Applications of Integration

Transcription

1 9 Chpter 9 Applictions of Integrtion 9 Applictions of Integrtion Ö ØÛ Ò ÙÖÚ º½ We hve seen how integrtion cn be used to find n re between curve nd the x-xis With very little chnge we cn find some res between curves; indeed, the re between curve nd the x-xis my be interpreted s the re between the curve nd second curve with eqution y = In the simplest of cses, the ide is quite esy to understnd EXAMPLE 9 Find the re below f(x) = x +4x+3 nd bove g(x) = x 3 + 7x x+5 over the intervl x In figure 9 we show the two curves together, with the desired re shded, then f lone with the re under f shded, nd then g lone with the re under g shded It is cler from the figure tht the re we wnt is the re under f minus the re under g, which is to sy f(x)dx g(x)dx = f(x) g(x)dx It doesn t mtter whether we compute the two integrls on the left nd then subtrct or compute the single integrl on the right In this cse, the ltter is perhps bit esier: f(x) g(x)dx = = x +4x+3 ( x 3 +7x x+5)dx x 3 8x +4x dx = x4 4 8x3 3 +7x x = ( ) = = 49 It is worth exmining this problem bit more We hve seen one wy to look t it, by viewing the desired re s big re minus smll re, which leds nturlly to the difference between two integrls But it is instructive to consider how we might find the desired re directly We cn pproximte the re by dividing the re into thin sections nd pproximting the re of ech section by rectngle, s indicted in figure 9 The re of typicl rectngle is x(f(x i ) g(x i )), so the totl re is pproximtely y y y n (f(x i ) g(x i )) x i= This is exctly the sort of sum tht turns into n integrl in the limit, nmely the integrl f(x) g(x)dx x x Figure 9 Are between curves s difference of res x Of course, this is the integrl we ctully computed bove, but we hve now rrived t it directly rther thn s modifiction of the difference between two other integrls In tht exmple it relly doesn t mtter which pproch we tke, but in some cses this second pproch is better 89

2 5 3 9 Are between curves 9 Figure 9 Approximting re between curves with rectngles EXAMPLE 9 Find the re below f(x) = x +4x+ nd bove g(x) = x 3 + 7x x+3 over the intervl x ; these re the sme curves s before but lowered by In figure 93 we show the two curves together Note tht the lower curve now dips below the x-xis This mkes it somewht tricky to view the desired re s big re minus smller re, but it is just s esy s before to think of pproximting the re by rectngles The height of typicl rectngle will still be f(x i ) g(x i ), even if g(x i ) is negtive Thus the re is x +4x+ ( x 3 +7x x+3)dx = x 3 8x +4x dx This is of course the sme integrl s before, becuse the region between the curves is identicl to the former region it hs just been moved down by y 9 Chpter 9 Applictions of Integrtion re in the usul sense, s necessrily positive quntity Since the two curves cross, we need to compute two res nd dd them First we find the intersection point of the curves: x +4x = x 6x+5 = x x+5 x = ± 4 4 = 5± 5 The intersection point we wnt is x = = (5 5)/ Then the totl re is x 6x+5 ( x +4x)dx+ fter bit of simplifiction y = x +4x (x 6x+5)dx x x+5dx+ x +x 5dx = x3 3 5x +5x + x3 3 +5x 5x = , 5 3 x x Figure 93 Are between curves Figure 94 Are between curves tht cross EXAMPLE 93 Find the re between f(x) = x +4x nd g(x) = x 6x+5 over the intervl x ; the curves re shown in figure 94 Generlly we should interpret EXAMPLE 94 Find the re between f(x) = x + 4x nd g(x) = x 6x + 5; the curves re shown in figure 95 Here we re not given specific intervl, so it must

3 9 Are between curves 93 be the cse tht there is nturl region involved Since the curves re both prbols, the only resonble interprettion is the region between the two intersection points, which we found in the previous exmple: 5± 5 If we let = (5 5)/ nd b = (5+ 5)/, the totl re is b fter bit of simplifiction x +4x (x 6x+5)dx = 5 b x +x 5dx = x3 3 +5x 5x = 5 5 b 94 Chpter 9 Applictions of Integrtion y = x 3/ nd y = x /3 y = x x nd y = x The following three exercises expnd on the geometric interprettion of the hyperbolic functions Refer to section 4 nd prticulrly to figure 4 nd exercise 6 in section 4 x 3 Compute dx using the substitution u = rccoshx, or x = coshu; use exercise 6 in section 4 4 Fix t > Sketch the region R in the right hlf plne bounded by the curves y = xtnht, y = xtnht, nd x y = Note well: t is fixed, the plne is the x-y plne 5 Prove tht the re of R is t We next recll generl principle tht will lter be pplied to distnce-velocity-ccelertion º¾ Ø Ò Î ÐÓ ØÝ Ð Ö Ø ÓÒ problems, mong other things If F(u) is n nti-derivtive of f(u), then b f(u)du = F(b) F() Suppose tht we wnt to let the upper limit of integrtion vry, ie, we replce b by some vrible x We think of s fixed strting vlue x In this new nottion the lst eqution (fter dding F() to both sides) becomes: F(x) = F(x )+ x x f(u) du 5 Figure 95 Are bounded by two curves Exercises 9 Find the re bounded by the curves y = x 4 x nd y = x (the prt to the right of the y-xis) x = y 3 nd x = y 3 x = y nd y = x 4 x = 3y y nd x+y = 3 5 y = cos(πx/) nd y = x (in the first qudrnt) 6 y = sin(πx/3) nd y = x (in the first qudrnt) 7 y = x nd y = x 8 y = x nd y = x+, x 4 9 x = nd x = 5 y y = sinxcosx nd y = sinx, x π (Here u is the vribleof integrtion, clled dummy vrible, since it is not the vrible in the function F(x) In generl, it is not good ide to use the sme letter s vrible of integrtion nd s limit of integrtion Tht is, x x f(x)dx is bd nottion, nd cn led to errors nd confusion) An importnt ppliction of this principle occurs when we re interested in the position of n object t time t (sy, on the x-xis) nd we know its position t time t Let s(t) denote the position of the object t time t (its distnce from reference point, such s the origin on the x-xis) Then the net chnge in position between t nd t is s(t) s(t ) Since s(t) is n nti-derivtive of the velocity function v(t), we cn write s(t) = s(t )+ t t v(u)du Similrly, since the velocity is n nti-derivtive of the ccelertion function (t), we hve v(t) = v(t )+ t t (u)du

4 9 Distnce, Velocity, Accelertion 95 EXAMPLE 9 Suppose n object is cted upon by constnt force F Find v(t) nd s(t) By Newton s lw F = m, so the ccelertion is F/m, where m is the mss of the object Then we first hve t v(t) = v(t )+ t F m du = v + F m u using the usul convention v = v(t ) Then s(t) = s(t )+ t t t t = v + F m (t t ), ( v + F ) m (u t ) du = s + (v u+ F m (u t ) ) = s +v (t t )+ F m (t t ) For instnce, when F/m = g is the constnt of grvittionl ccelertion, then this is the flling body formul (if we neglect ir resistnce) fmilir from elementry physics: or in the common cse tht t =, s +v (t t ) g (t t ), s +v t g t Recll tht the integrl of the velocity function gives the net distnce trveled, tht is, the displcement If you wnt to know the totl distnce trveled, you must find out where the velocity function crosses the t-xis, integrte seprtely over the time intervls when v(t) is positive nd when v(t) is negtive, nd dd up the bsolute vlues of the different integrls For exmple, if n object is thrown stright upwrd t 96 m/sec, its velocity function is v(t) = 98t + 96, using g = 98 m/sec for the force of grvity This is stright line which is positive for t < nd negtive for t > The net distnce trveled in the first 4 seconds is thus 4 ( 98t+96)dt =, while the totl distnce trveled in the first 4 seconds is 4 ( 98t+96)dt+ ( 98t + 96)dt = = 39 meters, 96 meters up nd 96 meters down t t 96 Chpter 9 Applictions of Integrtion EXAMPLE 9 The ccelertion of n object is given by (t) = cos(πt), nd its velocity t time t = is /(π) Find both the net nd the totl distnce trveled in the first 5 seconds We compute v(t) = v()+ t The net distnce trveled is then s(3/) s() = cos(πu)du = π + t π sin(πu) = ( π +sin(πt)) 3/ = π π ( ) +sin(πt) dt ( t ) 3/ π cos(πt) = 3 4π + 34 meters π To find the totl distnce trveled, we need to know when (5+sin(πt)) is positive nd when it is negtive This function is when sin(πt) is 5, ie, when πt = 7π/6, π/6, etc The vlue πt = 7π/6, ie, t = 7/6, is the only vlue in the rnge t 5 Since v(t) > for t < 7/6 nd v(t) < for t > 7/6, the totl distnce trveled is 7/6 Exercises 9 ( ) 3/ ( ) π +sin(πt) dt+ 7/6 π +sin(πt) dt = ( 7 π + π cos(7π/6)+ ) + 3 π π π cos(7π/6) ( = 7 π + ) 3 π π π π 49 meters For ech velocity function find both the net distnce nd the totl distnce trveled during the indicted time intervl (grph v(t) to determine when it s positive nd when it s negtive): v = cos(πt), t 5 v = 98t+49, t 3 v = 3(t 3)(t ), t 5 4 v = sin(πt/3) t, t 5 An object is shot upwrds from ground level with n initil velocity of meters per second; it is subject only to the force of grvity (no ir resistnce) Find its mximum ltitude nd the time t which it hits the ground 6 An object is shot upwrds from ground level with n initil velocity of 3 meters per second; it is subject only to the force of grvity (no ir resistnce) Find its mximum ltitude nd the time t which it hits the ground

5 93 Volume 97 7 An object is shot upwrds from ground level with n initil velocity of meters per second; it is subject only to the force of grvity (no ir resistnce) Find its mximum ltitude nd the time t which it hits the ground 8 An object moves long stright line with ccelertion given by (t) = cos(t), nd s() = nd v() = Findthe mximumdistnce the objecttrvels from zero, nd find its mximum speed Describe the motion of the object 9 An object moves long stright line with ccelertion given by (t) = sin(πt) Assume tht when t =, s(t) = v(t) = Find s(t), v(t), nd the mximum speed of the object Describe the motion of the object An object moves long stright line with ccelertion given by (t) = +sin(πt) Assume tht when t =, s(t) = v(t) = Find s(t) nd v(t) An object moves long stright line with ccelertion given by (t) = sin(πt) Assume tht when t =, s(t) = v(t) = Find s(t) nd v(t) ÎÓÐÙÑ We hve seen how to compute certin res by using integrtion; some volumes my lso be computed by evluting n integrl Generlly, the volumes tht we cn compute this wy hve cross-sections tht re esy to describe º 98 Chpter 9 Applictions of Integrtion the volume of the pyrmid, s shown in figure 93: on the left is cross-sectionl view, on the right is 3D view of prt of the pyrmid with some of the boxes used to pproximte the volume Ech box hs volume of the form (x i )(x i ) y Unfortuntely, there re two vribles here; fortuntely, we cn write x in terms of y: x = y/ or x i = y i / Then the totl volume is pproximtely n 4( y i /) y i= nd in the limit we get the volume s the vlue of n integrl: 4( y/) dy = ( y) dy = ( y)3 3 = = 8 3 As you my know, the volume of pyrmid is(/3)(height)(re of bse) = (/3)()(4), which grees with our nswer EXAMPLE 93 The bse of solid is the region between f(x) = x nd g(x) = x +, nd its cross-sections perpendiculr to the x-xis re equilterl tringles, s indicted in figure 93 The solid hs been truncted to show tringulr cross-section bove x = / Find the volume of the solid y i x i Figure 93 Volume of pyrmid pproximted by rectngulr prisms (AP) EXAMPLE 93 Find the volume of pyrmid with squre bse tht is meters tll nd meters on side t the bse Aswith most of our pplictions of integrtion, we begin by sking how we might pproximte the volume Since we cn esily compute the volume of rectngulr prism (tht is, box ), we will use some boxes to pproximte Figure 93 Solid with equilterl tringles s cross-sections (AP) A cross-section t vlue x i on the x-xis is tringle with bse ( x i) nd height 3( x i ), so the re of the cross-section is (bse)(height) = ( x i) 3( x i),

6 93 Volume 99 Chpter 9 Applictions of Integrtion nd the volume of thin slb is then ( x i ) 3( x i ) x Thus the totl volume is 3( x ) dx = One esy wy to get nice cross-sections is by rotting plne figure round line For exmple, in figure 933 we see plne region under curve nd between two verticl lines; then the result of rotting this round the x-xis, nd typicl circulr cross-section Figure 934 A region tht genertes cone; pproximting the volume by circulr disks (AP) Note tht we cn insted do the clcultion with generic height nd rdius: h π r h x dx = πr h h 3 3 = πr h 3, Figure 933 A solid of rottion (AP) Of course rel slice of this figure will not hve stright sides, but we cn pproximtethe volumeofthe slice by cylinder or disk withcirculr topnd bottom ndstright sides; the volume of this disk will hve the form πr x As long s we cn write r in terms of x we cn compute the volume by n integrl EXAMPLE 933 Find the volume of right circulr cone with bse rdius nd height (A right circulr cone is one with circulr bse nd with the tip of the cone directly over the center of the bse) We cn view this cone s produced by the rottion of the line y = x/ rotted bout the x-xis, s indicted in figure 934 At prticulr point on the x-xis, sy x i, the rdius of the resulting cone is the y-coordinte of the corresponding point on the line, nmely y i = x i / Thus the totl volume is pproximtely n π(x i /) dx nd the exct volume is i= π x 4 dx = π = π 3 giving us the usul formul for the volume of cone EXAMPLE 934 Find the volume of the object generted when the re between y = x nd y = x is rotted round the x-xis This solid hs hole in the middle; we cn compute the volume by subtrcting the volume of the hole from the volume enclosed by the outer surfce of the solid In figure 935 we show the region tht is rotted, the resulting solid with the front hlf cut wy, the cone tht forms the outer surfce, the horn-shped hole, nd cross-section perpendiculr to the x-xis We hve lredy computed the volume of cone; in this cse it is π/3 At prticulr vlue of x, sy x i, the cross-section of the horn is circle with rdius x i, so the volume of the horn is π(x ) dx = πx 4 dx = π 5, so the desired volume is π/3 π/5 = π/5 As with the re between curves, there is n lternte pproch tht computes the desired volume ll t once by pproximting the volume of the ctul solid We cn pproximte the volume of slice of the solid with wsher-shped volume, s indicted in figure 935 The volume of such wsher is the re of the fce times the thickness The thickness, s usul, is x, while the re of the fce is the re of the outer circle minus the re of

7 93 Volume Chpter 9 Applictions of Integrtion brek the problem into two prts nd compute two integrls: π (+ y) ( y) dy +π 4 (+ y) (y ) dy = 8 3 π π = 7 π If insted we consider typicl verticl rectngle, but still rotte round the y-xis, we get thin shell insted of thin wsher If we dd up the volume of such thin shells we will get n pproximtion to the true volume Wht is the volume of such shell? Consider the shell t x i Imgine tht we cut the shell verticlly in one plce nd unroll it into thin, flt sheet This sheet will be lmost rectngulr prism tht is x thick, f(x i ) g(x i ) tll, nd πx i wide (nmely, the circumference of the shell before it ws unrolled) The volume will then be pproximtely the volume of rectngulr prism with these dimensions: πx i (f(x i ) g(x i )) x If we dd these up nd tke the limit s usul, we get the integrl 3 πx(f(x) g(x))dx = 3 πx(x+ (x ) )dx = 7 π Not only does this ccomplish the tsk with only one integrl, the integrl is somewht esier thn those in the previous clcultion Things re not lwys so net, but it is often the cse tht one of the two methods will be simpler thn the other, so it is worth considering both before strting to do clcultions 4 4 Figure 935 Solid with hole, showing the outer cone nd the shpe to be removed to form the hole (AP) the inner circle, sy πr πr In the present exmple, t prticulr x i, the rdius R is x i nd r is x i Hence, the whole volume is ( ) x πx πx 4 3 dx = π 3 x5 ( = π 5 3 ) = π 5 5 Of course, wht we hve done here is exctly the sme clcultion s before, except we hve in effect recomputed the volume of the outer cone Suppose the region between f(x) = x+ nd g(x) = (x ) is rotted round the y-xis; see figure 936 It is possible, but inconvenient, to compute the volume of the resulting solid by the method we hve used so fr The problem is tht there re two kinds of typicl rectngles: those tht go from the line to the prbol nd those tht touch the prbol on both ends To compute the volume using this pproch, we need to Figure 936 Computing volumes with shells (AP) EXAMPLE 935 Suppose the re under y = x + between x = nd x = is rotted round the x-xis Find the volume by both methods Disk method: Shell method: π( x ) dx = 8 5 π πy ydy = 8 5 π

8 Exercises 93 Verify tht π Verify tht 3 Verify tht 4 Verify tht 3 (+ y) ( y) dy +π 4 πx(x+ (x ) )dx = 7 π π( x ) dx = 8 5 π πy ydy = 8 5 π 93 Volume 3 (+ y) (y ) = 8 3 π π = 7 π 5 Use integrtion to find the volume of the solid obtined by revolving the region bounded by x+y = nd the x nd y xes round the x-xis 6 Find the volume of the solid obtined by revolving the region bounded by y = x x nd the x-xis round the x-xis 7 Find the volume of the solid obtined by revolving the region bounded by y = sinx between x = nd x = π/, the y-xis, nd the line y = round the x-xis 8 Let S be the region of the xy-plne bounded bove by the curve x 3 y = 64, below by the line y =, on the left by the line x =, nd on the right by the line x = 4 Find the volume of the solid obtined by rotting S round () the x-xis, (b) the line y =, (c) the y-xis, (d) the line x = 9 The eqution x /9 + y /4 = describes n ellipse Find the volume of the solid obtined by rotting the ellipse round the x-xis nd lso round the y-xis These solids re clled ellipsoids; one is vguely rugby-bll shped, one is sort of flying-sucer shped, or perhps squished-bech-bll-shped Figure 937 Ellipsoids Use integrtion to compute the volume of sphere of rdius r You should of course get the well-known formul 4πr 3 /3 A hemispheric bowl of rdius r contins wter to depth h Find the volume of wter in the bowl The bse of tetrhedron ( tringulr pyrmid) of height h is n equilterl tringle of side s Its cross-sections perpendiculr to n ltitude re equilterl tringles Express its volume V s n integrl, nd find formul for V in terms of h nd s Verify tht your nswer is (/3)(re of bse)(height) 3 The bse of solid is the region between f(x) = cosx nd g(x) = cosx, π/ x π/, nd its cross-sections perpendiculr to the x-xis re squres Find the volume of the solid 4 Chpter 9 Applictions of Integrtion Ú Ö Ú ÐÙ Ó ÙÒØ ÓÒ º The verge of some finite set of vlues is fmilir concept If, for exmple, the clss scores on quiz re, 9,, 8, 7, 5, 7, 6, 3,, 7, 8, then the verge score is the sum of these numbers divided by the size of the clss: verge score = = Suppose tht between t = nd t = the speed of n object is sin(πt) Wht is the verge speed of the object over tht time? The question sounds s if it must mke sense, yet we cn t merely dd up some number of speeds nd divide, since the speed is chnging continuously over the time intervl To mke sense of verge in this context, we fll bck on the ide of pproximtion Consider the speed of the object t tenth of second intervls: sin, sin(π), sin(π), sin(3π),, sin(9π) The verge speed should be firly close to the verge of these ten speeds: 9 i= sin(πi/) 63 = 63 Of course, if we compute more speeds t more times, the verge of these speeds should be closer to the rel verge If we tke the verge of n speeds t evenly spced times, we get: n sin(πi/n) n Here the individul times re t i = i/n, so rewriting slightly we hve i= n sin(πt i ) n i= This is lmost the sort of sum tht we know turns into n integrl; wht s pprently missing is t but in fct, t = /n, the length of ech subintervl So rewriting gin: n sin(πt i ) n n = sin(πt i ) t i= Now this hs exctly the right form, so tht in the limit we get verge speed = sin(πt)dt = cos(πt) π = cos(π) + cos() = π π π It s not entirely obvious from this one simple exmple how to compute such n verge in generl Let s look t somewht more complicted cse Suppose tht the velocity i=

9 94 Averge vlue of function 5 of n object is 6t +5 feet per second Wht is the verge velocity between t = nd t = 3? Agin we set up n pproximtion to the verge: n 6t i +5, n i= where the vlues t i re evenly spced times between nd 3 Once gin we re missing t, nd this time /n is not the correct vlue Wht is t in generl? It is the length of subintervl; in this cse we tke the intervl [,3] nd divide it into n subintervls, so ech hs length (3 )/n = /n = t Now with the usul multiply nd divide by the sme thing trick we cn rewrite the sum: n n i= 6t i +5 = n 3 In the limit this becomes i= (6t i +5)3 n 3 = n (6t i +5) n = n (6t i +5) t i= 6t +5dt = = 3 3 Does this seem resonble? Let s picture it: in figure 94 is the velocity function together with the horizontl line y = 3/3 743 Certinly the height of the horizontl line looks t lest plusible for the verge height of the curve i= 6 Chpter 9 Applictions of Integrtion nd t = 3 If insted the object were to trvel t the verge speed over the sme time, it should go the sme distnce At n verge speed of 3/3 feet per second for two seconds the object would go 446/3 feet How fr does it ctully go? We know how to compute this: 3 3 v(t)dt = 6t +5dt = So now we see tht nother interprettion of the clcultion 3 6t +5dt = = 3 3 is: totl distnce trveled divided by the time in trnsit, nmely, the usul interprettion of verge speed In the cse of speed, or more properly velocity, we cn lwys interpret verge s totl (net) distnce divided by time But in the cse of different sort of quntity this interprettion does not obviously pply, while the pproximtion pproch lwys does We might interpret the sme problem geometriclly: wht is the verge height of 6x +5 on the intervl [,3]? We pproximte this in exctly the sme wy, by dding up mny smple heights nd dividing by the number of smples In the limit we get the sme result: lim n n 6x i n +5 = i= 3 6x +5dx = = We cn interpret this result in slightly different wy The re under y = 6x +5 bove [,3] is 3 6t +5dt = The re under y = 3/3 over the sme intervl [,3] is simply the re of rectngle tht is by 3/3 with re 446/3 So the verge height of function is the height of the horizontl line tht produces the sme re over the given intervl 5 3 Figure 94 Averge velocity Here s nother wy to interpret verge tht my mke our computtion pper even more resonble The object of our exmple goes certin distnce between t = Exercises 94 Find the verge height of cosx over the intervls [,π/], [ π/,π/], nd [,π] Find the verge height of x over the intervl [,] 3 Find the verge height of /x over the intervl [,A] 4 Find the verge height of x over the intervl [,] 5 An object moves with velocity v(t) = t + feet per second between t = nd t = Find the verge velocity nd the verge speed of the object between t = nd t =

10 95 Work 7 6 The observtion deck on the nd floor of the Empire Stte Building is,4 feet bove the ground If steel bll is dropped from the observtion deck its velocity t time t is pproximtely v(t) = 3t feet per second Find the verge speed between the time it is dropped nd the time it hits the ground, nd find its speed when it hits the ground ÏÓÖ A fundmentl concept in clssicl physics is work: If n object is moved in stright line ginst force F for distnce s the work done is W = Fs º EXAMPLE 95 How much work is done in lifting pound weight verticlly distnce of 5 feet? The force due to grvity on pound weight is pounds t the surfce of the erth, nd it does not chnge pprecibly over 5 feet The work done is W = 5 = 5 foot-pounds In relity few situtions re so simple The force might not be constnt over the rnge of motion, s in the next exmple EXAMPLE 95 How much work is done in lifting pound weight from the surfce oftheerth tonorbit milesbovethesurfce? Over milestheforce due togrvity does chnge significntly, so we need to tke this into ccount The force exerted on pound weight t distnce r from the center of the erth is F = k/r nd by definition it is when r is the rdius of the erth (we ssume the erth is sphere) How cn we pproximte the work done? We divide the pth from the surfce to orbit into n smll subpths On ech subpth the force due to grvity is roughly constnt, with vlue k/r i t distnce r i The work to rise the object from r i to r i+ is thus pproximtely k/r i r nd the totl work is pproximtely or in the limit n r i= i k r, r k W = r dr, where r is the rdius of the erth nd r is r plus miles The work is r W = r r k r dr = k r r r = k r + k r Using r = 9555 feet we hve r = The force on the pound weight t the surfce of the erth is pounds, so = k/9555, giving k = Chpter 9 Applictions of Integrtion Then k r + k r = foot-pounds Note tht if we ssume the force due to grvity is pounds over the whole distnce we would clculte the work s (r r ) = 58 = 58, somewht higher since we don t ccount for the wekening of the grvittionl force EXAMPLE 953 How much work is done in lifting kilogrm object from the surfce of the erth to distnce D from the center of the erth? This is the sme problem s before in different units, nd we re not specifying vlue for D As before D W = r k r dr = k r D r = k D + k r While weight inpounds ismesure offorce, weight inkilogrms ismesure ofmss To convert to force we need to use Newton s lw F = m At the surfce of the erth the ccelertion due to grvity is pproximtely 98 meters per second squred, so the force is F = 98 = 98 The units here re kilogrm-meters per second squred or kg m/s, lso known s Newton (N), so F = 98 N The rdius of the erth is pproximtely 6378 kilometers or 6378 meters Now the problem proceeds s before From F = k/r we compute k: 98 = k/6378, k = Then the work is: W = k D Newton-meters As D increses W of course gets lrger, since the quntity being subtrcted, k/d, gets smller But note tht the work W will never exceed , nd in fct will pproch this vlue s D gets lrger In short, with finite mount of work, nmely N-m, we cn lift the kilogrm object s fr s we wish from erth Next is n exmple in which the force is constnt, but there re mny objects moving different distnces EXAMPLE 954 Suppose tht wter tnk is shped like right circulr cone with the tip t the bottom, nd hs height meters nd rdius meters t the top If the tnk is full, how much work is required to pump ll the wter out over the top? Here we hve lrge number of toms of wter tht must be lifted different distnces to get to the top of the tnk Fortuntely, we don t relly hve to del with individul toms we cn consider ll the toms t given depth together

11 h Figure 95 Cross-section of conicl wter tnk 95 Work 9 To pproximte the work, we cn divide the wter in the tnk into horizontl sections, pproximte the volume of wter in section by thin disk, nd compute the mount of work required to lift ech disk to the top of the tnk As usul, we tke the limit s the sections get thinner nd thinner to get the totl work At depth h the circulr cross-section through the tnk hs rdius r = ( h)/5, by similr tringles, nd re π( h) /5 A section of the tnk t depth h thus hs volume pproximtely π( h) /5 h nd so contins σπ( h) /5 h kilogrms of wter, where σ is the density of wter in kilogrms per cubic meter; σ The force due to grvity on this much wter is 98σπ( h) /5 h, nd finlly, this section of wter must be lifted distnce h, which requires h98σπ( h) /5 h Newton-meters of work The totl work is therefore W = 98σπ 5 h( h) dh = 98 π 654 Newton-meters 3 Chpter 9 Applictions of Integrtion EXAMPLE 956 How much work is done in compressing the spring in the previous exmple from its nturl length to 8 meters? From 8 meters to 5 meters? How much work is done to stretch the spring from meters to 5 meters? We cn pproximte the work by dividing the distnce tht the spring is compressed (or stretched) into smll subintervls Then the force exerted by the spring is pproximtely constnt over the subintervl, so the work required to compress the spring from x i to x i+ is pproximtely 5(x i ) x The totl work is pproximtely nd in the limit W = 8 n 5(x i ) x i= 5(x )dx = 5(x ) 8 = 5(8 ) 5( ) = N-m The other vlues we seek simply use different limits To compress the spring from 8 meters to 5 meters tkes W = 5 8 5(x )dx = 5(x ) 5 8 = 5(5 ) 5(8 ) = 4 nd to stretch the spring from meters to 5 meters requires W = 5 5(x )dx = 5(x ) 5 = 5(5 ) 5( ) = 6 N-m N-m A spring hs nturl length, its length if nothing is stretching or compressing it If the spring is either stretched or compressed the spring provides n opposing force; ccording to Hooke s Lw the mgnitude of this force is proportionl to the distnce the spring hs been stretched or compressed: F = kx The constnt of proportionlity, k, of course depends on the spring Note tht x here represents the chnge in length from the nturl length EXAMPLE 955 Suppose k = 5 for given spring tht hs nturl length of meters Suppose force is pplied tht compresses the spring to length 8 Wht is the mgnitude of the force? Assuming tht the constnt k hs pproprite dimensions (nmely, kg/s ), the force is 5( 8) = 5() = Newtons Exercises 95 How much work is done in lifting kilogrm weight from the surfce of the erth to n orbit 35,786 kilometers bove the surfce of the erth? How much work is done in lifting kilogrm weight from n orbit kilometers bove the surfce of the erth to n orbit 35,786 kilometers bove the surfce of the erth? 3 A wter tnk hs the shpe of n upright cylinder with rdius r = meter nd height meters If the depth of the wter is 5 meters, how much work is required to pump ll the wter out the top of the tnk? 4 Suppose the tnk of the previous problem is lying on its side, so tht the circulr ends re verticl, nd tht it hs the sme mount of wter s before How much work is required to pump the wter out the top of the tnk (which is now meters bove the bottom of the tnk)?

12 96 Center of Mss 5 A wter tnk hs the shpe of the bottom hlf of sphere with rdius r = meter If the tnk is full, how much work is required to pump ll the wter out the top of the tnk? 6 A spring hs constnt k = kg/s How much work is done in compressing it / meter from its nturl length? 7 A force of Newtons will compress spring from meter (its nturl length) to 8 meters How much work is required to stretch the spring from meters to 5 meters? 8 A meter long steel cble hs density kilogrms per meter, nd is hnging stright down How much work is required to lift the entire cble to the height of its top end? 9 The cble in the previous problem hs kilogrm bucket of concrete ttched to its lower end How much work is required to lift the entire cble nd bucket to the height of its top end? Consider gin the cble nd bucket of the previous problem How much work is required to lift the bucket meters by rising the cble meters? (The top hlf of the cble ends up t the height of the top end of the cble, while the bottom hlf of the cble is lifted meters) ÒØ Ö Ó Å º 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 96 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 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 5 4 Chpter 9 Applictions of Integrtion (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 96 A solid bem EXAMPLE 96 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 of the bem s sequence of weights on bem For exmple, we cn think of the portion of the bem between x = nd x = s weight sitting t x =, the portion between x = ndx = sweight sittingt x =, nd so on, sindictedinfigure 96 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

13 96 Center of Mss 3 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 9 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 96 Suppose bem liesonthex-xisbetween nd 3, nd hsdensity 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 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 963 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

14 ( x,ȳ) x i 96 Center of Mss 5 m i x i Figure 963 Center of mss for two dimensionl plte of 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 963 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 963 EXAMPLE 964 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 6 Chpter 9 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 96 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 =

15 97 Kinetic energy; improper integrls 7 Ã Ò Ø Ò Ö Ý ÑÔÖÓÔ Ö ÒØ Ö Ð º Recll exmple 953 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 D 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, D 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 D 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 D x dx is the re under y = /x from x = to x = D Of course, s D increses this re increses But since D 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 8 Chpter 9 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 D 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 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

16 97 Kinetic energy; improper integrls 9 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 Chpter 9 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 97 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

17 98 Probbility 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: º 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 Chpter 9 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=

18 98 Probbility 3 DEFINITION 98 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 98 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 98 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 4 Chpter 9 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 983 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 984 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 98 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,