Integration. 148 Chapter 7 Integration

Save this PDF as:
Size: px
Start display at page:

Download "Integration. 148 Chapter 7 Integration"

Transcription

1 48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but only for tenth of second; during tht time the object would not move During the tenth of second from t = to t =, we suppose tht the object is trveling t 3 cm/sec, nmely, its ctul speed t t = In this cse the object would trvel (3)() = 3 centimeters: 3 cm/sec times seconds Similrly, between t = nd t = 3 the object would trvel (6)() = 6 centimeters Continuing, we get s n pproimtion tht the object trvels ()()+(3)()+(6)()+ +(7)() = 3 centimeters, ending up t position 3 This is better pproimtion thn, certinly, but is still just n pproimtion (We know in fct tht the object ends up t position, becuse we ve lredy done the problem using the first pproch) Presumbly, we will get better pproimtion if we divide the time into one hundred intervls of hundredth of second ech, nd repet the process: ÌÛÓ Ü ÑÔÐ º½ Up to now we hve been concerned with etrcting informtion bout how function chnges from the function itself Given knowledge bout n object s position, for emple, we wnt to know the object s speed Given informtion bout the height of curve we wnt to know its slope We now consider problems tht re, whether obviously or not, the reverse of such problems EXAMPLE 7 An object moves in stright line so tht its speed t time t is given by v(t) = 3t in, sy, cm/sec If the object is t position on the stright line when t =, where is the object t ny time t? There re two resonble wys to pproch this problem If s(t) is the position of the object t time t, we know tht s (t) = v(t) Becuse of our knowledge of derivtives, we know therefore tht s(t) = 3t /+k, nd becuse s() = we esily discover tht k =, so s(t) = 3t /+ For emple, t t = the object is t position 3/+ = This is certinly the esiest wy to del with this problem Not ll similr problems re so esy, s we will see; the second pproch to the problem is more difficult but lso more generl We strt by considering how we might pproimte solution We know tht t t = the object is t position How might we pproimte its position t, sy, t =? We know tht the speed of the object t time t = is ; if its speed were constnt then in the first second the object would not move nd its position would still be when t = In fct, the object will not be too fr from t t =, but certinly we cn do better Let s look t the times,, 3,,, nd try pproimting the loction of the object ()()+(3)()+(6)()+ +(97)() = 48 We thus pproimte the position s 48 Since we know the ect nswer, we cn see tht this is much closer, but if we did not lredy know the nswer, we wouldn t relly know how close We cnkeep thisup, but we llnever rellyknow theectnswer ifwe simplycompute more nd more emples Let s insted look t typicl pproimtion Suppose we divide the time into n equl intervls, nd imgine tht on ech of these the object trvels t constnt speed Over the first time intervl we pproimte the distnce trveled s ()(/n) =, s before During the second time intervl, from t = /n to t = /n, the object trvels pproimtely 3(/n)(/n) = 3/n centimeters During time intervl number i, the object trvels pproimtely (3(i )/n)(/n) = 3(i )/n centimeters, tht is, its speed t time (i )/n, 3(i )/n, times the length of time intervl number i, /n Adding these up s before, we pproimte the distnce trveled s () n +3 n +3() n +3(3) n + +3(n ) n centimeters Wht cn we sy bout this? At first it looks rther less useful thn the concrete clcultions we ve lredy done But in fct bit of lgebr revels it to be much 47

2 more useful We cn fctor out 3 nd /n to get 3 n ( (n )), 7 Two emples 49 tht is, 3/n times the sum of the first n positive integers Now we mke use of fct you my hve run cross before: k = In our cse we re interested in k = n, so k(k +) (n ) = (n )(n) This simplifies the pproimte distnce trveled to = n n 3 n n n = 3 n n n = 3 ( n n n ) n = 3 ( ) n Now this is quite esy to understnd: s n gets lrger nd lrger this pproimtion gets closer nd closer to (3/)( ) = 3/, so tht 3/ is the ect distnce trveled during one second, nd the finl position is So for t =, t lest, this rther cumbersome pproch gives the sme nswer s the first pproch But relly there s nothing specil bout t = ; let s just cll it t insted In this cse the pproimte distnce trveled during time intervl number i is 3(i )(t/n)(t/n) = 3(i )t /n, tht is, speed 3(i )(t/n) times time t/n, nd the totl distnce trveled is pproimtely Chpter 7 Integrtion but the second pproch works fine (Becuse the function y = 3 is so simple, there is nother pproch tht works here, but it is even more limited in potentil ppliction thn is pproch number one) How might we pproimte the desired re? We know how to compute res of rectngles, so we pproimte the re by rectngles Jumping stright to the generl cse, suppose we divide the intervl between nd into n equl subintervls, nd use rectngle bove ech subintervl to pproimte the re under the curve There re mny wys we might do this, but let s use the height of the curve t the left endpoint of the subintervl s the height of the rectngle, s in figure 7 The height of rectngle number i is then 3(i )(/n), the width is /n, nd the re is 3(i )( /n ) The totl re of the rectngles is () n +3() n +3() n +3(3) n + +3(n ) n By fctoring out 3 /n this simplifies to 3 3 n n (+++ +(n )) = n n = 3 ( ) n As n gets lrger this gets closer nd closer to 3 /, which must therefore be the true re under the curve As before we cn simplify this to () t n +3()t n +3()t n +3(3)t n + +3(n )t n 3t 3t n n (+++ +(n )) = = 3 ( n n t ) n In the limit, s n gets lrger, this gets closer nd closer to (3/)t nd the pproimted position of the object gets closer nd closer to (3/)t +, so the ctul position is (3/)t +, ectly the nswer given by the first pproch to the problem EXAMPLE 7 Findthereunder thecurvey = 3between = ndny positive vlue There is here no obvious nlogue to the first pproch in the previous emple, Figure 7 Approimting the re under y = 3 with rectngles Drg the slider to chnge the number of rectngles Wht you will hve noticed, of course, is tht while the problem in the second emple ppers to be much different thn the problem in the first emple, nd while the esy pproch to problem one does not pper to pply to problem two, the pproimtion pproch works in both, nd moreover the clcultions re identicl As we will see, there

3 7 The Fundmentl Theorem of Clculus re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we end up with mthemtics tht looks like the two emples, though of course the function involved will not lwys be so simple Even better, we now see tht while the second problem did not pper to be menble to pproch one, it cn in fct be solved in the sme wy The resoning is this: we know tht problem one cn be solved esily by finding function whose derivtive is 3t We lso know tht mthemticlly the two problems re the sme, becuse both cn be solved by tking limit of sum, nd the sums re identicl Therefore, we don t relly need to compute the limit of either sum becuse we know tht we will get the sme nswer by computing function with the derivtive 3t or, which is the sme thing, 3 It struethtthefirstproblemhdtheddedcomplictionofthe,ndwecertinly need to be ble to del with such minor vritions, but tht turns out to be quite simple The lesson then is this: whenever we cn solve problem by tking the limit of sum of certin form, we cn insted of computing the (often nsty) limit find new function with certin derivtive Eercises 7 Suppose n object moves in stright line so tht its speed t time t is given by v(t) = t+, nd tht t t = the object is t position Find the position of the object t t = Suppose n object moves in stright line so tht its speed t time t is given by v(t) = t +, nd tht t t = the object is t position Find the position of the object t t = 3 By method similr to tht in emple 7, find the re under y = between = nd ny positive vlue for 4 By method similr to tht in emple 7, find the re under y = 4 between = nd ny positive vlue for By method similr to tht in emple 7, find the re under y = 4 between = nd ny positive vlue for bigger thn 6 By method similr to tht in emple 7, find the re under y = 4 between ny two positive vlues for, sy < b 7 Let f() = Approimte the re under the curve between = nd = using 4 rectngles nd lso using 8 rectngles 8 Let f() = + 3 Approimte the re under the curve between = nd = 3 using 4 rectngles Ì ÙÒ Ñ ÒØ Ð Ì ÓÖ Ñ Ó ÐÙÐÙ º¾ Let s recst the first emple from the previous section Suppose tht the speed of the object is 3t t time t How fr does the object trvel between time t = nd time t = b? We re no longer ssuming tht we know where the object is t time t = or t ny other Chpter 7 Integrtion time Itiscertinlytruethtitissomewhere,solet ssupposethttt = thepositionisk Thenjustsintheemple, weknowthtthepositionoftheobjecttnytimeis3t /+k This mens tht t time t = the position is 3 /+k nd t time t = b the position is 3b /+k Therefore the chnge in position is 3b /+k (3 /+k) = 3b / 3 / Notice tht the k drops out; this mens tht it doesn t mtter tht we don t know k, it doesn t even mtter if we use the wrong k, we get the correct nswer In other words, to find the chnge in position between time nd time b we cn use ny ntiderivtive of the speed function 3t it need not be the one ntiderivtive tht ctully gives the loction of the object Wht bout the second pproch to this problem, in the new form? We now wnt to pproimte the chnge in position between time nd time b We tke the intervl of time between nd b, divide it into n subintervls, nd pproimte the distnce trveled during ech The strting time of subintervl number i is now +(i )(b )/n, which we bbrevite s t i, so tht t =, t = + (b )/n, nd so on The speed of the object is f(t) = 3t, nd ech subintervl is (b )/n = t seconds long The distnce trveled during subintervl number i is pproimtely f(t i ) t, nd the totl chnge in distnce is pproimtely f(t ) t+f(t ) t+ +f(t n ) t The ect chnge in position is the limit of this sum s n goes to infinity We bbrevite this sum using sigm nottion: n f(t i ) t = f(t ) t+f(t ) t+ +f(t n ) t i= The nottion on the left side of the equl sign uses lrge cpitl sigm, Greek letter, nd the left side is n bbrevition for the right side The nswer we seek is n lim f(t i ) t i= Since this must be the sme s the nswer we hve lredy obtined, we know tht n lim i= f(t i ) t = 3b 3 The significnce of 3t /, into which we substitute t = b nd t =, is of course tht it is function whose derivtive is f(t) As we hve discussed, by the time we know tht we

4 wnt to compute 7 The Fundmentl Theorem of Clculus 3 n lim f(t i ) t, i= it no longer mtters wht f(t) stnds for it could be speed, or the height of curve, or something else entirely We know tht the limit cn be computed by finding ny function with derivtive f(t), substituting nd b, nd subtrcting We summrize this in theorem First, we introduce some new nottion nd terms We write n f(t)dt = lim f(t i ) t i= if the limit eists Tht is, the left hnd side mens, or is n bbrevition for, the right hnd side The symbol is clled n integrl sign, nd the whole epression is red s the integrl of f(t) from to b Wht we hve lerned is tht this integrl cn be computed by finding function, sy F(t), with the property tht F (t) = f(t), nd then computing F(b) F() The function F(t) is clled n ntiderivtive of f(t) Now the theorem: THEOREM 7 Fundmentl Theorem of Clculus Suppose tht f() is continuous on the intervl [,b] If F() is ny ntiderivtive of f(), then Let s rewrite this slightly: f()d = F(b) F() f(t)dt = F() F() We ve replced the vrible by t nd b by These re just different nmes for quntities, so the substitution doesn t chnge the mening It does mke it esier to think of the two sides of the eqution s functions The epression is function: plug in vlue for, get out some other vlue The epression F() F() is of course lso function, nd it hs nice property: d d (F() F()) = F () = f(), 4 Chpter 7 Integrtion since F() is constnt nd hs derivtive zero In other words, by shifting our point of view slightly, we see tht the odd looking function G() = hs derivtive, nd tht in fct G () = f() This is relly just resttement of the Fundmentl Theorem of Clculus, nd indeed is often clled the Fundmentl Theorem of Clculus To void confusion, some people cll the two versions of the theorem The Fundmentl Theorem of Clculus, prt I nd The Fundmentl Theorem of Clculus, prt II, lthough unfortuntely there is no universl greement s to which is prt I nd which prt II Since it relly is the sme theorem, differently stted, some people simply cll them both The Fundmentl Theorem of Clculus THEOREM 7 Fundmentl Theorem of Clculus continuous on the intervl [,b] nd let Then G () = f() G() = Suppose tht f() is We hve not relly proved the Fundmentl Theorem In nutshell, we gve the following rgument to justify it: Suppose we wnt to know the vlue of n f(t)dt = lim f(t i ) t i= We cn interpret the right hnd side s the distnce trveled by n object whose speed is given by f(t) We know nother wy to compute the nswer to such problem: find the position of the object by finding n ntiderivtive of f(t), then substitute t = nd t = b nd subtrct to find the distnce trveled This must be the nswer to the originl problem s well, even if f(t) does not represent speed Wht s wrong with this? In some sense, nothing As prcticl mtter it is very convincing rgument, becuse our understnding of the reltionship between speed nd distnce seems to be quite solid From the point of view of mthemtics, however, it is unstisfctory to justify purely mthemticl reltionship by ppeling to our understnding of the physicl universe, which could, however unlikely it is in this cse, be wrong A complete proof is bit too involved to include here, but we will indicte how it goes First, if we cn prove the second version of the Fundmentl Theorem, theorem 7, then we cn prove the first version from tht:

5 Proof of Theorem 7 7 The Fundmentl Theorem of Clculus We know from theorem 7 tht G() = is n ntiderivtive of f(), nd therefore ny ntiderivtive F() of f() is of the form F() = G()+k Then It is not hrd to see tht F(b) F() = G(b)+k (G()+k) = G(b) G() = f(t)dt =, so this mens tht F(b) F() = which is ectly wht theorem 7 sys, f(t)dt So the rel job is to prove theorem 7 We will sketch the proof, using some fcts tht we do not prove First, the following identity is true of integrls: f(t)dt = c f(t)dt+ c This cn be proved directly from the definition of the integrl, tht is, using the limits of sums It is quite esy to see tht it must be true by thinking of either of the two pplictions of integrls tht we hve seen It turns out tht the identity is true no mtter wht c is, but it is esiest to think bout the mening when c b First, if f(t) represents speed, then we know tht the three integrls represent the distnce trveled between time nd time b; the distnce trveled between time nd time c; nd the distnce trveled between time c nd time b Clerly the sum of the ltter two is equl to the first of these Second, if f(t) represents the height of curve, the three integrls represent the re under the curve between nd b; the re under the curve between nd c; nd the re under the curve between c nd b Agin it is cler from the geometry tht the first is equl to the sum of the second nd third 6 Chpter 7 Integrtion Proof sketch for Theorem 7 We wnt to compute G (), so we strt with the definition of the derivtive in terms of limit: G G(+ ) G() () = lim ( + = lim f(t)dt ( = lim f(t)dt+ = lim + Now we need to know something bout + + ) f(t)dt when is smll; in fct, it is very close to f(), but we will not prove this Once gin, it is esy to believe this is true by thinking of our two pplictions: The integrl + cn be interpreted s the distnce trveled by n object over very short intervl of time Over sufficiently short period of time, the speed of the object will not chnge very much, so the distnce trveled will be pproimtely the length of time multiplied by the speed t the beginning of the intervl, nmely, f() Alterntely, the integrl my be interpreted s the re under the curve between nd + When is very smll, this will be very close to the re of the rectngle with bse nd height f(); gin this is f() If we ccept this, we my proceed: lim + which is wht we wnted to show f() f(t)dt = lim = f(), It is still true tht we re depending on n interprettion of the integrl to justify the rgument, but we hve isolted this prt of the rgument into two fcts tht re not too hrd to prove Once the lst reference to interprettion hs been removed from the proofs of these fcts, we will hve rel proof of the Fundmentl Theorem )

6 7 The Fundmentl Theorem of Clculus 7 Now we know tht to solve certin kinds of problems, those tht led to sum of certin form, we merely find n ntiderivtive nd substitute two vlues nd subtrct Unfortuntely, finding ntiderivtives cn be quite difficult While there re smll number of rules tht llow us to compute the derivtive of ny common function, there re no such rules for ntiderivtives There re some techniques tht frequently prove useful, but we will never be ble to reduce the problem to completely mechnicl process Becuse of the close reltionship between n integrl nd n ntiderivtive, the integrl sign is lso used to men ntiderivtive You cn tell which is intended by whether the limits of integrtion re included: d is n ordinry integrl, lso clled definite integrl, becuse it hs definite vlue, nmely We use d = = 7 3 d to denote the ntiderivtive of, lso clled n indefinite integrl So this is evluted s d = 3 3 +C It is customry to include the constnt C to indicte tht there re relly n infinite number of ntiderivtives We do not need this C to compute definite integrls, but in other circumstnces we will need to remember tht the C is there, so it is best to get into the hbit of writing the C When we compute definite integrl, we first find n ntiderivtive nd then substitute It is convenient to first disply the ntiderivtive nd then do the substitution; we need nottion indicting tht the substitution is yet to be done A typicl solution would look like this: d = 3 3 = = 7 3 The verticl line with subscript nd superscript is used to indicte the opertion substitute nd subtrct tht is needed to finish the evlution 8 Chpter 7 Integrtion Eercises 7 Find the ntiderivtives of the functions: 8 3t + 3 4/ 4 /z 7s 6 (+) 7 ( 6) 8 3/ 9 t 4 Compute the vlues of the integrls: t +3tdt d 4 3 d 6 7 Find the derivtive of G() = 8 Find the derivtive of G() = 9 Find the derivtive of G() = Find the derivtive of G() = Find the derivtive of G() = Find the derivtive of G() = t 3tdt t 3tdt e t dt e t dt tn(t )dt tn(t )dt π sintdt e d d ËÓÑ ÈÖÓÔ ÖØ Ó ÁÒØ Ö Ð º Suppose n object moves so tht its speed, or more properly velocity, is given by v(t) = t + t, s shown in figure 73 Let s emine the motion of this object crefully We know tht the velocity is the derivtive of position, so position is given by s(t) = t 3 /3 + t / + C Let s suppose tht t time t = the object is t position, so s(t) = t 3 /3+t /; this function is lso pictured in figure 73 Between t = nd t = the velocity is positive, so the object moves wy from the strting point, until it is bit pst position Then the velocity becomes negtive nd the object moves bck towrd its strting point The position of the object t t = is

7 Some Properties of Integrls Figure 73 The velocity of n object nd its position ectly s() = /6, nd t t = 6 it is s(6) = 8 The totl distnce trveled by the object is therefore /6+(/6 8) = 7/3 37 As we hve seen, we cn lso compute distnce trveled with n integrl; let s try it v(t)dt = t +tdt = t t = 8 Wht went wrong? Well, nothing relly, ecept tht it s not relly true fter ll tht we cn lso compute distnce trveled with n integrl Insted, s you might guess from this emple, the integrl ctully computes the net distnce trveled, tht is, the difference between the strting nd ending point As we hve lredy seen, v(t)dt = v(t)dt+ v(t) dt Computing the two integrls on the right (do it!) gives /6 nd 7/6, nd the sum of these is indeed 8 But wht does tht negtive sign men? It mens precisely wht you might think: it mens tht the object moves bckwrds To get the totl distnce trveled we cn dd /6+7/6 = 7/3, the sme nswer we got before Remember tht we cn lso interpret n integrl s mesuring n re, but now we see tht this too is little more complicted tht we hve suspected The re under the curve v(t) from to is given by nd the re from to 6 is v(t)dt = 6, v(t)dt = Chpter 7 Integrtion In other words, the re between the -is nd the curve, but under the -is, counts s negtive re So the integrl v(t)dt = 8 mesures net re, the re bove the is minus the (positive) re below the is If we recll tht the integrl is the limit of certin kind of sum, this behvior is not surprising Recll the sort of sum involved: n v(t i ) t i= In ech term v(t) t the t is positive, but if v(t i ) is negtive then the term is negtive If over n entire intervl, like to 6, the function is lwys negtive, then the entire sum is negtive In terms of re, v(t) t is then negtive height times positive width, giving negtive rectngle re So now we see tht when evluting v(t)dt = 7 6 by finding n ntiderivtive, substituting, nd subtrcting, we get surprising nswer, but one tht turns out to mke sense Let s now try something bit different: 6 v(t)dt = t3 3 + t 6 = = 7 6 Here we simply interchnged the limits nd 6, so of course when we substitute nd subtrct we re subtrcting in the opposite order nd we end up multiplying the nswer by This too mkes sense in terms of the underlying sum, though it tkes bit more thought Recll tht in the sum n v(t i ) t, i= the t is the length of ech little subintervl, but more precisely we could sy tht t = t i+ t i, the difference between two endpoints of subintervl We hve until now ssumed tht we were working left to right, but could s well number the subintervls from

8 73 Some Properties of Integrls 6 right to left, so tht t = b nd t n = Then t = t i+ t i is negtive nd in 6 n v(t)dt = v(t i ) t, the vlues v(t i ) re negtive but lso t is negtive, so ll terms re positive gin On the other hnd, in n v(t)dt = v(t i ) t, the vlues v(t i ) re positive but t is negtive,nd we get negtive result: v(t)dt = t3 3 + t i= i= Finlly we note one simple property of integrls: f()+g()d = = 3 3 = 6 f()d+ g() d This is esy to understnd once you recll tht (F()+G()) = F ()+G () Hence, if F () = f() nd G () = g(), then f()+g()d = (F()+G()) b = F(b)+G(b) F() G() = F(b) F()+G(b) G() 6 Chpter 7 Integrtion nd if < b nd f() on [,b] then nd in fct Eercises 73 f()d = f()d f() d An object moves so tht its velocity t time t is v(t) = 98t+ m/s Describe the motion of the object between t = nd t =, find the totl distnce trveled by the object during tht time, nd find the net distnce trveled An object moves so tht its velocity t time t is v(t) = sint Set up nd evlute single definite integrl to compute the net distnce trveled between t = nd t = π 3 An object moves so tht its velocity t time t is v(t) = +sint m/s Find the net distnce trveled by the object between t = nd t = π, nd find the totl distnce trveled during the sme period 4 Consider the function f() = ( + )( + )( )( ) on [,] Find the totl re between the curve nd the -is (mesuring ll re s positive) Consider the function f() = 3 + on [,4] Find the totl re between the curve nd the -is (mesuring ll re s positive) 6 Evlute the three integrls: A = 3 ( +9)d B = nd verify tht A = B +C 4 ( +9)d C = 3 4 ( +9)d, = F() b + G() b = f()d+ g() d In summry, we will frequently use these properties of integrls: f()d = c f()+g()d = f()d+ f()d = b c f()d+ f() d f() d g() d

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( ) Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +

More information

AREA OF A SURFACE OF REVOLUTION

AREA OF A SURFACE OF REVOLUTION AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7.

More information

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde

More information

Algebra Review. How well do you remember your algebra?

Algebra Review. How well do you remember your algebra? Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then

More information

SPECIAL PRODUCTS AND FACTORIZATION

SPECIAL PRODUCTS AND FACTORIZATION MODULE - Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come

More information

Operations with Polynomials

Operations with Polynomials 38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply

More information

5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.

5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one. 5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous rel-vlued

More information

Integration by Substitution

Integration by Substitution Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is

More information

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive

More information

Factoring Polynomials

Factoring Polynomials Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles

More information

Introduction to Integration Part 2: The Definite Integral

Introduction to Integration Part 2: The Definite Integral Mthemtics Lerning Centre Introduction to Integrtion Prt : The Definite Integrl Mr Brnes c 999 Universit of Sdne Contents Introduction. Objectives...... Finding Ares 3 Ares Under Curves 4 3. Wht is the

More information

Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.

Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions. Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd

More information

9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes

9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes The Sclr Product 9.3 Introduction There re two kinds of multipliction involving vectors. The first is known s the sclr product or dot product. This is so-clled becuse when the sclr product of two vectors

More information

Appendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:

Appendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered: Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you

More information

EQUATIONS OF LINES AND PLANES

EQUATIONS OF LINES AND PLANES EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in point-direction nd twopoint

More information

Reasoning to Solve Equations and Inequalities

Reasoning to Solve Equations and Inequalities Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing

More information

Graphs on Logarithmic and Semilogarithmic Paper

Graphs on Logarithmic and Semilogarithmic Paper 0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl

More information

P.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn

P.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn 33337_0P03.qp 2/27/06 24 9:3 AM Chpter P Pge 24 Prerequisites P.3 Polynomils nd Fctoring Wht you should lern Polynomils An lgeric epression is collection of vriles nd rel numers. The most common type of

More information

The Definite Integral

The Definite Integral Chpter 4 The Definite Integrl 4. Determining distnce trveled from velocity Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: If we know

More information

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100 hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by

More information

Math 135 Circles and Completing the Square Examples

Math 135 Circles and Completing the Square Examples Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for

More information

6.2 Volumes of Revolution: The Disk Method

6.2 Volumes of Revolution: The Disk Method mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine so-clled volumes of

More information

Section 7-4 Translation of Axes

Section 7-4 Translation of Axes 62 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 7-4 Trnsltion of Aes Trnsltion of Aes Stndrd Equtions of Trnslted Conics Grphing Equtions of the Form A 2 C 2 D E F 0 Finding Equtions of Conics In the

More information

Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.

Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers. 2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this

More information

LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES

LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of

More information

Binary Representation of Numbers Autar Kaw

Binary Representation of Numbers Autar Kaw Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse- rel number to its binry representtion,. convert binry number to n equivlent bse- number. In everydy

More information

Experiment 6: Friction

Experiment 6: Friction Experiment 6: Friction In previous lbs we studied Newton s lws in n idel setting, tht is, one where friction nd ir resistnce were ignored. However, from our everydy experience with motion, we know tht

More information

10.6 Applications of Quadratic Equations

10.6 Applications of Quadratic Equations 10.6 Applictions of Qudrtic Equtions In this section we wnt to look t the pplictions tht qudrtic equtions nd functions hve in the rel world. There re severl stndrd types: problems where the formul is given,

More information

1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator

1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator AP Clculus Finl Review Sheet When you see the words. This is wht you think of doing. Find the zeros Find roots. Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor.

More information

15.6. The mean value and the root-mean-square value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style

15.6. The mean value and the root-mean-square value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style The men vlue nd the root-men-squre vlue of function 5.6 Introduction Currents nd voltges often vry with time nd engineers my wish to know the verge vlue of such current or voltge over some prticulr time

More information

Module Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials

Module Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials MEI Mthemtics in Ection nd Instry Topic : Proof MEI Structured Mthemtics Mole Summry Sheets C, Methods for Anced Mthemtics (Version B reference to new book) Topic : Nturl Logrithms nd Eponentils Topic

More information

Bayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom

Bayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom Byesin Updting with Continuous Priors Clss 3, 8.05, Spring 04 Jeremy Orloff nd Jonthn Bloom Lerning Gols. Understnd prmeterized fmily of distriutions s representing continuous rnge of hypotheses for the

More information

Applications to Physics and Engineering

Applications to Physics and Engineering Section 7.5 Applictions to Physics nd Engineering Applictions to Physics nd Engineering Work The term work is used in everydy lnguge to men the totl mount of effort required to perform tsk. In physics

More information

9 CONTINUOUS DISTRIBUTIONS

9 CONTINUOUS DISTRIBUTIONS 9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete

More information

6 Energy Methods And The Energy of Waves MATH 22C

6 Energy Methods And The Energy of Waves MATH 22C 6 Energy Methods And The Energy of Wves MATH 22C. Conservtion of Energy We discuss the principle of conservtion of energy for ODE s, derive the energy ssocited with the hrmonic oscilltor, nd then use this

More information

Or more simply put, when adding or subtracting quantities, their uncertainties add.

Or more simply put, when adding or subtracting quantities, their uncertainties add. Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re

More information

2005-06 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration

2005-06 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration Source: http://www.mth.cuhk.edu.hk/~mt26/mt26b/notes/notes3.pdf 25-6 Second Term MAT26B 1 Supplementry Notes 3 Interchnge of Differentition nd Integrtion The theme of this course is bout vrious limiting

More information

Lecture 3 Gaussian Probability Distribution

Lecture 3 Gaussian Probability Distribution Lecture 3 Gussin Probbility Distribution Introduction l Gussin probbility distribution is perhps the most used distribution in ll of science. u lso clled bell shped curve or norml distribution l Unlike

More information

MATH 150 HOMEWORK 4 SOLUTIONS

MATH 150 HOMEWORK 4 SOLUTIONS MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers 65 1000 8 2001 + 3 177, 79 1212 9 2399 + 2 2001, nd 24 4493 5 8192 + 7 1777 is nonnegtive. Is your proof constructive

More information

www.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values)

www.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values) www.mthsbo.org.uk CORE SUMMARY NOTES Functions A function is rule which genertes ectl ONE OUTPUT for EVERY INPUT. To be defined full the function hs RULE tells ou how to clculte the output from the input

More information

Thinking out of the Box... Problem It s a richer problem than we ever imagined

Thinking out of the Box... Problem It s a richer problem than we ever imagined From the Mthemtics Techer, Vol. 95, No. 8, pges 568-574 Wlter Dodge (not pictured) nd Steve Viktor Thinking out of the Bo... Problem It s richer problem thn we ever imgined The bo problem hs been stndrd

More information

Section 5-4 Trigonometric Functions

Section 5-4 Trigonometric Functions 5- Trigonometric Functions Section 5- Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form

More information

Review guide for the final exam in Math 233

Review guide for the final exam in Math 233 Review guide for the finl exm in Mth 33 1 Bsic mteril. This review includes the reminder of the mteril for mth 33. The finl exm will be cumultive exm with mny of the problems coming from the mteril covered

More information

Helicopter Theme and Variations

Helicopter Theme and Variations Helicopter Theme nd Vritions Or, Some Experimentl Designs Employing Pper Helicopters Some possible explntory vribles re: Who drops the helicopter The length of the rotor bldes The height from which the

More information

Unit 6: Exponents and Radicals

Unit 6: Exponents and Radicals Eponents nd Rdicls -: The Rel Numer Sstem Unit : Eponents nd Rdicls Pure Mth 0 Notes Nturl Numers (N): - counting numers. {,,,,, } Whole Numers (W): - counting numers with 0. {0,,,,,, } Integers (I): -

More information

Physics 43 Homework Set 9 Chapter 40 Key

Physics 43 Homework Set 9 Chapter 40 Key Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nm-wide region t x

More information

A.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324

A.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324 A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................

More information

Review Problems for the Final of Math 121, Fall 2014

Review Problems for the Final of Math 121, Fall 2014 Review Problems for the Finl of Mth, Fll The following is collection of vrious types of smple problems covering sections.,.5, nd.7 6.6 of the text which constitute only prt of the common Mth Finl. Since

More information

RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS

RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS Known for over 500 yers is the fct tht the sum of the squres of the legs of right tringle equls the squre of the hypotenuse. Tht is +b c. A simple proof is

More information

Econ 4721 Money and Banking Problem Set 2 Answer Key

Econ 4721 Money and Banking Problem Set 2 Answer Key Econ 472 Money nd Bnking Problem Set 2 Answer Key Problem (35 points) Consider n overlpping genertions model in which consumers live for two periods. The number of people born in ech genertion grows in

More information

CURVES ANDRÉ NEVES. that is, the curve α has finite length. v = p q p q. a i.e., the curve of smallest length connecting p to q is a straight line.

CURVES ANDRÉ NEVES. that is, the curve α has finite length. v = p q p q. a i.e., the curve of smallest length connecting p to q is a straight line. CURVES ANDRÉ NEVES 1. Problems (1) (Ex 1 of 1.3 of Do Crmo) Show tht the tngent line to the curve α(t) (3t, 3t 2, 2t 3 ) mkes constnt ngle with the line z x, y. (2) (Ex 6 of 1.3 of Do Crmo) Let α(t) (e

More information

and thus, they are similar. If k = 3 then the Jordan form of both matrices is

and thus, they are similar. If k = 3 then the Jordan form of both matrices is Homework ssignment 11 Section 7. pp. 249-25 Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If

More information

3 The Utility Maximization Problem

3 The Utility Maximization Problem 3 The Utility Mxiiztion Proble We hve now discussed how to describe preferences in ters of utility functions nd how to forulte siple budget sets. The rtionl choice ssuption, tht consuers pick the best

More information

Basic Analysis of Autarky and Free Trade Models

Basic Analysis of Autarky and Free Trade Models Bsic Anlysis of Autrky nd Free Trde Models AUTARKY Autrky condition in prticulr commodity mrket refers to sitution in which country does not engge in ny trde in tht commodity with other countries. Consequently

More information

CHAPTER 11 Numerical Differentiation and Integration

CHAPTER 11 Numerical Differentiation and Integration CHAPTER 11 Numericl Differentition nd Integrtion Differentition nd integrtion re bsic mthemticl opertions with wide rnge of pplictions in mny res of science. It is therefore importnt to hve good methods

More information

DIFFERENTIATING UNDER THE INTEGRAL SIGN

DIFFERENTIATING UNDER THE INTEGRAL SIGN DIFFEENTIATING UNDE THE INTEGAL SIGN KEITH CONAD I hd lerned to do integrls by vrious methods shown in book tht my high school physics techer Mr. Bder hd given me. [It] showed how to differentite prmeters

More information

Regular Sets and Expressions

Regular Sets and Expressions Regulr Sets nd Expressions Finite utomt re importnt in science, mthemtics, nd engineering. Engineers like them ecuse they re super models for circuits (And, since the dvent of VLSI systems sometimes finite

More information

Vectors 2. 1. Recap of vectors

Vectors 2. 1. Recap of vectors Vectors 2. Recp of vectors Vectors re directed line segments - they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms

More information

Lectures 8 and 9 1 Rectangular waveguides

Lectures 8 and 9 1 Rectangular waveguides 1 Lectures 8 nd 9 1 Rectngulr wveguides y b x z Consider rectngulr wveguide with 0 < x b. There re two types of wves in hollow wveguide with only one conductor; Trnsverse electric wves

More information

Distributions. (corresponding to the cumulative distribution function for the discrete case).

Distributions. (corresponding to the cumulative distribution function for the discrete case). Distributions Recll tht n integrble function f : R [,] such tht R f()d = is clled probbility density function (pdf). The distribution function for the pdf is given by F() = (corresponding to the cumultive

More information

Warm-up for Differential Calculus

Warm-up for Differential Calculus Summer Assignment Wrm-up for Differentil Clculus Who should complete this pcket? Students who hve completed Functions or Honors Functions nd will be tking Differentil Clculus in the fll of 015. Due Dte:

More information

Treatment Spring Late Summer Fall 0.10 5.56 3.85 0.61 6.97 3.01 1.91 3.01 2.13 2.99 5.33 2.50 1.06 3.53 6.10 Mean = 1.33 Mean = 4.88 Mean = 3.

Treatment Spring Late Summer Fall 0.10 5.56 3.85 0.61 6.97 3.01 1.91 3.01 2.13 2.99 5.33 2.50 1.06 3.53 6.10 Mean = 1.33 Mean = 4.88 Mean = 3. The nlysis of vrince (ANOVA) Although the t-test is one of the most commonly used sttisticl hypothesis tests, it hs limittions. The mjor limittion is tht the t-test cn be used to compre the mens of only

More information

MODULE 3. 0, y = 0 for all y

MODULE 3. 0, y = 0 for all y Topics: Inner products MOULE 3 The inner product of two vectors: The inner product of two vectors x, y V, denoted by x, y is (in generl) complex vlued function which hs the following four properties: i)

More information

Numerical Methods of Approximating Definite Integrals

Numerical Methods of Approximating Definite Integrals 6 C H A P T E R Numericl Methods o Approimting Deinite Integrls 6. APPROXIMATING SUMS: L n, R n, T n, AND M n Introduction Not only cn we dierentite ll the bsic unctions we ve encountered, polynomils,

More information

Pure C4. Revision Notes

Pure C4. Revision Notes Pure C4 Revision Notes Mrch 0 Contents Core 4 Alger Prtil frctions Coordinte Geometry 5 Prmetric equtions 5 Conversion from prmetric to Crtesin form 6 Are under curve given prmetriclly 7 Sequences nd

More information

Derivatives and Rates of Change

Derivatives and Rates of Change Section 2.1 Derivtives nd Rtes of Cnge 2010 Kiryl Tsiscnk Derivtives nd Rtes of Cnge Te Tngent Problem EXAMPLE: Grp te prbol y = x 2 nd te tngent line t te point P(1,1). Solution: We ve: DEFINITION: Te

More information

4.11 Inner Product Spaces

4.11 Inner Product Spaces 314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible. 4.11 Inner Product Spces

More information

Rotating DC Motors Part II

Rotating DC Motors Part II Rotting Motors rt II II.1 Motor Equivlent Circuit The next step in our consiertion of motors is to evelop n equivlent circuit which cn be use to better unerstn motor opertion. The rmtures in rel motors

More information

Exponential and Logarithmic Functions

Exponential and Logarithmic Functions Nme Chpter Eponentil nd Logrithmic Functions Section. Eponentil Functions nd Their Grphs Objective: In this lesson ou lerned how to recognize, evlute, nd grph eponentil functions. Importnt Vocbulr Define

More information

MA 15800 Lesson 16 Notes Summer 2016 Properties of Logarithms. Remember: A logarithm is an exponent! It behaves like an exponent!

MA 15800 Lesson 16 Notes Summer 2016 Properties of Logarithms. Remember: A logarithm is an exponent! It behaves like an exponent! MA 5800 Lesson 6 otes Summer 06 Rememer: A logrithm is n eponent! It ehves like n eponent! In the lst lesson, we discussed four properties of logrithms. ) log 0 ) log ) log log 4) This lesson covers more

More information

All pay auctions with certain and uncertain prizes a comment

All pay auctions with certain and uncertain prizes a comment CENTER FOR RESEARC IN ECONOMICS AND MANAGEMENT CREAM Publiction No. 1-2015 All py uctions with certin nd uncertin prizes comment Christin Riis All py uctions with certin nd uncertin prizes comment Christin

More information

Solving BAMO Problems

Solving BAMO Problems Solving BAMO Problems Tom Dvis tomrdvis@erthlink.net http://www.geometer.org/mthcircles Februry 20, 2000 Abstrct Strtegies for solving problems in the BAMO contest (the By Are Mthemticl Olympid). Only

More information

FUNCTIONS AND EQUATIONS. xεs. The simplest way to represent a set is by listing its members. We use the notation

FUNCTIONS AND EQUATIONS. xεs. The simplest way to represent a set is by listing its members. We use the notation FUNCTIONS AND EQUATIONS. SETS AND SUBSETS.. Definition of set. A set is ny collection of objects which re clled its elements. If x is n element of the set S, we sy tht x belongs to S nd write If y does

More information

Radius of the Earth - Radii Used in Geodesy James R. Clynch February 2006

Radius of the Earth - Radii Used in Geodesy James R. Clynch February 2006 dius of the Erth - dii Used in Geodesy Jmes. Clynch Februry 006 I. Erth dii Uses There is only one rdius of sphere. The erth is pproximtely sphere nd therefore, for some cses, this pproximtion is dequte.

More information

Lecture 5. Inner Product

Lecture 5. Inner Product Lecture 5 Inner Product Let us strt with the following problem. Given point P R nd line L R, how cn we find the point on the line closest to P? Answer: Drw line segment from P meeting the line in right

More information

PROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1

PROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1 PROBLEMS - APPLICATIONS OF DERIVATIVES Pge ( ) Wter seeps out of conicl filter t the constnt rte of 5 cc / sec. When the height of wter level in the cone is 5 cm, find the rte t which the height decreses.

More information

19. The Fermat-Euler Prime Number Theorem

19. The Fermat-Euler Prime Number Theorem 19. The Fermt-Euler Prime Number Theorem Every prime number of the form 4n 1 cn be written s sum of two squres in only one wy (side from the order of the summnds). This fmous theorem ws discovered bout

More information

The Velocity Factor of an Insulated Two-Wire Transmission Line

The Velocity Factor of an Insulated Two-Wire Transmission Line The Velocity Fctor of n Insulted Two-Wire Trnsmission Line Problem Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Mrch 7, 008 Estimte the velocity fctor F = v/c nd the

More information

Physics 6010, Fall 2010 Symmetries and Conservation Laws: Energy, Momentum and Angular Momentum Relevant Sections in Text: 2.6, 2.

Physics 6010, Fall 2010 Symmetries and Conservation Laws: Energy, Momentum and Angular Momentum Relevant Sections in Text: 2.6, 2. Physics 6010, Fll 2010 Symmetries nd Conservtion Lws: Energy, Momentum nd Angulr Momentum Relevnt Sections in Text: 2.6, 2.7 Symmetries nd Conservtion Lws By conservtion lw we men quntity constructed from

More information

Version 001 Summer Review #03 tubman (IBII20142015) 1

Version 001 Summer Review #03 tubman (IBII20142015) 1 Version 001 Summer Reiew #03 tubmn (IBII20142015) 1 This print-out should he 35 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. Concept 20 P03

More information

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus Section 5.4 Te Funmentl Teorem of Clculus Kiryl Tsiscnk Te Funmentl Teorem of Clculus EXAMPLE: If f is function wose grp is sown below n g() = f(t)t, fin te vlues of g(), g(), g(), g(3), g(4), n g(5).

More information

5.6 POSITIVE INTEGRAL EXPONENTS

5.6 POSITIVE INTEGRAL EXPONENTS 54 (5 ) Chpter 5 Polynoils nd Eponents 5.6 POSITIVE INTEGRAL EXPONENTS In this section The product rule for positive integrl eponents ws presented in Section 5., nd the quotient rule ws presented in Section

More information

Euler Euler Everywhere Using the Euler-Lagrange Equation to Solve Calculus of Variation Problems

Euler Euler Everywhere Using the Euler-Lagrange Equation to Solve Calculus of Variation Problems Euler Euler Everywhere Using the Euler-Lgrnge Eqution to Solve Clculus of Vrition Problems Jenine Smllwood Principles of Anlysis Professor Flschk My 12, 1998 1 1. Introduction Clculus of vritions is brnch

More information

Small Business Cloud Services

Small Business Cloud Services Smll Business Cloud Services Summry. We re thick in the midst of historic se-chnge in computing. Like the emergence of personl computers, grphicl user interfces, nd mobile devices, the cloud is lredy profoundly

More information

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur Module Anlysis of Stticlly Indeterminte Structures by the Mtrix Force Method Version CE IIT, Khrgpur esson 9 The Force Method of Anlysis: Bems (Continued) Version CE IIT, Khrgpur Instructionl Objectives

More information

AAPT UNITED STATES PHYSICS TEAM AIP 2010

AAPT UNITED STATES PHYSICS TEAM AIP 2010 2010 F = m Exm 1 AAPT UNITED STATES PHYSICS TEAM AIP 2010 Enti non multiplicnd sunt preter necessittem 2010 F = m Contest 25 QUESTIONS - 75 MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD

More information

Babylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity

Babylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity Bbylonin Method of Computing the Squre Root: Justifictions Bsed on Fuzzy Techniques nd on Computtionl Complexity Olg Koshelev Deprtment of Mthemtics Eduction University of Texs t El Pso 500 W. University

More information

Small Business Networking

Small Business Networking Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd business. Introducing technology

More information

Small Business Networking

Small Business Networking Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd processes. Introducing technology

More information

Multiplication and Division - Left to Right. Addition and Subtraction - Left to Right.

Multiplication and Division - Left to Right. Addition and Subtraction - Left to Right. Order of Opertions r of Opertions Alger P lese Prenthesis - Do ll grouped opertions first. E cuse Eponents - Second M D er Multipliction nd Division - Left to Right. A unt S hniqu Addition nd Sutrction

More information

Linear Equations in Two Variables

Linear Equations in Two Variables Liner Equtions in Two Vribles In this chpter, we ll use the geometry of lines to help us solve equtions. Liner equtions in two vribles If, b, ndr re rel numbers (nd if nd b re not both equl to 0) then

More information

Small Business Networking

Small Business Networking Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd business. Introducing technology

More information

PHY 140A: Solid State Physics. Solution to Homework #2

PHY 140A: Solid State Physics. Solution to Homework #2 PHY 140A: Solid Stte Physics Solution to Homework # TA: Xun Ji 1 October 14, 006 1 Emil: jixun@physics.ucl.edu Problem #1 Prove tht the reciprocl lttice for the reciprocl lttice is the originl lttice.

More information

CS99S Laboratory 2 Preparation Copyright W. J. Dally 2001 October 1, 2001

CS99S Laboratory 2 Preparation Copyright W. J. Dally 2001 October 1, 2001 CS99S Lortory 2 Preprtion Copyright W. J. Dlly 2 Octoer, 2 Ojectives:. Understnd the principle of sttic CMOS gte circuits 2. Build simple logic gtes from MOS trnsistors 3. Evlute these gtes to oserve logic

More information

g(y(a), y(b)) = o, B a y(a)+b b y(b)=c, Boundary Value Problems Lecture Notes to Accompany

g(y(a), y(b)) = o, B a y(a)+b b y(b)=c, Boundary Value Problems Lecture Notes to Accompany Lecture Notes to Accompny Scientific Computing An Introductory Survey Second Edition by Michel T Heth Boundry Vlue Problems Side conditions prescribing solution or derivtive vlues t specified points required

More information

Chapter 2 The Number System (Integers and Rational Numbers)

Chapter 2 The Number System (Integers and Rational Numbers) Chpter 2 The Number System (Integers nd Rtionl Numbers) In this second chpter, students extend nd formlize their understnding of the number system, including negtive rtionl numbers. Students first develop

More information

1.2 The Integers and Rational Numbers

1.2 The Integers and Rational Numbers .2. THE INTEGERS AND RATIONAL NUMBERS.2 The Integers n Rtionl Numers The elements of the set of integers: consist of three types of numers: Z {..., 5, 4, 3, 2,, 0,, 2, 3, 4, 5,...} I. The (positive) nturl

More information

Small Business Networking

Small Business Networking Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd processes. Introducing technology

More information

Redistributing the Gains from Trade through Non-linear. Lump-sum Transfers

Redistributing the Gains from Trade through Non-linear. Lump-sum Transfers Redistributing the Gins from Trde through Non-liner Lump-sum Trnsfers Ysukzu Ichino Fculty of Economics, Konn University April 21, 214 Abstrct I exmine lump-sum trnsfer rules to redistribute the gins from

More information