f[a] x + f[a + x] x + f[a +2 x] x + + f[b x] x

Transcription

1 Bsic Integrtion This chpter contins the fundmentl theory of integrtion. We begin with some problems to motivte the min ide: pproximtion by sum of slices. The chpter confronts this squrely, nd Chpter 3 concentrtes on the bsic rules of clculus tht you use fter you hve found the integrnd. Definite integrls hve importnt uses in geometry nd physics. Both the geometric nd the physicl integrl formuls re derived in the following wy: First, find formul for the quntity (volume, re, length, distnce, etc.) in the cse of either constnt or liner function. Next, pproximte the nonliner quntity by sum of slices using the constnt or liner formul for ech slice. The primry difficulty is usully in expressing the vrible sizes of the pproximting pieces. In the geometric problems, this step is nlyticl geometry - finding the formul tht goes with the picture you wnt. The Fundmentl Theorem of Integrl Clculus.3 gives us simple wy to compute the limit of the sum pproximtions exctly once we hve the integrl formul. In order for sum pproximtions to tend to n integrl we need to write them in the form f[] x + f[ + x] x + f[ + x] x + + f[b x] x where x is the thickness of the slice nd f[x] is the vrible bse of the slice, so tht f[x] x is the vrible mount of the slice t x. Thissymbolicexpressionisnimportntprtofthe wy the formuls re expressed in integrtion; without the symbolic expression the more or less obvious pproximtions could not be computed exctly in common wy. The computer hs sum commnd tht computes this expression [f [x] x] Computer summtion is studied in detil in Section.4 below. The integrl is the limit of this sum s x tends to zero, f[x] dx = lim [f [x] x] x 0 The slicing pproximtion is the forest tht you must strive to see through tngle of technicl trees. The first problem is in finding the symbolic sum. 7

2 Chpter - BASIC INTEGRATION 7. Geometric Slice Approximtions The volume of right circulr cone with height h nd bse of rdius r is V = π 3 r h We wnt you to derive this formul by pproximting cone with sum of cylindricl disks. Once you understnd the step from constnt rdius to vrying rdius, you will be ble to find integrl formuls for generl solids of revolution. This importnt generliztion illustrtes the power of integrtion theory. The volume of right cylinder is V = Ah the re of the bse times the height. If the bse is circle of rdius r, s shown in Figure.:, the formul becomes V = πr h Figure.:: V = πr h Our next tsk is to let the rdius vry with x, r = R[x], nd not be the sme ll long cylinder... The Volume of Cone Now we think of cone s figure with circulr cross-sections tht vry linerly, strting with zero rdius nd incresing to rdius r when the distnce from the tip is h. Letthevriblex run down the xis of the cone with x =0t the tip s shown in Figure.:. The expression for the rdius of the cross-section cn be expressed s function of x.

3 Chpter - BASIC INTEGRATION Figure.:: A cone long the x-xis Exmple. The Vrying Rdius of the Cross-Section of Cone. We wnt formul R[x] =?for the rdius of the cross-section t x for cone with tip t x =0 nd bse rdius r t height x = h. Since the side of the cone is stright line, we wnt function such tht R[x] = mx+ b, liner function R[0] = 0, zero cross-section t x =0, so b =0 R[h] =r, or mx= r, when x = h, so m = r h So, the rdius of the cross-section t x of cone of height h nd bse rdius r is R[x] = r h x. Exmple. The Volume of Prticulr Cone Our pproximtion of the volume cn be obtined by slicing the cone in steps of x, using the formul for the volume of cylinder for ech slice, V slice = πr [x] x. For exmple, suppose we hve cone of height 0 nd rdius 5 t the bse (in the sme length units). In this cse the formul for the rdius x units below the pex is R[x] = x If we let x =,orweslicethecone5 times, we obtin 5 disks with rdii t x =, x =3,, x =9. The disks re generted by the horizontl segments shown in Figure.:3. These segments generte the rough pproximting pile of disks shown in Figure.:4, nd hve volumes given s follows:

4 Chpter - BASIC INTEGRATION Figure.:3: Generting lines for the cone pproximtion. () x from 0 to, R() =, thickness of cylindricl disk = x =, V slice = π (b) x from to, R(3) = 3, thickness of cylindricl disk = x =, V slice = π 3 (c) x from 4 to 6, R(5) = 5, thickness of cylindricl disk = x =, V slice3 = π 5 (d) x from 6 to 8, R(7) = 7, thickness of cylindricl disk = x =, V slice4 = π 7 (e) x from 8 to 0, R(9) = 9, thickness of cylindricl disk = x =, V slice5 = π 9 The totl volume pproximtion is π ( )= π Figure.:4: A cone pproximted by 5 disks

5 Chpter - BASIC INTEGRATION 75 We need to find the formul tht expresses the sum of the volumes of the slices in terms of vrible for the thickness of the slice, x. The previous exmples suggest n interesting specilpurpose lgebric pproch, but tht is likely to work only for this specific cone. We cn express the pproximtion in terms of the formul for the rdius R[x] of the cross-section t x nd thickness x s π[(r[ x/]) x +(R[3 x/]) x +(R[5 x/]) x + +(R[0 x/]) x] = π 0 x/ x= x/ (R[x]) x.. A Prbolic Nose Cone ThesimplelinerformulforR[x] in the cse of the cone is esily generlized to more complicted shpes. A prbolic rocket nose cone my be described by the volume swept out s we rotte the curve y = x for 0 x 0 bout the x-xis shown in Figure.: Figure.:5: A prbolic nose cone The formul for the rdius of cross-section is R[x] = x, nd the volume of slice becomes πr [x]h = π( x) x = πx x, mking the pproximting sum with the rdius t the left side of the slices..3 A Sper Tip π (0 x + x x + x x + +(0 x) x) = π 0 x x=0 [x x] A slender sper tip my be described by the volume swept out by the region between y = x 50 nd the x-xis s the region is revolved bout the x-xis. In this cse, the rdius of cross-section

6 Chpter - BASIC INTEGRATION 76 t x is R[x] = x 50, the volume of one slice is πr [x]h = π pproximting disks is h x 50 i x, nd the left rdius sum by π (0 x + x4 500 = π 500 ( x)4 x x x=0 [x 4 x] x + + (0 x)4 500 x) Figure.:6: A cusped sper point nd slice genertors..4 The Are Between Two Curves We need two formuls to build n pproximtion to the complicted res between curves. First, the distnce between points with rel coordintes y nd y,is y y. The signed quntity y y gives the directed distnce from y to y where minus sign mens moving opposite the direction from zero to one. (The bsolute vlue will cuse us technicl difficulties in clculus.) Second, the re of rectngle is the height times the width.

7 Chpter - BASIC INTEGRATION 77 Exmple.3 Are Between Sinusoids Slice the region between the sine nd cosine curves from π to π into strips of width x s shown in Figure.:7. Figure.:7: The re between the sine nd cosine Theheightoftheslicewithleftpointtx is given by Sin[x] Cos[x] = Cos[x] Sin[x]. The re of the slice is Cos[x] Sin[x] x. The sum of ll the pproximting slices is..5 More Slices ( Cos[ π] Sin[ π] x+ Cos[ π + x] Sin[ π + x] x+ Cos[ π + x] Sin[ π + x] x+ + Cos[π x] Sin[π x] x) = π x x= π [ Cos[x] Sin[x] x] In Chpter 4, we study geometric slicing pproximtions in greter detil. The first pproximtion in tht chpter, the length of the sine curve, shows why extr cre must sometimes be tken to find correct pproximtion.

8 Chpter - BASIC INTEGRATION 78 Exercise Set.. Show tht the volume pproximtion for the cone of height 0 nd bse rdius 5 when cut into 0 slices with thickness nd rdii t the midpoints,, 3,, 9 is π 4 ( )= π Begin by writing the volume pproximtion s sum of the volumes of pproximting disks s in the bove list for 5 slices. See the progrm Sums.. Verify tht your 0-slice pproximtion sum is specil cse of this formul b π (R[x]) x, {x, x/, 0 x/, x} in the cse where x =by writing out the steps of the sum commnd Figure.:8: A cone pproximted by 5 disks -rdii t left 3. () Use the progrm ConeVol to drw cone sliced into 0 slices. (b) Modify the progrm ConeVol to drw sphere sliced into 0 slices. If we mesured our rdii t the left end of the slice points insted of the midpoint, the pproximtion would be less ccurte, becuse the slices fit completely inside the cone but the formul would be simpler, π[(r[0]) x +(R[ x]) x +(R[ x]) x + +(R[0 x]) x] b = π (R[x]) x, {x, 0, 0 x, x}

9 Chpter - BASIC INTEGRATION 79 This hs the form f[] x + f[ + x] x + f[ + x] x + + f[b x] x where f[x] =πr [x], =0nd b =0. 4. The pproximtion for the re between the sine nd cosine curves shown on Figure.:7 is π x x= π [ Cos[x] Sin[x] x] Wht function f[x] nd numbers nd b mkes this sum hve the form [f [x] x]. Distnce When Speed Vries If we drive 50 mph for n hour nd hlf, we compute the distnce trveled by D = R T, distnce equlsthertetimesthetime, D =50 3 =75miles. If we vry our speed, we cnnot use this formul. We cn compute the distnce trveled s n integrl. This section shows how. Suppose we get on the highwy nd ccelerte. We speed up cutiously nd drive 5 mph for minute, 6 mph for minute, 7 mph for minute,, 49 mph for minute nd 50 mph for the reminder of the hour nd hlf s shown on Figure.:9. How fr do we go? The distnce trveled t 5 mph must be computed in the correct units, D = (mph hours = miles) 60 The distnce trveled t 6 mph is D =6 The distnce trveled t 7 mph is (miles) 60 D 3 = (miles) 60 Ech minute s distnce cn be computed for 5 minutes, giving sum of The lst prt of the trip is 50 (90 5) fortotlof69.59miles.

10 Chpter - BASIC INTEGRATION 80 There re severl questions:. () Wht hve we done symboliclly? (b) How cn we interpret the computtion geometriclly? (c) Why should we interpret the distnce computtion geometriclly? (d) How cn we extend this to continuously vrying speed? Symboliclly, we hve speed s function of time in minutes, S[] = 5, S[] = 6, S[3] = 7, nd 5 distnce trveled during ccelertion = [S[m] 60 ] wheres we hve x= step distnce trveled t 50 mph = Figure.:9: Speed vs. time in minutes We my lso think of this s sum with rte function definedtechminute,s[m] =50,for ll the minutes from 6 to 90. However, it is better to lso use the correct units, t = elpsed time in hours, nd give the speed by the function R[t] =5 for 0 t< 60 R[t] =6 for 60 t< 60 R[t] =7 for 60 t< 3 60 R[t] =8 for 3 60 t< R[t] =50 for 3 60 t 90 60

11 Chpter - BASIC INTEGRATION 8 In this cse, we cn combine the two pieces into one sum with t = 60, =0nd b = 3. distnce trveled = = 3 60 t=0 step 60 b t step t [R[t] 60 ] [R[t] t] Figure.:0: Speed vs. time in hours for five minutes The products 5 60, 6 60, 7 60, etc., cn be ssocited with rectngles of height 5, 6, 7, etc.,ndwidth 60 hours or minute. Distnce is not n re; but, in this cse, the distnce trveled t 5 mph is product nd tht product is lso n re of the rectngle under the segment on the grph of speed vs. time. This mens tht we cn represent the distnce trveled s the totl re between the speed curve nd the x-xis - provided our units re hours on the x-xis nd miles perhouronthey-xis. Figure.:: Distnce represented s re

12 Chpter - BASIC INTEGRATION 8 The re representtion helps us see wht hppens if we ccelerte continuously, insted of going exctly 5 mph for exctly minute, 6 mph for nother minute, nd so forth Suppose we ccelerte linerly from 5 mph to 50 mph in 5 minutes, so tht our speed is given by R[t] =5+60t for 0 t< 5 60 nd R[t] =50 for t< 60 We could clculte the distnce trveled in ech second using the speed t the beginning of the second, = µ = µ = µ = This computtion is no fun to do by hnd, but the computer gives.5 x=0 step /3600 [R [t] t] = when t =/3600. The computer computtion is wste in this cse, however, becuse we cn ssocite the distnce trveled with the re under the speed curve in Figure.: nd use the formul for trpezoid for 0 t<5/60 nd for rectngle for 5/60 t 90/60, re = (5 + 50) =

13 Chpter - BASIC INTEGRATION 83 Figure.:: Liner ccelertion nd distnce If speed vries by more complicted rule thn linerly in pieces, we cn still ssocite the re under the speed curve with the distnce trveled, but now we hve no formul to find the re. We will hve to find tht re nd distnce by definite integrtion. For short intervl of time, where speed does not chnge very much, the distnce is pproximtely the product R[t] t. distnce trveled from time t to time t + t = R[t] t + ε t The pproximtion is ctully close, even compred to t, which mkes totl distnce n integrl. We will see exctly wht the pproximtion is nd verify tht it holds s long s R[t] is continuous function. This should be plusible, becuse we could lso pproximte by speed trveled from time t to time t + t R[t 0 ] for ny t 0 between t nd t + t. ContinuityofR[t] mens tht R[t] R[t 0 ],sor[t] t = R[t 0 ] t + ε t, withε 0. Exercise Set.. The Shortcut to Grndmother s You re lte nd rush to get to Thnksgiving dinner t Grndmother s house in hours. Your speed is recorded on Figure.:3. How fr wy does Grndmother live? (Note: 0 minutes = hours.) How much ws the ticket?

14 Chpter - BASIC INTEGRATION 84 Figure.:3: The speed on trip to grndmother s. The Return from Grndmother s A grph of your trip odometer is shown on Figure.:4 on the return from Grndmother s to school. Grph your speed nd show the distnce s n re. Figure.:4: The speed on the trip bck from grndmother s.3 The Definite Integrl This section records the definition of the integrl tht ws motivted by ides like the problem of computing distnce from vrible speeds nd by the geometric re, length, nd volume slicing problems. Definition. The Definite Integrl in One Vrible Let f[x] be continuous function defined on the intervl [, b]. The definite integrl of f[x] over

15 Chpter - BASIC INTEGRATION 85 [, b] is given by the following limit: f[x] dx = lim [f [x] x] x 0 [f [x] δx], when δx, 0 We need to use the lgebric properties of the summtion function to deduce results bout the integrl. In either the limiting cse x 0 or when we extend the P b function to tiny increments, δx 0, lgebr of generl sums extends to integrls. The most bsic question is, Wht is the role of continuity of f[x] in the definition of R b f[x] dx? The nswer is, Continuity gurntees us tht the limit exists or, equivlently, tht we get nerly the sme vlue for ech tiny δx. Extension of the integrl to certin discontinuous functions is possible, but involves extr mthemticl complictions. Four steps in n exmple of this limit re shown in Figure.3:5. Figure.3:5: Approximtions to the Are ArcT n[].075 One of the most importnt lgebric properties of summtion, the Telescoping Sum Theorem., plus the differentil pproximtion (or microscope eqution) for smooth function will give us hlf of the Fundmentl Theorem of Integrl Clculus.5. This theorem tells us how to find exct (symbolic) integrls without summing or tking limit. We build lots of techniques of integrtion round tht theorem becuse it gives us these exct nswers. Summtion is still importnt, however, becuse the sums of little slices ide is where nerly ll pplictions of integrtion come from.

16 Chpter - BASIC INTEGRATION 86 Exercise Set.3. Run the computer progrm IntegrAprx. Compre the convergence of the slice pproximtions with the exct re of circle described there..4 Computer Summtion Suppose tht F [x] is function of x. To form the sum we my use single computer commnd This section studies this opertor. F []+F [ + x]+f [ + x]+ + F [b] b F [x] For exmple, if F [x] =, =0, b =nd x = b =F [0] + F [] = + = If we chnge to x =,then b b =F [0] + F + F [] = 3 The wy P b F [x] works on the computer is s follows: We strt with x = nd strt with P b = F [x] =F []. Next, we dd x to x nd sk whether x>b. If not, we dd F [x] =F [ + x] to P b,incrementx gin, check x>b,ddf [x] to P b, nd continue until x>b.

17 Chpter - BASIC INTEGRATION 87 Suppose we hve F [x] =x, =0, b =nd x =,then b F [x] =F [0] + F Keeping the other vlues, but chnging to x = 4 gives b x = F [0] + F + F [] = 0 + += 3 + F 4 = F 4 += F [] 4 Vlues of x tht do not exctly divide b relsollowed,sowhen x =.5, b x = = 3.5 In integrtion, we will tke x smller nd smller nd still rech limiting vlue in our sums by hving fctor x in the summnd. The next exercise shows you how this works. Exercise Set.4. Write out ll the terms of the sums P b increments, lwys using =0nd b = () F [x] =x, x =/5 (b) F [x] =x x, x =/ (c) F [x] =x x, x =/3 (d) F [x] =x x, x =/4 (e) F [x] =x x, x =/5 F [x] defined by the following functions nd. Use the computer progrm Sums to compre the summed vlues of P b x nd P b x x for x =, x = 3, x = 4, x = 5.

18 Chpter - BASIC INTEGRATION 88.5 The Algebr of Summtion This section writes simple properties of summtion in terms of the summtion opertor P b. These properties llow us to develop n lgebr of integrtion in the following section. Our first result is n lgebric method of computing exct sums. Theorem. The Telescoping Sum Theorem If F [x] is defined for x b, then the sum of differences below equls the lst vlue minus the first, b F [x + x] F [x] =F [b 0 ] F [] where b 0 is the lst vlue of x of the form + n x, which is less thn or equl to b. If x divides b exctly, b 0 = b. Proof: b F [x + x] F [x] = (F [ + x] F []) + (F [ + x] F [ + x]) +(F [ +3 x] F [ + x]) + (F [ +4 x] F [ +3 x]) + +(F[b 0 ] F [b 0 x]) The prt F [ + x] from the first term cncels the prt F [ + x] from the second term. The prt F [ + x] from the second term cncels the prt F [ + x] from the third term, nd so on. Ech term hs positive nd prt tht cncels the corresponding negtive prt from the next term. The negtive prt from the first term nd the positive prt from the lst term re never cnceled, so the sum telescopes to F [b 0 ] F []. Notice tht we sum to b x (in steps of x), so tht the lst term is F [b] when x divides b (rther thn (F [b + x] F [b]))

19 Chpter - BASIC INTEGRATION 89 Exmple.4 Finding the Difference The difficulty in using the Telescoping Sum Theorem is in finding n expression of the form F [x + x] F [x] for the summnd. Consider the sum b x [ x(x + x) ] Assuming tht neither x =0nor x + x =0,wecnwrite x x(x + x) = x x + x becuse putting the right-hnd side on common denomintor gives, x x + x = This mkes computtion of the sum esy, x + x x(x + x) = x + x x x(x + x) = x x(x + x) x x(x + x) x [ x(x + x) ]= [ x x + x ] = [ = F [b 0 ] F [] = µ x + x ] x µ b 0 = b 0

20 Chpter - BASIC INTEGRATION 90 Theorem. Superposition of Summtion Let α nd β be constnts, nd let F [x] nd G [x] be functions defined on [, b]. Then [αf [x]+βg[x]] = α [F [x]] + β [G [x]] Proof: [αf []+βg[]] + [αf [ + x]+βg[ + x]] +[αf [ + x]+βg[ + x]] + +[αf b 0 + βg b 0 ] = α[f []+F[ + x]+f [ + x]+ + F b 0 ] + β[g []+G[ + x]+g [ + x]+ + G b 0 ] Exmple.5 P b [x x] = b x b Given tht [ x] = [([x + x] x)] = b [[x + x] x ]=b nd [x + x] x =x x + x we cn use superposition to find [x x] =?

21 Chpter - BASIC INTEGRATION 9 We hve Solving for the unknown sum, b = b = = = Dividing by, we hve our formul: [[x + x] x ] [x x + x ] [x x]+ x [x x]+ x (b ) [ x] [x x] =(b ) x (b ) [x x] = (b ) x (b ) Notice tht if x is smll, this formul sys, [x x] (b )

22 Chpter - BASIC INTEGRATION 9 Theorem.3 The Tringle Inequlity For F [x] defined on [, b], [F [x]] [ F [x] ] Proof: Terms of opposite sign inside the sum F []+F [ + x] + + F [b] could cncel nd mke tht sum smller thn the sum with ll positive terms, F [] + F [ + x] + + F [b]. Theorem.4 Additivity of Summtion Suppose F [x] is defined on [, c], x divides b nd <b<c.then [F [x]] + c x x=b [F [x]] = c x [F [x]] Proof: [F []+F[ + x]+ + F [b x]] + [F [b]+f [b + x]+ + F c 0 ] =[F []+F[ + x]+ + F [b x]+f [b]+f [b + x]+ + F c 0 ] Theorem.5 Monotony of Summtion If F [x] G [x], then [F [x]] [G [x]] Proof: This is obvious becuse ech term in the left sum is smller thn the corresponding term in the right sum.

23 Chpter - BASIC INTEGRATION 93 Theorem.6 Orienttion of Summtion Suppose f[x] is defined on [, b] with <bnd x divides b, then x=b step x [f[x]( x)] = b [f[x] x] Proof: To go from b to, we must tke negtive steps. Exercise Set.5. Wht s the Difference? Use the Telescoping Sum Theorem to give exct nswers to the following sums in the cse where x divides b. () P [ x] (b) P [x x + x ] (c) P [3x x +3x x + x 3 ] The question relly is, Wht F [x + x] F [x] equls the expressions tht you re sked to sum? () Use telescoping sums, superposition, nd known sums to find Notice tht [ x] =(b ) [x x] = [x x] =? (b ) x (b ) nd (x + x) 3 x 3 =3x x +3x x + x 3

24 Chpter - BASIC INTEGRATION 94 (b) Use telescoping sums, superposition nd known sums to find [x 3 x] =?. Give n exmple of sum with the cncelltion described in the proof of the tringle inequlity. 3. Write out the three sums in the proof of Additivity for =0, b =, c =, x =,nd F [x] =x x. 4. Write out the two sums in the proof of Monotony for =0, b =, x =, F [x] =x x, nd G [x] =x x. 5. Write out the two sums in the proof of Orienttion for =0, b =, x =,ndf[x] =x x..6 The Algebr of Integrtion For fixed choice of the continuous rel function f[x], we know tht f[x] dx f [x] δx when δx 0. The sum properties of the previous section hold for ll x, so, in prticulr, they hold when x = δx 0 is smll. This shows tht integrls hve the following properties. Proofs of the properties of integrls re given in the Mthemticl Bckground. Theorem.7 Superposition [αf[x]+βg [x]] dx = α f[x] dx + β g [x] dx Theorem.8 Tringle Inequlity f[x] dx f[x] dx, ( <b)

25 Chpter - BASIC INTEGRATION 95 Theorem.9 Monotony f[x] dx g [x] dx, iff[x] g [x] Theorem.0 Additivity f[x] dx + Z c b f[x] dx = Z c f[x] dx Theorem. Orienttion Z b f[x] dx = f[x] dx Note: The dx in the reverse-oriented integrl cn be thought of s negtive to move us from b bck to. Exmple.6 Integrl Computtion the Hrd Wy We hve seen tht so, when δx 0, [x x] = (b ) x(b ) xdx [xδx] = (b ) δx(b ) b nd the two fixed quntities must be equl, xdx= b

26 Chpter - BASIC INTEGRATION 96 Theorem. Estimtion of Sums In prticulr, if ε 0 for ll x in [, b] when δx 0. [ F [x, δx] δx] F [x Mx,δx] [b ] [ε [x, δx] δx] 0 Proof: This is the lst technicl result we need from lgebr. We use the finite mximum function. Define function like the computer s mximum We know tht Mx[F [x, x], {x =, b x, x}] = mximum[f [, x],f [ + x, x],f [ + x, x],...,f b 0, x ] Mx[F [x, x], {, b x, x}] =F [x M, x] for some rel x M of the form x M = + n x, x M b x. Apply the Mx function to smll δx 0, Mx[F [x, δx], {, b δx, δx}] =F [x M,δx] for some x M of the form x M = + nδx, x M b. The formul we need is the forml sttement tht we cn estimte by mking ll the terms of sumlrger, [ F [x, δx] δx] Mx[ F [x, δx], {, b δx, δx}] = F [x M,δx] [δx] [δx] = F [x M,δx] (b )

27 Chpter - BASIC INTEGRATION 97 Exercise Set.6. () Compute R b x dx by extending the exct formul you computed erlier for [x x] to δx 0. (b) Compute R b x3 dx by extending the exct formul for [x 3 x] to δx 0 (c) Compute R b xn dx the hrd wy for integer n by showing with ε = ε[x, δx] 0. Note tht. Estimtion for R b dx x We showed bove tht (x + δx) n x n = nx n δx + ε δx [(x + δx) n x n ] b n n x [ x(x + x) ]= b 0 provided tht neither x =0nor x + x =0for ny term of the sum. Compute the integrl x dx = b by estimting the difference between the unknown sum [ δx x ] nd the sum bove, δx [ x(x + δx) ]= b 0

28 Chpter - BASIC INTEGRATION 98 HINT: Clculte the difference h i x x(x+δx) δx to estimte Is your computtion vlid if <0 <b? 3. Estimtion for R b Show tht x 3 dx [ δx x ] δx [ x(x + δx) ] x 3 dx = b s follows: First, by putting the difference on common denomintor, show tht x 3 = x (x δx ) + ε with ε 0 if δx 0. ε = x 3 x (x δx ) Second, show tht 4 xδx (x δx ) = (x δx) (x + δx) by putting the right-hnd side on common denomintor. Third, show tht µ 4 xδx (x δx ) = (x δx) + µ x (x + δx) + x nd tht 4 xδx [ (x δx ) ]= µ µ + ( δx) b + (b δx) 4. Estimtion for R b dx x? You cn sum. Wht integrl does tht tell you? (x δx) 3 (x+δx) 3 5. Estimtion for R b Show tht x n+ dx s follows. First, show tht x n+ dx = µ n n b n x n (x + δx) n = nxn δx x n (x + δx) n + ε δx

29 Chpter - BASIC INTEGRATION 99 with ε 0 when δx 0. Second, x n x n (x + δx) n = x n+ + ε with ε 0 when δx 0. Cn we hve <0 <b? 6. Estimtion for R b x dx You cn sum x + δx x in generl for x b. Multiply this expression by x + δx + x x + δx x = ( x + δx x)( x + δx + x) x + δx + x to find 7. Estimtion for R b xdx Sum p (x + δx) 3 x 3 to find x dx xdx.7 Fundmentl Theorem, Prt This section gives the theory to find integrls by ntiderivtives without tking limits or pproximting with tiny increments. (Chpter 3 gives techniques for using the theory.) The theory results from n estimte nd telescoping sum. The generl estimte is even simpler thn the specific telescoping sum exercises of the previous section. Recll the differentil pproximtion for smooth function from Chpter 5 t this time, becuse we re bout to use it. Theorem.3 First Hlf of the Fundmentl Theorem Given n integrnd f[x] dx, suppose tht we cn findfunctionf [x] such tht df [x] =f[x] dx (or df dx = f [x)) forll x b, then f[x] dx = F [b] F []

30 Chpter - BASIC INTEGRATION 300 We lso sometimes write f[x] dx = The nottion F [x] b simply mens F [b] F []. df [x] =F [x] b = F [b] F [] Proof: The differentil pproximtion (microscope eqution, see Chpter 5) for F [x] is F [x + δx] F [x] =f[x]δx + ε δx for ll x stisfying x b, whereε 0 when δx 0. (This forces f[x] to be continuous function for the ordinry derivtive defined in Chpter 5. See Section 5.5.) The telescoping sum nd superposition properties sy F [b] F [] = [F [x + δx] F [x]] = = [f[x]δx + ε δx] [f[x]δx]+ f[x] dx + We conclude the proof by using Theorem., which shows We hve shown tht the fixed quntities stisfy F [b] F [] [ε δx] 0 f[x] dx [ε δx] [ε δx] which forces them to be equl nd proves the theorem. Strnge though it sounds, we hve shown two things with the proof of the firsthlfofthe Fundmentl Theorem: ) the integrl exists! nd ) its vlue is F [b] F []. (Instructor Note: The usul continuity hypothesis is hidden in our bility to find function F [x] stisfying the uniform microscope eqution derivble with f[x] =F 0 [x] for ll x in [, b].)

31 Chpter - BASIC INTEGRATION 30 Exmple.7 Computtion of R b x dx, theesywy Compre the following use of the Fundmentl Theorem with your direct computtion from the lst section. We hve f[x] =x ;so,ifwetkef [x] =x 3 /3, thendf [x] =x dx nd x dx = 3 x3 b = b Exmple.8 Integrl of Cosine Cos [θ] dθ =Sin[θ] b =Sin[b] Sin [] It would be quite difficult to compute this integrl without the help of the Fundmentl Theorem. Exercise Set.7. Fundmentl Drill Show tht df [x] =f[x] dx for the given pirs of functions F [x] nd f[x]. Usethistocompute the integrls. () F [x] = 4 x4, f[x] =x 3, R 0 x3 dx = (b) F [x] = x, f[x] = x, R 9 4 (c) F [x] = 3 (d) F [x] = x, f[x] = n, R 3 n x n+ x dx = x 3, f[x] = x, R 9 4 xdx= x 3 dx =. Cn you use the formul F [x] = x implies df [x] = dx to find R 3 n x n+ x dx? Wht function F [x] hs df [x] = x dx? If f[x] is not continuous everywhere on [, b], the Fundmentl Theorem does not pply nd df we shll see tht using F [b] F [] nywy will led to errors, even if dx = f[x] t ll but one point of [, b]

32 Chpter - BASIC INTEGRATION Figure.7:6: Are under the curve y =/x 3. Is the following computtion correct? Z Z x dx = x dx = x = x = = 3 Note tht if F [x] = x,thendf [x] =x dx, but how could the re under positive curve be negtive number? Drw picture nd compute the res R.000 x dx, R.000 x dx dx. Wht do you think the re R.000 x.000 nd R + Incorrect progrms cn produce incorrect results. x dx should be? 4. Grbge In, Grbge Out Wht is wrong with the following computer computtion of R 0 In[] f := /(Cos[x]) ; Integrte[f,{x, 0, }] <Enter> Out[] Sin[] Cos[].8504 Cos [x] dx? Cn the integrl of positive function be negtive? Is the function F [x] = Sin[x]/ Cos[x] n ntiderivtive for the function f[x] =/ [Cos[x]] over the whole intervl 0 x? Tht is, does df [x] =f[x] dx for ll these vlues of x? Use the computer to plot the integrnd function over the intervl 0 x.

33 Chpter - BASIC INTEGRATION Fundmentl Theorem, Prt The second prt of the Fundmentl Theorem of Integrl Clculus sys tht the derivtive of n integrl of continuous function is the integrnd, Z d f[x] dx = f [] d The function A[] = R f[x] dx cn be thought of s the ccumulted re under the curve y = f[x] from to shown in Figure.8:7. The ccumultion function cn lso be thought of s the reding of your odometer t time for given speed function f[x] Figure.8:7: A [] = R f [x] dx When we cnnot find n ntiderivtive F [x] for given f[x], we sometimes still wnt to work directly with the definition of the integrl. A number of importnt functions re given by integrls of simple functions. The logrithm nd rctngent re elementry exmples. Some other functions like the probbility function of the bell-shped curve of probbility nd sttistics do not hve elementry formuls but do hve integrl formuls. The second hlf of the Fundmentl Theorem justifies the integrl formuls, which re useful s pproximtions. The NumIntAprx computer progrm shows us efficient wys to estimte the limit of sums directly. Continuity of f[x] is needed to show tht the limit ctully converges. With discontinuous integrnds, it is possible to mke the sums oscillte s x 0. In these cses, numericl integrtion by computer will likely give the wrong nswer. We need nked existence of the limit for the second hlf of the Fundmentl Theorem. The first hlf of the Fundmentl Theorem is used often, wheres the second hlf is seldom used. This is why we hve postponed generl proof of existence (without ntiderivtives). (Our proof of the First Hlf of the Fundmentl Theorem gve existence nd formul t the sme time.)

34 Chpter - BASIC INTEGRATION 304 Theorem.4 Existence of the Definite Integrl Let f[x] be continuous function on the intervl [, b]. Then there is rel number I such tht [f [x] x] I s x 0 or, equivlently, [f [x] δx] I for ny δx 0 Proof: First, by the Extreme Vlue Theorem for Continuous Functions, f[x] hs min, m, ndmx, M, on the intervl [, b]. Monotony of summtion tells us m [b ] [f [x] δx] M [b ] so tht P [f [x] δx] is finite number nd thus ner some fixed vlue I [δx] P [f [x] δx]. Wht we need to show is tht if we choose different increment, for exmple δu 0, then I [δx] =I [δu] or [f [x] δx] b δu [f [u] δu] Drw your own picture of two different rectngulr pproximtions to clrify this ide. If we superimpose the different prtitions nd consider two overlpping rectngles, the res only differ by n smll mount on scle of the increments becuse f[x] f[u] by continuity for x u. The proof tht we do get the sme number is tken up in detil in the book, Foundtions of Infinitesiml Clculus t

35 Chpter - BASIC INTEGRATION 305 Theorem.5 Second Hlf of the Fundmentl Theorem Suppose tht f[x] is continuous function on n intervl contining nd we define new function by ccumultion, Z A [] = f[x] dx Then A [] is smooth nd d da [] =f []; inotherwords, Z d f[x] dx = f [] d Proof: We show tht A [] stisfies the differentil pproximtion A [ + δ] A [] =f [] δ + ε δ with ε 0 when δ 0. Thisprovesthtf[] is the derivtive of A[]. Figure.8:8: R +δ f[x] dx By definition of A nd the dditivity property of integrls, we hve Z +δ Z Z +δ A [ + δ] = f[x] dx = f[x] dx + f[x] dx So, we need to show tht with ε 0, whenδ 0. = A []+ Z +δ Z +δ f[x] dx f[x] dx = f [] δ + ε δ

36 Chpter - BASIC INTEGRATION 306 Figure.8:9: R +δ f[x] dx This is when we use the continuity hypothesis bout f[x]. The Extreme Vlue Theorem for Continuous Functions.3 sys tht f[x] hs mx nd min on the intervl [, + δ], m = f [ m ] f[x] M = f [ M ] for ll x + δ. Monotony of the integrl gives us the estimtes Z +δ Z +δ Z +δ mδ = mdx f[x] dx Mdx= M δ Since both m nd M lie in the intervl [, + δ], we know tht m M when δ 0. Continuity of f[x] mens tht f [ m ] f [ M ] f [] in this cse, so Z +δ f[x] dx = f [] δ + ε δ using upper nd lower estimtes of the integrl bsed on the mx nd the min of the function over the smll subintervl. Figure.8:0: Upper nd lower estimtes Thisprovesthetheorembecusewehveverified the microscope eqution from Chpter 5, A [ + δ] =A []+A 0 [] δ + ε δ with A 0 [] =f [].

37 Chpter - BASIC INTEGRATION 307 Exercise Set.8. Use the Second Hlf of the Fundmentl Theorem to explin how YOU could compute Log [x] nd ArcTn [x] directly using numericl integrtion progrm. We know tht d Log [x] dx = Z x nd d d x dx = nd d ArcTn [x] dx = +x nd Z d d 0 +x dx = + Use the Numericl Integrtion progrm NumIntAprx to compute the vrious numericl integrl pproximtions for these functions, nd compre your results with the computer s built-in lgorithms for these functions. For exmple, how do the following compre: Log [ ] NIntegrte[/x, {x,, }] nd 4 ArcTn [] 4 NIntegrte[/( + x ), {x, 0, }] Explin in terms of the exct symbolic computtions why we hve chosen these prticulr numbers ( in log nd the 4 in rctngent). See the bsic left- right- nd midpointpproximtions in the progrm IntegrAprx. Sometimes it is not possible to find n ntiderivtive in terms of elementry functions, but even in these cses, the integrl defines function nd the numericl pproximtions give direct wy to compute these functions.. Non-elementry Integrls Use the computer s numericl integrl numericl integrtion commnd to pproximte the following integrls: ) R 0 e x dx = b) R 0 Sin[x] x dx = c) R 0 Sin[x ] dx = Also use the computer s symbolic integrtion to find expressions for these integrls. These expressions re not combintions of elementry functions. See the symbolic integrtion progrm SymbolicIntegr for detils.