (b )/n, it follows tht s n increses, ech rectngle is ssocited with smller portion of the shpe. The smller the portion is, the less tht the height of the portion cn vry. Therefore, the re of the portion cn be pproximted more ccurtely by the re of rectngle. This discussion suggests tht we cn obtin the exct re of the shpe by computing the limit of A n, s n becomes infinite, ssuming tht this limit

17 .. AREAS AND DISTANCES 7 exists. This is in fct the cse when the function f(x), which defines the top boundry of the shpe, is continuous on the intervl [, b]. This leds to the following definition. Definition Let f(x) be continuous function on n intervl [, b], nd ssume tht f(x) on [, b]. The re A of the region bounded by the verticl lines x = nd x = b, the horizonl line y =, nd the curve y = f(x), is A = lim n n f(x i ) x, (.7) where x = (b )/n nd x i = + i x, for i =,,..., n. i= We now illustrte how this definition of re cn be used to compute the exct re of non-rectngulr shpe. Exmple 5 We will use Definition to compute the re A of the shpe shown in Figure.. In this cse, we hve =, b =, nd f(x) = x +. We hve A = lim n = lim n = lim n = lim n = lim n = lim n n f(x i ) x i= n ( x i + ) n n [ ( + i x) + ] n [ n ( i ) + ] n n n [ ] 4i n + n n ( ) 8i i= i= i= i= n 3 + n i= [( 8 = lim n n 3 + ) ( 8 + n n 3 + n [ 8 = lim n n 3 ( n ) + ) ( 8 n + + ( n + + n )] n 3 + n )]. (.8)

18 8 CHAPTER. INTEGRALS To compute this limit, we first note tht /n is being dded n times, which yields n(/n) =. Second, we use the formul Using the Limit Lws n = n(n + )(n + ). (.9) 6 lim =, n lim cf(n) = c lim f(n), n n where nd c re constnts, we obtin A = lim n = lim n lim f(n)+g(n) = lim f(n)+ lim g(n), n n n (.) [ 8 n(n + )(n + ) n 3 6 n(n + )(n + ) n 3 = lim n + 3n + n n = + 4 ( 3 lim + 3n + n ) n = The Distnce Problem ] + = 4.6. (.) Just s the formul A = wh for the re of rectngle cnnot be used directly to compute the re of shpe whose height is not constnt, the formul d = vt cnnot be used directly to compute the distnce tht n object trvels during given intervl in time if the object s velocity is not constnt. However, we cn use this formul indirectly, just s we used A = wh indirectly to solve the re problem. Suppose tht n object is trveling t non-constnt velocity. If the object strts moving t time t = nd continues until t = b, then wht is the totl distnce trveled? For concreteness, we let v(t) be function tht indictes the object s velocity t time t, where t b. Then, we cn pproximte the distnce by ssuming tht in very smll intervls of time, the velocity is nerly constnt. This is resonble ssumption if the function v(t) tht represents the velocity is continuous function, which we will ssume in this cse.

19 .. AREAS AND DISTANCES 9 We then divide the intervl [, b] into n subintervls of width t = (b )/n, with endpoints [t i, t i ] where t i = + i t for i =,,,..., n. It follows tht between time t i nd time t i, for t =,,..., n, the distnce trveled is pproximtely v(t i ) t, nd the totl distnce d is therefore pproximted by the sum of these n distnces, which we denote by d n : d d n = v(t ) t + v(t ) t + + v(t n ) t. (.) Proceeding s in the re problem, we cn define the totl distnce trveled to be the limit of the pproximtion d n s n pproches infinity: d = lim n n v(t i ) t. (.3) i= Exmple 6 The velocity of bll thrown stright up with n initil velocity of ft/s is described by the function v(t) = 3t +, where t denotes the number of seconds tht hve elpsed since the bll hs been thrown, nd v(t) is the velocity t time t, mesured in ft/s. Using the bove definition of distnce trveled, we cn compute the distnce tht the bll trvels in the first two seconds fter it hs been thrown by dividing the intervl [, ] into n subintervls of width t = ( )/n, nd computing the sum d n = n v(t i ) n = i= n ( 3t i +) n, t i = +i t = i/n, i =,,..., n. i= (.4) It follows tht the totl distnce d tht the bll trvels in its first two seconds of flight is d = lim n d n = lim n n i= n = lim n i= [( = lim n = lim n [ ( 3 i ) n + n ) ( 8i n + n 8 n + n ) ( + 8 ( n) + n 8 n + ) n ( n + + n ( n n + n )] )]. (.5)

20 CHAPTER. INTEGRALS To compute this limit, we first note tht /n is being dded n times, which yields n(/n) =. Second, we use the formul Using the Limit Lws n = n(n + ). (.6) lim =, n lim cf(n) = c lim f(n), n n where nd c re constnts, we obtin d = lim n = 8 = 64 lim [ 8 n(n + ) n lim n n = 64 lim n = 64 lim f(n)+g(n) = lim f(n)+ lim g(n), n n n (.7) n(n + ) n n + ( n + n ] + ) = 36. (.8) We conclude tht in its first two seconds of flight, the bll trvels 36 feet up. Therefore, if the bll ws thrown from height of 6 feet, the ltitude of the bll fter two seconds would be 4 feet. Note tht the ltitude of the bll t prticulr time cnnot be determined exclusively from our knowledge of the velocity; the initil ltitude must lso be known. This point will be exmined in greter detil in upcoming sections..3 The Definite Integrl In the previous section, we pproximted the re under the grph of continuous, nonnegtive function y = f(x) between the verticl lines x = nd x = b by computing the sum of the res of n rectngles determined by the division of the intervl [, b] into n subintervls of equl width x = (b )/n. Intuitively, we concluded tht s n, the pproximte re A A n = n f(x i ) x, (.9) i=

21 .3. THE DEFINITE INTEGRAL where x i = + i x for i =,,..., n, converges to the exct re of the given region. More generlly, suppose tht for ech n =,,..., we define the quntity R n by choosing points = x < x < < x n = b, nd computing the sum R n = n f(x i ) x i, x i = x i x i, x i x i x i. (.3) i= The sum in eqution (.3) is known s Riemnn sum. Note tht the intervl [, b] need not be divided into subintervls of equl width, nd tht f(x) cn be evluted t rbitrry points belonging to ech subintervl. The Riemnn sum R n pproximtes the re under y = f(x), between x = nd x = b, by the sum of the re of n rectngles, where the width of the ith rectngle is x i, nd the height is f(x i ), for i =,,..., n. This is generliztion of our previous pproximtions of the re, in which we required tht ech rectngle hd the sme width x = (b )/n, nd, for i =,,..., n, the point x i belonging to the ith subintervl [x i, x i ] ws lwys chosen to be the left endpoint x i or the right endpoint x i. If f(x) on [, b], then, s n, the Riemnn sum R n converges to the re under the curve y = f(x), provided tht the widths of ll subintervls [x i, x i ], for i =,,..., n, lso pproch zero. If f ssumes negtive vlues on [, b], then under the sme conditions on the widths of the subintervls, R n converges to the net re between the grph of f nd the x-xis, where re below the x-xis is counted negtively. This is due to the fct tht the height of ech rectngle is determined by f(x), so when f(x) is negtive, rectngles will hve negtive height, nd therefore negtive re. This will be discussed in greter detil in the next section. Our definition of the re under the grph of f(x) between x = nd x = b, while precise, is lso, unfortuntely, rther cumbersome. This cn be llevited by introducing nottion tht concisely describes the whole process of computing the re in the mnner in which we hve described. To tht end, we stte the following definition. Definition Let the function f(x) be continuous on the intervl [, b]. Furthermore, let {R n } n= be sequence of Riemnn sums, s defined in eqution (.3), tht hs the property tht [ ] lim mx x i =. (.3) n i n

22 CHAPTER. INTEGRALS We define the definite integrl of f(x) from to b to be b f(x) dx = lim n R n. (.3) The function f(x) is clled the integrnd, nd the vlues nd b re clled the lower nd upper limits of integrtion, respectively. The process of computing the vlue of definite integrl is clled integrtion. The concept of the definite integrl embodies the entire process of pproximting the re of complicted region by reducing the problem to severl simpler problems of computing res of simple regions, nd then obtining the exct re through limiting process. By representing the result of such complex process using such simple nottion, we cn much more esily obtin solutions to wide vriety of problems. This is the primry dvntge offered by clculus: it provides us with simple mens of working mthemticlly with intuitive concepts such s distnce nd re..3. Evluting Integrls Now tht we hve defined the definite integrl for the purpose of computing quntities such s res nd distnces, we now discuss the problem of ctully evluting definite integrl of the form b f(x) dx, (.33) for given function f(x) nd limits of integrtion nd b. Certinly, one pproch is to use the definition directly. This entils pproximting the integrl (.33) using Riemnn sum of the res of n rectngles, nd then computing the limit s n. In the cse where f(x) is polynomil, it is helpful to use vrious well-known formuls for evluting summtions, such s n n(n + ) i = (.34) or i= n i = i= n(n + )(n + ). (.35) 6 Similr formuls for higher powers of i cn be derived by solving systems of liner equtions.

23 .3. THE DEFINITE INTEGRAL 3 The following summtion rules re lso useful for reducing the complexity of Riemnn sums: n c = nc (.36) i= n c i = c i= n i + b i = i= n i b i = i= n i (.37) i= n i + i= n i i= n b i (.38) i= n b i (.39) We now show how these rules cn be helpful for the tsk of evluting definite integrl using the definition. Exmple 7 Compute the re of the region bounded by the curve y = f(x), where f(x) = x + 3x +, (.4) s well s the horizontl line y =, the verticl line x =, nd the verticl line x = 3. Solution The re A of this region, which is shown in Figure.4, is given by the vlue of the definite integrl A = 3 f(x) dx = 3 i= x + 3x + dx. (.4) To compute this integrl, we pproximte the region by n rectngles of equl width, with height determined by f(x). Specificlly, we divide the intervl [, 3] into n subintervls of equl width x = (3 )/n = 3/n. These subintervls hve endpoints [x, x ], [x, x ],..., [x n, x n ] where x i = i x, for i =,..., n. These points re used to determine the rectngles tht pproximte the originl region. The ith rectngle is determined by the ith subintervl [x i, x i ] s follows: the width is the length of the subintervl, which is x = 3/n, nd the height is given by f(x i ) = x i + 3x i +. Figure.5 illustrtes these rectngles for the cse of n = 4. To pproximte the re of the region, we cn compute the res of the n rectngles nd dd them. The resulting sum is n exmple of Riemnn

24 4 CHAPTER. INTEGRALS Figure.4: Region bounded by the curve y = x + 3x + nd the lines y =, x = nd x = 3. sum. As n, the number of rectngles, increses, we cn expect tht the pproximte re will pproch the exct re. This is in fct the cse, which is why the definite integrl is defined to be the limit of Riemnn sum of the res of n rectngles, s n becomes infinite. In other words, b n A = f(x) dx = lim f(x i ) x, (.4) n where the sigm nottion n i= is used to denote the ddition of n expressions tht re indexed by the vrible i tht counts from to n. We now compute this limit of Riemnn sum to obtin the re of our region. Using the lws of sums, we hve A = 3 x + 3x + dx i=

25 .3. THE DEFINITE INTEGRAL 5 Figure.5: Approximting rectngles for the cse of n = 4. = lim n = lim n = lim n n (x i + 3x i + ) x i= n [(i x) + 3(i x) + ] x i= i= 3 = lim n n 3 = lim n n [ n (3i ) ( 3i + 3 n n [ n (3i ) ( 3i + 3 n n i= [ n ( ) 3i n + 3 n i= i= ) ] 3 + n ) + ( 3i n ] ) + ] n i=

26 6 CHAPTER. INTEGRALS [ n ] 3 9i n = lim n n n + 9i n n + i= i= i= [ ] 3 9 n = lim n n n i + 9 n n i + n i= i= i= [ 3 9 n(n + )(n + ) = lim n n n + 9 ] n(n + ) + n 6 n [ ] 7n(n + )(n + ) 7n(n + ) = lim n 6n 3 + n + 6 [ 7(n 3 + 3n ] + n) = lim n 6n 3 + 7(n + n) n + 6 [ 7 = lim ( + 3n n 6 ) + 7 ( + ) ] + 6 n = 7 ( 6 lim + 3n + n ) n + 7 ( lim + ) + 6 n n = = 8.5 (.43) Certinly, this process of evluting definite integrl using the definition is quite tedious. In the next section, we will see how diligent study of the behvior of the definite integrl cn led to much more efficient methods of evlution, t lest for certin types of integrnds such s the polynomil fetured in the preceding exmple..3. The Midpoint Rule Suppose tht we re computing Riemnn sum by dividing [, b] into n subintervls [x i, x i ], for i =,,..., n, where x = nd x n = b. By the definition of Riemnn sum in eqution (.3), we cn pproximte the underlying definite integrl (.33) by evluting the integrnd f(x) t ny point x i in ech subintervl [x i, x i ] to obtin the height of the corresponding rectngle. However, there is prticulr dvntge to choosing x i to be the midpoint of the subintervl, m i = (x i +x i )/. In the cse where f is liner function on [x i, x i ], the re of the corresponding rectngle is equl to the exct re under f(x), since the region bounded by the grph of f, the x-xis, nd the verticl lines x = x i nd x = x i is trpezoid with

27 .3. THE DEFINITE INTEGRAL 7 verge height m i. This technique of choosing x i to be the midpoint m i is clled the Midpoint Rule. For more generl functions, the Midpoint Rule tends to yield more ccurte pproximtion thn using either the left or right endpoint of ech subintervl..3.3 Properties of the Definite Integrl The properties (.36)-(.39) of summtions cn be used to prove nlogous properties for definite integrls. For exmple, by (.36), b c dx = c(b ). (.44) Similrly, the rules (.37)-(.39) cn be used to estblish the following dditionl properties of definite integrls. In stting these properties, we ssume tht f(x) nd g(x) re continuous on the intervl [, b], nd tht c is constnt. b b b cf(x) dx = c f(x) + g(x) dx = f(x) g(x) dx = b b b f(x) dx, (.45) f(x) dx + f(x) dx b b g(x) dx, (.46) g(x) dx. (.47) These properties show tht the definite integrl is liner function of the integrnd, just s the differentition opertor d/dx is liner opertor, in view of the Sum Rule, Difference Rule nd Constnt Multiple Rule for differentition. There re some useful inequlities pertining to definite integrls. First of ll, if f(x) g(x), then b f(x) dx It follows tht if m f(x) M, then m(b ) b b g(x) dx. (.48) f(x) dx M(b ). (.49) This property will be very useful in the next section, when we prove very importnt theoreticl result concerning definite integrls.

28 8 CHAPTER. INTEGRALS Intuitively, the re under the grph of f from to b, combined with the re under the grph of f from b to c, should equl the re under f from to c. More precisely, b f(x) dx + c b f(x) dx = c f(x) dx. (.5) A relted property is tht interchnging the limits of integrtion reverses the sign of the integrl; i.e., b f(x) dx = b f(x) dx. (.5) This is esy to see by exmining Riemnn sum, since it includes differences of x-vlues tht re negted by interchnging the limits. We illustrte this in the following exmple. Exmple 8 By the definition of the definite integrl, nd the definition of Riemnn sum given in eqution (.3), we cn define b f(x) dx = lim n R n, (.5) where the Riemnn sum R n is defined using eqully spced points = x, x, x,..., x n = b, where x i = + i x, for i =,,,..., n. Such Riemnn sum hs the form R n = n f(x i ) x, (.53) i= where x = (b )/n nd x i is ny point in the subintervl [x i, x i ]. This is the cse regrdless of whether b > or > b. In the ltter cse, x is negtive insted of positive. For exmple, if n = 4, =, nd b = 3, then we hve x = (3 )/4 = /, nd x i = + i/, for i =,,, 3, 4. It follows tht if we choose x i = x i for i =,, 3, 4, then R 4 = f(x ) x + f(x ) x + f(x 3 ) x + f(x 4 ) x = f(3/)(/) + f()(/) + f(5/)(/) + f(3)(/) = [ ( ) ( ) ] 3 3 f + f() + f + f(3). (.54)

29 .4. THE FUNDAMENTAL THEOREM OF CALCULUS 9 On the other hnd, suppose tht we interchnge the limits of integrtion so tht = 3 nd b =. Then, x = ( 3)/4 = /, nd x i = 3 i/, for i =,,, 3, 4. If we choose ech point x i in (.53) so tht x i = x i, for i =,, 3, 4, then we hve R 4 = f(x ) x + f(x ) x + f(x ) x + f(x 3 ) x = f(3)( /) + f(5/)( /) + f()( /) + f(3/)( /) = [ ( ) ( ) ] 3 3 f + f() + f + f(3). (.55) We see tht the two Riemnn sums re identicl, except tht they re of opposite sign. As n, this leds to the reltion from eqution (.5). One importnt consequence of the property (.5) is tht for ny function f(x) nd ny number, f(x) dx =. (.56) This cn be proved by setting b = in eqution (.5). We find tht the definite integrl of f(x) from to is equl to its own negtion, which implies tht it must be equl to zero..4 The Fundmentl Theorem of Clculus In order to evlute definite integrl, we must compute the limit of Riemnn sum, which, s we hve seen, cn be very tedious. In this section, we will discover much more efficient method for evluting definite integrls tht will prove to be useful in mny cses. Our discussion will be very similr to the derivtion of differentition rules. By the definition of derivtive, the derivtive of function f(x) t point x = x is number, denoted by f (x ), nd tht number represents the instntneous rte of chnge of y = f(x) with respect to x, t x = x. In order to compute the derivtive, using the definition, it is necessry to compute limit, which cn lso be rther tedious. To llevite this difficulty, we choose to view the derivtive of f(x) s function in its own right. Tht is, we cn define the function f (x) to be the function whose vlue t every point x in the domin of f is equl to the derivtive of f t x, ssuming tht the derivtive exists t x. Specificlly, f is defined by f f(x + h) f(x) (x) = lim, (.57) h h

30 3 CHAPTER. INTEGRALS for ech x in the domin of f. By llowing x to be vrible, insted of fixed number, we re ble to use the definition of the derivtive to obtin functions tht represent the derivtives of certin types of functions, such s polynomils or trigonometric functions, insted of hving to use the definition directly every time we wnt to compute the derivtive of function t specific point. Exmple 9 Consider the function f(x) = x. We cn compute the derivtive of f(x) t x = by using the definition of the derivtive to obtin f () = f( + h) f() lim h h = ( + h) 4 lim h h = 4 + 4h + h 4 lim h h = lim 4 + h h = 4. (.58) Alterntively, we cn use the definition of the derivtive in more bstrct mnner, llowing x to vry insted of fixing x =. This yields f (x) = f(x + h) f(x) lim h h = (x + h) x lim h h = x + hx + h x lim h h = lim x + h h = x. (.59) Then, we cn esily compute f (x) t ny vlue of x. By being even more bstrct, nd using the definition of the derivtive with types of functions insted of specific functions, we obtin the vrious differentition rules with which we re fmilir. We will use the sme pproch to simplify the process of computing definite integrls. First, we will view the definite integrl of function f(x) over n intervl [, b] s function in its own right, nd try to understnd this function s behvior. It turns out tht this perspective will led to remrkble discovery concerning the reltionship between integrls nd derivtives.

31 .4. THE FUNDAMENTAL THEOREM OF CALCULUS 3.4. Prt : Differentition Undoes Integrtion Let f(x) be function tht is continuous for ll x. How does the definite integrl of f chnge s the limits of integrtion chnge? It is our hope tht the nswer to this question will provide us with more efficient method for computing definite integrls. To nswer this question, we view the definite integrl s function of the limits of integrtion, which mens tht the limits re now vribles, insted of fixed numbers. For simplicity, we will only llow the upper limit to vry, nd we will keep the lower limit fixed t. We define g(x) = x f(t) dt, x >. (.6) Then, the vlue of g(x) t ny point x > is the definite integrl of f(x) from to x. How does g(x) chnge s x chnges? We cn nswer this question by ttempting to compute g (x), which is the instntneous rte of chnge of y = g(x) with respect to x. Using the definition of the derivtive, nd the properties of definite integrls introduced in Section.3, we obtin g g(x + h) g(x) (x) = lim h h [ x+h = lim f(t) dt h h = lim h h x+h x x ] f(t) dt f(t) dt. (.6) To compute this limit, we note tht if h is sufficiently smll, then f is continuous on the intervl [x, x + h]. We my ssume tht h is sufficiently smll, becuse we re computing limit s h pproches. By the Extreme Vlue Theorem, f ssumes mximum vlue nd minimum vlue on the intervl [x, x + h]. Let the mximum nd minimum vlues of f on [x, x + h] be ttined t the points M(h) nd m(h) respectively. Tht is, f(m(h)) f(t) f(m(h)), x t x + h. (.6) Then, by the properties of definite integrls, f(m(h))h x+h x f(t) dt f(m(h))h, (.63)

32 3 CHAPTER. INTEGRALS since h is the width of the intervl [x, x + h]. Becuse we re computing the limit s h pproches, we re not concerned with the cse of h =. Therefore, we cn divide through by h nd obtin f(m(h)) h x+h x f(t) dt f(m(h)). (.64) Now, we let h pproch. Becuse x m(h) x + h, nd x M(h) x + h, it follows from the Squeeze Theorem tht lim m(h) = x, lim h M(h) = x. (.65) h Becuse f is continuous on [x, x + h], it follows from the definition of continuity tht lim f(m(h)) = f(x), lim f(m(h)) = f(x). (.66) h h By the Squeeze Theorem, it follows from equtions (.64) nd (.66) tht We conclude tht lim h h x+h x g (x) = lim h h f(t) dt = f(x). (.67) x+h x f(t) dt = f(x). (.68) In words, this eqution sttes tht the instntneous rte of chnge of the re of region with respect to its width is equl to the height of the region t the point where the width is chnging. This is esy to see in the simple cse of rectngle of width w nd height h. Since the re A of the rectngle is given by the formul A = wh, we cn esily determine tht the rte of chnge of the re with respect to the width is given by da dw = d(wh) = h, (.69) dw since the height of rectngle is constnt. Now, however, we cn compute the rte of chnge of the re of region with respect to its width, even if the height is not constnt. More generlly, suppose tht we define the function F (x) by F (x) = x f(t) dt, (.7)

33 .4. THE FUNDAMENTAL THEOREM OF CALCULUS 33 where f(t) is continuous on [, b]. Then, Prt of the Fundmentl Theorem of Clculus sttes tht for ny x in (, b), F is continuous nd differentible t x, nd F (x) = f(x). The function F (x) is clled n ntiderivtive of f(x). Lter in this section, we will see how ntiderivtives cn be used to esily evlute definite integrls. Exmple Differentite the function F (x) = x 3 t + dt. (.7) Solution Recll Prt of the Fundmentl Theorem of Clculus, which sttes tht if f(x) is continuous, nd the function F (x) is defined by the integrl F (x) = x f(t) dt, (.7) then F (x) = f(x). Applying this result to the given function F (x), we hve F (x) = d [ x ] t + dt = x +. (.73) dx Exmple Differentite the function 3 F (x) = x t + dt. (.74) Solution We use the following property of definite integrls: It follows tht b f(x) dx = F (x) = d [ dx x [ = d dx = d dx b f(x) dx. (.75) ] t + dt x ] t + dt [ x ] t + dt = (x + ). (.76)

34 34 CHAPTER. INTEGRALS It should be noted tht except for the minus sign, the derivtive is the sme s in the previous exmple. In other words, the integrl from to x hs the sme derivtive s integrl from 3 to x. This mkes sense becuse the rte of chnge of the re under sin(t + ), with respect to the width x, should only depend on the height of the region, t the point where the region is chnging, which is t t = x. Exmple Differentite the function F (x) = x 3 sin(t + ) dt. (.77) Solution Recll Prt of the Fundmentl Theorem of Clculus, which sttes tht if f(x) is continuous, nd the function F (x) is defined by the integrl F (x) = x f(t) dt, (.78) then F (x) = f(x). Note tht the vlue of F (x) t ny x is the re under the curve y = f(t), between t = nd t = x. Applying this result to the given function F (x), we hve F (x) = d [ x ] sin(t + ) dt = sin(x + ). (.79) dx 3 Exmple 3 Differentite the function F (x) = x Solution Let g(x) be the function defined by g(x) = 3 x 3 sin(t + ) dt. (.8) sin(t + ) dt. (.8) From the previous exmple, we know tht g (x) = sin(x + ). The function F (x) in this exmple cn be written s g(x ), since F (x) nd g(x) re identicl, except tht the upper limit of integrtion for g(x) is x nd the upper limit for F (x) is x. Therefore, F (x) cn be obtined from g(x) simply by replcing x with x in the definition of g(x).

35 .4. THE FUNDAMENTAL THEOREM OF CALCULUS 35 We cn differentite F (x) = g(x ) using the Chin Rule, which yields F (x) = d dx [g(x )] = g (x )(x) = sin((x ) + )(x) = x sin(x 4 + ). (.8) Exmple 4 Differentite the function F (x) = Solution Let g(x) be defined by x x /x sin(t + ) dt. (.83) sin(t + ) dt. (.84) Then, s in the previous exmple, g (x) = sin(x + ). Using the properties of definite integrls, we cn write F (x) s F (x) = = = x /x /x /x sin(t + ) dt x sin(t + ) dt + sin(t + ) dt + sin(t + ) dt x sin(t + ) dt = g(/x) + g(x ). (.85) Therefore, we cn use the Chin Rule to obtin F (x) = d dx [g(x )] d dx [g(/x)] = g (x )(x) g (/x)( /x ) = sin((x ) + )(x) + sin((/x) + )(/x ) = x sin(x 4 + ) + sin ( + ) x x. (.86) Exmple 5 Differentite the function F (x) = sin x 3 t + dt. (.87)

36 36 CHAPTER. INTEGRALS Solution If we define g(x) = x 3 t + dt, (.88) then it follows tht F (x) = g(sin x). From Exmple, we know tht Using the Chin Rule, we obtin g (x) = x +. (.89) F (x) = d [g(sin x)] dx = g (sin x) cos x Exmple 6 Differentite the function F (x) = = [(sin x) + ] cos x. (.9) sin x x t + dt. (.9) Solution We use the following property of definite integrls, b f(x) dx = c f(x) dx + b with = x, b = sin x, nd c =. Furthermore, we define g(x) = x c f(x) dx, (.9) t + dt, (.93) nd note tht by Prt of the Fundmentl Theorem of Clculus, It follows tht F (x) = d dx = d dx = d dx [ sin x x [ sin x g (x) = x +. (.94) ] t + dt t + dt + [ sin x t + dt ] t + dt x ] x t + dt

37 .4. THE FUNDAMENTAL THEOREM OF CALCULUS 37 = d [ sin x ] [ ] t + dt d x t + dt dx dx = d d [ [g(sin x)] g(x ) ] dx dx = g (sin x) cos x g (x )(x) = [(sin x) + ] cos x [(x ) + ](x) = (sin x + ) cos x x(x 4 + ). (.95) Exmple 7 Differentite the function F (x) = g(x) h(x) f(t) dt. (.96) Solution We define the function G(x) by G(x) = x f(t) dt. (.97) By Prt of the Fundmentl Theorem of Clculus, G (x) = f(x). Using the sme pproch s in the previous exmple, we rewrite F (x) using the properties of definite integrls nd obtin F (x) = = h(x) h(x) Using the Chin Rule, we obtin f(t) dt + g(x) f(t) dt + g(x) f(t) dt f(t) dt = G(h(x)) + G(g(x)). (.98) F (x) = d dx [G(g(x))] d dx [G(h(x))] = G (g(x))g (x) G (h(x))h (x) = f(g(x))g (x) f(h(x))h (x). (.99)

38 38 CHAPTER. INTEGRALS.4. Prt : Integrtion Undoes Differentition Previously, we lerned tht the derivtive of the integrl of function f(x) with respect to the integrl s upper limit ws equl to f(x). Specificlly, if g(x) = x f(t) dt, (.) then, by Prt of the Fundmentl Theorem of Clculus, g (x) = f(x). This result, however, is only one prt of this theorem. The second prt, which follows from the first prt described in eqution (.), tells us how we cn evlute the definite integrl b f(t) dt. (.) Suppose tht F (x) is ny ntiderivtive of f; tht is, F (x) = f(x). Then the ntiderivtive g(x) defined in eqution (.) stisfies g(x) = F (x) + C for some constnt C, since g (x) F (x) implies g(x) F (x) must be constnt. By the properties of definite integrls nd the definition of g(x), we hve g() =. It follows tht b f(t) dt = g(b) = g(b) g() = F (b) + C [F () + C] = F (b) F (). (.) In summry, we cn compute definite integrl of f simply by finding n ntiderivtive of f, nd then evluting the ntiderivtive t the limits of integrtion..4.3 Sttement of the Fundmentl Theorem of Clculus We re now redy to summrize these results nd formlly stte the most importnt theoreticl result in ll of differentil nd integrl clculus. In the sttement of the following theorem, note tht we ssume tht functions re differentible on closed intervl. A function f(x) is differentible on closed intervl [, b] if it is differentible on (, b) nd f is differentible from right t nd differentible from the left t b; tht is, the one-sided limits f(x) f() lim x + x nd f(x) f(b) lim x b x b (.3) both exist. This is similr to the definition of continuity on closed intervl.

39 .4. THE FUNDAMENTAL THEOREM OF CALCULUS 39 Theorem (The Fundmentl Theorem of Clculus) Let f(x) be continuous on [, b].. If the function g(x) is defined by the definite integrl g(x) = x f(t) dt, (.4) then g(x) is continuous on [, b] nd differentible on [, b], nd g (x) = f(x), x b. (.5). If F (x) is ny ntiderivtive of f(x) on [, b]; tht is, if F (x) = f(x), x b, (.6) then b f(x) dx = F (b) F (). (.7) Exmple 8 Let f(x) = x. Then, if we define g(x) = x f(t) dt = x t dt, (.8) then, by Prt of the Fundmentl Theorem of Clculus, g (x) = x. Furthermore, if F (x) = x 3 /3, then F (x) is n ntiderivtive of f(x), since F (x) = x. It follows from Prt of the Fundmentl Theorem of Clculus tht 4 x dx = F (4) F () = = (.9) This implies tht the re under the grph of x from x = to x = 4 is equl to 64/ Differentition nd Integrtion s Inverse Processes In words, the Fundmentl Theorem of Clculus sttes tht the derivtive of n integrl of function is the function itself, nd tht the integrl of derivtive of function is the function, up to n dditive constnt. In other words, differentition nd integrtion re inverse processes, just s division nd multipliction re inverse opertions. This prllel is even closer thn it would pper on the surfce. Consider the following:

40 4 CHAPTER. INTEGRALS When speed is constnt over time, distnce is the product of speed nd time. In generl, distnce is the integrl of speed over time. If speed is constnt over time, then it is equl to the rtio of distnce to time. In generl, the speed t ny given time is the derivtive, or rte of chnge, of distnce trveled with respect to time..4.5 Computing Are Using the Fundmentl Theorem of Clculus Exmple 9 Let h nd b be positive constnts. Use the Fundmentl Theorem of Clculus to compute the re of the region bounded by the curve y = f(x), where f(x) = hx/b, the horizontl line y =, nd the verticl lines x = nd x = b. Solution The region is shown in Figure.6. It is tringle with height h Figure.6: Region bounded by the curve y = f(x) = hx/b, the horizontl line y =, nd the verticl lines x = nd x = b.

41 .4. THE FUNDAMENTAL THEOREM OF CALCULUS 4 nd bse of length b. By the definition of the definite integrl, the re A of this region is given by A = b f(x) dx = b hx b dx. (.) We use Prt of the Fundmentl Theorem of Clculus, which sttes tht for ny function f tht is continuous on [, b], b f(x) dx = F (b) F (), (.) where F (x) is ny ntiderivtive of f(x); tht is, F (x) = f(x). We therefore need to find n ntiderivtive of f(x) = hx/b. To ccomplish this, we use two nti-differentition rules: first, if F is n ntiderivtive of f, then, for ny constnt c, cf is n ntiderivtive of cf. Second, the ntiderivtive of x n, for ny integer n, is x n /(n + ). It follows tht if f(x) = hx/b, then F (x) = h b x + + = hx b (.) is n ntiderivtive of f(x). This cn be verified by differentiting F (x), which yields F (x) = d dx [ ] hx = h d b b dx [x ] = h b (x) = hx b = f(x). (.3) Now tht we hve found n ntiderivtive of f(x), we pply the Fundmentl Theorem of Clculus nd obtin b f(x) dx = F (b) F () = h(b ) b h( ) b = hb b = bh, (.4) which is the fmilir formul for the re of tringle of height h nd bse of length b. Exmple Compute the re of the region bounded by the grph of y = f(x) = y y x + y, (.5) h the line y =, nd the lines x = nd x = h, where y, y, nd h re ll constnts.

42 4 CHAPTER. INTEGRALS Solution This region is trpezoid of width h nd heights y nd y. The function f(x) is liner function, defining line tht psses through the points (, y ) nd (h, y ). To compute the re of this region, we gin use Prt of the Fundmentl Theorem of Clculus. Using the rule tht n ntiderivtive of x n is x n+ /(n + ), provided n, we cn determine tht the function F (x), defined by F (x) = y y x h + y x, (.6) is n ntiderivtive of f(x). This cn be verified by differentiting F (x), s in the previous exmple. From Prt of the Fundmentl Theorem of Clculus, we find tht the re A of the given region is A = h h f(x) dx y y = x + y dx h = F (h) F () [ y y h ] = h + y h = y y h h + y h = (y y ) h + y h [ y y ] h + y () = h y h y + hy = h y + h y = h y + y, (.7) which is the well-known formul for the re of trpezoid of width h nd heights y nd y. Exmple Compute the re of the region bounded by the curve y = sin x, where x π, nd the line y =. Solution The region is shown in Figure.7. To compute the re A of this region, we need to evlute the definite integrl A = π sin x dx. (.8)

43 .4. THE FUNDAMENTAL THEOREM OF CALCULUS 43 Figure.7: Region bounded by y = sin x nd y = To ccomplish this, we first obtin n ntiderivtive of f(x) = sin x. Since d [cos x] = sin x, (.9) dx it follows tht F (x) = cos x is n ntiderivtive of sin x, s cn be verified by noting tht F (x) = d d [ cos x] = [cos x] = ( sin x) = sin x. (.) dx dx Therefore, by Prt of the Fundmentl Theorem of Clculus, the re A of the region is given by A = π sin x dx = F (π) F () = cos π ( cos )

44 44 CHAPTER. INTEGRALS = ( ) ( ) = + =. (.).4.6 Net Are vs. Totl Are We know tht the definite integrl b f(x) dx cn be used to compute the re of the region bounded by the verticl lines x = nd x = b, the horizontl line y =, nd the curve y = f(x), when f(x) on [, b]. If f(x) < on [, b], then the integrl represents the negtive of the re of the region, becuse it is the limit of sum of res of rectngles whose heights re negtive. We now consider the interprettion of the integrl when the integrnd f(x) cn be either positive or negtive on [, b]. Exmple Evlute the definite integrl π Wht is the geometric interprettion of this integrl? sin x dx. (.) Solution Using the Fundmentl Theorem of Clculus, nd the fct tht n ntiderivtive of f(x) = sin x is F (x) = cos x, we obtin π sin x dx = F (π) F () = cos(π) ( cos()) = ( ) = + =. (.3) The grph of f(x) = sin x on the intervl [, π] is shown in Figure.8. Wht is the geometric significnce of the integrl being equl to? Note tht in the previous exmple, we determined tht the integrl of the sme

45 .4. THE FUNDAMENTAL THEOREM OF CALCULUS 45 Figure.8: Grph of y = sin x, x π function from x = to x = π is equl to, nd tht is the re of the region bounded by the curve y = sin x nd the lines y =, x = nd x = π. From Figure.8, we cn visully determine tht the re of the region bounded by the curve y = sin x nd the lines y =, x = π nd x = π is lso equl to. Given this informtion, why is the integrl from x = to x = π equl to? We cn nswer this question by computing π π sin x dx = F (π) F (π) = cos(π) ( cos(π)) = ( ( )) = =. (.4) This exmple illustrtes tht b f(x) dx is not lwys equl to the re of the region bounded by the curve y = f(x) nd the lines y =, x = nd

46 46 CHAPTER. INTEGRALS x = b. Rther, it is the net re bove the x-xis; tht is, re bove the x- xis is counted positively, while re below the x-xis is counted negtively. Therefore, the re between x = nd x = π, which is positive, cncels with the re between x = π nd x = π, which is negtive, thus explining why the integrl is equl to zero. In generl, to compute the gross or totl re, rther thn the net re, of the region bounded by y = f(x), y =, x = nd x = b, it is necessry to evlute the definite integrl b f(x) dx. (.5) This is best ccomplished by dividing the intervl [, b] into subintervls bsed on where f(x) is positive or negtive, nd then integrting over ech of these subintervls, using the fct tht x = x when x < in order to eliminte the bsolute vlue from the problem. Then, the integrl from x = to x = b is the sum of the integrls over the subintervls. Exmple To compute the totl re under y = sin x from x = to x = π, we need to compute A = π sin x dx. (.6) We cn compute this integrl s follows: we note tht sin x on [, π], while sin x on [π, π]. It follows tht sin x = sin x on [, π], while sin x = sin x on [π, π]. We then hve A = = = = π π π π sin x dx π sin x dx + sin x dx + sin x dx π π π π π sin x dx sin x dx sin x dx = cos x π [ cos x] π π = [ cos(π) ( cos())] [ cos(π) ( cos(π))] = [ ( ) ( )] [ ( ( ))]

47 .4. THE FUNDAMENTAL THEOREM OF CALCULUS 47 = [ + ] [ ] = ( ) = + = 4. (.7) This vlue is the totl re of the region enclosed by the curve y = sin x nd the lines y =, x = nd x = π. Note tht in this cse we used the nottion f(x) b = f(b) f(). (.8).4.7 Displcement vs. Distnce We now revisit the problem of computing the distnce tht n object hs trveled long stright line, given its velocity v(t), for t b. If the velocity chnges sign from positive, then it follows tht the object hs chnged direction from right to left, nd positive distnce previously trveled to the right is offset by negtive distnce trveled to the left. Just s the definite integrl of function f(x) represents the net re under the curve y = f(x), the definite integrl of function v(t) represents the displcement, or net distnce, trveled by n object long stright line with velocity v(t). In order to compute the totl distnce trveled by the object, it is necessry to integrte v(t) insted. Exmple 3 A flling object hs constnt ccelertion of (t) = 3 ft/s. Since ccelertion is the derivtive of the velocity v(t), it follows tht v(t) is n ntiderivtive of (t). From Prt of the Fundmentl Theorem of Clculus, we hve t (s) ds = v(t) v(). Therefore, if the initil velocity is v = v() ft/s, then the velocity is given by rerrnging the bove eqution: v(t) = v() + t (s) ds = v + t ( 3) ds = v + ( 3s) t = v 3t. To compute the displcement d(t), whose derivtive is the velocity v(t), we gin pply Prt of the Fundmentl Theorem of Clculus to obtin t v(s) ds = d(t) d(),

48 48 CHAPTER. INTEGRALS nd if we denote the initil position by d = d() ft, we hve d(t) = d + = d + t t v(s) ds (v 3s) ds = d + ( v s 6s ) t = d + v t 6t. Now, suppose tht d = 6 ft nd v = 8 ft/s. Then, to compute the displcement between t = s nd t = 5 s, we use the fct tht displcement is n ntiderivtive of velocity to obtin 5 v(s) ds = d(5) d() = [6 + 8(5) 6(5) ] [6 + 8() 6() ] = 46 6 = 4 ft. On the other hnd, to compute the distnce trveled by the object between t = s nd t = 5 s, we must compute 5 v(s) ds = = s ds (8 3s) ds (3s 8) ds = ( 8s 6s ) 4 + ( 6s 8s ) 5 4 = 56 + ( 4 ( 56)) = 7 ft. The integrl from s = to s = 5 is divided into two integrls t s = 4, becuse v(s) on [, 4], while v(s) on [4, 5]. By integrting v(s) from s = to s = 4, nd v(s) from s = 4 to s = 5, we obtin the integrl of v(s) from s = to s = 5, which represents the distnce trveled between t = s nd t = 5 s..4.8 Averge Vlues Suppose tht we wish to compute the verge vlue of tht function f(x) ssumes over the intervl [, b]. If f(x) is constnt function f(x) = c, then

49 .4. THE FUNDAMENTAL THEOREM OF CALCULUS 49 we know tht the verge vlue is simply c. For more generl function f, we cn pproximte the verge by dividing [, b] into subintervls of width x = (b )/n, nd ssuming tht f is pproximtely constnt over ech subintervl. Tking the verge over the ith subintervl to be f(x i ), where is the midpoint of the ith subintervl, we obtin the pproximte verge x i which cn be rewritten s Avg f n Avg f b n f(x i ) (.9) i= n f(x i ) x. (.3) Letting n, we obtin the exct verge s definite integrl Avg f = b i= b f(x) dx. (.3) Let f be differentible over the intervl [, b]. Recll from differentil clculus tht by the Men Vlue Theorem for derivtives, there exists some point c in [, b] such tht f (c) is equl to the verge of f (x) over [, b]. It is nturl to sk if there is some point c in [, b] such tht the totl chnge in f from to b is equl to f(c)(b ), the re of the rectngle tht hs height f(c) nd width (b ). The following theorem nswers this question. Theorem (Men Vlue Theorem for Integrls) If f is continuous on [, b], then for some c in (, b). b f(x) dx = f(c)(b ) (.3) Proof Since f is continuous on [, b], it follows from Prt of the Fundmentl Theorem of Clculus tht the function F (x) defined by F (x) = x f(t) dt (.33) is continuous on [, b] nd differentible on (, b), nd F (x) = f(x) on (, b). By the Men Vlue Theorem for derivtives, F (b) F () b = F (c) (.34)

50 5 CHAPTER. INTEGRALS for some c in (, b). However, by Prt of the Fundmentl Theorem of Clculus, F (b) F () = b f(x) dx. (.35) Substituting the right side of eqution (.35) into eqution (.34) yields b b f(x) dx = F (c), (.36) nd since F (x) = f(x) for ny x in (, b), it follows tht b b f(x) dx = f(c), (.37) nd multiplying both sides by b yields the conclusion of the theorem, stted in eqution (.3). Remrk From eqution (.37), we obtin n intuitive interprettion of the Men Vlue Theorem for Integrls: function f tht is continuous on [, b] ssumes its verge vlue t some point in (, b). Exmple 4 Find b such tht the verge vlue of f(x) = x + b on the intervl [, ] is equl to 8. Solution The verge vlue of f(x) = x + b on [, ] is given by x + b dx = ( ) x bx = ( ) b = 4 + b. (.38) 3 It follows tht in order for the verge vlue of f(x) on [, ] to be equl to 8, we must hve b = = 3. (.39) Exmple 5 Compute the verge vlue of f(x) = x on the intervl [, ]. Solution We use the formul for the verge vlue of f(x) on n intervl [, b], Avg f = b f(x) dx. (.4) b This yields Avg f = x dx = x 3 3 = 8 3 = 4 3. (.4)

51 .5. INDEFINITE INTEGRALS 5 Exmple 6 Find the vlue of b so tht the verge vlue of f(x) = x 3 on the intervl [, b] is equl to 4. Solution The verge vlue of f(x) on [, b] is given by b b x 3 dx = b x 4 4 b = b 4 b 4 = b3 4. (.4) Therefore, in order for the verge vlue to be equl to 4, we must hve b 3 /4 = 4. It follows tht b 3 = 6, which yields the solution b = Indefinite Integrls As evluting integrls cn be ccomplished by computing ntiderivtives, it is desirble to know the ntiderivtives of commonly used functions. For ese of exposition, the nottion of n integrl sign without limits of integrtion is used to denote the set of ll ntiderivtives of function. We now define this nottion precisely. Definition 3 (Indefinite Integrl) Let f(x) be continuous function. The indefinite integrl of f(x), denoted by the symbol f(x) dx, (.43) is the set of ll ntiderivtives of f(x). It is importnt to note tht the indefinite integrl of f(x) does not represent specific function. Insted, it represents fmily of infinitely mny functions, ll of which hve the common property tht their derivtives re equl to f(x). Fortuntely, this fmily of functions is very esy to describe. If we dd constnt to function, its derivtive does not chnge, since the derivtive of constnt is zero. Furthermore, by the Men Vlue Theorem, ny two functions tht hve the sme derivtive differ from one nother by constnt. It follows tht if F (x) is ny ntiderivtive of f(x), then function G(x) is lso n ntiderivtive of f(x) if nd only if G(x) = F (x) + C, (.44)

52 5 CHAPTER. INTEGRALS where C is constnt. Since ll ntiderivtives of f(x) hve the form F (x)+ C, we write f(x) dx = F (x) + C, (.45) where F (x) is ny ntiderivtive of f(x), nd C represents n rbitrry constnt. This nottion is used in order to specify the most generl ntiderivtive of f(x), becuse this generlity my be needed if the ntiderivtive is used in nother problem tht my specify vlue for C, s in the following exmple. Exmple 7 We cn determine tht n ntiderivtive of the function f(t) = 3 is F (t) = 3t, since the derivtive of 3t with respect to t is 3. However, the most generl ntiderivtive of this function is F (t) = 3t+C, where C is n rbitrry constnt. If we interpret f(t) s the ccelertion of n bll thrown in to the ir, then the velocity of the bll is function of the form F (t) = 3t + C, where the constnt C is the initil velocity of the bll. Therefore, the ntiderivtive 3t, without the constnt C, is of little use, becuse it is not generl enough to describe the velocity. Severl indefinite integrls cn be obtined esily from differentition rules, s shown in the following tble. We ssume tht F (x) nd G(x) re ntiderivtives of f(x) nd g(x), respectively; tht is, F (x) = f(x) nd G (x) = g(x). With ech indefinite integrl, we dd the rbitrry constnt C in order to specify the most generl indefinite integrl possible.

53 .5. INDEFINITE INTEGRALS 53 Differentition Rule d dx [f(x) + g(x)] = f (x) + g (x) d dx [f(x) g(x)] = f (x) g (x) d dx [cf(x)] = cf (x) d dx [xn ] = nx n d dx [sin x] = cos x d dx [cos x] = sin x d dx [tn x] = sec x d dx [cot x] = csc x d dx [sec x] = sec x tn x d dx [csc x] = csc x cot x d dx [sin x] = x d dx [tn x] = d dx [ln x ] = x d dx [ex ] = e x d dx [x ] = x ln d dx [f(g(x))] = f (g(x))g (x) d +x x dx [f(x)g(x)] = g(x)f (x) + f(x)g (x) Indefinite Integrl f(x) + g(x) dx = F (x) + G(x) + C f(x) + g(x) dx = F (x) G(x) + C cf(x) dx = cf (x) + C x n dx = xn+ n+ + C, n cos x dx = sin x + C sin x dx = cos x + C sec x dx = tn x + C csc x dx = cot x + C sec x tn x dx = sec x + C csc x cot x dx = csc x + C dx = x sin x + C dx = tn x + C +x dx = ln x + C e x dx = e x + C x dx = x ln + C f(g(x))g (x) dx = F (g(x)) + C F (x)g(x) dx = F (x)g(x) G(x)f(x) dx In the lst indefinite integrl, which is ctully sttement of the technique of integrtion by prts, the rbitrry constnt C is not included becuse n indefinite integrl ppers on both sides of the eqution, nd therefore it is not necessry to include C in order to specify the most generl indefinite integrl. Exmple 8 Compute x 3 6x + 5x 4 dx. (.46) Solution We use the following nti-differentition rules, which re nmed ccording to the differentition rules from which they re obtined. In describing these rules, we ssume tht F (x) = f(x) nd G (x) = g(x). f(x) + g(x) dx = F (x) + G(x) + C (Sum Rule) (.47) f(x) g(x) dx = F (x) G(x) + C (Difference Rule) (.48) cf(x) dx = cf (x) + C (Constnt Multiple Rule)(.49)

54 54 CHAPTER. INTEGRALS We proceed s follows: x 3 6x + 5x 4 dx = x n dx = xn+, n (Power Rule) (.5) n + = x 3 dx 6x dx + 5x dx 4 dx x 3 dx 6 x dx + 5 x dx 4 dx = x4 4 6x x 4x + C = x4 4 x3 + 5x 4x + C. (.5) Note tht even though we evluted ech term s seprte indefinite integrl, it is only necessry to include one term tht is n rbitrry constnt C, becuse it cn be equl to ny constnt. Exmple 9 Compute sec x tn x cos x dx. (.5) Solution Using the Difference Rule s well s the rules sec x tn x dx = sec x + C, (.53) cos x dx = sin x + C, (.54) we obtin sec x tn x cos x dx = sec x tn x dx cos x dx = sec x sin x + C. (.55) Exmple 3 Compute 8x 7 + sec x dx. (.56) Solution Using the Sum Rule, the Constnt Multiple Rule, the Power Rule nd the rule sec x dx = tn x + C, (.57)

55 .5. INDEFINITE INTEGRALS 55 we obtin 8x 7 +sec x dx = 8 x 7 dx+ sec x dx = 8 x8 8 +tn x+c = x8 +tn x+c. (.58) Exmple 3 Given tht the ccelertion (t) of n object t time t is given by (t) = 7 cos t + 5, (.59) nd its initil velocity is v() = 4 nd its initil position is s() =, compute the object s velocity function v(t) nd the position function s(t). Solution Since v (t) = (t), it follows tht v(t) must be n ntiderivtive of (t). Therefore, v(t) hs the form v(t) = F (t) + C, (.6) where F (t) is ny ntiderivtive of (t) nd the constnt C is to be determined so tht v() = 4. To compute n ntiderivtive of (t), we use the Sum Rule, the Constnt Multiple Rule, nd the rule cos x dx = sin x + C (.6) nd obtin v(t) = (t) dt = 7 cos t dt + 5 dt = 7 sin t + 5t + C. (.6) We now specify the constnt C so tht v() = 4. Substituting t = into the eqution v(t) = 7 sin t + 5t + C yields v() = 7 sin + 5() + C, (.63) which simplifies to 4 = C since v() = 4 nd sin =. Therefore, the velocity of the object is given by v(t) = 7 sin t + 5t 4. (.64)

56 56 CHAPTER. INTEGRALS To obtin the position function, we use the fct tht s (t) = v(t); tht is, s(t) is n ntiderivtive of v(t). We hve v(t) dt = 7 sin t + 5t 4 dt = 7 sin t dt + 5 t dt 4 dt = 7 cos t + 5t 4t + C. (.65) We need to determine the vlue of C so tht s() =. Substituting t = into the eqution s(t) = 7 cos t + 5t 4t + C (.66) yields s() = 7 cos + C, (.67) but since cos = nd s() =, we hve = 7 + C, which implies C = 7. We conclude tht s(t) = 7 cos t + 5t is the position of the object t time t. 4t + 7 (.68).5. The Net Chnge Theorem Recll the second prt of the Fundmentl Theorem of Clculus, F (x) = f(x) = By rewriting this reltion s b f(x) dx = F (b) F (). (.69) b F (x) dx = F (b) F (), (.7) we obtin the interprettion tht the integrl of rte of chnge of quntity from to b represents the net chnge in tht quntity from to b. This result is known s the Net Chnge Theorem. Exmple 3 Let v(t) represent the velocity of n object tht is moving in stright line, nd s(t) represent its position. Then s (t) = v(t), nd, by the

57 .6. THE SUBSTITUTION RULE 57 Net Chnge Theorem, the totl displcement of the object from the time t = until the time t = 5 is given by s(5) s() = 5 s(t) dt = 5 v(t) dt. (.7) Alterntively, if the position s() is known, nd the velocity v(t) is known, then the position t ny time t = b is given by the following rerrngement of the Net Chnge Theorem, s(b) = s() + b v(t) dt. (.7).6 The Substitution Rule In the lst section, it ws mentioned tht indefinite integrls cn be obtined from differentition rules. This includes the Chin Rule for differentition, which sttes tht d dx [f(g(x))] = f (g(x))g (x). (.73) Reversing this process, suppose tht F is n ntiderivtive of continuous function f, nd tht g is differentible function. It follows from (.73) tht d dx [F (g(x))] = f(g(x))g (x) (.74) which yields f(g(x))g (x) dx = F (g(x)) + C, (.75) where C is n rbitrry constnt. If we mke chnge of vrible u = g(x) on the right side of eqution (.75), then from the reltion f(x) dx = F (x) + C (.76) we obtin the Substitution Rule f(g(x))g (x) dx = f(u) du. (.77) Note tht the constnt C is dropped becuse both sides of the eqution (.77) re indefinite integrls.

58 58 CHAPTER. INTEGRALS The Substitution Rule is used to compute indefinite integrls by ttempting to recognize whether the integrnd is of the form f(g(x))g (x) for some choice of f nd g. This illustrtes the primry symmetry of differentition nd integrtion: differentition tends to be much more strightforwrd process, involving the ppliction of sequence of rules, where the sequence is esily determined. Integrtion, on the other hnd, requires the determintion of whether the integrnd cn be viewed s the result of some differentition rule, nd such determintion consists of difficult pttern-mtching process. This contrst shows in the fct tht differentition is much esier to utomte by computer progrm thn integrtion. Exmple 33 Evlute the indefinite integrl x sin(x ) dx. (.78) Solution Consider the portion of the integrnd sin(x ). This is composition of two functions, with f(u) = sin u being the outer function nd g(x) = x being the inner function. Since the derivtive of the inner function g(x) = x is g (x) = x, nd since this derivtive ppers s fctor in the integrnd, we cn recognize tht the integrnd cn be written s x sin(x ) = x sin(x ) = g (x)f(g(x)). (.79) Therefore, we cn pply the Substitution Rule, f(g(x))g (x) dx = f(u) du, (.8) where the substitution is u = g(x). To pply this rule, we first note tht the indefinite integrl, or most generl ntiderivtive, of f(x) = sin x is sin x dx = cos x + C. (.8) Using this result, we obtin x sin(x ) dx = x sin(x ) dx = x sin(x ) dx

59 .6. THE SUBSTITUTION RULE 59 = = = g (x) sin(g(x)) dx g (x)f(g(x)) dx f(u) du = [ cos u + C] = [ cos(g(x)) + C] = [ cos(x ) + C] = cos(x ) + C, (.8) since / multiplied by n rbitrry constnt is still n rbitrry constnt. Note tht the Substitution Rule only pplies on integrnds tht involve composition of functions tht hs the form f(g(x)), where g (x) ppers s fctor in the integrnd. Therefore, when evluting n integrl, it is wise to begin by exmining the integrnd to determine whether it hs such form, by trying to recognize the outer function f(x) nd the inner function g(x). If such composition cn be recognized, then it is necessry to compute g (x) nd determine whether it lso ppers in the integrnd. Exmple 34 Compute tn x dx. (.83) Solution Writing tn x = sin x/ cos x nd letting u = cos x, we hve du = sin x dx. It follows tht sin x tn x dx = cos x dx du = u = ln u + C = ln cos x + C = ln cos x + C = ln cos x + C = ln sec x + C. (.84)

60 6 CHAPTER. INTEGRALS As will be seen in lter sections, similr integrtion rules cn be obtined for cot x, sec x, nd csc x. Exmple 35 Evlute e 3x dx. (.85) Solution Using the substitution u = 3x, we hve du = 3 dx, or dx = du/3, which yields e 3x dx = e u du = 3 3 eu + C = 3 e3x + C. (.86) Exmple 36 Evlute e x cos(e x ) dx. (.87) Solution Using the substitution u = e x, we hve du = e x dx which yields e x cos(e x ) dx = cos u du = sin u + C = sin(e x ) + C. (.88) Exmple 37 Evlute e x dx. (.89) + ex Solution Using the substitution u = +e x, we hve du = e x dx which yields e x + e x dx = u du = ln u + C = ln + ex + C. (.9) Exmple 38 Evlute (sin x)e cos x dx. (.9) Solution Using the substitution u = cos x, we hve du = sin x dx which yields (sin x)e cos x dx = e u du = e u + C = e cos x + C. (.9)

61 .6. THE SUBSTITUTION RULE 6 Exmple 39 Evlute x dx x 4. (.93) Solution Using the substitution u = x, we hve du = x dx nd therefore x dx = du = x 4 u sin u + C = sin x + C. (.94) Exmple 4 Evlute the indefinite integrl x 7 ( x 4 ) /3 dx. (.95) Solution This integrnd includes fctor ( x 4 ) /3, which cn be viewed s composition of two functions. Tht is, ( x 4 ) /3 = f(g(x)) where f(u) = u /3 nd g(x) = x 4. We hve g (x) = 4x 3, which ppers s fctor in the integrnd; tht is, we cn write x 7 ( x 4 ) /3 = x 4 (x 3 )( x 4 ) /3 = 4 x4 (4x 3 )( x 4 ) /3. (.96) Becuse of the dditionl fctor of x 4, however, it cn be difficult to pply the Substitution Rule directly. We therefore use n explicit substitution u = x 4, which yields du = 4x 3, nd obtin x 7 ( x 4 ) /3 dx = 4 x4 (4x 3 )( x 4 ) /3 dx = ( u)u /3 du, 4 (.97) since u = x 4 implies x 4 = u. We cn now evlute this integrl using the Power Rule. We hve ( u)u /3 du = u /3 u 4/3 du 4 4 [ ] = u 4/3 4 4/3 u7/3 + C 7/3 = 3 6 u4/ u7/3 + C. (.98) However, we need to compute the indefinite integrl of x 7 ( x 4 ) /3. In order to ccomplish this, we note tht by the substitution u = g(x), we hve trnsformed the originl problem of integrting f(g(x))g (x) dx = F (g(x)) + C, (.99)

62 6 CHAPTER. INTEGRALS where F is ny ntiderivtive of f, to the simpler problem f(u) du = F (u) + C. (.) In other words, we hve computed the function F (u) = 3 6 u4/ u7/3. (.) Therefore, the solution to our originl problem is x 7 ( x 4 ) /3 dx = F (g(x)) + C = 3 6 [g(x)]4/ [g(x)]7/3 + C = 3 6 ( x4 ) 4/ ( x4 ) 7/3 + C (.) In summry, we cn pply the Substitution Rule by choosing n pproprite substitution u = g(x) (where the integrnd is composition f(g(x)), with g being the inner function) to obtin new integrl with respect to u, where the integrnd is simply f(u). After evluting tht integrl, we use the substitution u = g(x) to obtin our finl nswer s function of x..6. Definite Integrls We now consider the evlution of definite integrls using substitution. Suppose tht g is continuously differentible on [, b] nd tht F is n ntiderivtive of function f tht is continuous on the imge of [, b] under g. From the second prt of the Fundmentl Theorem of Clculus nd the reltion (.74), it follows tht b f(g(x))g (x) dx = F (g(b)) F (g()) = = g(b) g() g(b) g() F (u) du f(u) du. (.3) We see tht the intervl x b is mpped by the substitution u = g(x) to the intervl g() u g(b).

63 .6. THE SUBSTITUTION RULE 63 Exmple 4 Evlute the definite integrl π x sin(x ) dx. (.4) Solution We will discuss two pproches to solving this problem. In previous exmple, we noted tht the integrnd cn be written in the form f(g(x))g (x), where f(u) = sin u nd g(x) = x. Becuse n ntiderivtive of f(u) = sin u is F (u) = cos u, it follows tht x sin(x ) dx = cos(x ) + C, (.5) in view of the Substitution Rule f(g(x))g (x) dx = F (g(x)) + C. (.6) We cn then use the Fundmentl Theorem of Clculus to obtin π x sin(x ) dx = F (g(π)) F (g()) = cos((π) ) ( cos( )) = cos(4π ) ( cos( )) (.7) ( ) (.7) Alterntively, we cn use the substitution u = x, since x is the inner function of the composition ppering in the integrnd. This llows us to rewrite the integrl s π x sin(x ) dx = 4π sin u du, (.8) since u = x implies du = x dx. The new limits u = nd u = π re obtined by pplying the substitution u = x to the originl limits x = nd x = π. We cn now evlute simpler definite integrl nd obtin π x sin(x ) dx = 4π sin u du = cos u 4π

64 64 CHAPTER. INTEGRALS = cos(4π ) ( cos()) (.7) ( ) (.9) Exmple 4 Evlute the integrl dx. (.) + 4x Solution Using the substitution u = x, we hve du = dx nd therefore + 4x dx = 4 du (.) + u since u = when x = nd u = () = 4 when x =. We then hve 4 + u du = tn u 4 = (tn 4 tn ) = tn Symmetry (.) The Substitution Rule cn be used to esily evlute certin integrls by wy of the simple substitution u = x. Specificlly, suppose tht f is continuous on the intervl [, ]. If f is n even function (i.e., f( x) = f(x)), then the re under f from to is exctly the sme s the re under f from to, nd it follows tht f(x) dx = f(x) dx. (.3) On the other hnd, if f is n odd function (i.e., f( x) = f(x)), then the net re under f from to is equl in mgnitude to the net re under f from to, but of opposite sign. Therefore f(x) dx =. (.4) These rules re prticulrly useful for trigonometric functions, s sin x is n odd function nd cos x is n even function.

65 .6. THE SUBSTITUTION RULE 65 Exmple 43 Since sin x is n odd function, we cn immeditely conclude, without using ny nti-differentition rules, tht π π sin x dx =. (.5) Exmple 44 The function f(x) = x is even, since f( x) = ( x) = x = f(x). It follows tht 4 4 x dx = 4 = x3 3 x dx 4 = 43 3 = 64 3 = 8 3. (.6)

66 66 CHAPTER. INTEGRALS

67 Chpter Techniques of Integrtion. Integrtion by Prts Recll the Product Rule for differentition, d dx [f(x)g(x)] = f(x)g (x) + g(x)f (x). (.) Like other differentition rules, this rule cn be reversed to obtin n integrtion rule. Specificlly, the product f(x)g(x) is n ntiderivtive of f(x)g (x) + g(x)f (x); tht is, f(x)g (x) + g(x)f (x) dx = f(x)g(x) + C. (.) Rerrnging lgebriclly, we obtin the eqution f(x)g (x) dx = f(x)g(x) g(x)f (x) dx, (.3) where the rbitrry constnt C cn be dropped becuse n indefinite integrl ppers on both sides of the eqution. If we let u = f(x) nd v = g(x), then we hve du = f (x) dx nd dv = g (x) dx. It follows tht eqution (.3) cn be rewritten s u dv = uv v du. (.4) This rule is known s the rule of integrtion by prts. To pply this rule properly to n indefinite integrl of the form h(x) dx, (.5) 67

68 68 CHAPTER. TECHNIQUES OF INTEGRATION it is necessry to identify the functions tht will ply the role of u nd v in the bove rule. In other words, we must be ble to write h(x) dx in the form u dv, where u nd v re chosen so tht v du cn be evluted more esily thn the originl integrl. Typiclly, one chooses u = u(x) to be some portion of h(x), nd then sets dv = h(x)/u(x) dx. Then, v nd du cn esily be determined. Exmple 45 Evlute ln x dx. (.6) Solution We use integrtion by prts with u = ln x, du = dx, v = x, dv = dx (.7) x to obtin ln x dx = x ln x x x dx = x ln x dx = x ln x x + C. (.8) Exmple 46 Evlute tn x dx. (.9) Solution We use integrtion by prts with u = tn x, du = dx, v = x, dv = dx (.) + x s well s substitution u = x + to obtin tn x dx = x tn x = x tn x x + x dx u du = x tn x ln u + C = x tn x ln x + + C. (.)

69 .. INTEGRATION BY PARTS 69 Exmple 47 Evlute e x sin x dx. (.) Solution We use integrtion by prts with u = e x, du = e x dx, v = cos x, dv = sin x dx (.3) to obtin e x sin x dx = e x cos x dx + e x cos x dx. (.4) We then use integrtion by prts gin with u = e x, du = e x dx, v = sin x, dv = cos x dx (.5) to obtin e x sin x dx = e x cos x dx + e x sin x e x sin x dx. (.6) We hve obtined the sme integrl tht we strted with, but with the opposite sign. We cn therefore dd this integrl to both sides to obtin e x sin x dx = e x sin x e x cos x + C (.7) which yields e x sin x dx = ex (sin x cos x) + C. (.8) We now exmine how integrtion by prts cn be used to evlute definite integrl. If we integrte both sides of the Product Rule for differentition from to b, we hve b d [f(x)g(x)] dx = dx b f(x)g (x) + g(x)f (x) dx. (.9) It follows from Prt of the Fundmentl Theorem of Clculus tht or, rerrnging, b b f(x)g(x) b = f(x)g (x) + g(x)f (x) dx, (.) b f(x)g (x) dx = f(x)g(x) b g(x)f (x) dx. (.)

70 7 CHAPTER. TECHNIQUES OF INTEGRATION Therefore, to pply integrtion prts to definite integrl of the form b h(x) dx, (.) we identify the functions u nd v s in the cse of the indefinite integrl. Then, we cn simply compute u(b)v(b) u()v(), nd evlute the integrl of v(x)u (x) from to b. Exmple 48 Compute π/ x cos x dx. (.3) Solution We use integrtion by prts with to obtin u = x, du = dx, v = sin x, dv = cos x dx (.4) π/ x cos x dx = x sin x π/ π/ = x sin x + cos x π/ sin x dx = π sin π + cos π cos = π. (.5) A common usge of integrtion by prts is to evlute integrls of the form x m f(x) dx, (.6) where the function f(x) cn esily be nti-differentited repetedly, s is the cse with vrious exponentil or trigonometric functions. For such n integrnd, one cn choose u = x m, which yields du = mx m dx. By pplying integrtion by prts in this wy m times, the power of x in the integrnd cn be reduced ll the wy to zero, hopefully yielding simpler integrnd. Exmple 49 Evlute x e x dx. (.7)

71 .. INTEGRATION BY PARTS 7 Solution We use integrtion by prts with to obtin u = x, du = x dx, v = e x, dv = e x dx (.8) x e x dx = x e x We then use integrtion by prts gin with to obtin xe x dx. (.9) u = x, du = dx, v = e x, dv = e x dx (.3) x e x dx = x e x xe x dx [ = x e x xe x ] e x dx = x e x [xe x e x + C] = x e x xe x + e x + C = (x x + )e x + C. (.3) Exmple 5 Derive formul for sin n x dx (.3) where n is n integer, nd n. Solution We use integrtion by prts with u = sin n x, du = (n ) sin n x cos x dx, v = cos x, dv = sin x dx (.33) to obtin sin n x dx = sin n x cos x + (n ) sin n x cos x dx. (.34) Writing cos x = sin x yields sin n x dx = sin n x cos x + (n ) = sin n x cos x + (n ) = sin n x cos x + (n ) = sin n x cos x + (n ) sin n x cos x dx sin n x( sin x) dx sin n x sin n x dx sin n x dx (n ) sin n (.35) x dx.

72 7 CHAPTER. TECHNIQUES OF INTEGRATION Moving the lst integrl on the right side to the left side of the eqution yields n sin n x dx = sin n x cos x + (n ) sin n x dx (.36) or sin n x dx = n sinn x cos x + n n sin n x dx. (.37). Trigonometric Integrls So fr we hve obtined severl integrtion rules by reversing vrious differentition rules. Additionl integrtion rules cn be obtined by employing trigonometric identities to convert integrnds involving trigonometric functions into form in which other integrtion rules cn be pplied. For such integrnds, the resulting rules cn then be used directly, insted of continuing to rely on the trigonometric identities used to estblish them. For exmple, suppose we wish to evlute n integrl of the form sin m x cos n x dx. (.38) If n is odd, then we cn use the identity sin x + cos x = to express n of the powers of cosine in terms of sine. Then, we cn use the substitution u = sin x. Becuse du = cos x dx, the resulting integrl with respect to u will be simple to evlute, since the integrnd will consist of severl terms of the form u m+k, for k =,..., (n )/, which cn esily be integrted using the power rule. An integrl of the form (.38) where m is odd cn be hndled similrly, using the sme identity. Exmple 5 Evlute sin 3 x cos x dx. (.39) Solution Since the power of sine is odd, we write ll but one power of sine in terms of cosines: sin 3 x cos x dx = sin x( cos x) cos x dx = sin x cos x dx sin x cos 4 x dx. (.4)

73 .. TRIGONOMETRIC INTEGRALS 73 Using the substitution u = cos x, with du = sin x dx, yields sin x cos x dx sin x cos 4 x dx = u du+ u 4 du = u5 5 u3 3 +C = cos5 x cos3 x +C. 5 3 (.4) If both m nd n re even, then the hlf-ngle formuls sin x = ( cos x), cos x = ( + cos x) (.4) cn be used to reduce the powers of sine nd cosine until every term in the trnsformed integrnd either hs n odd power of sine or cosine, or until simpler integrtion rule cn be used. If term of the form sin x cos x rises, this cn be hndled using the substitution u = cos x or u = sin x, but one my find it esier to use the double-ngle formul sin x cos x = sin x. (.43) Exmple 5 Evlute sin x cos x dx. (.44) Solution Using the double-ngle formul yields sin x = sin x cos x (.45) ( ) sin x sin x cos x dx = dx = 4 sin x dx. (.46) We cn then use hlf-ngle formul sin cos x x = (.47) to obtin sin x cos x dx = cos 4x dx 4 = cos 4x dx 8 = ( ) sin 4x x 8 4 = x sin 4x + C. (.48) 8 3

74 74 CHAPTER. TECHNIQUES OF INTEGRATION Integrls of the form tn m x sec n x dx (.49) cn be hndled using similr strtegy. In this cse, the useful identity is sec x = + tn x. If the power of secnt is even, then ll but two of the powers of secnt cn be expressed in terms of tn x using this identity, nd then the substitution u = tn x cn be used to obtin simpler integrnd. If, on the other hnd, the power of tngent is odd nd n >, then ll but one of the powers of tngent cn be expressed in terms of sec x, nd then the substitution u = sec x cn be used, since we would hve du = sec x tn x. Exmple 53 Evlute sec 4 x tn 3 x dx. (.5) Solution We write ll but two powers of secnt s powers of tngents, using the identity sec x = tn x +. (.5) This yields sec 4 x tn 3 x dx = sec x(tn x+) tn 3 x dx = Using the substitution u = tn x, with du = sec x dx, yields sec 4 x tn 3 x dx = = sec x tn 5 x dx + u 5 du + u 3 du = u6 6 + u4 4 + C = tn6 x 6 + tn4 x 4 sec x tn 5 x dx+ sec x tn 3 x dx (.5) + C. (.53) An lterntive pproch is to write ll but one power of tngent s powers of secnts, using the identity tn x = sec x. (.54) sec x tn 3 x dx.

75 .. TRIGONOMETRIC INTEGRALS 75 This yields sec 4 x tn 3 x dx = = sec 4 x(sec x ) tn x dx sec 6 x tn x dx sec 4 x tn x dx. (.55) Using the substitution u = sec x, with du = sec x tn x dx, yields sec 4 x tn 3 x dx = sec 6 x tn x dx sec 4 x tn x dx = sec 5 x sec x tn x dx sec 3 x sec x tn x dx = u 5 du u 3 du = u6 6 u4 4 + C = sec6 x 6 sec4 x 4 + C. (.56) Other cses re not s strightforwrd. It my be necessry to use other trigonometric identities, integrtion by prts, or the rules for integrting tn x nd sec x: tn x dx = ln sec x + C, (.57) sec x dx = ln sec x + tn x + C. (.58) Exmple 54 Evlute sec 3 x tn x dx. (.59) Solution Since the power of tngent is even nd the power of secnt is odd, it is not possible to rewrite powers of one in terms of powers of the other nd use simple substitution s in the previous exmple. Insted, we cn proceed using integrtion by prts with which implies u = sec x tn x, dv = sec x, (.6) du = ( sec 3 x tn x + sec x tn 3 x) dx, v = tn x. (.6)

76 76 CHAPTER. TECHNIQUES OF INTEGRATION Before pplying integrtion by prts, we use the identity tn x = sec x to write du = ( sec 3 x tn x + sec x(sec x ) tn x) dx = ( sec 3 x tn x + sec x sec x tn x sec x tn x) dx = (3 sec 3 x tn x sec x tn x) dx. (.6) Integrting by prts, we obtin sec 3 x tn x dx = sec x tn 3 x 3 Rerrnging lgebriclly yields 4 sec 3 x tn x dx = sec x tn 3 x + sec 3 x tn x dx + sec x tn x dx. (.63) sec x tn x dx (.64) or sec 3 x tn x dx = 4 sec x tn3 x + 4 sec x tn x dx. (.65) We now focus on the remining integrl sec x tn x dx. (.66) Rewriting the powers of tngent s powers of secnt yields sec x tn x dx = sec x(sec x ) dx = sec 3 x sec x dx. (.67) To integrte sec x, we proceed s follows, using the substitution u = sec x + tn x with du = (sec x tn x + sec x) dx: sec x + tn x sec x dx = sec x sec x + tn x dx sec x + sec x tn x = dx sec x + tn x du = u = ln u + C = ln sec x + tn x + C. (.68) To integrte sec 3 x, we use integrtion by prts with u = sec x, dv = sec x dx (.69)

77 .. TRIGONOMETRIC INTEGRALS 77 which yields It follows tht du = sec x tn x dx, v = tn x. (.7) sec 3 x dx = sec x tn x sec x tn x dx. (.7) Using the identity tn x = sec x yields sec 3 x dx = sec x tn x sec x(sec x ) dx = sec x tn x Rerrnging lgebriclly yields sec 3 x dx = sec x tn x + sec 3 x dx+ (.7) sec x dx (.73) sec x dx. or sec 3 x dx = sec x tn x + sec x dx. (.74) Applying our erlier result of integrting sec x, we obtin sec 3 x dx = sec x tn x + ln sec x + tn x. (.75) Finlly, we put ll of the pieces together nd obtin sec 3 x tn x dx = 4 sec x tn3 x + sec x tn x dx 4 = 4 sec x tn3 x + [ ] sec 3 x dx sec x dx 4 = 4 sec x tn3 x + [ 4 sec x tn x + ] ln sec x + tn x ln sec x + tn x = 4 sec x tn3 x + [ 4 sec x tn x ] ln sec x + tn x = 4 sec x tn3 x + 8 sec x tn x ln sec x + tn x. (.76) 8 Trigonometric identities cn lso be helpful for integrls of the forms sin mx cos nx dx, (.77)

78 78 CHAPTER. TECHNIQUES OF INTEGRATION or The identities sin mx sin nx dx, (.78) cos mx cos nx dx. (.79) sin A cos B = [sin(a B) + sin(a + B)], (.8) sin A sin B = [cos(a B) cos(a + B)], (.8) cos A cos B = [cos(a B) + cos(a + B)], (.8) which cn be derived from the identities for cosines nd sines of sums nd differences, cn be used to convert these integrnds into much simpler ones involving only sine or cosine, in which cse the rules for integrting these functions cn be employed. Exmple 55 Prove tht for ny nonnegtive integers m nd n, π π Solution Using the identity we obtin π π sin mx cos nx dx =. (.83) sin A cos B = [sin(a B) + sin(a + B)], (.84) sin mx cos nx dx = π π sin(m n)x + sin(m + n)x dx. (.85) Since sin x is n odd function, it follows tht sin kx is odd for ny integer k. Since the integrl of ny odd function over n intervl of the form [, ] is equl to, we must hve π π sin(m n)x dx =, nd therefore we must hve π π π π sin(m + n)x dx =, (.86) sin mx cos nx dx =. (.87)

79 .. TRIGONOMETRIC INTEGRALS 79 By now it should be cler tht there re gret mny integrls tht cnnot be evluted using the Fundmentl Theorem of Clculus without some ingenuity. Although we hve gretly expnded the rnge of integrnds for which integrtion rules cn be developed, there re still mny integrnds for which no such rules exist, nd therefore the definition of the definite integrl must be used insted. Exmple 56 Evlute π π Solution We use the product sum identity sin 3x cos 7x dx. (.88) sin A cos B = [sin(a + B) + sin(a B)] (.89) with A = 3x nd B = 7x. We then hve π π sin 3x cos 7x dx = π π π [sin(3x + 7x) + sin(3x 7x)] dx = sin x + sin( 4x) dx π = ( cos x + cos( 4x) ) π 4 π ( cos x = + cos( 4x) ) π 8 π ( cos π = + cos( 4π) ) ( cos( π) 8 ( = + ) ( 8 + ) 8 = cos(4π) ) 8 =. (.9) The ntiderivtive in the third step cn be obtined using seprte substitutions u = x nd u = 4x in ech term, or by noting tht in generl, cos x sin x dx = + C. (.9)

80 8 CHAPTER. TECHNIQUES OF INTEGRATION.3 Trigonometric Substitution Recll the substitution rule, f(g(x))g (x) dx = f(u) du. (.9) In pplying this rule, one mkes the substitution u = g(x), which yields du = g (x) dx. Using this reltion, the originl integrnd f(g(x))g (x) is then rewritten s f(u). In rewriting the integrnd in this fshion, one cn write the substitution in such wy tht x is expressed s function of u, insted of the other wy round: x = g (u). Then, we hve dx = du/g (g (u)). This is not lwys the esiest pproch to pplying the substitution rule, but cn be very effective when the substitution to be employed is more complicted, or if the function g is simpler to work with thn g. This pproch to substitution is known s inverse substitution, but it should be recognized tht it is nothing more thn n lterntive pproch to u-substitution. This is precisely the cse when evluting n integrl of the form x dx. (.93) Suppose tht we use the substitution x = sin θ. Clerly, it is preferble to write the substitution in this mnner thn expressing u s function of x. From the reltion dx = cos θ dθ, we obtin the new integrl sin θ( cos θ) dθ = cos θ dθ, (.94) which cn esily be evluted using hlf-ngle formul. For more complicted integrnds involving the expression x, the sme substitution cn sometimes be used. Exmple 57 Evlute x 4 x dx. (.95) Solution One pproch is to use the substitution u = 4 x, which implies du = x dx. The new limits cn be obtined by substituting the limits x = nd x = into the reltion u = 4 x, which yields u = 4 nd u =.

81 .3. TRIGONOMETRIC SUBSTITUTION 8 We then hve x 4 x dx = = = = 4 4 u 3/ 3/ u / du u / du u3/ = 3 43/ = 3 (4/ ) 3 = 3 3 = 8 3. (.96) Alterntively, we cn use n inverse substitution x = sin θ, which implies dx = cos θ dθ. We then hve 4 x = 4 ( sin θ) = 4 4 sin θ = sin θ = cos θ = cos θ. (.97) It follows tht x 4 x dx = = 8 π/ = 8 = 8 π/ = 8 u3 3 ( sin θ)( cos θ)( cos θ) dθ sin θ cos θ dθ u du, u = cos θ, du = sin θ dθ u du = 8 3. (.98)

82 8 CHAPTER. TECHNIQUES OF INTEGRATION Similr substitutions cn be used for integrnds contining expressions of the form + x or x. In the first cse, the substitution x = tn θ cn be used in conjunction with the identity + tn θ = sec θ. In the second cse, the substitution x = sec θ cn be used in conjunction with the identity sec θ = tn θ. Exmple 58 Evlute dx x + 4. (.99) Solution We use the inverse substitution x = tn θ, which implies dx = sec θ dθ. Furthermore, x + 4 = ( tn θ) + 4 = 4 tn θ + 4 = tn θ + = sec θ = sec θ. (.) Therefore dx sec θ dθ = x + 4 sec θ = sec θ dθ = ln sec θ + tn θ + C x = ln x + C = ln x x + C since C represents n rbitrry constnt. = ln x x + ln + C = ln x x ln + C = ln x x + C (.) Often, these substitutions must be used in conjunction with other integrtion techniques. We illustrte this with simple exmple. Suppose we wish to evlute + x dx. (.) From the discussion bove, we should use the substitution x = tn θ. This yields + x dx = sec 3 θ dθ. (.3)

83 Exmple 59 Evlute x x dx. (.7).3. TRIGONOMETRIC SUBSTITUTION 83 However, in order to evlute the new integrl, we must resort to integrtion by prts, which yields sec 3 θ dθ = sec θ tn θ sec θ tn θ dθ. (.4) Using the identity tn θ = sec θ, we obtin sec 3 θ dθ = sec θ tn θ sec 3 θ dθ + sec θ dθ. (.5) Rerrnging lgebriclly nd using the integrtion rule for sec θ, we finlly obtin + x dx = sec 3 θ dθ = [sec θ tn θ + ln sec θ + tn θ ] + C = x x + + ( ) ln x + + x + C.(.6) The key lesson to be lerned from this exmple is tht one should be prepred to use ny integrtion technique t his or her disposl, since, for mny integrnds, some combintion of such techniques is necessry to evlute the given integrl. Unlike the more strightforwrd process of differentition, nti-differentition often requires substntil ingenuity nd openmindedness. Solution Before we cn use trigonometric substitution, we need to write the expression x x s sum or difference of two squres. Completing the squre, we hve x + x = ( x + ) ( 4 = x + ) 5 4 (.8) nd therefore x x dx = ( 5 4 x + dx. (.9) )

84 84 CHAPTER. TECHNIQUES OF INTEGRATION Since we now hve difference of two squres under the rdicl sign, nd since the first term is constnt, we try to write the rdicl in the form sin θ for some choice of nd some substitution involving sin θ. Mtching sin θ to 5/4 (x + /), we find tht we must hve = 5/4 = 5/, nd lso x + = 5 sin θ. (.) Using this inverse substitution, we hve 5 dx = cos θ dθ (.) nd ( 5 4 x + = ) It follows tht 5 4 x x dx = Using the hlf-ngle formul ( ) 5 5 sin θ = 4 54 sin θ = ( ) 5 cos θ dθ = sin θ = cos θ. (.) cos θ dθ. (.3) yields cos θ = + cos θ + cos θ cos θ dθ = dθ = ( θ + (.4) ) sin θ + C (.5) nd therefore x x dx = 5 8 θ + 5 sin θ + C. (.6) 6 From our originl substitution, we obtin ( ( θ = sin 5 x + )), (.7)

85 .3. TRIGONOMETRIC SUBSTITUTION 85 nd using the double-ngle formul long with the reltions nd yields sin θ = sin θ cos θ (.8) x + = 5 sin θ (.9) ( 5 4 x + 5 = cos θ (.) ) sin θ = sin θ cos θ ( ( = 5 x + )) = 8 5 ( x + ) ( x + ) x x. (.) Now we cn ssemble our finl nswer, x x dx = 5 ( ( 8 sin 5 x + )) + ( x + x x ) +C. (.) Exmple 6 Evlute x x dx (.3) where is positive constnt. Solution We use the inverse substitution which implies x = sec θ (.4) dx = sec θ tn θ dθ (.5) nd x = ( sec θ) = sec θ = sec θ = tn θ = tn θ. (.6)

86 86 CHAPTER. TECHNIQUES OF INTEGRATION Therefore, x x dx = ( sec θ) ( tn θ)( sec θ tn θ) dθ = 4 sec 3 θ tn θ dθ. (.7) This integrl ws worked out in the previous section. Using tht result, we obtin x x dx = 4 [ 4 sec θ tn3 θ + 8 sec θ tn θ 8 ln sec θ + tn θ ]+C. Using the reltions (.8) x = sec θ (.9) nd x = tn θ, (.3) we obtin x x dx = 4 x(x ) 3/ + 8 x x 4 8 ln x x + + C x + x + C = 4 x(x ) 3/ + 8 x x 4 8 ln = 4 x(x ) 3/ + 8 x x 4 8 ln x + x 4 8 ln + C = 4 x(x ) 3/ + 8 x x 4 8 ln x + x + C (.3) since is constnt nd C is n rbitrry constnt..4 Integrtion of Rtionl Functions In this section we consider the integrtion of rtionl functions. A function f(x) is sid to be rtionl if it cn be written s f(x) = P (x) Q(x), (.3) where P (x) nd Q(x) re polynomils. If the degree of P is less thn tht of Q, then f is sid to be proper rtionl function; otherwise it is improper. We shll only consider proper rtionl functions, becuse if f is improper,

87 .4. INTEGRATION OF RATIONAL FUNCTIONS 87 then we cn divide P by Q using long division of polynomils to obtin the representtion f(x) = S(x) + R(x) Q(x), (.33) where S(x) nd R(x) re both polynomils, nd the rtionl function g(x) = R(x)/Q(x) is proper. Therefore we cn integrte f(x) by integrting ech term of S(x) using the power rule nd then hndling g(x) seprtely. Exmple 6 Evlute x 3 + 3x + 7x + 4 dx. (.34) x + Solution Dividing the numertor nd denomintor, we hve x 3 +3x +7x+4 = x (x+)+x +7x+4 = (x +x)(x+)+6x+4 = (x +x+3)(x+)+ (.35) nd therefore x 3 + 3x + 7x + 4 x + = x + x x +, (.36) which yields, by the substitution u = x +, x 3 + 3x + 7x + 4 dx = x + x x + x + dx = x3 3 + x + 3x + u du.4. Simple Proper Rtionl Functions = x3 3 + x + 3x + ln x + + C. (.37) We know how to integrte some proper rtionl functions using techniques we hve lredy lerned. To review: dx = ln x + + C, (.38) x + x + dx = ( x ) tn + C. (.39)

88 88 CHAPTER. TECHNIQUES OF INTEGRATION Some slightly more complex proper rtionl functions cn be hndled using these rules in conjunction with other techniques. For exmple, using the substitution u = x, we hve x x + dx = du u + = ln u + + C = ln x + + C. (.4) To integrte more generl rtionl function, we cn rewrite it s sum of simpler rtionl functions tht cn be integrted using simpler rules nd/or techniques such s those illustrted bove. Such sum cn be obtined using the method of prtil frctions..4. The Method of Prtil Frctions The method of prtil frctions involves the decomposition of generl proper rtionl function f(x) = R(x) (.4) Q(x) into sum of simpler rtionl functions tht cn be integrted using known integrtion techniques such s the rules bove. The method is bsed on the fct tht Q(x) cn be fctored into product of liner fctors of the form (x + b) or n irreducible qudrtic fctor of the form (x + bx + c), where b 4c <. This is simply consequence of the Fundmentl Theorem of Algebr, which sttes tht ny polynomil of degree n hs exctly n roots, which cn be rel numbers or complex numbers. The first step in the method of prtil frctions is to fctor Q(x) into these liner nd qudrtic fctors. The second step is to write f(x) s sum of proper rtionl functions tht hve these fctors s denomintors. This sum is clled the prtil frction decomposition of f(x). Finlly, ech of term in this decomposition, which hs much simpler form thn f(x), cn be integrted individully. Depending on the fctoriztion of Q(x), there re four distinct scenrios:. Q(x) is product of liner fctors of the form (x+b), where ll of the fctors hve distinct roots (in other words, the rtio b/ is different in ll fctors). Then, since we cn write Q(x) s Q(x) = ( x + b ) ( k x + b k ) = k ( j x + b j ), (.4) j=

89 .4. INTEGRATION OF RATIONAL FUNCTIONS 89 we cn then write f(x) s f(x) = R(x) Q(x) = A A k + +. (.43) x + b k x + b k The constnts A,..., A k cn be determined by solving the eqution [ A R(x) = + + A ] k Q(x). (.44) x + b x + b k This cn be ccomplished by cncelling the denomintors with fctors of Q(x) where possible, resulting in polynomil on the right side. Then, by mtching powers of x on both sides of the eqution, we obtin system of k liner equtions for the constnts A,..., A k. Then, we cn integrte ech term individully s follows: A j dx = A j ln j x + b j j x j + b j + C, j =,..., k. (.45) j Exmple 6 Evlute A dx. (.46) x + Solution Using the substitution u = x +, we hve du = dx nd A x + dx = A du = A ln u + C = A ln x + + C. (.47) u Exmple 63 Evlute x 4 + x dx. (.48) Solution Dividing the numertor by the denomintor yields x 4 + = x (x ) + x + = (x + )(x ) + 3 (.49) nd therefore x 4 + x = x x. (.5)

90 9 CHAPTER. TECHNIQUES OF INTEGRATION To evlute the integrl, we need to compute the prtil frction decomposition of x = (x + )(x ) = A x + + B x. (.5) Multiplying through this eqution by x yields [ A = x + + B ] A(x + )(x ) B(x + )(x ) (x+)(x ) = + x x + x (.5) Collecting terms with like powers of x yields = Ax A + Bx + B = (A + B)x + B A. (.53) Mtching like powers of x on both sides of the eqution yields the system of equtions A + B =, B A =. (.54) From the first eqution, A = B. Substituting this reltion into the second eqution yields A A = A =, nd therefore we must hve A = / nd B = /. It follows tht x 4 + x dx = x x dx = x3 3 + x + 3 / x + + / x dx = x3 3 + x 3 3 ln x ln x + C.(.55) = A(x )+B(x+) Exmple 64 Evlute x 7 dx. (.56) (x )(x + 3) Solution We need to compute the prtil frction decomposition x 7 (x )(x + 3) = A x + B x + 3, (.57)

91 .4. INTEGRATION OF RATIONAL FUNCTIONS 9 where the constnts A nd B re to be determined. through this eqution by (x )(x + 3) yields x 7 = Multiplying [ A x + B ] (x )(x+3) = A(x+3)+B(x ). (.58) x + 3 Collecting terms with like powers of x yields x 7 = Ax + 3A + Bx B = (A + B)x + 3A B. (.59) Mtching like powers of x on both sides of the eqution yields the system of equtions A + B = 3A B = 7 (.6) for the constnts A nd B. From the first eqution, A = B. Substituting this reltion into the second eqution yields 3( B) B = 3 5B = 7, which yields B = nd therefore we must hve A =. It follows tht x 7 (x )(x + 3) dx = x + x + 3 dx = ln x + ln x C.(.6) Exmple 65 Evlute x + 3 (x )(x dx. (.6) ) Solution The denomintor fctors s (x )(x ) = (x )(x + )(x ) (.63) nd therefore we must compute prtil frction decomposition of the form x + 3 (x )(x ) = A x + B x + + C x, (.64)

92 9 CHAPTER. TECHNIQUES OF INTEGRATION where the constnts A, B, nd C re to be determined. Multiplying through this eqution by (x )(x ) yields x + 3 = A(x )(x ) + B(x )(x ) + C(x )(x ) x x + x = A(x ) + B(x )(x ) + C(x )(x + ). (.65) To determine A, B nd C we use the method of creting zeros. Substituting x = into the bove eqution yields 5 = A(() ) + B( )( ) + C( )( + ) = 3A (.66) nd therefore A = 5/3. Similrly, substituting x = yields 4 = A(() ) + B( )( ) + C( )( + ) = C( )() = C (.67) nd therefore C =. Finlly, substituting x = yields = A(( ) )+B( )( )+C( )( +) = B( 3)( ) = 6B (.68) nd therefore B = /3. It follows tht x + 3 (x )(x ) dx = 5/3 x + /3 x + x dx = 5 3 ln x + 3 ln x + ln x (.69) + C.. Q(x) is product of liner fctors, where some of the fctors repet. In this cse, we cn write Q(x) s Q(x) = ( x + b ) m ( k x + b k ) m k = k ( j x + b j ) m j, (.7) where k is the number of distinct liner fctors, nd, for ech j =,..., k, m j is clled the multiplicity of the fctor j x + b j. Since t lest one fctor repets in this scenrio, this mens m j > for t lest one j. For ech distinct fctor j x + b j of Q(x), the prtil frction decomposition of f(x) includes the terms A j A j + j x + b j ( j x + b j ) + + A mj j ( j x + b j ) m. (.7) j j=

93 .4. INTEGRATION OF RATIONAL FUNCTIONS 93 Then, we cn solve for the constnts {A ij } s in the previous scenrio, nd integrte ech term seprtely. The terms with denomintors of the form (x + b) j, where j >, cn be integrted using the power rule. It follows tht f(x) cn be written s f(x) = R(x) Q(x) = m k j j= i= A ij ( j x + b j ) i, (.7) which yields f(x) dx = = = = m k j A ij j= i= m k j j= i= k j= k j= [ A j j [ A j j A ij i j dx ( j x + b j ) i dx (x + b j / j ) i dx (x + b j / j ) + ln x j + b mj j j i= m j i= A ij i j A ij j ( j x + b j ) i ] dx (x + b j / j ) i ] + (.73) C. Exmple 66 Evlute 3 dx. (.74) (x + 4) 8 Solution Using the substitution u = x + 4, we hve du = dx, so by the Power Rule, 3 (x + 4) 8 dx = 3 u 8 du = 3 u 8 du = 3 u 7 7 +C = 3 7(x + 4) 7 +C. Exmple 67 Evlute (.75) 3x x 3 (x + )x dx. (.76)

94 94 CHAPTER. TECHNIQUES OF INTEGRATION Solution We must compute prtil frction decomposition of the form 3x x 3 (x + )x = A x + + B x + C x (.77) due to the repeted liner fctor of x in the denomintor. Multiplying through this eqution by (x + )x yields [ A 3x x 3 = x + + B x + C ] x (x + )x A(x + )x B(x + )x = + + x + x = Ax + B(x + )x + C(x + ) = Ax + Bx + Bx + Cx + C C(x + )x x = (A + B)x + (B + C)x + C. (.78) Mtching like powers of x on both sides of the eqution, we obtin the system of equtions A + B = 3, B + C =, C = 3. (.79) Since C = 3, it follows tht B = ( 3) =. Therefore we must hve A = 3 B = 3 =. We then hve 3x x 3 (x + )x dx = x + + x 3 x dx = ln x + + ln x 3 x = ln x + + ln x + 3 x. (.8) 3. Q(x) is product of liner nd irreducible qudrtic fctors, where none of the qudrtic fctors repet. Then, for ech qudrtic fctor (x + bx + c) of Q(x), where b 4c <, the prtil frction decomposition of f(x) includes term of the form A(x + b) + B x + bx + c, (.8)

95 .4. INTEGRATION OF RATIONAL FUNCTIONS 95 where the constnts A nd B re to be determined s in the previous scenrios. This term cn then be integrted independently of the other terms in the decomposition by completing the squre, which yields A(x + b) + B x + bx + c x + b dx = A x + bx + c dx + B ( ) x + b dx + 4c b 4 du = A u + B ( ) x + b dx, u = x + bx + c + 4c b 4 = A ln x + bx + c + B 4 4c b tn x + b + C 4c b = A ln x + bx + c + 4 ( ) B x + b 4c b tn (.8) + C. 4c b Exmple 68 Evlute B x dx. (.83) + z Solution Using the substitution u = x/z, we hve du = dx/z nd B x + z = B x + z dx = zb (zu) + z dx = zb z u + dx = B z tn u + C = B z tn ( x z ) + C. (.84) The sme nswer cn be obtined immeditely by using eqution (.8), with A =, b =, = nd c = z. 4. Q(x) is product of liner nd irreducible qudrtic fctors which cn repet. In this cse, for ech fctor of Q(x) tht hs the form (x + bx + c) j, the prtil frction decomposition of f(x) cn be

96 96 CHAPTER. TECHNIQUES OF INTEGRATION written in such wy tht it includes terms of the form A (x + b) + B x + bx + c + j i= C i (x + bx + c) i +(x + b)[d i (i )C i x] (x + bx + c) i. (.85) where the constnts re to be determined s in the previous scenrios. The first term cn be hndled s described bove. The other terms cn be integrted s follows: D i (x + b) du (x + bx + c) i dx = D D i i u i = (i )(x + C, + bx + c) i (.86) where u = x + bx + c, nd [ ] (i )(x + b)x C i x C i (x + bx + c) i (x + bx + c) i dx = (x + bx + c) (.87) This rule is obtined by pplying the Quotient Rule for differentition in reverse. It should be noted tht the prtil frction decompositions used in the cse of qudrtic fctors, whether repeted or not, differ substntilly from the decompositions presented in the textbook. The reson for this discrepncy is tht the decompositions presented here, while somewht more complicted, mke the ctul integrtion process fr simpler, since terms corresponding to ech constnt in the numertor cn be integrted independently without hving to employ lgebric tricks tht re difficult to conceive nd pply correctly..5 Approximte Integrtion The Fundmentl Theorem of Clculus provides simple method for evluting mny definite integrls of the form b f(x) dx. (.88) However, the theorem cn only be used under the following conditions: The integrnd f(x) must be known continuous function. This is not lwys the cse, since knowledge of f(x) my be limited to smll set of vlues t specific smpling points in the intervl [, b], where such vlues re typiclly obtined from mesurements. i +C.

97 .5. APPROXIMATE INTEGRATION 97 Even if formul for f(x) is known, we must still be ble to ntidifferentite f(x). If no ntiderivtive is known, then we cnnot pply the Fundmentl Theorem of Clculus. In rel pplictions, integrls for which the Fundmentl Theorem of Clculus is of no use rise quite frequently. Such integrls cnnot be evluted exctly, nd therefore we must pproximte their vlues insted. This leds to the study of numericl integrtion, which is lso known s numericl qudrture..5. Riemnn Sums Since the definite integrl is defined to be the limit of sequence of Riemnn sums, certinly ny vlid Riemnn sum cn be used to pproximte the vlue of the integrl. we hve lredy used few prticulr types of Riemnn sums to obtin such n pproximtion: We cn divide the intervl [, b] into n subintervls of equl width x = (b )/n, nd pproximte the re under the grph of f by dding the res of n rectngles of width x nd height determined by the vlue of f(x) t the left endpoint of ech subintervl. The resulting Riemnn sum is b f(x) dx n f(x i ) x, x i = + i x. (.89) i= Similrly, we cn determine the height of ech rectngle using the right endpoint of ech subintervl, resulting in the pproximtion b f(x) dx n f(x i ) x. (.9) i= We hve seen tht more ccurte pproximtion cn be obtined by setting the height of ech rectngle equl to the vlue of f(x) t the midpoint of the corresponding subintervl. The resulting pproximtion is known s the Midpoint Rule: b f(x) dx n f(x i / ) x, i= x i / = x i + x i. (.9)

98 98 CHAPTER. TECHNIQUES OF INTEGRATION In ddition, we cn pproximte the integrl by verging the pproximtions obtined by using left endpoints nd right endpoints. The resulting rule, clled the Trpezoidl Rule, effectively pproximtes the re under the grph of f by n trpezoids of width x nd heights f(x i ) nd f(x i ), for i =,..., n. Intuitively, it mkes sense tht n pproximtion using trpezoids is more ccurte thn using rectngles. This is true in tht the Trpezoidl Rule is fr more ccurte thn using left or right endpoints. In fct, not only is it more ccurte, but the pproximtion obtined using the Trpezoidl Rule converges to the exct vlue of the integrl t much fster rte thn the pproximtion obtined using left or right endpoints. Specificlly, the error in the pproximtion by left or right endpoints decreses by fctor of when the number of rectngles is doubled. In other words, the error is proportionl to x, nd therefore we sy tht these pproximtions converge linerly to the exct vlue s n, the number of rectngles, tends to. The error in the Trpezoidl Rule, however, decreses by fctor of 4 when the number of rectngles is doubled. It follows tht the error is proportionl to x, nd therefore we sy tht the pproximtion obtined using the Trpezoidl Rule converges qudrticlly to the exct vlue s n, the number of trpezoids, tends to. Surprisingly, it turns out tht the Midpoint Rule, even though it pproximtes the re using rectngles, is even more ccurte thn the Trpezoidl Rule. Its pproximtions converge qudrticlly to the exct vlue, nd the error with n rectngles is pproximtely hlf of the error of the Trpezoidl Rule using n trpezoids. The reson for this greter ccurcy cn be seen by exmining prticulr subintervl [x i, x i ] of [, b]. Let x M be the midpoint of this subintervl. Then the re of the rectngle of height f(x M ) nd width x is equl to the re of the trpezoid whose upper side is tngent to the grph of f t x M, since such trpezoid, by the eqution of the tngent line, would hve heights f(x M ) f (x M ) x/ nd f(x M ) + f (x M ) x/. The fct tht the upper side of this trpezoid is tngent to the grph of f llows this side to more closely pproximte the grph of f on this subintervl thn the line pssing through f(x i ) nd f(x i ), which is the upper side of the trpezoid used by the Trpezoidl Rule..5. Simpson s Rule nd Other Approximtion Methods The left nd right endpoint pproximtions to the definite integrl pproximte the integrnd f(x) by constnt function on ech subintervl of [, b] to determine the height of ech rectngle. The Midpoint Rule nd Trpezoidl

99 .5. APPROXIMATE INTEGRATION 99 Rule pproximte f(x) by liner function on ech subintervl, which helps explin why they re more ccurte. It is nturl to sk if even more ccurcy cn be obtined by pproximting f(x) by higher-degree polynomil on ech subintervl. The nswer to this question is yes, nd this fct is the bsis of Simpson s Rule. This rule pproximtes f(x) by qudrtic function on ech consecutive pir of subintervls, where the number of subintervls is ssumed to be even. For ech pir of subintervls [x i, x i ] nd [x i, x i+ ], Simpson s Rule implicitly constructs the unique qudrtic function tht psses through the three endpoints x i, x i, nd x i+. This qudrtic function is then integrted exctly on the intervl [x i, x i+ ] using the Fundmentl Theorem of Clculus. This integrl turns out to be equl to x 3 [f(x i ) + 4f(x i ) + f(x i+ )], (.9) where x = (b )/n is the width of ech subintervl. Adding the results from ll pirs of subintervls, we obtin the pproximtion b f(x) dx x 3 [f(x )+4f(x )+f(x )+4f(x 3 )+ +f(x n )+4f(x n )+f(x n )]. (.93) This rule exhibits fourth-order convergence; tht is, doubling the number of subintervls reduces the error by fctor of 6. All of the rules we hve discussed re exmples of interpoltory qudrture rules. These rules hve the property tht they pproximte the integrnd by polynomils on subintervls of [, b], nd then integrte ech polynomil exctly over the corresponding subintervl. Effectively, these rules compute the exct integrl of function tht interpoltes the vlues of the integrnd f(x) on [, b] bsed on the vlues of f t the endpoints of ech subintervl. Interpoltory qudrture rules tht use k eqully spced smpling points for ech polynomil pproximtion hve the property tht they obtin the exct result if the integrnd is polynomil of degree k or less if k is odd, nd degree k or less if k is even. For exmple, Simpson s Rule is exct for cubic polynomils, since it uses three nodes in ech pproximtion. However, much higher ccurcy cn be obtined if we do not require the smpling points to be eqully spced. This is the bsic ide behind Gussin qudrture, in which the smpling points re chosen so s to obtin s much ccurcy s possible. Using this pproch with k smpling points for ech polynomil pproximtion yields exct results for n integrnd tht is

100 CHAPTER. TECHNIQUES OF INTEGRATION polynomil of degree k or less nerly twice the degree tht cn be obtined using eqully spced smpling points. It is for this reson tht Gussin qudrture is often the preferred pproch for pproximting integrls numericlly. For more informtion on numericl methods for pproximting integrls s well s derivtives, see jlmbers/mth5/. Exmple 69 The following tble describes function r(t) tht indictes the rte of snowfll t time t, where t is mesured in hours since some initil time. t / 3/ 5/ 3 r(t) Estimte the totl snowfll over the three-hour period for which dt hs been gthered. Solution We will estimte the totl snowfll, which is given by the definite integrl 3 r(t) dt, (.94) using five different methods: left endpoints, right endpoints, the Midpoint Rule, the Trpezoidl Rule, nd Simpson s Rule.. Left endpoints: the integrl cn be interpreted not only s totl snowfll, but lso the re under the curve y = r(t) from t = to t = 3. This re cn be estimted by pproximting the region under the curve by six rectngles tht re determined s follows: the intervl [, 3] is divided into six subintervls of equl with t = (3 )/6 = /. Ech subintervl hs endpoints [t i, t i ] where t i = i t for i =,..., 6. Then, we hve 3 6 ti r(t) dt = r(t) dt, (.95) t i i= nd ech integrl in the bove sum cn be pproximted s follows: ti t i r(t) dt r(t i )(t i t i ) = r(t i ) t. (.96) In other words, the integrl of the subintervl [t i, t i ] is pproximted by computing the re of rectngle with width t nd height equl to the vlue of r(t) t the left endpoint of the subintervl. Since there

101 .5. APPROXIMATE INTEGRATION re 6 subintervls, ech of width /, it follows tht the pproximtion L 6 to the totl snowfll obtined using left endpoints is given by L 6 = = = 6 r(t i ) t i= 6 r((i ) t) t i= 6 r((i )/)/ i= = [r() + r(/) + r() + r(3/) + r() + r(3/)] = [ ] = (7.5) = (.97) The pproximting rectngles re shown in Figure... Right endpoints: we proceed exctly s with left endpoints, but this time we pproximte the integrl over ech subintervl by the re of rectngle whose height is obtined by evluting r(t) t the right endpoint, t i, of ech subintervl [t i, t i ]. The pproximtion R 6 obtined using these right endpoints is R 6 = = = 6 r(t i ) t i= 6 r(i t) t i= 6 r(i/)/ i= = [r(/) + r() + r(3/) + r() + r(3/) + r(3)] = [ ] = (7) = 3.5. (.98)

102 CHAPTER. TECHNIQUES OF INTEGRATION Figure.: Approximting rectngles from using left endpoints The pproximting rectngles re shown in Figure.. 3. Midpoint Rule: the re under y = r(t) is once gin pproximted by computing the res of rectngles of equl width, but the height of ech rectngle is determined by evluting r(t) t the midpoint of ech subintervl. Becuse we only know the vlues of r(t) t select points, we cn use t most 3 subintervls of width t = (3 )/3 =. Our subintervls re [, ], [, ], nd [, 3], with midpoints /, 3/, nd 5/, respectively. The pproximtion M 3 is given by M 3 = = 3 r(t i / ) t i= 3 r((i /) t) t i=

103 .5. APPROXIMATE INTEGRATION 3 Figure.: Approximting rectngles from using right endpoints 3 = r(i /) i= = r(/) + r(3/) + r(5/) = = 3.4. (.99) The pproximting rectngles re shown in Figure.3. Alterntively, the Midpoint Rule cn be viewed s pproximting the region under r(t) on ech subintervl [t i, t i ] by trpezoid of width t nd heights r(t M ) + ( t/)r (t M ) nd r(t M ) ( t/)r (t M ), where t M = (t i + t i )/ is the midpoint of the subintervl. Of course, we do not know

104 4 CHAPTER. TECHNIQUES OF INTEGRATION Figure.3: Approximting rectngles from Midpoint Rule r (t M ), but we cn pproximte it using the finite difference r (t M ) r(t i) r(t i ). (.) t The pproximting trpezoids re shown in Figure Trpezoidl Rule: The intervl [, 3] is once gin divided into six subintervls of equl width t = /. On ech subintervl [t i, t i ], the region under y = r(t) is pproximted by trpezoid of width t nd heights determined by the vlue of r(t) t the left endpoint t i nd t the right endpoint t i. The pproximtion T 6 obtined using this pproch is T 6 = 6 i= t r(t i ) + r(t i )

105 .5. APPROXIMATE INTEGRATION 5 Figure.4: Approximting trpezoids from Midpoint Rule = t 6 r(t i ) + r(t i ) i= = t [(r(t ) + r(t )) + (r(t ) + r(t )) + (r(t ) + r(t 3 ))+ (r(t 3 ) + r(t 4 )) + (r(t 4 ) + r(t 5 )) + (r(t 5 ) + r(t 6 ))] = t [r(t ) + r(t ) + r(t ) + r(t 3 ) + r(t 4 ) + r(t 5 ) + r(t 6 )] = [.7 + (.) + (.3) + (.7) + (.) + (.5) +.] 4 = [ ] 4 = 4 (4.5)

106 6 CHAPTER. TECHNIQUES OF INTEGRATION = (.) The pproximting trpezoids re shown in Figure.5. Figure.5: Trpezoidl Rule 5. Simpson s Rule: wheres the pproch of using left endpoints or right endpoints pproximtes the integrnd r(t) by constnt function on ech subintervl, nd the Midpoint Rule nd Trpezoidl Rule pproximte r(t) by liner function on ech subintervl, Simpson s Rule pproximtes r(t) by qudrtic function on ech pir of djcent subintervls nd integrtes ech qudrtic function exctly using the Fundmentl Theorem of Clculus. The resulting pproximtion S 6 is S 6 = x 3 [r(t ) + 4r(t ) + r(t ) + 4r(t 3 ) + r(t 4 ) + 4r(t 5 ) + r(t 6 ) + r(t 7 )]

107 .5. APPROXIMATE INTEGRATION 7 = / [.7 + 4(.) + (.3) + 4(.7) + (.) + 4(.5) +.] 3 = [ ] 6 = 6 (.3) = (.) The pproximting qudrtic functions re shown in Figure.6. Note tht on the third nd fourth subintervls, the qudrtic pproximtion is ctully liner function, since the points (,.3), (3/,.7), nd (,.) ll lie on the sme line. Figure.6: Approximting qudrtic functions from Simpson s Rule

108 8 CHAPTER. TECHNIQUES OF INTEGRATION Exmple 7 Approximte the integrl using the following methods: 3 x dx (.3). Left endpoints with 3 nd 6 subintervls,. Right endpoints with 3 nd 6 subintervls, 3. The Midpoint Rule with 3 nd 6 subintervls, 4. The Trpezoidl Rule with 3 nd 6 subintervls, 5. Simpson s Rule with nd 4 subintervls. Compre to the exct nswer 3 x dx = x3 3 3 = 7 3 = 9. (.4) Solution In the cse of 3 subintervls, we hve width x = (3 )/3 =, nd for 6 subintervls, we hve x = (3 )/ = /.. Left endpoints: We hve L 3 = x [ x + x + x ] = ( + + ) = ( + + 4) = 5 (.5) nd L 6 = x [ x + x + x + x 3 + x 4 + x ] 5 = [ + (/) + + (3/) + + (5/) ]. Right endpoints: We hve = [ + / / /4] = 55 4 = (.6) R 3 = x [ x + x + x 3] = ( ) = ( ) = 4 (.7)

109 .5. APPROXIMATE INTEGRATION 9 nd R 6 = x [ x + x + x 3 + x 4 + x 5 + x ] 6 = [ (/) + + (3/) + + (5/) + 3 ] = [/ / /4 + 9] = 9 4 =.375. (.8) 3. The Midpoint Rule: We hve [ ] M 3 = x x / + x 3/ + x 5/ = [ (.5) + (.5) + (.5) ] = 8.75 (.9) nd [ ] M 6 = x x / + x 3/4 + 5/ +x 7/ + x 9/ + x / = [ (.5) + (.75) + (.5) + (.75) + (.5) + (.75) ] = = (.) 4. The Trpezoidl Rule: We hve T 3 = x [ x + x + x + x [ 3] = + ( ) + ( ) + 3 ] = (++8+9) = 9.5 (.) nd T 6 = x [ x + x + x + x 3 + x 4 + x 5 + x 6] = / [ + (.5 ) + ( ) + (.5 ) + ( ) + (.5 ) + 3 ] = = 9.5. (.) 5. Simpson s Rule: We hve S = x 3 [ x + 4x + x ] =.5 3 [ + 4(.5 ) + 3 ] = [ ] = 9 (.3)

110 CHAPTER. TECHNIQUES OF INTEGRATION nd S 4 = x [ x 3 + 4x + x + 4x 3 + x 4] = 3/4 [ + 4(.75 ) + (.5 ) + 4(.5 ) + 3 ] 3 = 9. (.4) Becuse Simpson s Rule pproximtes the integrnd by qudrtic function on ech pir of subintervls, nd the integrnd is qudrtic function, its pproximtion is exct no mtter how few subintervls re used. The following tble summrizes the error of the other pproximtions. In ll cses the error is obtined by tking the bsolute vlue of the difference between the pproximtion nd the exct nswer of 9. Rule Error, n = 3 Error, n = 6 Left endpoints 4.5 Right endpoints Midpoint Rule.5.65 Trpezoidl Rule.5.5 We mke the following observtions: The Midpoint Rule is the most ccurte, followed by the Trpezoidl Rule. Both of these rules re much more ccurte thn using either left or right endpoints. For the Midpoint Rule nd the Trpezoidl Rule, the error is reduced by fctor of 4 when the number of subintervls is doubled. When using the left or right endpoints, the error is reduced by fctor of only when the number of subintervls is doubled..6 Improper Integrls So fr, we hve only considered integrls of the form b f(x) dx (.5) in which the intervl [, b] is finite, nd the integrnd f(x) is either continuous on [, b], or, if it is discontinuous nywhere in [, b], the discontinuity is

111 .6. IMPROPER INTEGRALS not infinite. In this section, we consider integrls tht do not fll into this ctegory, either becuse the intervl [, b] is infinite, or the integrnd becomes infinite t some point in [, b]. Such integrls re known s improper integrls..6. Infinite Intervls In mny pplictions such s in sttistics or the study of differentil equtions on unbounded domins, it is necessry to compute integrls over n intervl [, b] where either endpoint or b is infinite. In this cse, we cn define such integrls in terms of integrls over finite intervls. Definition 4 If t f(x) dx exists for t, then we define f(x) dx = lim t t f(x) dx, (.6) provided this limit exists nd is finite. Similrly, if b t f(x) dx exists for ll t b, then we define b f(x) dx = lim t b t f(x) dx, (.7) provided the limit exists nd is finite. If n integrl over n infinite intervl exists, we sy tht the integrl is convergent; otherwise, we sy tht it is divergent. Exmple 7 Evlute dx. (.8) x Solution We hve dx x = b lim b x dx = lim ln x b b = lim ln b ln b = lim ln b b = (.9) nd therefore the integrl is divergent.

112 CHAPTER. TECHNIQUES OF INTEGRATION Exmple 7 Evlute dx. (.) x Solution We hve b dx = lim x b x dx = lim b b x = lim b b ( ) = lim b b =. (.) Exmple 73 Determine the vlues of p for which is convergent. Assume p >. dx (.) xp Solution We know tht the integrl diverges when p =. If p, then we hve, by the Power Rule, dx = lim xp b = lim b b = lim b x p dx x p+ b p + p = lim b p = b x p ( b p p p p lim b ). (.3) bp In order for /b p to tend to zero s b, the exponent p must be positive, nd therefore the integrl converges if p >. In this cse, its vlue is /(p ).

113 .6. IMPROPER INTEGRALS 3 Exmple 74 Determine the volume of the solid of revolution obtined by revolving the re under /x, for x, round the x-xis. Solution This solid hs circulr cross-sections of rdius /x, for ll x. It follows tht the cross-sectionl re is A(x) = π(/x) nd therefore the volume is given by ( ) V = π dx = π dx = π, (.4) x x bsed on the result from the previous exmple. Exmple 75 Evlute cos x dx. (.5) Solution We hve cos x dx = lim b b cos x dx = lim sin x b b = lim sin b sin b = lim sin b. b (.6) However, lim b sin b does not exist becuse sine oscilltes between nd. Therefore the integrl is divergent. We cn use the integrls defined bove to define the integrl of function f(x) over the entire rel line. If both f(x) dx nd f(x) dx re convergent, then we define f(x) dx = f(x) dx + f(x) dx. (.7) The process of computing n integrl over n infinite intervl is suggested by the definition. For exmple, to compute f(x) dx, one cn compute b f(x) dx where the upper limit b is vrible insted of fixed constnt, nd then compute the limit of the integrl s b. Exmple 76 Evlute e x dx. (.8)

114 4 CHAPTER. TECHNIQUES OF INTEGRATION Solution We hve e x dx = = lim e x dx + = lim e x dx e x dx + lim e x dx + lim b b b b = lim ex + lim b e x b e x dx e x dx = lim e e + lim b e b ( e ) = e + e lim e lim =. (.9) b e b.6. Discontinuous Integrnds An integrl cn be improper even if it is over finite intervl [, b]. If the integrnd f(x) hs verticl symptote t some point in [, b], then the usul definition of definite integrl cnnot pply, we cn define the integrl s limit of integrls computed over n intervl on which f(x) is continuous. Definition 5 Suppose tht f(x) hs verticl symptote t x =. Then we define b f(x) = lim t + b t f(x) dx, (.3) provided the limit exists nd is finite. Similrly, if f(x) hs verticl symptote t x = b, then we define b f(x) = lim t b t f(x) dx, (.3) provided the limit exists nd is finite. Finlly, if f(x) hs verticl symptote t x = c, where < c < b, then we define b f(x) dx = c f(x) dx + b c f(x) dx, (.3)

115 .6. IMPROPER INTEGRALS 5 provided tht the two (improper) integrls on the right side of the eqution exist. We sy tht n integrl over [, b] where f(x) hs verticl symptote in [, b] is convergent if it exists nd is finite, nd we sy tht it is divergent otherwise. As with integrls over infinite intervls, computing n integrl where the integrnd hs verticl symptote on [, b] cn proceed using method suggested by the definition. For exmple, to compute b hs verticl symptote t x = b, one cn compute f(x) dx where f(x) c f(x) dx where c is vrible upper limit ssumed to be between nd b, nd then tke the limit s c b from the left. Exmple 77 Evlute dx x. (.33) Solution The integrnd hs n infinite discontinuity t x =. Therefore, we hve, by the substitution u = x/, dx x b dx = lim b x = lim b = lim b = lim b b = lim b b b b dx x / dx (x/ ) du u du u = lim b sin u b = lim b sin b sin = lim b sin b = π. (.34)

116 6 CHAPTER. TECHNIQUES OF INTEGRATION Exmple 78 Assume p >. Determine the vlues of p for which dx (.35) xp is convergent, nd, when it is convergent, determine the integrl s vlue. Solution We hve dx = lim xp + x p dx x p+ = lim + p + = lim + p = lim + p x p ( p p ). (.36) For / p to tend to zero s, the exponent p must be negtive, nd therefore we must hve < p <. In this cse, the integrl is convergent to the vlue /( p). Exmple 79 Determine whether the integrl is convergent. 3 dx x (.37) Solution This integrnd hs discontinuity t , which lies in the intervl [, 3]. Therefore we hve 3 dx x = dx x + 3 dx x. (.38) However, becuse /(x ) hs the prtil frction decomposition x = A x + + B x (.39) for some constnts A nd B, it follows tht the ntiderivtive is given by dx x = A ln x + + B ln x + C. (.4) Since lim x ln x =, it follows tht the integrl from to is divergent, s is the integrl from to 3. Therefore the integrl from to 3 does not exist.

117 .6. IMPROPER INTEGRALS A Comprison Test for Improper Integrls If one only wnts to know whether n improper integrl over n infinite intervl is convergent or divergent, without necessrily knowing its exct vlue, then the Comprison Theorem cn be useful. Theorem 3 (Comprison Test) If f(x) g(x) for x, then the following sttements re true: If If f(x) dx is convergent, then g(x) dx is convergent. g(x) dx is divergent, then f(x) dx is divergent. This theorem not esy to prove, but, intuitively, it cn be seen tht it is true by compring the re under the grphs of f nd g for x, since the region under g is entirely contined within the region under f. Exmple 8 Use the Comprison Theorem to determine whether is convergent. Solution For x >, we hve dx x 3 + 7x + x + (.4) nd therefore x 3 + 7x + x + > x 3 (.4) x 3 + 7x + x + < x 3. (.43) Since b dx = lim x3 b x 3 dx =, (.44) it follows from the Comprison Theorem tht is convergent. dx x 3 + 7x + x + (.45)

118 8 CHAPTER. TECHNIQUES OF INTEGRATION

119 Chpter 3 Applictions of Integrtion In the next few sections, we begin to expnd the pplicbility of the definite integrl. In prticulr, we will lern how the definite integrl cn be used to compute res of more generl regions thn in previous discussion, nd even volumes of certin three-dimensionl solids. 3. Ares Between Curves We hve lredy lerned tht the definite integrl b f(x) dx (3.) cn be used to compute the re of the region bounded by the curve y = f(x), the horizontl line y =, nd the verticl lines x = nd x = b. Now, suppose we need to compute the re of somewht more complicted region: one tht is bounded bove by the curve y = f(x) nd bounded below by the curve y = g(x), between the verticl lines x = nd x = b. We ssume tht f(x) nd g(x) re continuous on [, b], nd tht f(x) g(x) on [, b]. As before, we cn pproximte this region using rectngles. We divide the intervl [, b] into n subintervls of width x = (b )/n. These subintervls re [x, x ], [x, x ],..., [x n, x n ], where x i = + i x for i =,..., n. We then pproximte the region by n rectngles of width x nd height f(x i ) g(x i ), where x i is ny point in the ith subintervl [x i, x i ]. Note tht the height the ith rectngle hppens to be the verticl distnce between the curves y = f(x) nd y = g(x) t the point x i. 9

120 CHAPTER 3. APPLICATIONS OF INTEGRATION Using these rectngles, we cn pproximte the re A of the region using the Riemnn sum A R n = n [f(x i ) g(x i )] x i. (3.) i= As the number of rectngles, n, pproches infinity, we obtin the exct re of the region between the curves y = f(x) nd y = g(x), which is given the definite integrl A = lim n R n = b f(x) g(x) dx. (3.3) We cn therefore use the definite integrl to compute the re of the region bounded bove nd below by ny two curves, not just regions bounded bove by the curve y = f(x) nd bounded below by the horizontl line y =. A similr pproch cn be used to compute the re of the region bounded on the right by curve of the form x = f(y), on the left by curve x = g(y), below by the horizontl line y = c nd bove by the horizontl line y = d. If f(y) g(y) on the intervl [c, d], then the re A of such region is given by the definite integrl A = d c f(y) g(y) dy. (3.4) On stndrd grph, x increses to the right, so in order to ensure tht the re hs the correct sign, the function whose grph is the left boundry of the region, in this cse g(y), is subtrcted from the function whose grph is the right boundry, which is f(y) in this cse. For this reson, it is helpful to remember right left when integrting with respect to y to compute the re between to curves. Similrly, it is helpful to remember top bottom when integrting with respect to x, since the integrl represents the correct re when the function whose grph is the bottom boundry is subtrcted from the function whose grph is the top boundry. The ssumption tht f(x) g(x), or tht f(y) g(y), should not be ignored. In generl, the bove integrls represent the net re of regions between two curves. If g(x) f(x) on ny subintervl of [, b], then on tht subintervl, the re between the curves is counted negtively towrd the vlue of the integrl of f(x) g(x) from to b. In generl, the re A of the region between the grphs of f(x) nd g(x) from x = to x = b is A = b f(x) g(x) dx, (3.5)

121 3.. AREAS BETWEEN CURVES where the bsolute vlue ensures tht the re between the two curves is lwys counted positively. A similr integrl cn be used to compute the re between the curves x = f(y) nd x = g(y) from y = c to y = d. Exmple 8 Use clculus to compute the re of the tringle with vertices ( 4, ), (, 7) nd (5, 3). Solution The tringle is displyed in Figure 3.. We will compute its re Figure 3.: Tringle with vertices ( 4, ), (, 7) nd (5, 3) by computing the re enclosed by the three lines tht define the edges of the tringle. We begin by computing the equtions of these lines. First, we consider the line tht psses through the vertices ( 4, ) nd (, 7). The slope is given by 7 ( 4) = 5 =, (3.6) 5 which yields the eqution y 7 = (x ) (3.7)

122 CHAPTER 3. APPLICATIONS OF INTEGRATION or y = x + 6. (3.8) Proceeding in the sme mnner with the other two lines, we obtin slopes nd the corresponding equtions which simplify to = = 5 4, 3 5 ( 4) = 5 9, (3.9) y 7 = 5 (x ), y = 5 (x ( 4)) (3.) 9 y = 5 x + 9, y = 5 9 x 9. (3.) We re now redy to compute the re of the tringle using definite integrls. We divide the tringle long the dshed line shown in Figure 3.. This yields two smller tringles, the re of which cn be computed using single definite integrl in ech cse. The left tringle is the region bounded by the lines x =, y = x + 6, nd y = 5x/9 /9. The line y = x + 6 defines the top of the tringle, while the line y = 5x/9 /9 defines the bottom. The limits of integrtion re dictted by the right boundry x = nd the fct tht the two lines y = x + 6 nd y = 5x/9 /9 intersect t x = 4. It follows tht its re, which we denote by A, is ( A = (x + 6) 5 9 x ) dx 9 = = = = 4 9 = 4 9 = x x + 9 dx 9 9 x x + 9 dx 4 9 x dx 4 x + 4 dx [ ] x + 4x 4 {[ ] [ ( 4) + 4 ]} + 4( 4)

123 3.. AREAS BETWEEN CURVES 3 = 4 { } 9 [8 6] 9 = 4 ( ) = 4 ( ) 5 9 = 35 8 = (3.) We use the sme pproch to compute the re of the tringle to the right of the dshed line. This tringle is the region bounded by the lines x =, y = 5x/ + 9/, nd y = 5x/9 /9. The line y = 5x/ + 9/ defines the top of the tringle, while the line y = 5x/9 /9 defines the bottom. The limits of integrtion re dictted by the right boundry x = nd the fct tht the two lines y = 5x/ + 9/ nd y = 5x/9 /9 intersect t x = 5. It follows tht its re, which we denote by A, is A = = = = ( 5 x + 9 ) ( 5 9 x 9 5 x x + 9 dx 45 8 x x dx 35 8 x dx 5 ) dx = 35x + 75 dx 8 = ] [ 35 x x = ] [ {[ (5) 35 ]} + 75 = { 875 } = { 84 } = ( 4 + 7) 8

124 4 CHAPTER 3. APPLICATIONS OF INTEGRATION = 8 8 = 4 9. (3.3) We conclude tht the re A of the entire tringle is A = A + A = = 35 = 35. (3.4) 9 Exmple 8 Compute the re of the region bounded by the curves y = sin x nd y = cos x, s well s the lines x = nd x = π. Solution The re tht is to be computed is shown in Figure 3.. The Figure 3.: Region bounded by y = sin x, y = cos x, x = nd x = π, shded region whose re we will compute is shded in the figure. We cn see tht this region cn be divided into three regions:

125 3.. AREAS BETWEEN CURVES 5 the region bounded bove by y = cos x, below by y = sin x, to the left by x =, nd to the right by x = π/4, which is where y = cos x nd y = sin x intersect the region bounded bove by y = sin x, below by y = cos x, to the left by x = π/4, nd to the right by x = 5π/4, which is nother point t which y = cos x nd y = sin x intersect the region bounded bove by y = cos x, below by y = sin x, to the left by x = 3π/4, nd to the right by x = π. The re of these regions cn be obtined by evluting the definite integrls A = π/4 cos x sin x dx, A = 5π/4 π/4 sin x cos x dx, A 3 = π (3.5) Using nti-differentition rules, we obtin cos x sin x dx = sin x + cos x + C. (3.6) It follows from the Fundmentl Theorem of Clculus tht A = A = π/4 cos x sin x dx = sin x + cos x π/4 = [sin(π/4) + cos(π/4)] [sin + cos ] [ ] = + [ + ] 5π/4 π/4 5π/4 =, (3.7) sin x cos x dx = cos x sin x 5π/4 π/4 = [ cos(5π/4) sin(5π/4)] [ cos(π/4) sin(π/4)] = = [ ( ) ( )] [ [ ] [ ] ] =, (3.8) cos x sin x dx.

126 6 CHAPTER 3. APPLICATIONS OF INTEGRATION nd A 3 = π 5π/4 cos x sin x dx = sin x + cos x π 5π/4 = [sin(π) + cos(π)] [sin(5π/4) + cos(5π/4)] [ ] = [ + ] = +. (3.9) We conclude tht the re A of the entire region between the curves is given by A = A + A + A 3 = ( ) + + ( + ) = 4. (3.) It should be noted tht this re is given by the definite integrl π cos x sin x dx, (3.) which we computed by dividing the intervl [, π] into subintervls on which cos x sin x is either positive or negtive, but not both. Exmple 83 Compute the re of the region bounded by the line y = x nd the prbol x = y. Solution This region is shown in Figure 3.3. The curve x = y cnnot be described by n eqution of the form y = f(x), becuse it does not pss the verticl line test. Therefore, it is difficult to compute the re of the shded region in the figure by integrting with respect to x. Insted, we will integrte with respect to y, with our integrnd being the distnce between the right curve nd the left curve, s function of y. This distnce is given by y (y ), since, for ech y-vlue in the shded region, the line x = y is to the right of the curve x = y. The limits of integrtion re given by the y-vlues t which these two curves intersect, since we re integrting with respect to y. These y-vlues stisfy the eqution y = y, or y y =. The polynomil y y fctors into (y )(y + ), so the curves intersect t y = nd y =. Therefore, the re A of the shded region is given by the definite integrl A = y (y ) dy. (3.)

127 3.. AREAS BETWEEN CURVES 7 Figure 3.3: Are between x = y nd x = y, shded We hve A = = = y = = y (y ) dy y y + dy y3 3 + y [ ] () [ 83 ] + 4 [ ( ) ] [ + 3 ( )3 3 ] + ( ) =

128 8 CHAPTER 3. APPLICATIONS OF INTEGRATION = = 9. (3.3) 3. Volume by Slices Suppose tht S is solid tht lies between the plnes x = nd x = b. If A(x) represents the re of the cross-section, or slice, obtined by intersecting S with the plne tht crosses the x-xis t x nd is perpendiculr to the x-xis, then the volume of S is V = b A(x) dx. (3.4) This formul is generliztion of the formul for the volume of cylinder of rdius r nd height (b ) plced prllel to the x-xis, which is V = πr (b ). In this cse, the solid hs constnt rdius, independent of x, nd therefore ech cross-section hs constnt re A(x) πr. The more generl formul (3.4) cn be obtined by dividing [, b] into subintervls [x, x ],..., [x n, x n ] where x = nd x n = b. On ech subintervl [x i, x i ], we then pproximte the portion of S between the plnes x = x i nd x = x i by cylinder of height x i x i nd rdius A(x i ). Just s the definite integrl llows us to compute the re of region with non-constnt height, it llows us to compute the volume of solid whose cross-sectionl re is non-constnt. Some solids whose volumes cn be computed using the formul (3.4) re known s solids of revolution, s they cn be obtined by revolving the re of some two-dimensionl region round line. For such solids, the cross-sectionl re A(x) is typiclly very esy to determine. To illustrte, we discuss some exmples: Suppose tht the region in question is the re below the curve y = f(x), where f(x), from x = to x = b. Furthermore, suppose tht this region is revolved round the x-xis to obtin the solid S. Then, for ech x in [, b], the cross-section t x is disc of rdius f(x), which implies A(x) = π[f(x)]. The volume cn then be computed by using eqution (3.4) with this choice of A(x). This pproch to computing the volume of such solid is clled the disc method.

129 3.. VOLUME BY SLICES 9 Suppose tht the region is the re between two curves y = f(x) nd y = g(x), where f(x) g(x), from x = to x = b. As before, the region is revolved round the x-xis. Then, for ech x between nd b, the cross-section of the solid t x is the region between two concentric circles. The function f(x) determines the outer rdius of the cross-section, nd g(x) determines the inner rdius. It follows tht A(x) = π[f(x) g(x) ]; tht is, A(x) is the difference of the res of two concentric circles. Becuse ech cross-section resembles wsher, the method of computing the volume of such solid by integrting A(x) is clled the wsher method. Mny solids, however, re not solids of revolution. In such cses, the cross-sectionl re A(x) my still be esy to determine. For exmple, the cross-section my be tringle or rectngle. If the cross-section is not shpe whose re is esy to compute for ech x, then it my be necessry to intersect the solid with plnes tht re perpendiculr to the y-xis or the z-xis nd determine the cross-sectionl re s function of y or z insted of s function of x. It is importnt to recognize the prllel between computing res nd computing volumes using definite integrls. In both cses, n object is pproximted by n simpler objects, such s rectngles or cylinders, so tht the desired quntity be obtined by computing it esily for ech of the simpler objects, nd dding the results to obtin n pproximtion R n, which is Riemnn sum. Then, by determining the overll result s function of n nd computing lim n R n, the exct result is obtined. It is interesting exercise to inquire s to wht other simple formuls cn be generlized to more difficult problems using this pproch. Exmple 84 We will compute the volume of conicl bem, crosssection of which is shown in Figure 3.4. For ech x between nd 5, the cross-section of the bem t x cn be viewed s rectngle of length 3x nd height x, with hlf-disc of rdius x removed from ech side. It follows tht the cross-sectionl re is given by A(x) = 6x πx. Using the formul in eqution (3.4), we cn compute volume of this solid s follows: 5 6x πx dx = x 3 π 3 x3 5 ( = 5 π ). (3.5) 3 Exmple 85 Suppose tht we form solid by revolving the region under y = x, from x = to x =, round the x-xis. The resulting solid is

130 3 CHAPTER 3. APPLICATIONS OF INTEGRATION Figure 3.4: Cross-section of conicl bem for some x between nd 5 curved funnel with cross-sectionl re A(x) = π(x ) = πx 4. It follows tht its volume is πx 4 dx = π x5 5 = π 5. (3.6) Exmple 86 Suppose tht we form solid by revolving the sme region s in Exmple 85, except tht in this cse, insted of revolving it round the x-xis, we revolve it round line y =. Then, the cross-section of x is disc with rdius x +, insted of x. It follows tht the volume of this solid is ( ) x π(x +) dx = π x 4 +4x 5 +4 dx = π 5 + 4x3 ( 3 + 4x = π ) (3.7) = 83π 5.

131 3.. VOLUME BY SLICES 3 Exmple 87 Suppose tht we form solid by revolving the region between y = x nd y = x round x-xis. Then, for ech x between nd (where the curves y = x nd y = x intersect), the cross-section of the solid t x is wsher with outer rdius x nd inner rdius x, since x x on [, ]. It follows tht the volume of the solid, obtined by the wsher method, is ( ) x πx πx 4 3 dx = π 3 x5 5 ( = π 3 ) = π 5 5. (3.8) Exmple 88 Suppose tht we form solid by revolving the region under y = x, from x = to x =, round line y =. This region is illustrted in Figure 3.5. For ech x between nd, the cross-section of the solid t x is wsher. The outer rdius of the wsher is, the distnce between the line y = nd the line y =. The inner rdius is the distnce between the curve y = x nd the line y =, which is ( x) = x. It follows tht the volume of the solid is π( ) x ) dx = π 4 x dx = π (4x x ( = π 4 ) = 7π. Exmple 89 Suppose tht we form solid of revolution using the sme region s in Exmple 88, (shown in Figure 3.5) except tht this time, we revolve the region round the y-xis, which is the verticl line x =. In this cse, we cn still use the formul in eqution (3.4), except tht the cross-sectionl re is now function of y. For ech y between nd, the cross-section t y is disc. For y, the rdius of the disc is equl to, nd for y, the rdius is equl to the horizontl distnce between the line x = nd the curve y = x. Rewriting the eqution of the curve s function of y, we obtin x = ( y), nd it follows tht the rdius of the disc is ( y), for y. Becuse two functions re needed to describe the cross-sectionl re A(y), one for y nd one for y, it is necessry to brek up the integrl of A(y) from to into two integrls, one for ech of these subintervls. It follows tht the volume V of the solid is V = π( ) dy + π[( y) ] dy

132 3 CHAPTER 3. APPLICATIONS OF INTEGRATION Figure 3.5: The region bounded by y = x, y =, x =, nd x = is to be revolved round the line y =. = π dy + = πy π u 4 du = π + π = π + π u5 5 = 6π 5 u 4 du π( y) 4 dy In the third step, we used the substitution u = y to simplify the integrnd in the second integrl. In the next section, we will compute the volume of this sme solid using nother method, clled the shell method, in which we

133 3.3. VOLUME BY SHELLS 33 will not need to brek up the integrl. 3.3 Volume by Shells In the previous section we lerned how we could compute the volumes of solid S tht ly between the plnes x = nd x = b by integrting its crosssectionl A(x) re over the intervl [, b]. Unfortuntely, this technique of computing volume by slices, where ech slice is cylinder of infinitely smll height, is only prcticl if the cross-sectionl re is reltively esy to determine. If this is not the cse, then we cn insted try to determine whether the solid cn be pproximted by severl concentric cylindricl shells. For exmple, suppose tht b >, nd we hve solid tht cn be obtined by rotting the region bounded by y = f(x) (where f(x) ), x =, x = b, nd y =. Then, the volume of resulting solid cn be pproximted by first dividing the intervl [, b] into n subintervls of equl width x = (b )/n, with the ith subintervl hving endpoints x i nd x i, where x i = i x for i =,..., n. For ech subintervl, we denote the midpoint x i by (x i + x i )/. Then, we compute the volumes of n shells of thickness x = (b )/n, height f(x i ), inner rdius x i, nd outer rdius x i, for i =,..., n. The volume of the ith shell is V i = πx i f(x i ) x, nd therefore the volume V of the entire solid is pproximted by V n V i = i= n πx i f(x i ) x. (3.9) i= Letting n, this pproximtion converges to the exct volume, with the summtion converging to the definite integrl V = b πxf(x) dx. (3.3) More generlly, if the solid cn be viewed s collection of concentric cylindricl shells of rdius r(x) nd height f(x), for x b, the volume of the solid is given by V = b πr(x)f(x) dx. (3.3) If the solid is formed by rotting region round the y-xis, then r(x) = x. However, different r(x) must be used if the solid is obtined by rotting

134 34 CHAPTER 3. APPLICATIONS OF INTEGRATION region round different line. For exmple, if the line round which the region is rotted is x = c, where c must be outside the intervl [, b], then r(x) = x c if c, while r(x) = c x if c b. Exmple 9 In the previous section, we used the disc method to compute the volume of the solid obtined by revolving the region bounded by y = x, y =, x = nd x = round the y-xis. Now, we will compute the volume of the sme solid using the shell method. The verge rdius of the shell t x is equl to x, nd the height of this shell is equl to the height of the region t x, which is x. It follows tht the volume of the solid is πx( x) dx = π x x 3/ dx = π ( x x5/ 5 ) ( = π ) = 6π 5 5. (3.3) Exmple 9 Let f(x) = x. Consider the region bounded by the curve y = f(x), the horizontl line y =, nd the verticl lines x = nd x =. Compute the volume of the solid obtined by revolving this region round the y-xis. Solution We use the method of cylindricl shells. The given region is shown in Figure 3.6, while the solid obtined by revolving the region round the y- xis is shown in Figure 3.7. This solid cn be pproximted by collection of concentric cylindricl shells. The pproximtion proceeds s follows: first, we divide the intervl [, ] into n subintervls of equl width x = /n. These intervls hve endpoints [x, x ], [x, x ],..., [x n, x n ] where x i = + i x, for i =,,,..., n. Then, we pproximte the region below y = f(x) by rectngles of width x nd height f(x i ), where x i is ny point in the intervl [x i, x i ] for i =,..., n. Then, by revolving ech rectngle round the y-xis, we obtin n cylindricl shells tht pproximte the solid, just s the rectngles pproximte the region. This process of revolving ech rectngle round the y-xis to obtin shell is illustrted in Figure 3.8. For ech i =,..., n, the ith shell hs thickness x, height f(x i ), inner rdius x i nd outer rdius x i. The volume V i of this shell is given by ( ) xi + x i V i = π f(x i ) x. (3.33) By dding the volume of ll of these shells, we obtin Riemnn sum tht yields n pproximtion to the volume of the solid. As the number of subin-

135 3.3. VOLUME BY SHELLS 35 Figure 3.6: Region bounded by y = x, y =, x = nd x =, shded tervls, n, becomes infinite, this pproximtion converges to the exct volume. The volume V of the solid is therefore given by V = lim n = lim n = lim n = n i= i= i= V i n ( xi + x i π n ( π x i x ) f(x i ) x ) f(x i ) x πxf(x) dx. (3.34)

136 36 CHAPTER 3. APPLICATIONS OF INTEGRATION Figure 3.7: Solid obtined by revolving region shown in Figure.4 round the y-xis Evluting the resulting definite integrl yields V = = π πx(x ) dx x 3 dx = π x4 4 ( ) 4 = π ( = π 4 ) 4

137 3.3. VOLUME BY SHELLS 37 Figure 3.8: Cylindricl shell obtined by revolving rectngle round the y- xis. The rectngle is one of severl tht is used to pproximte the region bounded by y = x, y =, x =, x =, whose outline is shown. = 5π. (3.35) Exmple 9 Let f(x) = x. Consider the region from the previous exmple, tht is bounded by the curve y = f(x), the horizontl line y =, nd the verticl lines x = nd x =. Compute the volume of the solid obtined by revolving this region round the verticl line x =. Solution In this cse, we cn pproximte the solid by cylindricl shells s before, but the center nd rdii of the shells is different. Becuse the center of the solid is the line x =, the inner rdii of the ith shell, corresponding

138 38 CHAPTER 3. APPLICATIONS OF INTEGRATION to the subintervl [x i, x i ], is x i +, since tht is the distnce between the inner boundry of the shell nd the center. Similrly, the outer rdius of the ith shell is x i +. Proceeding in the previous exmple, we cn determine tht the volume V of the solid is given by n V = lim V i n = lim n = lim n = i= n ( (xi + ) + (x i + ) π i= n ( π x i + x ) f(x i ) x i= ) f(x i ) x π(x + )f(x) dx. (3.36) Evluting the resulting definite integrl yields V = = π π(x + )x dx x 3 + x dx ( ) x 4 = π 4 + x3 3 ( 4 = π ) 3 3 ( = π ) 3 ( 5 = π ) 3 = π 6. (3.37) Exmple 93 Let f(x) = x. Consider the region from the previous exmple, tht is bounded by the curve y = f(x), the horizontl line y =, nd the verticl lines x = nd x =. Compute the volume of the solid obtined by revolving this region round the horizontl line y =. Solution We will compute the volume of this solid using both the wsher method (tht is, volume by slices) nd the shell method. Using the wsher

139 3.3. VOLUME BY SHELLS 39 method, we see tht for ech x in the intervl [, ], the corresponding wsher hs inner rdius nd outer rdius x +. It follows tht the volume V of the solid is given by V = = π = π π(x + ) π dx x 4 + x + dx x 4 + x dx ( ) x 5 = π 5 + x3 3 ( 5 = π ) 3 3 ( 3 = π ) 3 ( 3 = π ) 3 = 63π 5. (3.38) A smple wsher is illustrted in Figure 3.9. Using the shell method, we integrte with respect to y becuse we re revolving the region round horizontl line. For ech y in the intervl [, 4], we hve cylindricl shell centered t the line y = with thickness dy, verge rdius y +, nd height y, since tht is the horizontl distnce between the line x = nd the curve y = x, or, equivlently, x = y. Furthermore, for ech y in the intervl [, ], we hve cylindricl shell centered t the line y = with thickness dy, verge rdius y +, nd height. It follows tht the volume V is given by 4 V = π(y + )() dy + π(y + )( y) dy [ 4 ] = π y + dy + y y 3/ y / + dy ( ) ( ) = π y 4 + y + y y5/ 5/ y3/ 3/ + y

140 4 CHAPTER 3. APPLICATIONS OF INTEGRATION Figure 3.9: The region bounded by y = x, y =, x = nd x = is to be revolved round the line y =. The wsher corresponding to x =.4 is shown. [ ( ) ( ) ( = π / 43/ + 4 5/ [( ) ( = π ) ( 5 3 )] + [ ( 3 = π ) ( )] 3 [ 3 64 = π ] 3 [ 45 = π ] 3 3/ 3 + )]

141 3.3. VOLUME BY SHELLS 4 [ ] 63 = π 3 = 63π 5. (3.39) A smple shell is illustrted in Figure 3.. Figure 3.: The region bounded by y = x, y =, x = nd x = is to be revolved round the line y =. The shell corresponding to y =.3 is shown. Exmple 94 Let f(x) = x. Consider the region tht is bounded by the curve y = f(x), the horizontl line y =, nd the verticl lines x = nd x =. This region in shown in Figure 3.6. Compute the volume of the solid obtined by revolving this region round the verticl line x = 3.

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