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


 Sybil Clark
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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 xxis nd the curve, but under the xxis, 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 xxis (mesuring ll re s positive). 5. Consider the function f(x) = x 3x + on [,4]. Find the totl re between the curve nd the xxis (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
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
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