1 The fundamental theorems of calculus.


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1 The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Tody we provide the connection between the two min ides of the course. The integrl nd the derivtive. Theorem Suppose f is continuous function on [, b]. (FTC I) If F is n ntiderivtive of f, then b f(t) dt = F (b) F (). (FTC II) If g(x) = x f(t) dt, then g = f. Compute Compute d x dx t dt. 3 x 3 dx. Proof. An ide of the proofs. FTC I: We let F be n ntiderivtive of f nd let P = { = x < x < x 2 <... < x n = b}. We will express the chnge of F, F (b) F (), s Riemnn sum on the prtition. Letting the size of the lrgest intervl in the prtition tend to zero, we obtin the integrl is equl to the chnge in F. We begin by writing F (b) F () = F (x n ) F (x n )+F (x n )...+F (x i ) F (x i )+...+F (x ) F (x ). We recll tht F is n ntiderivtive of f nd pply the men vlue theorem on ech intervl [x i, x i ] nd find vlue c i so tht F (x i ) F (x i ) = f(c i )(x i x i ). Thus, we hve F (b) F () = f(c i )(x i x i ). i= Since the righthnd side is Riemnn sum for the integrl, we my let the width of the lrgest subintervl tend to zero nd obtin F (b) F () = b f(s) ds.
2 FTC II: Write g(x + h) g(x) h = h x+h x f(t) dt. 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. Indefinite integrls. We use the symbol f(x) dx to denote the indefinite integrl or ntiderivtive of f. The indefinite integrl is function. The definite integrl is number. According FTC I, 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 (). x b x= nd x y =x Solution. b 2 xy x 2
3 According to FTC II, ntiderivtives 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 ntiderivtives in Chpter 4.) Verify x cos(x 2 ) dx = 2 sin(x2 ). Solution. According to the definition of ntiderivtive, we need to see if d dx 2 sin(x2 ) = x cos(x 2 ). This holds, by the chin rule..2 Computing integrls. The min use of FTC I is to simplify the evlution of integrls. We give few exmples. ) Compute π sin(x) dx. b) Compute 4 2x 2 + x Solution. ) Since d ( cos(x)) = sin(x), we hve cos(x) is n ntiderivtive 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 ntiderivtive. As the indefinite integrl is liner, we write 2x 2 + dx = 2x 3/2 + x /2 dx = 2 x 3/2 dx + x /2 dx = 4 x 5 x5/2 + 2x /2 + C. dx. With this ntiderivtive, we my then use FTC I to find 4 2x 2 + x dx = 4 5 x5/2 + 2x /2 4 x= = / /2 ( ) = 28/5 + 2/5 (4/5 + /5) = 34/5.
4 π 2x cos(x 2 ) dx. Solution. We recognize tht sin(x 2 ) is n ntiderivtive of 2x cos(x 2 ), 2x cos(x 2 ) dx = sin(x 2 ) + C. Thus, π 2x cos(x 2 ) dx = sin(x 2 ) π x= =. Here, is more involved exmple tht illustrtes the progress we hve mde. lim n n sin(k/n). k= Solution. We recognize tht sin(k/n) n k= is Riemnn sum for n integrl. The points x k, k =,..., n divide the intervl [, ] into n equl subintervls of length /n. Thus, we my write the limit s n integrl lim sin(k/n) = sin(x) dx. n n k= To evlute the resulting integrl, we use FTC I. An ntiderivtive of sin(x) is cos(x), thus sin(x) dx = cos(x) x= = cos()..3 Differentiting integrls. FTC II shows tht ny continuous function hs n ntiderivtive nd cn be used to find the derivtives of integrls. d dx x sin(t 2 ) dt d x sin(t 2 ) dt dx x 2 d x dx t dt Is the function L(x) = x t dt incresing or decresing? Is the grph of L concve up or concve down?
5 .4 The net chnge theorem Since F is lwys n ntiderivtive 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 (2) 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 (2) 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 4, 22 m(x) = x dm dx dx.