More Applications of Integration

Transcription

1 4 Chpter More Applictions of Integrtion More Applictions of Integrtion fulcrum to the left To clculte exctly where the fulcrum should be, we let x denote the loction of the fulcrum when the bem is in blnce The totl torque on the bem is then (3 x)+(6 x)5+(8 x)4 = 9 9 x Since the bem blnces t x it must be tht 9 9 x = or x = 9/9 484, tht is, the fulcrum should be plced t x = 9/9 to blnce the bem Now suppose tht we hve bem with vrying density some portions of the bem contin more mss thn other portions of the sme size We wnt to figure out where to put the fulcrum so tht the bem blnces m m m m 3 m 4 m 5 m 6 m 7 m 8 m 9 Figure A solid bem ÒØ Ö Ó Å ½½º½ Suppose bem is meters long, nd tht there re three weights on the bem: kilogrm weight 3 meters from the left end, 5 kilogrm weight 6 meters from the left end, nd 4 kilogrm weight 8 meters from the left end Where should fulcrum be plced so tht the bem blnces? Let s ssign scle to the bem, from t the left end to t the right, so tht we cn denote loctions on the bem simply s x coordintes; the weights re t x = 3, x = 6, nd x = 8, s in figure Figure A bem with three msses Suppose to begin with tht the fulcrum is plced t x = 5 Wht will hppen? Ech weight pplies force to the bem tht tends to rotte it round the fulcrum; this effect is mesured by quntity clled torque, proportionl to the mss times the distnce from the fulcrum Of course, weights on different sides of the fulcrum rotte the bem in opposite directions We cn distinguish this by using signed distnce in the formul for torque So with the fulcrum t 5, the torques induced by the three weights will be proportionl to (3 5) =, (6 5)5 = 5, nd (8 5)4 = For the bem to blnce, the sum of the torques must be zero; since the sum is = 3, the bem rottes counter-clockwise, nd to get the bem to blnce we need to move the 5 4 EXAMPLE Suppose the bem is meters long nd tht the density is +x kilogrms per meter t loction x on the bem To pproximte the solution, we cn think ofthebemssequenceofweights on bem Forexmple, wecnthinkoftheportion of the bem between x = nd x = s weight sitting t x =, the portion between x = nd x = s weight sitting t x =, nd so on, s indicted in figure We then pproximte the mss of the weights by ssuming tht ech portion of the bem hs constnt density So the mss of the first weight is pproximtely m = (+) = kilogrms, nmely, ( + ) kilogrms per meter times meter The second weight is m = ( + ) = kilogrms, nd so on to the tenth weight with m 9 = ( + 9) = kilogrms So in this cse the totl torque is ( x)m +( x)m + +(9 x)m 9 = ( x)+( x)+ +(9 x) If we set this to zero nd solve for x we get x = 6 In generl, if we divide the bem into n portions, the mss of weight number i will be m i = ( + x i )(x i+ x i ) = ( + x i ) x nd the torque induced by weight number i will be (x i x)m i = (x i x)(+x i ) x The totl torque is then (x x)(+x ) x+(x x)(+x ) x+ +(x n x)(+x n ) x n = i= n = i= n x i (+x i ) x i= n x i (+x i ) x x i= x(+x i ) x (+x i ) x 3

2 Center of Mss 5 If we set this equl to zero nd solve for x we get n pproximtion to the blnce point of the bem: n x n n = x i (+x i ) x x (+x i ) x i= n (+x i ) x = x i (+x i ) x i= x = i= n x i (+x i ) x i= n (+x i ) x i= The denomintor of this frction hs very fmilir interprettion Consider one term of the sum in the denomintor: (+x i ) x This is the density ner x i times short length, x, which in other words is pproximtely the mss of the bem between x i nd x i+ When we dd these up we get pproximtely the mss of the bem Now ech of the sums in the frction hs the right form to turn into n integrl, which in turn gives us the exct vlue of x: x = x(+x)dx (+x)dx The numertor of this frction is clled the moment of the system round zero: x(+x)dx = nd the denomintor is the mss of the bem: (+x)dx = 6, i= x+x dx = 5 3, nd the blnce point, officilly clled the center of mss, is x = = Chpter More Applictions of Integrtion It should be pprent tht there ws nothing specil bout the density function σ(x) = + x or the length of the bem, or even tht the left end of the bem is t the origin In generl, if the density of the bem is σ(x) nd the bem covers the intervl [,b], the moment of the bem round zero is nd the totl mss of the bem is nd the center of mss is t M = M = b b xσ(x) dx σ(x) dx x = M M EXAMPLE Suppose bem lies on the x-xis between nd 3, nd hs density function σ(x) = x 9 Find the center of mss This is the sme s the previous exmple except tht the bem hs been moved Note tht the density t the left end is 9 = nd t the right end is 3 9 =, s before Hence the center of mss must be t pproximtely +639 = 639 Let s see how the clcultion works out M = M = M x(x 9)dx = x 9xdx = x3 3 9x x 9dx = x 3 9x = 6 M = = = EXAMPLE 3 Suppose flt plte of uniform density hs the shpe contined by y = x, y =, nd x =, in the first qudrnt Find the center of mss (Since the density is constnt, the center of mss depends only on the shpe of the plte, not the density, or in other words, this is purely geometric quntity In such cse the center of mss is clled the centroid) Thisistwodimensionlproblem, butitcnbesolvedsifitweretwoonedimensionl problems: we need to find the x nd y coordintes of the center of mss, x nd ȳ, nd fortuntely we cn do these independently Imgine looking t the plte edge on, from below the x-xis The plte will pper to be bem, nd the mss of short section of

3 ( x,ȳ) x i Center of Mss 7 m i x i Figure 3 Center of mss for two dimensionl plte the bem, sy between x i nd x i+, is the mss of strip of the plte between x i nd x i+ See figure 3 showing the plte from bove nd s it ppers edge on Since the plte hs uniform density we my s well ssume tht σ = Then the mss of the plte between x i nd x i+ is pproximtely m i = σ( x i ) x = ( x i ) x Now we cn compute the moment round the y-xis: nd the totl mss nd finlly M y = M = x( x )dx = 4 ( x )dx = 3 x = 3 4 = 3 8 Next we do the sme thing to find ȳ The mss of the plte between y i nd y i+ is pproximtely n i = y y, so nd M x = y ydy = 5 ȳ = 3 5 = 3 5, since the totl mss M is the sme The center of mss is shown in figure 3 EXAMPLE 4 Find the center of mss of thin, uniform plte whose shpe is the region between y = cosx nd the x-xis between x = π/ nd x = π/ It is cler 8 Chpter More Applictions of Integrtion tht x =, but for prctice let s compute it nywy We will need the totl mss, so we compute it first: π/ M = cosxdx = sinx π/ = π/ π/ The moment round the y-xis is π/ M y = xcosxdx = cosx+xsinx π/ π/ π/ nd the moment round the x-xis is M x = y rccosydy = y rccosy y y + rcsiny = π 4 Thus Exercises x =, ȳ = π A bem meters long hs density σ(x) = x t distnce x from the left end of the bem Find the center of mss x A bem meters long hs density σ(x) = sin(πx/) t distnce x from the left end of the bem Find the center of mss x 3 A bem 4 meters long hs density σ(x) = x 3 t distnce x from the left end of the bem Find the center of mss x 4 Verify tht xrccosxdx = x rccosx x x + rcsinx +C 5 A thin plte lies in the region between y = x nd the x-xis between x = nd x = Find the centroid 6 A thin plte fills the upper hlf of the unit circle x +y = Find the centroid 7 A thin plte lies in the region contined by y = x nd y = x Find the centroid 8 A thin plte lies in the region contined by y = 4 x nd the x-xis Find the centroid 9 A thin plte lies in the region contined by y = x /3 nd the x-xis between x = nd x = Find the centroid A thin plte lies in the region contined by x+ y = nd the xes in the first qudrnt Find the centroid A thin plte lies in the region between the circle x + y = 4 nd the circle x + y =, bove the x-xis Find the centroid A thin plte lies in the region between the circle x +y = 4 nd the circle x +y = in the first qudrnt Find the centroid 3 A thin plte lies in the region between the circle x + y = 5 nd the circle x + y = 6 bove the x-xis Find the centroid =

4 Kinetic energy; improper integrls 9 Ã Ò Ø Ò Ö Ý ÑÔÖÓÔ Ö ÒØ Ö Ð ½½º¾ Recll exmple 853 in which we computed the work required to lift n object from the surfce of the erth to some lrge distnce D wy Since F = k/x we computed r k x dx = k D + k r We noticed tht sd increses, k/d decreses to zero so tht the mount of work increses to k/r More precisely, k lim dx = lim D x k D D + k = k r r r We might resonbly describe this clcultion s computing the mount of work required to lift the object to infinity, nd bbrevite the limit s lim D r k x dx = r k x dx Such n integrl, with limit of infinity, is clled n improper integrl This is bit unfortunte, since it s not relly improper to do this, nor is it relly n integrl it is n bbrevition for the limit of prticulr sort of integrl Nevertheless, we re stuck with the term, nd the opertion itself is perfectly legitimte It my t first seem odd tht finite mount of work is sufficient to lift n object to infinity, but sometimes surprising things re nevertheless true, nd this is such cse If the vlue of n improper integrl is finite number, s in this exmple, we sy tht the integrl converges, nd if not we sy tht the integrl diverges Here s nother wy, perhps even more surprising, to interpret this clcultion We know tht one interprettion of x dx is the re under y = /x from x = to x = D Of course, s D increses this re increses But since x dx = D +, while the re increses, it never exceeds, tht is dx = x The re of the infinite region under y = /x from x = to infinity is finite 3 Chpter More Applictions of Integrtion Consider slightly different sort of improper integrl: xe x dx There re two wys we might try to compute this First, we could brek it up into two more fmilir integrls: xe x dx = xe x dx+ xe x dx Now we do these s before: nd so Alterntely, we might try xe x dx = lim xe x dx = lim xe x dx = lim xe x dx = lim D D D e x D e x D D xe x dx = + = D e x D D =, =, = lim D e D + e D So we get the sme nswer either wy This does not lwys hppen; sometimes the second pproch gives finite number, while the first pproch does not; the exercises provide exmples In generl, we interpret the integrl both integrls f(x)dx nd = f(x) dx ccording to the first method: f(x)dx must converge for the originl integrl to converge The second pproch does turn out to be useful; when lim nd L is finite, then L is clled the Cuchy Principl Vlue of D D f(x) dx Here s more concrete ppliction of these ides We know tht in generl W = x x F dx f(x)dx = L, is the work done ginst the force F in moving from x to x In the cse tht F is the force of grvity exerted by the erth, it is customry to mke F < since the force is

5 Kinetic energy; improper integrls 3 downwrd This mkes the work W negtive when it should be positive, so typiclly the work in this cse is defined s W = x x F dx Also, by Newton s Lw, F = m(t) This mens tht W = x x m(t) dx Unfortuntely this integrl is bit problemtic: (t) is in terms of t, while the limits nd the dx re in terms of x But x nd t re certinly relted here: x = x(t) is the function tht gives the position of the object t time t, so v = v(t) = dx/dt = x (t) is its velocity nd (t) = v (t) = x (t) We cn use v = x (t) s substitution to convert the integrl from dx to dv in the usul wy, with bit of cleverness long the wy: Substituting in the integrl: W = x x dv = x (t)dt = (t)dt = (t) dt dx dx dx dv = (t)dx dt vdv = (t)dx m(t)dx = v v mvdv = mv v v = mv + mv You my recll seeing the expression mv / in physics course it is clled the kinetic energy of the object We hve shown here tht the work done in moving the object from one plce to nother is the sme s the chnge in kinetic energy We know tht the work required to move n object from the surfce of the erth to infinity is W = r k r dr = k r At the surfce of the erth the ccelertion due to grvity is pproximtely 98 meters per second squred, so the force on n object of mss m is F = 98m The rdius of the erth is pproximtely 6378 kilometers or 6378 meters Since the force due to grvity obeys n inverse squre lw, F = k/r nd 98m = k/6378, k = m nd W = 65538m 3 Chpter More Applictions of Integrtion Now suppose tht the initil velocity of the object, v, is just enough to get it to infinity, tht is, just enough so tht the object never slows to stop, but so tht its speed decreses to zero, ie, so tht v = Then so 65538m = W = mv + mv = mv v = meters per second, or bout 45 kilometers per hour This speed is clled the escpe velocity Notice tht the mss of the object, m, cnceled out t the lst step; the escpe velocity is the sme for ll objects Of course, it tkes considerbly more energy to get lrge object up to 45 kph thn smll one, so it is certinly more difficult to get lrge object into deep spce thn smll one Also, note tht while we hve computed the escpe velocity for the erth, this speed would not in fct get n object to infinity becuse of the lrge mss in our neighborhood clled the sun Escpe velocity for the sun strting t the distnce of the erth from the sun is nerly 4 times the escpe velocity we hve clculted Exercises Is the re under y = /x from to infinity finite or infinite? If finite, compute the re Is the re under y = /x 3 from to infinity finite or infinite? If finite, compute the re 3 Does 4 Does 5 Does 6 / x +x dx converge or diverge? If it converges, find the vlue / xdx converge or diverge? If it converges, find the vlue e x dx converge or diverge? If it converges, find the vlue (x ) 3 dx is n improper integrl of slightly different sort Express it s limit nd determine whether it converges or diverges; if it converges, find the vlue 7 Does 8 Does 9 Does Does π/ / xdx converge or diverge? If it converges, find the vlue sec xdx converge or diverge? If it converges, find the vlue x dx converge or diverge? If it converges, find the vlue 4+x6 xdx converge or diverge? If it converges, find the vlue Also find the Cuchy Principl Vlue, if it exists Does sinxdx converge or diverge? If it converges, find the vlue Also find the Cuchy Principl Vlue, if it exists

6 3 Probbility 33 Does cosxdx converge or diverge? If it converges, find the vlue Also find the Cuchy Principl Vlue, if it exists 3 Suppose the curve y = /x is rotted round the x-xis generting sort of funnel or horn shpe, clled Gbriel s horn or Toricelli s trumpet Is the volume of this funnel from x = to infinity finite or infinite? If finite, compute the volume 4 An officilly snctioned bsebll must be between 4 nd 49 grms How much work, in Newton-meters, does it tke to throw bll t 8 miles per hour? At 9 mph? At 9 mph? (According to the Guinness Book of World Records, t The gretest relibly recorded speed t which bsebll hs been pitched is 9 mph by Lynn Noln Ryn (Cliforni Angels) t Anheim Stdium in Cliforni on August, 974 ) ÈÖÓ Ð ØÝ You perhps hve t lest rudimentry understnding of discrete probbility, which mesures the likelihood of n event when there re finite number of possibilities For exmple, when n ordinry six-sided die is rolled, the probbility of getting ny prticulr number is /6 In generl, the probbility of n event is the number of wys the event cn hppen divided by the number of wys tht nything cn hppen For slightly more complicted exmple, consider the cse of two six-sided dice The dice re physiclly distinct, which mens tht rolling 5 is different thn rolling 5 ; ech is n eqully likely event out of totl of 36 wys the dice cn lnd, so ech hs probbility of /36 Most interesting events re not so simple More interesting is the probbility of rolling certin sum out of the possibilities through It is clerly not true tht ll sums re eqully likely: the only wy to roll is to roll, while there re mny wys to roll 7 Becuse the number of possibilities is quite smll, nd becuse pttern quickly becomes evident, it is esy to see tht the probbilities of the vrious sums re: ½½º P() = P() = /36 P(3) = P() = /36 P(4) = P() = 3/36 P(5) = P(9) = 4/36 P(6) = P(8) = 5/36 P(7) = 6/36 Here we use P(n) to men the probbility of rolling n n Since we hve correctly ccounted for ll possibilities, the sum of ll these probbilities is 36/36 = ; the probbility tht the sum is one of through is, becuse there re no other possibilities 34 Chpter More Applictions of Integrtion The study of probbility is concerned with more difficult questions s well; for exmple, suppose the two dice re rolled mny times On the verge, wht sum will come up? In the lnguge of probbility, this verge is clled the expected vlue of the sum This is t first little misleding, s it does not tell us wht to expect when the two dice re rolled, but wht we expect the long term verge will be Suppose tht two dice re rolled 36 million times Bsed on the probbilities, we would expect bout million rolls to be, bout million to be 3, nd so on, with roll of 7 topping the list t bout 6 million The sum of ll rolls would be million times plus million times 3, nd so on, nd dividing by 36 million we would get the verge: x = ( 6 +3( 6 )+ +7(6 6 )+ + 6 ) 36 6 = = P()+3P(3)+ +7P(7)+ +P() = ip(i) = 7 i= There is nothing specil bout the 36 million in this clcultion No mtter wht the number of rolls, once we simplify the verge, we get the sme ip(i) While the ctul verge vlue of lrge number of rolls will not be exctly 7, the verge should be close to 7 when the number of rolls is lrge Turning this round, if the verge is not close to 7, we should suspect tht the dice re not fir A vrible, sy X, tht cn tke certin vlues, ech with corresponding probbility, is clled rndom vrible; in the exmple bove, the rndom vrible ws the sum of the two dice If the possible vlues for X re x, x,,x n, then the expected vlue of the n rndom vrible is E(X) = x i P(x i ) The expected vlue is lso clled the men i= When the number of possible vlues for X is finite, we sy tht X is discrete rndom vrible In mny pplictions of probbility, the number of possible vlues of rndom vrible is very lrge, perhps even infinite To del with the infinite cse we need different pproch, nd since there is sum involved, it should not be wholly surprising tht integrtion turns out to be useful tool It then turns out tht even when the number of possibilities is lrge but finite, it is frequently esier to pretend tht the number is infinite Suppose, for exmple, tht drt is thrown t drt bord Since the drt bord consists of finite number of toms, there re in some sense only finite number of plces for the drt to lnd, but it is esier to explore the probbilities involved by pretending tht the drt cn lnd on ny point in the usul x-y plne i=

7 3 Probbility 35 DEFINITION 3 Let f : R R be function If f(x) for every x nd f(x)dx = then f is probbility density function We ssocite probbility density function with rndom vrible X by stipulting tht the probbility tht X is between nd b is b f(x) dx Becuse of the requirement tht the integrl from to be, ll probbilities re less thn or equl to, nd the probbility tht X tkes on some vlue between nd is, s it should be EXAMPLE 3 Consider gin the two dice exmple; we cn view it in wy tht more resembles the probbility density function pproch Consider rndom vrible X tht tkes on ny rel vlue with probbilities given by the probbility density function in figure 3 The function f consists of just the top edges of the rectngles, with verticl sides drwn for clrity; the function is zero below 5 nd bove 5 The re of ech rectngle is the probbility of rolling the sum in the middle of the bottom of the rectngle, or P(n) = The probbility of rolling 4, 5, or 6 is P(n) = n+/ n / 3/ 7/ f(x) dx f(x) dx Of course, we could lso compute probbilities tht don t mke sense in the context of the dice, such s the probbility tht X is between 4 nd 58 6/36 36 Chpter More Applictions of Integrtion The function x F(x) = P(X x) = f(t)dt is clled the cumultive distribution function or simply (probbility) distribution EXAMPLE 33 Suppose tht < b nd { f(x) = if x b b otherwise Then f(x) is the uniform probbility density function on [, b] nd the corresponding distribution is the uniform distribution on [, b] EXAMPLE 34 Consider the function f(x) = e x / Wht cn we sy bout e x / dx? We cnnot find n ntiderivtive of f, but we cn see tht this integrl is some finite number Notice tht < f(x) = e x / e x/ for x > This implies tht the re under e x / is less thn the re under e x/, over the intervl [, ) It is esy to compute the ltter re, nmely so e x/ dx = e, e x / dx is some finite number smller thn / e Becuse f is symmetric round the y-xis, 5/36 4/36 3/36 /36 / Figure 3 A probbility density function for two dice This mens tht e x / dx = e x / dx = e x / dx e x / dx+ e x / dx+ for some finite positive number A Now if we let g(x) = f(x)/a, g(x)dx = A e x / dx = A A =, e x / dx = A so g is probbility density function It turns out to be very useful, nd is clled the stndrd norml probbility density function or more informlly the bell curve,

8 3 Probbility Chpter More Applictions of Integrtion Figure 3 The bell curve giving rise to the stndrd norml distribution See figure 3 for the grph of the bell curve We hve shown tht A issome finitenumber without computing it; we cnnot compute it with the techniques we hve vilble By using some techniques from multivrible clculus, it cn be shown tht A = π EXAMPLE 35 The exponentil distribution hs probbility density function where c is positive constnt { x < f(x) = ce cx x The men or expected vlue of rndom vrible is quite useful, s hinted t in our discussion of dice Recll tht the men for discrete rndom vrible is E(X) = n x i P(x i ) In the more generl context we use n integrl in plce of the sum i= DEFINITION 36 The men of rndom vrible X with probbility density function f is µ = E(X) = xf(x) dx, provided the integrl converges When the men exists it is unique, since it is the result of n explicit clcultion The men does not lwys exist The men might look fmilir; it is essentilly identicl to the center of mss of one-dimensionl bem, s discussed in section The probbility density function f plys the role of the physicl density function, but now the bem hs infinite length If we consider only finite portion of the bem, sy between nd b, then the center of mss is x = b b xf(x) dx f(x) dx If we extend the bem to infinity, we get becuse x = xf(x) dx f(x) dx = xf(x)dx = E(X), f(x)dx = In the center of mss interprettion, this integrl is the totl mss of the bem, which is lwys when f is probbility density function EXAMPLE 37 The men of the stndrd norml distribution is We compute the two hlves: nd x e x / π dx = lim x e x / π dx = lim The sum of these is, which is the men x e x / π dx / D e x π D / D D e x π = π = π While the men is very useful, it typiclly is not enough informtion to properly evlute sitution For exmple, suppose we could mnufcture n -sided die, with the fces numbered through so tht ech fce is eqully likely to be down when the die is rolled The vlue of roll is the vlue on this lower fce Rolling the die gives the sme rnge of vlues s rolling two ordinry dice, but now ech vlue occurs with probbility / The expected vlue of roll is = 7 The men does not distinguish the two cses, though of course they re quite different If f is probbility density function for rndom vrible X, with men µ, we would like to mesure how fr typicl vlue of X is from µ One wy to mesure this distnce

9 3 Probbility 39 is (X µ) ; we squre the difference so s to mesure ll distnces s positive To get the typicl such squred distnce, we compute the men For two dice, for exmple, we get ( 7) 36 +(3 7) (7 7) ( 7) 36 +( 7) 36 = Becuse we squred the differences this does not directly mesure the typicl distnce we seek; if we tke the squre root of this we do get such mesure, 35/36 4 Doing the computtion for the strnge -sided die we get ( 7) +(3 7) + +(7 7) + ( 7) +( 7) =, with squre root pproximtely 36 Compring 4 to 36 tells us tht the two-dice rolls clump somewht more closely ner 7 thn the rolls of the weird die, which of course we lredy knew becuse these exmples re quite simple To perform the sme computtion for probbility density function the sum is replced by n integrl, just s in the computtion of the men The expected vlue of the squred distnces is V(X) = (x µ) f(x)dx, clled the vrince The squre root of the vrince is the stndrd devition, denoted σ EXAMPLE 38 We compute the stndrd devition of the stndrd norml distrubution The vrince is x e x / dx π To compute the ntiderivtive, use integrtion by prts, with u = x nd dv = xe x / dx This gives x e x / dx = xe x / + e x / dx We cnnot do the new integrl, but we know its vlue when the limits re to, from our discussion of the stndrd norml distribution Thus π x e x / dx = xe x / π The stndrd devition is then = + π e x / dx = + π π = 4 Chpter More Applictions of Integrtion EXAMPLE 39 Here is simple exmple showing how these ides cn be useful Suppose it is known tht, in the long run, out of every computer memory chips produced by certin mnufcturing plnt is defective when the mnufcturing process is running correctly Suppose chips re selected t rndom nd 5 of them re defective This is more thn the expected number (), but is it so mny tht we should suspect tht something hs gone wrong in the mnufcturing process? We re interested in the probbility tht vrious numbers of defective chips rise; the probbility distribution is discrete: there cn only be whole number of defective chips But (under resonble ssumptions) the distribution is very close to norml distribution, nmely this one: f(x) = ( ) (x ) exp, π ()(99) ()()(99) which is pictured in figure 33 (recll tht exp(x) = e x ) Figure 33 Norml density function for the defective chips exmple Now how do we mesure how unlikely it is tht under norml circumstnces we would see 5 defective chips? We cn t compute the probbility of exctly 5 defective chips, s this would be 5 5 f(x)dx = We could compute f(x)dx 36; this mens there is only 36% chnce tht the number of defective chips is 5 (We cnnot compute these integrls exctly; computer softwre hs been used to pproximte the integrl vlues in this discussion) But this is misleding: 5 95 f(x) dx 6, which is lrger, certinly, but still smll, even for the most likely outcome The most useful question, in most circumstnces, is this: how likely is it tht the number of defective chips is fr from the men? For exmple, how likely, or unlikely, is it tht the number of defective chips is different by 5 or more from the expected vlue of? This is the probbility tht the number of defective chips is less thn 5 or lrger thn 5, nmely 5 f(x)dx+ 5 f(x)dx So there is n % chnce tht this hppens not lrge, but not tiny Hence the 5 defective chips does not pper to be cuse for lrm: bout one time in nine we would

10 3 Probbility 4 expect to see the number of defective chips 5 or more wy from the expected How bout? Here we compute f(x)dx+ f(x)dx 5 So there is only 5% chnce tht the number of defective chips is more thn wy from the men; this would typiclly be interpreted s too suspicious to ignore it shouldn t hppen if the process is running normlly The big question, of course, is wht level of improbbility should trigger concern? It depends to some degree on the ppliction, nd in prticulr on the consequences of getting it wrong in one direction or the other If we re wrong, do we lose little money? A lot of money? Do people die? In generl, the stndrd choices re 5% nd % So wht we should do is find the number of defective chips tht hs only, let us sy, % chnce of occurring under norml circumstnces, nd use tht s the relevnt number In other words, we wnt to know when r f(x)dx+ f(x)dx < +r A bit of tril nd error shows tht with r = 8 the vlue is bout, nd with r = 9 it is bout 4, so if the number of defective chips is 9 or more, or or fewer, we should look for problems If the number is high, we worry tht the mnufcturing process hs problem, or conceivbly tht the process tht tests for defective chips is not working correctly nd is flgging good chips s defective If the number is too low, we suspect tht the testing procedure is broken, nd is not detecting defective chips Exercises 3 Verify tht e x/ dx = / e Show tht the function in exmple 35 is probbility density function Compute the men nd stndrd devition 3 Compute the men nd stndrd devition of the uniform distribution on [, b] (See exmple 33) 4 Wht is the expected vlue of one roll of fir six-sided die? 5 Wht is the expected sum of one roll of three fir six-sided dice? 6 Let µ nd σ be rel numbers with σ > Show tht N(x) = e (x µ) σ πσ is probbility density function You will not be ble to compute this integrl directly; use substitution to convert the integrl into the one from exmple 34 The function N 4 Chpter More Applictions of Integrtion is the probbility density function of the norml distribution with men µ nd stndrd devition σ Show tht the men of the norml distribution is µ nd the stndrd devition is σ 7 Let f(x) = { x x x < Show tht f is probbility density function, nd tht the distribution hs no men 8 Let { x x f(x) = < x otherwise Show tht f(x)dx = Is f probbility density function? Justify your nswer 9 If you hve ccess to pproprite softwre, find r so tht r f(x)dx+ +r f(x)dx 5, using the function of exmple 39 Discuss the impct of using this new vlue of r to decide whether to investigte the chip mnufcturing process Ö Ä Ò Ø ½½º Here is nother geometric ppliction of the integrl: find the length of portion of curve As usul, we need to think bout how we might pproximte the length, nd turn the pproximtion into n integrl We lredy know how to compute one simple rc length, tht of line segment If the endpoints re P (x,y ) nd P (x,y ) then the length of the segment is the distnce between the points, (x x ) +(y y ), from the Pythgoren theorem, s illustrted in figure 4 (x x ) + (y y ) (x,y ) x x (x,y ) y y Figure 4 The length of line segment

11 4 Arc Length Chpter More Applictions of Integrtion EXAMPLE 4 Let f(x) = r x, the upper hlf circle of rdius r The length of this curve is hlf the circumference, nmely πr Let s compute this with the rc length formul The derivtive f is x/ r x so the integrl is r r + x r x dx = r r r r x dx = r r r r x dx Figure 4 Approximting rc length with line segments Now if the grph of f is nice (sy, differentible) it ppers tht we cn pproximte the length of portion of the curve with line segments, nd tht s the number of segments increses, nd their lengths decrese, the sum of the lengths of the line segments will pproch the true rc length; see figure 4 Now we need to write formul for the sum of the lengths of the line segments, in form tht we know becomes n integrl in the limit So we suppose we hve divided the intervl [,b] into n subintervls s usul, ech with length x = (b )/n, nd endpoints = x, x, x,, x n = b The length of typicl line segment, joining (x i,f(x i )) to (x i+,f(x i+ )), is ( x) +(f(x i+ ) f(x i )) By the Men Vlue Theorem (65), there is number t i in (x i,x i+ ) such tht f (t i ) x = f(x i+ ) f(x i ), so the length of the line segment cn be written s ( x) +(f (t i )) x = +(f (t i )) x The rc length is then n b lim +(f (t i )) x = +(f (x)) dx n i= Note tht the sum looks bit different thn others we hve encountered, becuse the pproximtioncontinst i instedofnx i Inthepst wehvelwysused left endpoints (nmely, x i ) to get representtive vlue of f on [x i,x i+ ]; now we re using different point, but the principle is the sme To summrize, to compute the length of curve on the intervl [,b], we compute the integrl b +(f (x)) dx Unfortuntely, integrls of this form re typiclly difficult or impossible to compute exctly, becuse usully none of our methods for finding ntiderivtives will work In prctice this mens tht the integrl will usully hve to be pproximted Using trigonometric substitution, we find the ntiderivtive, nmely rcsin(x/r) Notice tht the integrl is improper t both endpoints, s the function /(r x ) is undefined when x = ±r So we need to compute lim D r + D r dx+ lim x D r r x dx This is not difficult, nd hs vlue π, so the originl integrl, with the extr r in front, hs vlue πr s expected Exercises 4 Find the rc length of f(x) = x 3/ on [,] Find the rc length of f(x) = x /8 lnx on [,] 3 Find the rc length of f(x) = (/3)(x +) 3/ on the intervl [,] 4 Find the rc length of f(x) = ln(sinx) on the intervl [π/4,π/3] 5 Let > Show tht the length of y = coshx on [,] is equl to 6 Find the rc length of f(x) = coshx on [,ln] coshxdx 7 Set up the integrl to find the rc length of sinx on the intervl [,π]; do not evlute the integrl If you hve ccess to pproprite softwre, pproximte the vlue of the integrl 8 Set up the integrl to find the rc length of y = xe x on the intervl [,3]; do not evlute the integrl If you hve ccess to pproprite softwre, pproximte the vlue of the integrl 9 Find the rc length of y = e x on the intervl [,] (This cn be done exctly; it is bit tricky nd bit long) ËÙÖ Ö ½½º Another geometric question tht rises nturlly is: Wht is the surfce re of volume? For exmple, wht is the surfce re of sphere? More dvnced techniques re required to pproch this question in generl, but we cn compute the res of some volumes generted by revolution

12 5 Surfce Are Chpter More Applictions of Integrtion As usul, the question is: how might we pproximte the surfce re? For surfce obtined by rotting curve round n xis, we cn tke polygonl pproximtion to the curve, s in the lst section, nd rotte it round the sme xis This gives surfce composed of mny truncted cones; truncted cone is clled frustum of cone Figure 5 illustrtes this pproximtion h πr πr/h r h similr tringles, Figure 5 The re of cone h r = h +h r With bit of lgebr this becomes (r r )h = r h; substitution into the re gives Figure 5 Approximting surfce (left) by portions of cones (right) So we need to be ble to compute the re of frustum of cone Since the frustum cn be formed by removing smll cone from the top of lrger one, we cn compute the desired re if we know the surfce re of cone Suppose right circulr cone hs bse rdius r nd slnt height h If we cut the cone from the vertex to the bse circle nd fltten it out, we obtin sector of circle with rdius h nd rc length πr, s in figure 5 The ngle t the center, in rdins, is then πr/h, nd the re of the cone is equl to the re of the sector of the circle Let A be the re of the sector; since the re of the entire circle is πh, we hve π((r r )h +r h) = π(r h+r h) = πh(r +r ) = π r +r h = πrh The finl form is prticulrly esy to remember, with r equl to the verge of r nd r, s it is lso the formul for the re of cylinder (Think of cylinder of rdius r nd height h s the frustum of cone of infinite height) r h A πh = πr/h π A = πrh r h Now suppose we hve frustum of cone with slnt height h nd rdii r nd r, s in figure 53 The re of the entire cone is πr (h +h), nd the re of the smll cone is πr h ; thus, the re of the frustum is πr (h +h) πr h = π((r r )h +r h) By Figure 53 The re of frustum Now we re redy to pproximte the re of surfce of revolution On one subintervl, the sitution is s shown in figure 54 When the line joining two points on the

13 5 Surfce Are 47 curve is rotted round the x-xis, it forms frustum of cone The re is πrh = π f(x i)+f(x i+ ) +(f (t i )) x Here +(f (t i )) x is the length of the line segment, s we found in the previous section Assuming f is continuous function, there must be some x i in [x i,x i+ ] such tht (f(x i )+f(x i+ ))/ = f(x i ), so the pproximtion for the surfce re is n πf(x i ) +(f (t i )) x i= This is not quite the sort of sum we hve seen before, s it contins two different vlues in the intervl [x i,x i+ ], nmely x i nd t i Nevertheless, using more dvnced techniques thn we hve vilble here, it turns out tht n lim πf(x i ) +(f (t i )) x = n i= b πf(x) +(f (x)) dx is the surfce re we seek (Roughly speking, this is becuse while x i nd t i re distinct vlues in [x i,x i+ ], they get closer nd closer to ech other s the length of the intervl shrinks) (x i,f(x i )) x i x i x i+ (x i+,f(x i+ )) Figure 54 One subintervl EXAMPLE 5 We compute thesurfce re ofsphere of rdius r The sphere cn be obtined by rotting the grph of f(x) = r x bout the x-xis The derivtive 48 Chpter More Applictions of Integrtion f is x/ r x, so the surfce re is given by r A = π r x + x r r x dx r r = π r x r x dx = π r r r rdx = πr r r dx = 4πr If the curve is rotted round the y xis, the formul is nerly identicl, becuse the length of the line segment we use to pproximte portion of the curve doesn t chnge Insted of the rdius f(x i ), we use the new rdius x i = (x i +x i+ )/, nd the surfce re integrl becomes b πx +(f (x)) dx EXAMPLE 5 Compute the re of the surfce formed when f(x) = x between nd is rotted round the y-xis We compute f (x) = x, nd then by simple substitution Exercises 5 π x +4x dx = π 6 (73/ ), Compute the re of the surfce formed when f(x) = x between nd is rotted round the x-xis Compute the surfce re of exmple 5 by rotting f(x) = x round the x-xis 3 Compute the re of the surfce formed when f(x) = x 3 between nd 3 is rotted round the x-xis 4 Compute the re of the surfce formed when f(x) = +cosh(x) between nd is rotted round the x-xis 5 Consider the surfce obtined by rotting the grph of f(x) = /x, x, round the x-xis This surfce is clled Gbriel s horn or Toricelli s trumpet In exercise 3 in section we sw tht Gbriel s horn hs finite volume Show tht Gbriel s horn hs infinite surfce re 6 Consider the circle (x ) + y = Sketch the surfce obtined by rotting this circle bout the y-xis (The surfce is clled torus) Wht is the surfce re?

14 5 Surfce Are 49 7 Consider the ellipse with eqution x /4+y = If the ellipse is rotted round the x-xis it forms n ellipsoid Compute the surfce re 8 Generlize the preceding result: rotte the ellipse given by x / + y /b = bout the x-xis nd find the surfce re of the resulting ellipsoid You should consider two cses, when > b nd when < b Compre to the re of sphere