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 end up with mthemtics tht looks like the two emples, though of course the function involved will not lwys be so simple. Even better, we now see tht while the second problem did not pper to be menble to pproch one, it cn in fct be solved in the sme wy. The resoning is this: we know tht problem one cn be solved esily by finding function whose derivtive is 3t. We lso know tht mthemticlly the two problems re the sme, becuse both cn be solved by tking it of sum, nd the sums re identicl. Therefore, we don t relly need to compute the it of either sum becuse we know tht we will get the sme nswer by computing function with the derivtive 3t or, which is the sme thing, 3. It struethtthefirstproblemhdtheddedcomplictionofthe 0,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 it of sum of certin form, we cn insted of computing the (often nsty) it find new function with certin derivtive. Eercises 7... Suppose n object moves in stright line so tht its speed t time t is given by v(t) = 2t+2, nd tht t t = the object is t position 5. Find the position of the object t t = 2. 2. Suppose n object moves in stright line so tht its speed t time t is given by v(t) = t 2 +2, nd tht t t = 0 the object is t position 5. Find the position of the object t t = 2. 3. By method similr to tht in emple 7.2, find the re under y = 2 between = 0 nd ny positive vlue for. 4. By method similr to tht in emple 7.2, find the re under y = 4 between = 0 nd ny positive vlue for. 5. By method similr to tht in emple 7.2, find the re under y = 4 between = 2 nd ny positive vlue for bigger thn 2. 6. By method similr to tht in emple 7.2, find the re under y = 4 between ny two positive vlues for, sy < b. 7. Let f() = 2 + 3 + 2. Approimte the re under the curve between = 0 nd = 2 using 4 rectngles nd lso using 8 rectngles. 8. Let f() = 2 2 + 3. Approimte the re under the curve between = nd = 3 using 4 rectngles. º¾ Ì ÙÒ Ñ ÒØ Ð Ì ÓÖ Ñ Ó ÐÙÐÙ Let s recst the first emple from the previous section. Suppose tht the speed of the object is 3t t time t. How fr does the object trvel between time t = nd time t = b? We re no longer ssuming tht we know where the object is t time t = 0 or t ny other Pge 5 of 6
50 Chpter 7 Integrtion time. Itiscertinlytruethtitissomewhere,solet ssupposethttt = 0thepositionisk. Thenjustsintheemple, weknowthtthepositionoftheobjecttnytimeis3t 2 /2+k. This mens tht t time t = the position is 3 2 /2+k nd t time t = b the position is 3b 2 /2+k. Therefore the chnge in position is 3b 2 /2+k (3 2 /2+k) = 3b 2 /2 3 2 /2. Notice tht the k drops out; this mens tht it doesn t mtter tht we don t know k, it doesn t even mtter if we use the wrong k, we get the correct nswer. In other words, to find the chnge in position between time nd time b we cn use ny ntiderivtive of the speed function 3t it need not be the one ntiderivtive tht ctully gives the loction of the object. Wht bout the second pproch to this problem, in the new form? We now wnt to pproimte the chnge in position between time nd time b. We tke the intervl of time between nd b, divide it into n subintervls, nd pproimte the distnce trveled during ech. The strting time of subintervl number i is now +(i )(b )/n, which we bbrevite s t i, so tht t 0 =, t = + (b )/n, nd so on. The speed of the object is f(t) = 3t, nd ech subintervl is (b )/n = t seconds long. The distnce trveled during subintervl number i is pproimtely f(t i ) t, nd the totl chnge in distnce is pproimtely f(t 0 ) t+f(t ) t+ +f(t ) t. The ect chnge in position is the it of this sum s n goes to infinity. We bbrevite this sum using sigm nottion: f(t i ) t = f(t 0 ) t+f(t ) t+ +f(t ) 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 f(t i ) t. Since this must be the sme s the nswer we hve lredy obtined, we know tht f(t i ) t = 3b2 2 32 2. The significnce of 3t 2 /2, 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 Pge 6 of 6
wnt to compute 7.2 The Fundmentl Theorem of Clculus 5 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 it 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 = f(t i ) t if the it eists. Tht is, the left hnd side mens, or is n bbrevition for, the right hnd side. The symbol is clled n integrl sign, nd the whole epression is red s the integrl of f(t) from to b. Wht we hve lerned is tht this integrl cn be computed by finding function, sy F(t), with the property tht F (t) = f(t), nd then computing F(b) F(). The function F(t) is clled n ntiderivtive of f(t). Now the theorem: THEOREM 7.3 Fundmentl Theorem of Clculus Suppose tht f() is continuous on the intervl [,b]. If F() is ny ntiderivtive of f(), then f()d = F(b) F(). Let s rewrite this slightly: = F() F(). We ve replced the vrible by t nd b by. These re just different nmes for quntities, so the substitution doesn t chnge the mening. It does mke it esier to think of the two sides of the eqution s functions. The epression is function: plug in vlue for, get out some other vlue. The epression F() F() is of course lso function, nd it hs nice property: d d (F() F()) = F () = f(), Pge 7 of 6
52 Chpter 7 Integrtion since F() is constnt nd hs derivtive zero. In other words, by shifting our point of view slightly, we see tht the odd looking function G() = hs derivtive, nd tht in fct G () = f(). This is relly just resttement of the Fundmentl Theorem of Clculus, nd indeed is often clled the Fundmentl Theorem of Clculus. To void confusion, some people cll the two versions of the theorem The Fundmentl Theorem of Clculus, prt I nd The Fundmentl Theorem of Clculus, prt II, lthough unfortuntely there is no universl greement s to which is prt I nd which prt II. Since it relly is the sme theorem, differently stted, some people simply cll them both The Fundmentl Theorem of Clculus. Suppose tht f() is con- THEOREM 7.4 Fundmentl Theorem of Clculus tinuous on the intervl [,b] nd let G() =. Then G () = f(). 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 = f(t i ) t. 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 7.4, then we cn prove the first version from tht: Pge 8 of 6
Proof of Theorem 7.3. We know from theorem 7.4 tht 7.2 The Fundmentl Theorem of Clculus 53 G() = is n ntiderivtive of f(), nd therefore ny ntiderivtive F() of f() is of the form F() = G()+k. Then F(b) F() = G(b)+k (G()+k) = G(b) G() =. It is not hrd to see tht = 0, so this mens tht F(b) F() = which is ectly wht theorem 7.3 sys., So the rel job is to prove theorem 7.4. We will sketch the proof, using some fcts tht we do not prove. First, the following identity is true of integrls: = c + c. This cn be proved directly from the definition of the integrl, tht is, using the its 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. Pge 9 of 6
54 Chpter 7 Integrtion Proof sketch for Theorem 7.4. We wnt to compute G (), so we strt with the definition of the derivtive in terms of it: G G(+ ) G() () = 0 ( + = 0 ( = + 0 = 0 + Now we need to know something bout. + ) ) + when is smll; in fct, it is very close to f(), but we will not prove this. Once gin, it is esy to believe this is true by thinking of our two pplictions: The integrl + cn be interpreted s the distnce trveled by n object over very short intervl of time. Over sufficiently short period of time, the speed of the object will not chnge very much, so the distnce trveled will be pproimtely the length of time multiplied by the speed t the beginning of the intervl, nmely, f(). Alterntely, the integrl my be interpreted s the re under the curve between nd +. When is very smll, this will be very close to the re of the rectngle with bse nd height f(); gin this is f(). If we ccept this, we my proceed: 0 which is wht we wnted to show. + f() = 0 = f(), It is still true tht we re depending on n interprettion of the integrl to justify the rgument, but we hve isolted this prt of the rgument into two fcts tht re not too hrd to prove. Once the lst reference to interprettion hs been removed from the proofs of these fcts, we will hve rel proof of the Fundmentl Theorem. Pge 0 of 6
7.2 The Fundmentl Theorem of Clculus 55 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 its of integrtion re included: 2 2 d is n ordinry integrl, lso clled definite integrl, becuse it hs definite vlue, nmely We use 2 2 d = 23 3 3 3 = 7 3. 2 d to denote the ntiderivtive of 2, lso clled n indefinite integrl. So this is evluted s 2 d = 3 3 +C. It is customry to include the constnt C to indicte tht there re relly n infinite number of ntiderivtives. We do not need this C to compute definite integrls, but in other circumstnces we will need to remember tht the C is there, so it is best to get into the hbit of writing the C. When we compute definite integrl, we first find n ntiderivtive nd then substitute. It is convenient to first disply the ntiderivtive nd then do the substitution; we need nottion indicting tht the substitution is yet to be done. A typicl solution would look like this: 2 2 d = 3 3 2 = 23 3 3 3 = 7 3. The verticl line with subscript nd superscript is used to indicte the opertion substitute nd subtrct tht is needed to finish the evlution. Pge of 6