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

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,

and that at t = 0 the object is at position 5. Find the position of the object at t = 2.

and that at t = 0 the object is at position 5. Find the position of the object at t = 2. 7.2 The Fundmentl Theorem of Clculus 49 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

More information

Integration. 148 Chapter 7 Integration

Integration. 148 Chapter 7 Integration 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

More information

7.2 The Definite Integral

7.2 The Definite Integral 7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where

More information

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019 ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil

More information

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph

More information

Riemann Sums and Riemann Integrals

Riemann Sums and Riemann Integrals Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties

More information

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a). The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples

More information

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1 The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the

More information

Riemann Sums and Riemann Integrals

Riemann Sums and Riemann Integrals Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct

More information

INTRODUCTION TO INTEGRATION

INTRODUCTION TO INTEGRATION INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide

More information

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl

More information

Chapter 0. What is the Lebesgue integral about?

Chapter 0. What is the Lebesgue integral about? Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous

More information

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

More information

The Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.

The Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve. Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F

More information

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

More information

The Regulated and Riemann Integrals

The Regulated and Riemann Integrals Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue

More information

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

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

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

Review of Calculus, cont d

Review of Calculus, cont d Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

More information

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,

More information

Integrals - Motivation

Integrals - Motivation Integrls - Motivtion When we looked t function s rte of chnge If f(x) is liner, the nswer is esy slope If f(x) is non-liner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but

More information

x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b

x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick

More information

Chapters 4 & 5 Integrals & Applications

Chapters 4 & 5 Integrals & Applications Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO - Ares Under Functions............................................ 3.2 VIDEO - Applictions

More information

MA 124 January 18, Derivatives are. Integrals are.

MA 124 January 18, Derivatives are. Integrals are. MA 124 Jnury 18, 2018 Prof PB s one-minute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,

More information

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite

More information

x = b a N. (13-1) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is

x = b a N. (13-1) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is Jnury 28, 2002 13. The Integrl The concept of integrtion, nd the motivtion for developing this concept, were described in the previous chpter. Now we must define the integrl, crefully nd completely. According

More information

4.4 Areas, Integrals and Antiderivatives

4.4 Areas, Integrals and Antiderivatives . res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order

More information

1 The Riemann Integral

1 The Riemann Integral The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re

More information

Improper Integrals, and Differential Equations

Improper Integrals, and Differential Equations Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted

More information

Lecture 1: Introduction to integration theory and bounded variation

Lecture 1: Introduction to integration theory and bounded variation Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You

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

Overview of Calculus I

Overview of Calculus I Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,

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

p(t) dt + i 1 re it ireit dt =

p(t) dt + i 1 re it ireit dt = Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)

More information

Section 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40

Section 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40 Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since

More information

MATH , Calculus 2, Fall 2018

MATH , Calculus 2, Fall 2018 MATH 36-2, 36-3 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly

More information

Section Areas and Distances. Example 1: Suppose a car travels at a constant 50 miles per hour for 2 hours. What is the total distance traveled?

Section Areas and Distances. Example 1: Suppose a car travels at a constant 50 miles per hour for 2 hours. What is the total distance traveled? Section 5. - Ares nd Distnces Exmple : Suppose cr trvels t constnt 5 miles per hour for 2 hours. Wht is the totl distnce trveled? Exmple 2: Suppose cr trvels 75 miles per hour for the first hour, 7 miles

More information

Big idea in Calculus: approximation

Big idea in Calculus: approximation Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:

More information

Section 6.1 INTRO to LAPLACE TRANSFORMS

Section 6.1 INTRO to LAPLACE TRANSFORMS Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform

More information

MAT 168: Calculus II with Analytic Geometry. James V. Lambers

MAT 168: Calculus II with Analytic Geometry. James V. Lambers MAT 68: Clculus II with Anlytic Geometry Jmes V. Lmbers Februry 7, Contents Integrls 5. Introduction............................ 5.. Differentil Clculus nd Quotient Formuls...... 5.. Integrl Clculus nd

More information

a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1

a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1 Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the

More information

Reversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b

Reversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b Mth 32 Substitution Method Stewrt 4.5 Reversing the Chin Rule. As we hve seen from the Second Fundmentl Theorem ( 4.3), the esiest wy to evlute n integrl b f(x) dx is to find n ntiderivtive, the indefinite

More information

Math 1B, lecture 4: Error bounds for numerical methods

Math 1B, lecture 4: Error bounds for numerical methods Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the

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

Chapter 6 Notes, Larson/Hostetler 3e

Chapter 6 Notes, Larson/Hostetler 3e Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn

More information

Math& 152 Section Integration by Parts

Math& 152 Section Integration by Parts Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

More information

dt. However, we might also be curious about dy

dt. However, we might also be curious about dy Section 0. The Clculus of Prmetric Curves Even though curve defined prmetricly my not be function, we cn still consider concepts such s rtes of chnge. However, the concepts will need specil tretment. For

More information

Chapter 5. Numerical Integration

Chapter 5. Numerical Integration Chpter 5. Numericl Integrtion These re just summries of the lecture notes, nd few detils re included. Most of wht we include here is to be found in more detil in Anton. 5. Remrk. There re two topics with

More information

Interpreting Integrals and the Fundamental Theorem

Interpreting Integrals and the Fundamental Theorem Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of

More information

Riemann Integrals and the Fundamental Theorem of Calculus

Riemann Integrals and the Fundamental Theorem of Calculus Riemnn Integrls nd the Fundmentl Theorem of Clculus Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University September 16, 2013 Outline Grphing Riemnn Sums

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 practical version

The practical version Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht

More information

1 The fundamental theorems of calculus.

1 The fundamental theorems of calculus. The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl- new nme for nti-derivtive. Differentiting integrls. Tody we provide the connection

More information

Week 10: Line Integrals

Week 10: Line Integrals Week 10: Line Integrls Introduction In this finl week we return to prmetrised curves nd consider integrtion long such curves. We lredy sw this in Week 2 when we integrted long curve to find its length.

More information

Review of basic calculus

Review of basic calculus Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below

More information

5.7 Improper Integrals

5.7 Improper Integrals 458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the

More information

Section 6: Area, Volume, and Average Value

Section 6: Area, Volume, and Average Value Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find

More information

Definite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +

Definite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 + Definite Integrls --5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the left-hnd

More information

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b

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

Chapter 8.2: The Integral

Chapter 8.2: The Integral Chpter 8.: The Integrl You cn think of Clculus s doule-wide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms 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

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus The Fundmentl Theorem of Clculus Professor Richrd Blecksmith richrd@mth.niu.edu Dept. of Mthemticl Sciences Northern Illinois University http://mth.niu.edu/ richrd/mth229. The Definite Integrl We define

More information

Math Calculus with Analytic Geometry II

Math Calculus with Analytic Geometry II orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove x-xis) ( bove f under x-xis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem

More information

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS. THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem

More information

The Evaluation Theorem

The Evaluation Theorem These notes closely follow the presenttion of the mteril given in Jmes Stewrt s textook Clculus, Concepts nd Contexts (2nd edition) These notes re intended primrily for in-clss presenttion nd should not

More information

ROB EBY Blinn College Mathematics Department

ROB EBY Blinn College Mathematics Department ROB EBY Blinn College Mthemtics Deprtment Mthemtics Deprtment 5.1, 5.2 Are, Definite Integrls MATH 2413 Rob Eby-Fll 26 Weknowthtwhengiventhedistncefunction, wecnfindthevelocitytnypointbyfindingthederivtiveorinstntneous

More information

Recitation 3: More Applications of the Derivative

Recitation 3: More Applications of the Derivative Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech

More information

Fundamental Theorem of Calculus

Fundamental Theorem of Calculus Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under

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

Infinite Geometric Series

Infinite Geometric Series Infinite Geometric Series Finite Geometric Series ( finite SUM) Let 0 < r < 1, nd let n be positive integer. Consider the finite sum It turns out there is simple lgebric expression tht is equivlent to

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

We divide the interval [a, b] into subintervals of equal length x = b a n

We divide the interval [a, b] into subintervals of equal length x = b a n Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:

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

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique? XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk bout solving systems of liner equtions. These re problems tht give couple of equtions with couple of unknowns, like: 6 2 3 7 4

More information

Anti-derivatives/Indefinite Integrals of Basic Functions

Anti-derivatives/Indefinite Integrals of Basic Functions Anti-derivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second

More information

different methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).

different methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s). Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different

More information

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O 1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the x-xis etween nd is denoted y f(x) dx nd clled the

More information

MAA 4212 Improper Integrals

MAA 4212 Improper Integrals Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly well-defined, is too restrictive for mny purposes; there re functions which

More information

Stuff You Need to Know From Calculus

Stuff You Need to Know From Calculus Stuff You Need to Know From Clculus For the first time in the semester, the stuff we re doing is finlly going to look like clculus (with vector slnt, of course). This mens tht in order to succeed, you

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

ODE: Existence and Uniqueness of a Solution

ODE: Existence and Uniqueness of a Solution Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =

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

1 The fundamental theorems of calculus.

1 The fundamental theorems of calculus. The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl- new nme for nti-derivtive. Differentiting integrls. Theorem Suppose f is continuous

More information

Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions

Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner

More information

Math 116 Calculus II

Math 116 Calculus II Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................

More information

Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim

Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)

More information

Z b. f(x)dx. Yet in the above two cases we know what f(x) is. Sometimes, engineers want to calculate an area by computing I, but...

Z b. f(x)dx. Yet in the above two cases we know what f(x) is. Sometimes, engineers want to calculate an area by computing I, but... Chpter 7 Numericl Methods 7. Introduction In mny cses the integrl f(x)dx cn be found by finding function F (x) such tht F 0 (x) =f(x), nd using f(x)dx = F (b) F () which is known s the nlyticl (exct) solution.

More information

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.) MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give

More information

Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1

Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1 Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH-115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixed-point itertion to converge when solving the eqution

More information

38 Riemann sums and existence of the definite integral.

38 Riemann sums and existence of the definite integral. 38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These

More information

SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus

SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is

More information

Taylor Polynomial Inequalities

Taylor Polynomial Inequalities Tylor Polynomil Inequlities Ben Glin September 17, 24 Abstrct There re instnces where we my wish to pproximte the vlue of complicted function round given point by constructing simpler function such s polynomil

More information

5 Accumulated Change: The Definite Integral

5 Accumulated Change: The Definite Integral 5 Accumulted Chnge: The Definite Integrl 5.1 Distnce nd Accumulted Chnge * How To Mesure Distnce Trveled nd Visulize Distnce on the Velocity Grph Distnce = Velocity Time Exmple 1 Suppose tht you trvel

More information

Section 6.1 INTRO to LAPLACE TRANSFORMS

Section 6.1 INTRO to LAPLACE TRANSFORMS Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform

More information

Indefinite Integral. Chapter Integration - reverse of differentiation

Indefinite Integral. Chapter Integration - reverse of differentiation Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the

More information

Main topics for the First Midterm

Main topics for the First Midterm Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the

More information

APPROXIMATE INTEGRATION

APPROXIMATE INTEGRATION APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose nti-derivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be

More information