4.3 The Fundamental Theorems of Calculus

Transcription

1 4.3 The Fundmentl Theorems of Clculus Mny students think of the Fundmentl Theorems of Clculus s being formuls, such s the following ones: 1. Define F : [, b] R by F (x) := F (x) = f (x) for ll x (, b). f (t) dt. Then F is differentible nd 2. If there exists F differentible such tht F = f on [, b], then f (t) dt = F (t) dt = F (b) F (). But these require the right hypotheses. We re given f s dt. Wht hypotheses on f will mke the first sttement bove true? Wht hypotheses on f mke the second sttement bove true? Tht is wht we study in this section. 1 / 10

2 Recll tht of the conditions f R[, b] nd f C[, b], we hve f C[, b] f R[, b] f R[, b] f C[, b] Do you recll how to prove the first of these, nd wht is counterexmple for the second? 2 / 10

3 Hypotheses of the first fundmentl theorem of clculus 1. Define F : [, b] R by F (x) := F (x) = f (x) for ll x (, b). f (t) dt. Then F is differentible nd This sttement would t lest mke sense if f were merely in R[, b]. Actully, is tht true sttement? It presupposes tht if f is Riemnn integrble on n intervl [, b], then it is lso Riemnn integrble on ny subintervl [c, d] of [, b]. Actully tht is true sttement. Exercise. Let f R[, b]. Let [c, d] be subintervl of [, b]. Prove tht f R[c, d]. 3 / 10

4 Hypotheses of the first fundmentl theorem of clculus 1. Define F : [, b] R by F (x) := F (x) = f (x) for ll x (, b). f (t) dt. Then F is differentible nd Is it true tht if f is merely Riemnn integrble on [, b], then the bove function F is differentible nd F = f on [, b]? Not ccording to the following counterexmple. Exercise. Let f (t) := ) Clculte F. { 0 if 0 t 1 1 if 1 < t 2 b) Is F differentible on [0, 2]?. Let F (x) := f (t) dt. 4 / 10

5 Hypotheses of the first fundmentl theorem of clculus 1. Define F : [, b] R by F (x) := F (x) = f (x) for ll x (, b). f (t) dt. Then F is differentible nd So the bove exercise shows tht in the boxed sttement bove, it is not sufficient to merely put the hypothesis tht f R[, b]. So let s view this s the following problem: Problem: Sy we hve this unknown function f (x), nd wht we know bout it re the vlues of ll of the integrls f (t) dt for ll x [, b]. Cn we determine ech f (x) from this informtion? Wht condition on f is sufficient to chieve this? 5 / 10

6 Problem: Sy we hve this unknown function f (x), nd wht we know bout it re the vlues of ll of the integrls f (t) dt for ll x [, b]. Cn we determine ech f (x) from this informtion? Wht condition on f is sufficient to chieve this? Well the ide is tht if we know f (t) dt for ech x, then we lso know x (, b) nd h sufficiently close to 0. Do you see why? +h x f (t) dt for ech +h If f is merely ssumed to be continuous, then for h close to 0, we expect the integrl f (t) dt to x x+h f (t) dt x be close to f (x) h, nd so we would expect to be ble to prove tht lim = f (x). h 0 h This solves the bove boxed problem, nd then to finish the proof of the theorem on the previous slide (it becomes theorem if we ssume in ddition tht f is continuous), it is enough to recognize +h f (t) dt x s difference quotient of F. h In fct it is quite esy to use the continuity of f to mke the bove rigorous. We do so on the next slide. I wnted to mention tht our textbook nd just bout every clculus book proves it by mking use of the so-clled men vlue theorem for integrls. In order to prove tht theorem you hve to mke use of the Extreme Vlue Theorem nd the Intermedite Vlue Theorem. This is pretty hevy rtillery, but the proof we give doesn t require ny of this, nd is relly quite simple. 6 / 10

7 Theorem (First Fundmentl Theorem of Clculus) Let f : [, b] R be continuous. Define F : [, b] R by F (x) := differentible nd F (x) = f (x) for ll x (, b). f (t) dt. Then F is Exercise. Give the proof. Do it by proving directly from the definition of continuity of f tht for ech x (, b) we hve +h f (t) dt x lim f (x) h 0 h = 0. 7 / 10

8 Hypotheses of the second fundmentl theorem of clculus Let s now discuss the other of the fundmentl theorems of clculus. It s something like: If there exists F differentible such tht F = f on [, b], then f (t) dt = F (t) dt = F (b) F () So wht bout the correct hypotheses on f? Exercise. In the bove sttement, let s sy we merely ssume tht f is function for which there exists F such tht F is differentible nd F = f. Wht else would we hve to implicitly be sserting in order for the conclusion to hve chnce to be true? We show next tht this necessry hypothesis is lso sufficient. 8 / 10

9 Theorem (Second Fundmentl Theorem of Clculus) Let f : [, b] R be Riemnn integrble. Suppose there exists F : [, b] R such tht F (x) = f (x) for ll x [, b]. Then f (x) dx = F (b) F (). Some comments nd steps of the proof: Notice tht this theorem is stronger thn the one tht ppers in most clculus books where the hypothesis is the more stringent one tht f should be continuous. Becuse we hve developed the theory of integrtion bit differently from the text, we hve to modify the text s proof bit, but it still works out nicely. Give yourself ε > 0; we ll try to prove tht f (x) dx (F (b) F ()) < ε. Since ε is rbitrry, tht would imply the conclusion of the theorem. Begin by writing down wht the Prtition Chrcteriztion of Riemnn Integrbility Theorem sys bout f reltive to the given ε. It produces specil prtition P. Note tht ny number tht lies between U(f, P) nd L(f, P) must be within ε of f (x) dx. Apply the Men Vlue Theorem to F on ech of the intervls [x i 1, x i ] of P. This produces specil points c i in ech subintervl [x i 1, x i ] of the prtition. Using these c i s, write down the ssocited Riemnn sum, nd note tht it must lie between U(f, P) nd L(f, P). Then mke use of the fct tht f (c i ) = F (c i ). This cuses the Riemnn sum to be telescoping sum, nd the result follows. 9 / 10

10 Theorem (Second Fundmentl Theorem of Clculus) Let f : [, b] R be Riemnn integrble. Suppose there exists F : [, b] R such tht F (x) = f (x) for ll x [, b]. Then f (x) dx = F (b) F (). Exercise. Write the proof of the second fundmentl theorem of clculus. 10 / 10