1 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 function on [, b]. (FTC I) If g(x) = x f(t) dt, then g = f. (FTC II) If F is n nti-derivtive of f, then b f(t) dt = F (b) F (). Exmple. Compute Compute d x dx t dt. 3 x 3 dx. Proof. An ide of the proofs. FTC I: Write g(x + h) g(x) = x+h f(t) dt. h h x We will show x+h lim f(t) dt = f(x). h + h x The reder should write out similr rgument for the limit from the below. If f is continuous, then f hs mximum nd minimum vlues M h nd m h on the intervl [x, x + h]. Using the order property of the integrl, m h h x+h x f(t) dt M h. As h tends to, we hve lim h + M h = lim h + m h = f(x) since f is continuous. It follows tht x+h lim f(t) dt = f(x). h + h x FTC II:
2 We know from FTCI tht f hs one nti-derivtive, x f(t) dt. We let G(x) = x f(t) dt F (x) where F is some nti-derivtive s in FTC II. The derivtive of G, G (x) = f(x) f(x) = for ll x in (, b). This uses FTC I nd the hypothesis tht F is n ntiderivtive of f. Since the derivtive of G is identiclly zero, we cn conclude tht G is constnt. If we set x = in the definition of G, we find G() = F () so tht we cn conclude the constnt is F (). If we set x = b in the definition of G, then we conclud F () = b f(t) dt F (b). Adding F (b) to both sides give the conclusion of FTC II.. Indefinite integrls. We use the symbol f(x) dx to denote the indefinite integrl or nti-derivtive of f. The indefinite integrl is function. The definite integrl is number. According FTC II, we cn find the (numericl) vlue of definite integrl by evluting the indefinite integrl t the endpoints of the integrl. Since this procedure hppens so often, we hve specil nottion for this evlution. F (x) b x= = F (b) F (). Exmple. x b x= nd x y =x Solution. b 2 xy x 2 According to FTC I, nti-derivtives exist provided f is continuous. The box on pge 35 should be memorized. (In fct, you should lredy hve memorized this informtion when we studied derivtives in Chpter 3 nd when we studied nti-derivtives in Chpter 4.) Exmple. Verify x cos(x 2 ) dx = 2 sin(x2 ).
3 Solution. According to the definition of nti-derivtive, we need to see if This holds, by the chin rule..2 Computing integrls. d dx 2 sin(x2 ) = x cos(x 2 ). The min use of FTC II is to simplify the evlution of integrls. We give few exmples. Exmple. ) Compute π sin(x) dx. b) Compute 4 2x 2 + x dx. Solution. ) Since d ( cos(x)) = sin(x), we hve cos(x) is n nti-derivtive of dx sin(x). Using the second prt of the fundmentl theorem of clculus gives, π sin(x) dx = cos(x) π x= = 2. b) We first find n nti-derivtive. As the indefinite integrl is liner, we write 2x 2 + dx = x 2x 3/2 + x /2 dx = 2 With this nti-derivtive, we my then use FTC II to find 4 2x 2 + x dx = 4 5 x5/2 + 2x /2 4 x 3/2 dx + x /2 dx = 4 5 x5/2 + 2x /2 + C. x= = / /2 ( ) = 28/5 + 2/5 (4/5 + /5) = 34/5. Here, is more involved exmple tht illustrtes the progress we hve mde. Exmple. lim n n sin(k/n). k=
4 Solution. We recognize tht n sin(k/n) k= is Riemnn sum for n integrl. The points x k, k =,..., n divide the intervl [, ] into n equl sub-intervls of length /n. Thus, we my write the limit s n integrl lim n n sin(k/n) = sin(x) dx. k= To evlute the resulting integrl, we use FTCII. An nti-derivtive of sin(x) is cos(x), thus sin(x) dx = cos(x) x= = cos()..3 Differentiting integrls. FTC I plys n importnt role in the proof of FTC II. It is lso used to find the derivtives of integrls. Exmple. d x sin(t 2 ) dt dx d x sin(t 2 ) dt dx x 2 d x dx t dt Is the function L(x) = x dt incresing or decresing? Is the grph of L concve t up or concve down?.4 The net chnge theorem Since F is lwys n nti-derivtive of F, one consequence of prt II of the fundmentl theorem of clculus is tht if F is continuous on the intervl [, b], then b F (t) dt = F (b) F (). This helps us to understnd some common physicl interprettions of the integrl. For exmple, if p(t) denotes the position of n object. More precisely, if n object is moving long line nd p gives the number of meters the object lies to the right of reference point, then p = v is the velocity of the object. The definite integrl p(b) p() = b v(t) dt ()
5 denotes the net chnge in position of the object during the intervl [, b]. Note tht if v is mesured in meters/second, then units on v(t)dt would be meters/second seconds so the eqution () is sophisticted version of the fmilir fct tht distnce = rte time. To give less fmilir exmple, suppose we hve rope whose thickness vries long its length. Fix one end of the rope to mesure from nd let m(x) denote the mss in kilogrms of the rope from to x meters long the rope. If we tke the derivtive, dm = lim dx h m(x + h) m(x)h, then this represents n verge mss of the rope ner x whose units re kilogrms/meter. If we integrte this liner density nd observe tht m() =, then we recover the mss November 6, 26 m(x) = x dm dx dx.