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 of these theorems in this set of notes. The net set of notes will consider some pplictions of these theorems. The First Fundmentl Theorem of Clculus. If f() is continuous on [, b nd F () is ny ntiderivtive of f() on [, b, then f() d = F (b) F (). In words: In order to compute definite integrl f(), it suffices to find n ntiderivtive of f(), then compute the difference of this ntiderivtive. The Second Fundmentl Theorem of Clculus. If f() is continuous on n intervl nd is ny number in this intervl, then the function is n ntiderivtive of f(). In other words, A() = d d. In words: In order to constuct n ntiderivtive of function f(), it is enough to be ble to compute the definite integrl of f(). Both of these theorems estblish reltionship between ntiderivtives (i.e., indefinite integrls) nd net res (i.e., definite integrls.) In effect, the First Fundmentl Theorem sys tht definite integrtion reverses differentition, wheres the Second Fundmentl Theorem sys tht differentition reverses definite integrtion. Thus, the Fundmentl Theorems sy tht definite integrtion is ctully sort of ntidifferentition. This eplins why we use integrtion nottion nd terminology (indefinite integrls) for ntiderivtives. From mthemticl point of view, the Second Fundmentl Theorem is the true fundmentl theorem of clculus. The First Fundmentl Theorem is just logicl consequence of Second Fundmentl Theorem. However, the First Fundmentl Theorem is wht most non-mthemticins think of s the fundmentl theorem. This is becuse the First Fundmentl Theorem is the one tht ppers in pplictions in lots of different subjects. Thus, the First Fundmentl Theorem is of prcticl interest wheres the Second Fundmentl Theorem is primrily of theoreticl interest, lthough it does hve some prcticl pplictions. Emple : Find. Using the First Fundmentl Theorem of Clculus sin θ dθ. Solution: We lredy know this integrl is 0 by the previous set of notes. But lets compute this using the First Fundmentl Theorem. First, we need to find n ntiderivtive of sin θ. But sin θ dθ = cos θ + C,
so we ll use F (θ) = cos θ in the fundmentl theorem. (The fundmentl theorem only requires tht we find some ntiderivtive, not the most generl ntiderivtive, so we don t need to use the +C.) So, sin θ dθ = F ( 3 ) F ( ) = cos (3 ) ( cos ( )) = 0 + 0 = 0, consistent with the nswer we obtined in the lst set of notes. We mke two comments on the bove emple. First, wht if we used different ntiderivtive? Could tht hve chnged our nswer? Fortuntely the nswer is no. The most generl ntiderivtive is F (θ) = cos θ + C, so if we used this we would hve hd sin θ dθ = F ( 3 ) F ( ) = ( cos (3 ) + C) ( cos ( ) + C) = 0 + C + 0 C = 0. The C s will lwys cncel off like this, so we don t need the +C when computing definite integrls. Second, we will find it convenient to write Thus, we could write Emple : Find Solution: First we find n ntiderivtive: d = We ll use s our ntiderivtive. Thus Emple 3: Find 4 F () b = F (b) F (). sin θ dθ = F (θ) 3 4 d. =.... / d = / + C. d = 4 = 4 =. ( 3 + 5) d. Solution: Often times, we will just compute the ntiderivtives in our heds. Thus ( 3 + 5) d = ( 4 4 + 5 ) = ( 4 4 + 5 ) ( 4 ( )4 + 5 ( ) ) =.5 Emple 4: Things cn go wrong if the integrnd hs verticl symptote. (Plese note the Fundmentl Theorem does require f() to be defined nd continuous on the whole intervl of integrtion.) For emple, the following clcultion is wrong. d = = () ( ( ) ) =. We cn tell immeditely tht something is wrong since d is the net re of figure tht is lwys bove the is, nd therefore the integrl better be postive. Furthermore, using pproimting sums, it looks s though the integrl is infinite. Thus, while the fundmentl theorem gives us very powerful
method to compute definite integrls, we do need to be creful nd check tht our function stisfies the hypotheses of the fundmentl theorem before using the conclusion of tht theorem. Hving sid tht, I should point out tht, lthough s stted the fundmentl theorem requires the function to be continuous, in ctulity the fundmentl theorem holds for clss of functions more generl thn the continuous functions. In prticulr, if the integrnd hs finite number of jump discontinuities then the Fundmentl Theorem still works. Integrls whose integrnds contin symptotes re referred to s improper integrls nd re studied in detil in Clculus. A lot of integrls tht rise in prctice (especilly in Probbility Theory nd Quntum Mechnics) re improper integrls.. Using the Second Fundmentl Theorem of Clculus A lrge prt of the difficulty in understnding the Second Fundmentl Theorem of Clculus is getting grsp on the function. As definite integrl, we should think of A() s giving the net re of geometric figure. The function A() depends on three different things. First, it depends on the integrnd f(t), different integrnd gives rise to different A(). In terms of res, the integrnd determines the height of the figure. Net, A() depends on. tells us where the firgure strts. Finlly, A() depends on, n tells us where the figure stops. Now, we usully think of f(t) nd s being fied, n is our vrible. For emple, if f(t) = t nd =, then A() gives us the re of trpezoid, but the bse of this trpezoid is not necessrily given. See the figures below. Thus, A() = 0 since the trpezpoid degenertes into verticl line. A() is the re of trpezoid with bse = nd heights nd, thus its re is ( ) + = 3. Generlly, A() is trpezoid with bse nd heights n, so A() = ( ) + Now we cn esily compute the derivtive of A(): A () = =. =. This required quite bit of work. First, we hd to write down formul for A(), then we hd to tke the derivtive. The Second Fundmentl Theorem tells us tht we didn t ctully need to find n eplicit formul for A(), tht we could immeditely write down A () =. We remind ourselves of the Second Fundmentl Theorem. The Second Fundmentl Theorem of Clculus. If f() is continuous on n intervl nd is ny number in tht intervl, then = f(). d In words: Integrting f(t) up to, then computing the derivtive of this with respect to is the sme s evluting the integrnd t, i.e., the sme s plugging into f(t). We turn to some more emples. 3 Emple 5: Find (t + 3t) dt. d Solution: By the second Fundmentl Theorem of Clculus, we need only evlute the integrnd, t + 3t, t. Hence (t + 3t) dt = + 3. d
4 (You could just s well hve first computed the definite integrl using the first Fundmentl Theorem, then computed the derivtive. Doing this should gin give + 3. But doing so misses the whole point of the Second Fundmentl Theorem.) Emple 6: Find sin (θ ) dθ. d 0 Solution: A few notes bck we mentioned tht sin (θ ) hs no ntiderivtive in terms of simple functions, so it is impossible to compute this by first computing the definite integrl, then computing the derivtive. But this doesn t bother us since we know the Second Fundmentl Theorem: sin (θ ) dθ = sin ( ). d 0 Mtters re bit more complicted if the limits of integrtion re not simply. The most generl sitution llows both the upper nd lower limits of integrtion to be functions of. The Second Fundmentl Theorem of Clculus with rbitrry limits of integrtion. If f() is continuous nd u() nd l() re differentible functions, then [ d u() = u ()f(u()) l ()f(l()). d l() Emple 7: According to this form of the Fundmentl Theorem, e t dt = () e ( ) e = e + e. d This hs nice geometric interprettion. As increses, the re represented by et dt epnds both to the left nd to the right. On the right we dd little rectngle of height e, on the left we dd little rectngle of height e. Emple 8: Emple 9: Emple 0: Let G() = ln 0 Solution: [ cos t dt = ( ) cos ( ) ( ) cos ( ) = cos ( ), d d e t 3 dt = (e ) (e ) 3 (3) (3) 3 = e 4 8 3. d 3 e t / dt. Find G (). G () = d ln e t / dt = (ln ) e (ln ) / = d 0 e (ln ) /. The lower limit of integrtion is constnt, so the (0) e 0 / term is just 0, so we did not include tht term. We ll see lots of pplictions of the First Fundmentl Theorem of Clculus in the net set of notes. Direct pplictions of the Second Fundmentl Theorem re bit hrder to come by, but pplictions do eist for it. In prticulr, the lst emple is typicl sort of clcultion you might do in probbility clss (specificlly, this emple reltes the probbility densities of the norml distribution nd the log-norml distribution.)
3. Theory We look t some of the more theoreticl spects of the Fundmentl Theorems in this section. We begin by outlining proof of the fundmentl theorems. Actully, this rgument should be very fmilir to you by now. We ve gone through this rgument in severl specific cses lredy. Even though I m tking the time to go through the proofs of these theorems, you will not be responsible for either of these proofs. Being ble to do clculutions like the emples t the beginning of the set of notes on Ares nd Antiderivtives is fr more importnt thn memorizing this proof. After you ve done enough emples like those you should see tht the proof of the fundmentl theorem is the ect sme rgument just written out in bstrct generlity. We begin with the Second Fundmentl Theorem: We suppose f() is continuous on n intervl, is in the intervl, nd we need to show tht the function A() = is n ntiderivtive of f(t). In other words, we need to show tht A () = f(), or = f(). d We don t hve n eplicit formul for the integrl, so our only hope for computing the derivtive is to write out the definition of derivtive: +h = lim. d The numertor is the difference between two net res, the first from to + h, the second from to. Thus, the difference is the net re from to + h, i.e., +h = +h. If h is very smll then this lst integrl is the net re of figure tht is essentilly rectngle with width h nd height f(), so +h Piecing together everything, = lim d = lim f() h. +h +h f() h lim = f(). This only shows f(), d so techniclly this is not proof of the second Fundmentl Theorem. To get rigorous proof we would need to quntify just how good the pproimtion +h f() h is. Requiring f() to be continuous is enough to gurntee tht this pproimtion is good enough. You should look bck t the first two emples in the the notes on Ares nd Antiderivtives. The proof of the Second Fundmentl Theorem of Clculus is ectly the sme rgument we used in those specific emples. 5
6 Now for the first Fundmentl Theorem. Honestly, the proof I m going to give is bit confusing the first time you see it. The problem is tht the First Fundmentl Theorem of Clculus is relly just sying the sme thing s the Second Fundmentl Theorem, but just in different wy, so it my pper tht we re not doing much in the proof. (There ctully is something going on. To go from the Second to the First, we need to pply the +C Theorem, nd the +C Theorem is consequnce of the Men Vlue Theorem) Thus, we ssume f() is continuous on [, b nd we let F () be n rbitrry ntiderivtive of f(). We need to show = F (b) F (). (I ve swithced the vrible of integrtion from n to t only becuse I wnt to use for something else.) Do we know n ntiderivtive of f()? Yes, the Second Fundmentl Theorem tells us tht A() = is n ntiderivtive of f(). Furthermore, since both F () nd A() re ntiderivtives of f() then there is constnt C so tht F () = A() + C. Also, A(b) A() = = ; = 0 since this integrl is the net re of rectngle of height f() nd width 0. But Therefore F (b) F () = (A(b) + C) (A() + C) = (A(b) A()) + (C C) = A(b) A(). F (b) F () = where F () is ny ntiderivtive of f() on [, b. It is possible to prove the First Fundmentl Theorem of Clculus directly, tht is, without recourse to the Second Fundmentl Theorem. Such proof requires doing creful estimtes with Riemnn sums. The dvntge of the bove proof of the Fundmentl Theorems is tht we did not hve to del directly with the Riemnn Sums. We close by writting down ech of the Fundmentl Theorems in yet nother form. The First Fundmentl Theorem of Clculus. If F () is continuous on [, b then F () d = F (b) F (). The originl sttement involved two functions, f() nd F (), nd F () ws n ntiderivtive of f(). But this just mens tht F () = f(), so we replce f() with F (). This resttement fits in nicely with the Second Fundmentl Theorem. Wheres the Second Fundmentl Theorem sys tht tking the derivtive of n integrl brings you bck to the originl function, this form of the First Fundmentl Theorem sys tht tking the integrl of derivtive brings you bck to the originl function. The Second Fundmentl Theorem of Clculus. If f() is continuous on n intervl contining, then +h lim = f(). The originl sttement sys d d = +h using +h = f(). But writting out the definition of the derivtive then gives this lterntive form. Corollry: If f() is continous on [, b then f() hs n ntiderivtive. This follows directly from the Second Fundmentl Theorem. ntiderivtive of f(). Tht theorem eplicitly describes n