(0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+ +(2.7)(0.1) = 1.35

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "(0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+ +(2.7)(0.1) = 1.35"

Transcription

1 7 Integrtion º½ ÌÛÓ Ü ÑÔÐ Up to now we hve been concerned with extrcting informtion bout how function chnges from the function itself. Given knowledge bout n object s position, for exmple, 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 exmple, t t = the object is t position 3/+ =.5. 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 pproximte solution. We know tht t t = the object is t position. How might we pproximte 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 pproximting the loction of the object 47

2 48 Chpter 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 pproximtion tht the object trvels (.)(.)+(.3)(.)+(.6)(.)+ +(.7)(.) =.35 centimeters, ending up t position.35. This is better pproximtion thn, certinly, but is still just n pproximtion. (We know in fct tht the object ends up t position.5, becuse we ve lredy done the problem using the first pproch.) Presumbly, we will get better pproximtion if we divide the time into one hundred intervls of hundredth of second ech, nd repet the process: (.)(.)+(.3)(.)+(.6)(.)+ +(.97)(.) =.485. We thus pproximte the position s.485. Since we know the exct 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 theexctnswer ifwe simplycompute more nd more exmples. Let s insted look t typicl pproximtion. 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 pproximte the distnce trveled s (.)(/n) =, s before. During the second time intervl, from t = /n to t = /n, the object trvels pproximtely 3(/n)(/n) = 3/n centimeters. During time intervl number i, the object trvels pproximtely (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 pproximte 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

3 7. Two exmples 49 more useful. We cn fctor out 3 nd /n to get 3 n ( (n )), 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 = k(k +). In our cse we re interested in k = n, so (n ) = (n )(n) = n n. This simplifies the pproximte distnce trveled to 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 pproximtion gets closer nd closer to (3/)( ) = 3/, so tht 3/ is the exct distnce trveled during one second, nd the finl position is.5. 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 pproximte 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 pproximtely 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 )) = n n = 3 ( t ). n In the limit, s n gets lrger, this gets closer nd closer to (3/)t nd the pproximted position of the object gets closer nd closer to (3/)t +, so the ctul position is (3/)t +, exctly the nswer given by the first pproch to the problem. EXAMPLE 7.. Find the re under the curve y = 3x between x = nd ny positive vlue x. There is here no obvious nlogue to the first pproch in the previous exmple,

4 5 Chpter 7 Integrtion but the second pproch works fine. (Becuse the function y = 3x 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 pproximte the desired re? We know how to compute res of rectngles, so we pproximte the re by rectngles. Jumping stright to the generl cse, suppose we divide the intervl between nd x into n equl subintervls, nd use rectngle bove ech subintervl to pproximte 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 )(x/n), the width is x/n, nd the re is 3(i )(x /n ). The totl re of the rectngles is () x n +3()x n +3()x n +3(3)x n + +3(n )x n. By fctoring out 3x /n this simplifies to 3x 3x n n (+++ +(n )) = n n = 3 ( x ). n As n gets lrger this gets closer nd closer to 3x /, which must therefore be the true re under the curve..... Figure 7.. Approximting the re under y = 3x 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 exmple ppers to be much different thn the problem in the first exmple, nd while the esy pproch to problem one does not pper to pply to problem two, the pproximtion pproch works in both, nd moreover the clcultions re identicl. As we will see, there

5 7. The Fundmentl Theorem of Clculus 5 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 pproximte solutions we end up with mthemtics tht looks like the two exmples, 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, 3x. 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. Exercises 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 5. 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 5. Find the position of the object t t =. 3. By method similr to tht in exmple 7.., find the re under y = x between x = nd ny positive vlue for x. 4. By method similr to tht in exmple 7.., find the re under y = 4x between x = nd ny positive vlue for x. 5. By method similr to tht in exmple 7.., find the re under y = 4x between x = nd ny positive vlue for x bigger thn. 6. By method similr to tht in exmple 7.., find the re under y = 4x between ny two positive vlues for x, sy < b. 7. Let f(x) = x + 3x +. Approximte the re under the curve between x = nd x = using 4 rectngles nd lso using 8 rectngles. 8. Let f(x) = x x + 3. Approximte the re under the curve between x = nd x = 3 using 4 rectngles. º¾ Ì ÙÒ Ñ ÒØ Ð Ì ÓÖ Ñ Ó ÐÙÐÙ Let s recst the first exmple 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

6 5 Chpter 7 Integrtion time. Itiscertinlytruethtitissomewhere,solet ssupposethttt = thepositionisk. Thenjustsintheexmple, 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 pproximte the chnge in position between time nd time b. We tke the intervl of time between nd b, divide it into n subintervls, nd pproximte 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 pproximtely f(t i ) t, nd the totl chnge in distnce is pproximtely f(t ) t+f(t ) t+ +f(t n ) t. The exct chnge in position is the limit of this sum s n goes to infinity. We bbrevite this sum using sigm nottion: n i= f(t i ) t = f(t ) t+f(t ) t+ +f(t n ) t. 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 n i= f(t i ) t. Since this must be the sme s the nswer we hve lredy obtined, we know tht n lim n 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

7 wnt to compute 7. The Fundmentl Theorem of Clculus 53 n lim n i= f(t i ) t, 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 n if the limit exists. 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 expression 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(x) is continuous on the intervl [,b]. If F(x) is ny ntiderivtive of f(x), then i= f(x)dx = F(b) F(). Let s rewrite this slightly: f(t)dt = F(x) F(). We ve replced the vrible x by t nd b by x. 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 expression f(t)dt is function: plug in vlue for x, get out some other vlue. The expression F(x) F() is of course lso function, nd it hs nice property: d dx (F(x) F()) = F (x) = f(x),

8 54 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(x) = f(t)dt hs derivtive, nd tht in fct G (x) = f(x). 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 Suppose tht f(x) is G(x) = f(t)dt. Then G (x) = f(x). 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. n 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: i=

9 7. The Fundmentl Theorem of Clculus 55 Proof of Theorem 7... We know from theorem 7.. tht G(x) = f(t)dt is n ntiderivtive of f(x), nd therefore ny ntiderivtive F(x) of f(x) is of the form F(x) = G(x)+k. Then F(b) F() = G(b)+k (G()+k) = G(b) G() = f(t)dt f(t)dt. It is not hrd to see tht f(t)dt =, so this mens tht F(b) F() = which is exctly 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 f(t)dt. 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.

10 56 Chpter 7 Integrtion Proof sketch for Theorem 7... We wnt to compute G (x), so we strt with the definition of the derivtive in terms of limit: G G(x+ x) G(x) (x) = lim x x ( + x = lim x x f(t)dt ( x = lim f(t)dt+ x x = lim x x + x Now we need to know something bout x f(t)dt. + x x f(t)dt ) f(t)dt f(t)dt ) + x x f(t)dt when x is smll; in fct, it is very close to xf(x), but we will not prove this. Once gin, it is esy to believe this is true by thinking of our two pplictions: The integrl + x x f(t)dt 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 pproximtely the length of time multiplied by the speed t the beginning of the intervl, nmely, xf(x). Alterntely, the integrl my be interpreted s the re under the curve between x nd x+ x. When x is very smll, this will be very close to the re of the rectngle with bse x nd height f(x); gin this is xf(x). If we ccept this, we my proceed: lim x which is wht we wnted to show. x+ x xf(x) f(t)dt = lim x x x x = f(x), 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.

11 7. The Fundmentl Theorem of Clculus 57 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: x dx is n ordinry integrl, lso clled definite integrl, becuse it hs definite vlue, nmely We use x dx = = 7 3. x dx to denote the ntiderivtive of x, lso clled n indefinite integrl. So this is evluted s x dx = x3 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: x dx = x3 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.

12 58 Chpter 7 Integrtion Exercises 7.. Find the ntiderivtives of the functions:. 8 x. 3t / x 4. /z 5. 7s 6. (5x+) 7. (x 6) 8. x 3/ 9. x x. t 4 Compute the vlues of the integrls: t +3tdt. dx 4. x x 3 dx Find the derivtive of G(x) = 8. Find the derivtive of G(x) = 9. Find the derivtive of G(x) =. Find the derivtive of G(x) =. Find the derivtive of G(x) =. Find the derivtive of G(x) = t 3tdt t 3tdt e t dt e t dt tn(t )dt tn(t )dt π 5 sintdt e x dx x 5 dx º ËÓÑ ÈÖÓÔ ÖØ Ó ÁÒØ Ö Ð Suppose n object moves so tht its speed, or more properly velocity, is given by v(t) = t + 5t, s shown in figure Let s exmine 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 + 5t / + C. Let s suppose tht t time t = the object is t position, so s(t) = t 3 /3+5t /; this function is lso pictured in figure Between t = nd t = 5 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 = 5 is

13 7.3 Some Properties of Integrls Figure 7.3. The velocity of n object nd its position. exctly s(5) = 5/6, nd t t = 6 it is s(6) = 8. The totl distnce trveled by the object is therefore 5/6+(5/6 8) = 7/ As we hve seen, we cn lso compute distnce trveled with n integrl; let s try it. 6 v(t)dt = 6 t +5tdt = t t 6 = 8. Wht went wrong? Well, nothing relly, except 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 exmple, the integrl ctully computes the net distnce trveled, tht is, the difference between the strting nd ending point. As we hve lredy seen, 6 v(t)dt = 5 v(t)dt+ 6 5 v(t) dt. Computing the two integrls on the right (do it!) gives 5/6 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 5/6+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 5 is given by 5 v(t)dt = 5 6, nd the re from 5 to 6 is 6 5 v(t)dt = 7 6.

14 6 Chpter 7 Integrtion In other words, the re between the x-xis nd the curve, but under the x-xis, counts s negtive re. So the integrl 6 v(t)dt = 8 mesures net re, the re bove the xis minus the (positive) re below the xis. 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 5 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 6 5 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: 5 6 v(t)dt = t t 5 6 = = 7 6. Here we simply interchnged the limits 5 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

15 7.3 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 5 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, 5 the vlues v(t i ) re positive but t is negtive,nd we get negtive result: 5 v(t)dt = t t 5 i= i= Finlly we note one simple property of integrls: = = 5 6. f(x)+g(x)dx = f(x)dx+ g(x) dx. This is esy to understnd once you recll tht (F(x)+G(x)) = F (x)+g (x). Hence, if F (x) = f(x) nd G (x) = g(x), then f(x)+g(x)dx = (F(x)+G(x)) b = F(b)+G(b) F() G() = F(b) F()+G(b) G() = F(x) b + G(x) b = f(x)dx+ g(x) dx. In summry, we will frequently use these properties of integrls: f(x)dx = f(x)+g(x)dx = c f(x)dx+ f(x)dx = f(x)dx+ b c f(x)dx f(x)dx g(x) dx

16 6 Chpter 7 Integrtion nd if < b nd f(x) on [,b] then f(x)dx nd in fct f(x)dx = f(x) dx. Exercises An object moves so tht its velocity t time t is v(t) = 9.8t+ m/s. Describe the motion of the object between t = nd t = 5, 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(x) = (x + )(x + )(x )(x ) on [,]. Find the totl re between the curve nd the x-xis (mesuring ll re s positive). 5. Consider the function f(x) = x 3x + on [,4]. Find the totl re between the curve nd the x-xis (mesuring ll re s positive). 6. Evlute the three integrls: A = 3 ( x +9)dx B = nd verify tht A = B +C. 4 ( x +9)dx C = 3 4 ( x +9)dx,

Math 8 Winter 2015 Applications of Integration

Math 8 Winter 2015 Applications of Integration Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl

More information

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp. MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

More information

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows: Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl

More information

We know that if f is a continuous nonnegative function on the interval [a, b], then b

We know that if f is a continuous nonnegative function on the interval [a, b], then b 1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going

More information

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by. NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with

More information

The Wave Equation I. MA 436 Kurt Bryan

The Wave Equation I. MA 436 Kurt Bryan 1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string

More information

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1 3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =

More information

5.5 The Substitution Rule

5.5 The Substitution Rule 5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n nti-derivtive is not esily recognizble, then we re in

More information

5: The Definite Integral

5: The Definite Integral 5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce

More information

Section 4.8. D v(t j 1 ) t. (4.8.1) j=1

Section 4.8. D v(t j 1 ) t. (4.8.1) j=1 Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions

More information

10. AREAS BETWEEN CURVES

10. AREAS BETWEEN CURVES . AREAS BETWEEN CURVES.. Ares etween curves So res ove the x-xis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in

More information

Line and Surface Integrals: An Intuitive Understanding

Line and Surface Integrals: An Intuitive Understanding Line nd Surfce Integrls: An Intuitive Understnding Joseph Breen Introduction Multivrible clculus is ll bout bstrcting the ides of differentition nd integrtion from the fmilir single vrible cse to tht of

More information

STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.

STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0. STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t

More information

The Definite Integral

The Definite Integral CHAPTER 3 The Definite Integrl Key Words nd Concepts: Definite Integrl Questions to Consider: How do we use slicing to turn problem sttement into definite integrl? How re definite nd indefinite integrls

More information

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230 Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given

More information

Lecture 14: Quadrature

Lecture 14: Quadrature Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be rel-vlues nd smooth The pproximtion of n integrl by numericl

More information

Lecture 1. Functional series. Pointwise and uniform convergence.

Lecture 1. Functional series. Pointwise and uniform convergence. 1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is

More information

Line Integrals. Partitioning the Curve. Estimating the Mass

Line Integrals. Partitioning the Curve. Estimating the Mass Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to

More information

Anonymous Math 361: Homework 5. x i = 1 (1 u i )

Anonymous Math 361: Homework 5. x i = 1 (1 u i ) Anonymous Mth 36: Homewor 5 Rudin. Let I be the set of ll u (u,..., u ) R with u i for ll i; let Q be the set of ll x (x,..., x ) R with x i, x i. (I is the unit cube; Q is the stndrd simplex in R ). Define

More information

1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.

1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q. Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the

More information

Numerical integration

Numerical integration 2 Numericl integrtion This is pge i Printer: Opque this 2. Introduction Numericl integrtion is problem tht is prt of mny problems in the economics nd econometrics literture. The orgniztion of this chpter

More information

Section 14.3 Arc Length and Curvature

Section 14.3 Arc Length and Curvature Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in

More information

20 MATHEMATICS POLYNOMIALS

20 MATHEMATICS POLYNOMIALS 0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of

More information

Section 6.1 Definite Integral

Section 6.1 Definite Integral Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined

More information

Distance And Velocity

Distance And Velocity Unit #8 - The Integrl Some problems nd solutions selected or dpted from Hughes-Hllett Clculus. Distnce And Velocity. The grph below shows the velocity, v, of n object (in meters/sec). Estimte the totl

More information

Not for reproduction

Not for reproduction AREA OF A SURFACE OF REVOLUTION cut h FIGURE FIGURE πr r r l h FIGURE A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundry of solid of revolution of the type

More information

Lecture 3. Limits of Functions and Continuity

Lecture 3. Limits of Functions and Continuity Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live

More information

If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du

If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du Integrtion by Substitution: The Fundmentl Theorem of Clculus demonstrted the importnce of being ble to find nti-derivtives. We now introduce some methods for finding ntiderivtives: If u = g(x) is differentible

More information

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence

More information

PART 1 MULTIPLE CHOICE Circle the appropriate response to each of the questions below. Each question has a value of 1 point.

PART 1 MULTIPLE CHOICE Circle the appropriate response to each of the questions below. Each question has a value of 1 point. PART MULTIPLE CHOICE Circle the pproprite response to ech of the questions below. Ech question hs vlue of point.. If in sequence the second level difference is constnt, thn the sequence is:. rithmetic

More information

Best Approximation. Chapter The General Case

Best Approximation. Chapter The General Case Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given

More information

31.2. Numerical Integration. Introduction. Prerequisites. Learning Outcomes

31.2. Numerical Integration. Introduction. Prerequisites. Learning Outcomes Numericl Integrtion 3. Introduction In this Section we will present some methods tht cn be used to pproximte integrls. Attention will be pid to how we ensure tht such pproximtions cn be gurnteed to be

More information

Unit #10 De+inite Integration & The Fundamental Theorem Of Calculus

Unit #10 De+inite Integration & The Fundamental Theorem Of Calculus Unit # De+inite Integrtion & The Fundmentl Theorem Of Clculus. Find the re of the shded region ove nd explin the mening of your nswer. (squres re y units) ) The grph to the right is f(x) = -x + 8x )Use

More information

Notes on Calculus II Integral Calculus. Miguel A. Lerma

Notes on Calculus II Integral Calculus. Miguel A. Lerma Notes on Clculus II Integrl Clculus Miguel A. Lerm November 22, 22 Contents Introduction 5 Chpter. Integrls 6.. Ares nd Distnces. The Definite Integrl 6.2. The Evlution Theorem.3. The Fundmentl Theorem

More information

l 2 p2 n 4n 2, the total surface area of the

l 2 p2 n 4n 2, the total surface area of the Week 6 Lectures Sections 7.5, 7.6 Section 7.5: Surfce re of Revolution Surfce re of Cone: Let C be circle of rdius r. Let P n be n n-sided regulr polygon of perimeter p n with vertices on C. Form cone

More information

Summary of Elementary Calculus

Summary of Elementary Calculus Summry of Elementry Clculus Notes by Wlter Noll (1971) 1 The rel numbers The set of rel numbers is denoted by R. The set R is often visulized geometriclly s number-line nd its elements re often referred

More information

Math 360: A primitive integral and elementary functions

Math 360: A primitive integral and elementary functions Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:

More information

Lecture 17. Integration: Gauss Quadrature. David Semeraro. University of Illinois at Urbana-Champaign. March 20, 2014

Lecture 17. Integration: Gauss Quadrature. David Semeraro. University of Illinois at Urbana-Champaign. March 20, 2014 Lecture 17 Integrtion: Guss Qudrture Dvid Semerro University of Illinois t Urbn-Chmpign Mrch 0, 014 Dvid Semerro (NCSA) CS 57 Mrch 0, 014 1 / 9 Tody: Objectives identify the most widely used qudrture method

More information

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz

More information

Numerical Integration

Numerical Integration Chpter 1 Numericl Integrtion Numericl differentition methods compute pproximtions to the derivtive of function from known vlues of the function. Numericl integrtion uses the sme informtion to compute numericl

More information

Math 113 Exam 2 Practice

Math 113 Exam 2 Practice Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.-7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number

More information

a a a a a a a a a a a a a a a a a a a a a a a a In this section, we introduce a general formula for computing determinants.

a a a a a a a a a a a a a a a a a a a a a a a a In this section, we introduce a general formula for computing determinants. Section 9 The Lplce Expnsion In the lst section, we defined the determinnt of (3 3) mtrix A 12 to be 22 12 21 22 2231 22 12 21. In this section, we introduce generl formul for computing determinnts. Rewriting

More information

3.4 Numerical integration

3.4 Numerical integration 3.4. Numericl integrtion 63 3.4 Numericl integrtion In mny economic pplictions it is necessry to compute the definite integrl of relvlued function f with respect to "weight" function w over n intervl [,

More information

III. Lecture on Numerical Integration. File faclib/dattab/lecture-notes/numerical-inter03.tex /by EC, 3/14/2008 at 15:11, version 9

III. Lecture on Numerical Integration. File faclib/dattab/lecture-notes/numerical-inter03.tex /by EC, 3/14/2008 at 15:11, version 9 III Lecture on Numericl Integrtion File fclib/dttb/lecture-notes/numerical-inter03.tex /by EC, 3/14/008 t 15:11, version 9 1 Sttement of the Numericl Integrtion Problem In this lecture we consider the

More information

Continuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom

Continuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive

More information

MTH 122 Fall 2008 Essex County College Division of Mathematics Handout Version 10 1 October 14, 2008

MTH 122 Fall 2008 Essex County College Division of Mathematics Handout Version 10 1 October 14, 2008 MTH 22 Fll 28 Essex County College Division of Mthemtics Hndout Version October 4, 28 Arc Length Everyone should be fmilir with the distnce formul tht ws introduced in elementry lgebr. It is bsic formul

More information

The final exam will take place on Friday May 11th from 8am 11am in Evans room 60.

The final exam will take place on Friday May 11th from 8am 11am in Evans room 60. Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23

More information

HOW TO USE INTEGRALS E. L. Lady (December 21, 1998) Consider the following set of formulas from high-school geometry and physics:

HOW TO USE INTEGRALS E. L. Lady (December 21, 1998) Consider the following set of formulas from high-school geometry and physics: HOW TO USE INTEGRALS E L Ldy (December 21, 1998) Consider the following set of formuls from high-school geometry nd physics: Are = Width Length Are of Rectngle Distnce = Velocity Time Distnce Trveled by

More information

ECO 317 Economics of Uncertainty Fall Term 2007 Notes for lectures 4. Stochastic Dominance

ECO 317 Economics of Uncertainty Fall Term 2007 Notes for lectures 4. Stochastic Dominance Generl structure ECO 37 Economics of Uncertinty Fll Term 007 Notes for lectures 4. Stochstic Dominnce Here we suppose tht the consequences re welth mounts denoted by W, which cn tke on ny vlue between

More information

The Bernoulli Numbers John C. Baez, December 23, x k. x e x 1 = n 0. B k n = n 2 (n + 1) 2

The Bernoulli Numbers John C. Baez, December 23, x k. x e x 1 = n 0. B k n = n 2 (n + 1) 2 The Bernoulli Numbers John C. Bez, December 23, 2003 The numbers re defined by the eqution e 1 n 0 k. They re clled the Bernoulli numbers becuse they were first studied by Johnn Fulhber in book published

More information

Topic 1 Notes Jeremy Orloff

Topic 1 Notes Jeremy Orloff Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble

More information

Mapping the delta function and other Radon measures

Mapping the delta function and other Radon measures Mpping the delt function nd other Rdon mesures Notes for Mth583A, Fll 2008 November 25, 2008 Rdon mesures Consider continuous function f on the rel line with sclr vlues. It is sid to hve bounded support

More information

Math 554 Integration

Math 554 Integration Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we

More information

2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).

2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ). AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following

More information

Week 10: Riemann integral and its properties

Week 10: Riemann integral and its properties Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the

More information

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral. Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:

More information

Orthogonal Polynomials and Least-Squares Approximations to Functions

Orthogonal Polynomials and Least-Squares Approximations to Functions Chpter Orthogonl Polynomils nd Lest-Squres Approximtions to Functions **4/5/3 ET. Discrete Lest-Squres Approximtions Given set of dt points (x,y ), (x,y ),..., (x m,y m ), norml nd useful prctice in mny

More information

Physics 9 Fall 2011 Homework 2 - Solutions Friday September 2, 2011

Physics 9 Fall 2011 Homework 2 - Solutions Friday September 2, 2011 Physics 9 Fll 0 Homework - s Fridy September, 0 Mke sure your nme is on your homework, nd plese box your finl nswer. Becuse we will be giving prtil credit, be sure to ttempt ll the problems, even if you

More information

MATH 222 Second Semester Calculus. Fall 2015

MATH 222 Second Semester Calculus. Fall 2015 MATH Second Semester Clculus Fll 5 Typeset:August, 5 Mth nd Semester Clculus Lecture notes version. (Fll 5) This is self contined set of lecture notes for Mth. The notes were written by Sigurd Angenent,

More information

Linear Systems with Constant Coefficients

Linear Systems with Constant Coefficients Liner Systems with Constnt Coefficients 4-3-05 Here is system of n differentil equtions in n unknowns: x x + + n x n, x x + + n x n, x n n x + + nn x n This is constnt coefficient liner homogeneous system

More information

1 Error Analysis of Simple Rules for Numerical Integration

1 Error Analysis of Simple Rules for Numerical Integration cs41: introduction to numericl nlysis 11/16/10 Lecture 19: Numericl Integrtion II Instructor: Professor Amos Ron Scries: Mrk Cowlishw, Nthnel Fillmore 1 Error Anlysis of Simple Rules for Numericl Integrtion

More information

Synoptic Meteorology I: Finite Differences September Partial Derivatives (or, Why Do We Care About Finite Differences?

Synoptic Meteorology I: Finite Differences September Partial Derivatives (or, Why Do We Care About Finite Differences? Synoptic Meteorology I: Finite Differences 16-18 September 2014 Prtil Derivtives (or, Why Do We Cre About Finite Differences?) With the exception of the idel gs lw, the equtions tht govern the evolution

More information

THE HANKEL MATRIX METHOD FOR GAUSSIAN QUADRATURE IN 1 AND 2 DIMENSIONS

THE HANKEL MATRIX METHOD FOR GAUSSIAN QUADRATURE IN 1 AND 2 DIMENSIONS THE HANKEL MATRIX METHOD FOR GAUSSIAN QUADRATURE IN 1 AND 2 DIMENSIONS CARLOS SUERO, MAURICIO ALMANZAR CONTENTS 1 Introduction 1 2 Proof of Gussin Qudrture 6 3 Iterted 2-Dimensionl Gussin Qudrture 20 4

More information

Test , 8.2, 8.4 (density only), 8.5 (work only), 9.1, 9.2 and 9.3 related test 1 material and material from prior classes

Test , 8.2, 8.4 (density only), 8.5 (work only), 9.1, 9.2 and 9.3 related test 1 material and material from prior classes Test 2 8., 8.2, 8.4 (density only), 8.5 (work only), 9., 9.2 nd 9.3 relted test mteril nd mteril from prior clsses Locl to Globl Perspectives Anlyze smll pieces to understnd the big picture. Exmples: numericl

More information

Math 0230 Calculus 2 Lectures

Math 0230 Calculus 2 Lectures Mth Clculus Lectures Chpter 7 Applictions of Integrtion Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition. Section 7. Ares Between Curves Two

More information

Practice final exam solutions

Practice final exam solutions University of Pennsylvni Deprtment of Mthemtics Mth 26 Honors Clculus II Spring Semester 29 Prof. Grssi, T.A. Asher Auel Prctice finl exm solutions 1. Let F : 2 2 be defined by F (x, y (x + y, x y. If

More information

Calculus 2: Integration. Differentiation. Integration

Calculus 2: Integration. Differentiation. Integration Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is

More information

Lecture 6: Singular Integrals, Open Quadrature rules, and Gauss Quadrature

Lecture 6: Singular Integrals, Open Quadrature rules, and Gauss Quadrature Lecture notes on Vritionl nd Approximte Methods in Applied Mthemtics - A Peirce UBC Lecture 6: Singulr Integrls, Open Qudrture rules, nd Guss Qudrture (Compiled 6 August 7) In this lecture we discuss the

More information

Continuous Random Variables

Continuous Random Variables STAT/MATH 395 A - PROBABILITY II UW Winter Qurter 217 Néhémy Lim Continuous Rndom Vribles Nottion. The indictor function of set S is rel-vlued function defined by : { 1 if x S 1 S (x) if x S Suppose tht

More information

Integrals along Curves.

Integrals along Curves. Integrls long Curves. 1. Pth integrls. Let : [, b] R n be continuous function nd let be the imge ([, b]) of. We refer to both nd s curve. If we need to distinguish between the two we cll the function the

More information

USA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year

USA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year 1/1/21. Fill in the circles in the picture t right with the digits 1-8, one digit in ech circle with no digit repeted, so tht no two circles tht re connected by line segment contin consecutive digits.

More information

Lecture 7 notes Nodal Analysis

Lecture 7 notes Nodal Analysis Lecture 7 notes Nodl Anlysis Generl Network Anlysis In mny cses you hve multiple unknowns in circuit, sy the voltges cross multiple resistors. Network nlysis is systemtic wy to generte multiple equtions

More information

Introduction to Numerical Analysis

Introduction to Numerical Analysis Introduction to Numericl Anlysis Doron Levy Deprtment of Mthemtics nd Center for Scientific Computtion nd Mthemticl Modeling (CSCAMM) University of Mrylnd June 14, 2012 D. Levy CONTENTS Contents 1 Introduction

More information

u(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C.

u(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C. Lecture 4 Complex Integrtion MATH-GA 2451.001 Complex Vriles 1 Construction 1.1 Integrting complex function over curve in C A nturl wy to construct the integrl of complex function over curve in the complex

More information

Czechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction

Czechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction Czechoslovk Mthemticl Journl, 55 (130) (2005), 933 940 ESTIMATES OF THE REMAINDER IN TAYLOR S THEOREM USING THE HENSTOCK-KURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When rel-vlued

More information

7.6 The Use of Definite Integrals in Physics and Engineering

7.6 The Use of Definite Integrals in Physics and Engineering Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 7.6 The Use of Definite Integrls in Physics nd Engineering It hs been shown how clculus cn be pplied to find solutions to geometric problems

More information

Math 211A Homework. Edward Burkard. = tan (2x + z)

Math 211A Homework. Edward Burkard. = tan (2x + z) Mth A Homework Ewr Burkr Eercises 5-C Eercise 8 Show tht the utonomous system: 5 Plne Autonomous Systems = e sin 3y + sin cos + e z, y = sin ( + 3y, z = tn ( + z hs n unstble criticl point t = y = z =

More information

C1M14. Integrals as Area Accumulators

C1M14. Integrals as Area Accumulators CM Integrls s Are Accumultors Most tetbooks do good job of developing the integrl nd this is not the plce to provide tht development. We will show how Mple presents Riemnn Sums nd the ccompnying digrms

More information

Pre-Session Review. Part 1: Basic Algebra; Linear Functions and Graphs

Pre-Session Review. Part 1: Basic Algebra; Linear Functions and Graphs Pre-Session Review Prt 1: Bsic Algebr; Liner Functions nd Grphs A. Generl Review nd Introduction to Algebr Hierrchy of Arithmetic Opertions Opertions in ny expression re performed in the following order:

More information

11.1 Exponential Functions

11.1 Exponential Functions . Eponentil Functions In this chpter we wnt to look t specific type of function tht hs mny very useful pplictions, the eponentil function. Definition: Eponentil Function An eponentil function is function

More information

Discrete Least-squares Approximations

Discrete Least-squares Approximations Discrete Lest-squres Approximtions Given set of dt points (x, y ), (x, y ),, (x m, y m ), norml nd useful prctice in mny pplictions in sttistics, engineering nd other pplied sciences is to construct curve

More information

UniversitaireWiskundeCompetitie. Problem 2005/4-A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that

UniversitaireWiskundeCompetitie. Problem 2005/4-A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that Problemen/UWC NAW 5/7 nr juni 006 47 Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de

More information

3.1 Review of Sine, Cosine and Tangent for Right Angles

3.1 Review of Sine, Cosine and Tangent for Right Angles Foundtions of Mth 11 Section 3.1 Review of Sine, osine nd Tngent for Right Tringles 125 3.1 Review of Sine, osine nd Tngent for Right ngles The word trigonometry is derived from the Greek words trigon,

More information

Problem Set 9. Figure 1: Diagram. This picture is a rough sketch of the 4 parabolas that give us the area that we need to find. The equations are:

Problem Set 9. Figure 1: Diagram. This picture is a rough sketch of the 4 parabolas that give us the area that we need to find. The equations are: (x + y ) = y + (x + y ) = x + Problem Set 9 Discussion: Nov., Nov. 8, Nov. (on probbility nd binomil coefficients) The nme fter the problem is the designted writer of the solution of tht problem. (No one

More information

Energy Bands Energy Bands and Band Gap. Phys463.nb Phenomenon

Energy Bands Energy Bands and Band Gap. Phys463.nb Phenomenon Phys463.nb 49 7 Energy Bnds Ref: textbook, Chpter 7 Q: Why re there insultors nd conductors? Q: Wht will hppen when n electron moves in crystl? In the previous chpter, we discussed free electron gses,

More information

Mathematics notation and review

Mathematics notation and review Appendix A Mthemtics nottion nd review This ppendix gives brief coverge of the mthemticl nottion nd concepts tht you ll encounter in this book. In the spce of few pges it is of course impossible to do

More information

Week 7 Riemann Stieltjes Integration: Lectures 19-21

Week 7 Riemann Stieltjes Integration: Lectures 19-21 Week 7 Riemnn Stieltjes Integrtion: Lectures 19-21 Lecture 19 Throughout this section α will denote monotoniclly incresing function on n intervl [, b]. Let f be bounded function on [, b]. Let P = { = 0

More information

13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS

13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS 33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in

More information

Math 324 Course Notes: Brief description

Math 324 Course Notes: Brief description Brief description These re notes for Mth 324, n introductory course in Mesure nd Integrtion. Students re dvised to go through ll sections in detil nd ttempt ll problems. These notes will be modified nd

More information

Homework Solution - Set 5 Due: Friday 10/03/08

Homework Solution - Set 5 Due: Friday 10/03/08 CE 96 Introduction to the Theory of Computtion ll 2008 Homework olution - et 5 Due: ridy 10/0/08 1. Textook, Pge 86, Exercise 1.21. () 1 2 Add new strt stte nd finl stte. Mke originl finl stte non-finl.

More information

MATH Summary of Chapter 13

MATH Summary of Chapter 13 MATH 21-259 ummry of hpter 13 1. Vector Fields re vector functions of two or three vribles. Typiclly, vector field is denoted by F(x, y, z) = P (x, y, z)i+q(x, y, z)j+r(x, y, z)k where P, Q, R re clled

More information

KOÇ UNIVERSITY MATH 106 FINAL EXAM JANUARY 6, 2013

KOÇ UNIVERSITY MATH 106 FINAL EXAM JANUARY 6, 2013 KOÇ UNIVERSITY MATH 6 FINAL EXAM JANUARY 6, 23 Durtion of Exm: 2 minutes INSTRUCTIONS: No clcultors nd no cell phones my be used on the test. No questions, nd tlking llowed. You must lwys explin your nswers

More information

Using integration tables

Using integration tables Using integrtion tbles Integrtion tbles re inclue in most mth tetbooks, n vilble on the Internet. Using them is nother wy to evlute integrls. Sometimes the use is strightforwr; sometimes it tkes severl

More information

MATH STUDENT BOOK. 10th Grade Unit 5

MATH STUDENT BOOK. 10th Grade Unit 5 MATH STUDENT BOOK 10th Grde Unit 5 Unit 5 Similr Polygons MATH 1005 Similr Polygons INTRODUCTION 3 1. PRINCIPLES OF ALGEBRA 5 RATIOS AND PROPORTIONS 5 PROPERTIES OF PROPORTIONS 11 SELF TEST 1 16 2. SIMILARITY

More information

PDE Notes. Paul Carnig. January ODE s vs PDE s 1

PDE Notes. Paul Carnig. January ODE s vs PDE s 1 PDE Notes Pul Crnig Jnury 2014 Contents 1 ODE s vs PDE s 1 2 Section 1.2 Het diffusion Eqution 1 2.1 Fourier s w of Het Conduction............................. 2 2.2 Energy Conservtion.....................................

More information

ESCI 343 Atmospheric Dynamics II Lesson 14 Inertial/slantwise Instability

ESCI 343 Atmospheric Dynamics II Lesson 14 Inertial/slantwise Instability ESCI 343 Atmospheric Dynmics II Lesson 14 Inertil/slntwise Instbility Reference: An Introduction to Dynmic Meteorology (3 rd edition), J.R. Holton Atmosphere-Ocen Dynmics, A.E. Gill Mesoscle Meteorology

More information

SCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics

SCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics SCHOOL OF ENGINEERING & BUIL ENVIRONMEN Mthemtics An Introduction to Mtrices Definition of Mtri Size of Mtri Rows nd Columns of Mtri Mtri Addition Sclr Multipliction of Mtri Mtri Multipliction 7 rnspose

More information

u( t) + K 2 ( ) = 1 t > 0 Analyzing Damped Oscillations Problem (Meador, example 2-18, pp 44-48): Determine the equation of the following graph.

u( t) + K 2 ( ) = 1 t > 0 Analyzing Damped Oscillations Problem (Meador, example 2-18, pp 44-48): Determine the equation of the following graph. nlyzing Dmped Oscilltions Prolem (Medor, exmple 2-18, pp 44-48): Determine the eqution of the following grph. The eqution is ssumed to e of the following form f ( t) = K 1 u( t) + K 2 e!"t sin (#t + $

More information

MTH 505: Number Theory Spring 2017

MTH 505: Number Theory Spring 2017 MTH 505: Numer Theory Spring 207 Homework 2 Drew Armstrong The Froenius Coin Prolem. Consider the eqution x ` y c where,, c, x, y re nturl numers. We cn think of $ nd $ s two denomintions of coins nd $c

More information