Mth 70 Topics for third exm Chpter 3: Applictions of Derivtives x7: Liner pproximtion nd diæerentils Ide: The tngent line to grph of function mkes good pproximtion to the function, ner the point of tngency. Tngent line to y = fèxè tèx 0 ;fèx 0 è : Lèxè =fèx 0 è+f 0 èx 0 èèx, x 0 è fèxè ç Lèxè for x ner x 0 Ex.: p 7 ç 5+ æ 5 è7, 5è, using fèxè =p x è + xè k ç +kx, using x 0 =0 æf = fèx 0 +æxè, fèx 0 è, then fèx 0 +æxè ç Lèx 0 +æxè trnsltes to æf ç f 0 èx 0 è æ æx diæerentil nottion: df = f 0 èx 0 èdx So æf ç df, when æx = dx is smll In fct, æf, df = èdiærnce quot,f 0 èx 0 èèæx = èsmllèæèsmllè = relly smll, goes like èæxè x8: Newton's method A relly fst wy to pproximte roots of function. Ide: tngent line to the grph of function ëpoints towrds" root of the function oots of ètngentè lines re esy to ænd! Lèxè =fèx 0 è+f 0 èx 0 èèx, x 0 è ; root is x = x 0, fèx 0è f 0 èx 0 è Now use x s strting point for new tngent line; keep repeting! x n+ = x n, fèx nè f 0 èx n è Bsic fct: if x n pproximtes root to k deciml plces, then x n+ tends to pproximte it to k deciml plces! BUT: Newton's method might ænd the ëwrong" root: Int Vlue Thm might ænd one, ut N.M. ænds diæerent one! Newton's method might crsh: if f 0 èx n è = 0, then we cn't ænd x n+ èhorizontl lines don't hve roots!è Newton's method might wnder oæ to inænity, if f hs horizontl symptote; n initil guess too fr out the line will generte numers even frther out. Newton's method cn't ænd wht doesn't exist! If f hs no roots, Newton's method will try to ëænd" the function's closest pproch to the x-xis; ut everytime it gets close, nerly horizontl tngent line sends it zooming oæ gin! Chpter 4: Integrtion x: Antiderivtives Integrl clculus is ll out ænding res of things, e.g. the re etween the grph of function f nd the x-xis. This will, in the end, involve ænding function F whose derivtive is f.
F is n ntiderivtive èor èindeæniteè integrlè off if F 0 èxè =fèxè. Nottion: F èxè = ; it mens F 0 èxè=fèxè ëthe integrl of f of x dee x" Bsic list: xn dx = xn+ + C èprovided n 6=,è n +, cosèkxè sinèkxè dx = + C k sinèkxè cosèkxè dx = + C k sec x dx = tn x + C csc x dx =, cot x + C sec x tn x dx = sec x + C csc x cot x dx =, csc x + C Most diæerentition rules cn e turned into integrtion rules èlthough some re hrder thn others; some will even wit until Clc II!è Bsic integrtion rules: sum nd constnt multiple rules re esy to reverse k=constnt k æ = k èfèxè æ gèxè dx = æ gèxè dx x3: Integrtion y sustiution The ide: reverse the chin rule! if gèxè =u, then d dx fègèxèè= d dx fèuè =f 0 èuè du dx so f 0 èuè du dx dx = f 0 èuè du = fèuè+c fègèxèèg 0 èxè dx ; set u = gèxè then du = g 0 èxè dx, so fègèxèèg 0 èxè dx = intfèuè du, where u = gèxè Exmple: xèx +, 3è 4 dx ; set u = x, 3, so du=x dx. Then xèx +, 3è 4 dx = èx +, 3è4 x dx = u 4 du j u=x,3 = u 5 5 + c j u=x,3 = èx, 3è 5 + c 0 The three most importnt points:. Mke sure tht you clculte ènd then set sideè your du efore doing step!. Mke sure everything gets chnged from x's to u's 3. Don't push x's through the integrl sign! They're not constnts! x4: Estimting things with sums Ide: lot of things cn estimted y dding up lot of tiny pieces. Sigm nottion: forml properties: k i = k i i = + æææ n ; just dd the numers up
è i æ i è= i æ i Some things worth dding up: length of curve: pproximte curve y collection of stright line segments y=f(x) length of curve ç P èlength of line segmentsè distnce trvelled = èverge velocityèètime of trvelè over short periods of time, vg. vel. ç instntneous vel. so distnce trvelled ç P èinst. vel.èèshort time intervlsè E.g., sètè=position, vètè=velocity, use velocity 4 times per second dist. trvelled = sè0è, sè5è ç verge vlue of function X0 vè5 + i 4 èè 4 è verge of n numers: dd the numers, divide y n for function, dd up lots of vlues of f, divide y numer of vlues vg. vlue of f ç n fèc i è x5: Deænite integrls The most importnt thing to pproximte y sums: re under curve. Ide: pproximte region èw curve nd x-xis y things whose res we cn esily clculte: rectngles! 3
y=f(x) Are etween grph nd x-xis ç P ères of the rectnglesè = fèc i èæx i We deæne the re to e the limit of these sums s the numer of rectngles goes to èi.e., the width of the rectngles goes to 0è, nd cll this the deænite integrl of f from to : = lim n! When do such limits exist? fèc i èæx i Theorem If f is continuous on the intervl ë; ë, then exists. èi.e., the re under the grph is pproximted y rectngles.è x6: Properties of deænite integrls Fisrt note: the sum used to deæne deænite integrl does need to hve fèxè ç 0; the limit still mkes sense. When f is igger thn 0, we interpret the integrl s re under the grph. Bsic properties of deænite integrls: Z Z =0 =, k = k fèxè æ gèxè dx = + Z c = æ Z c If m ç fèxè ç M for ll x in ë; ë, then mè, è ç ç Mè, è gèxè dx More generlly, iffèxè ç gèxè for ll x in ë; ë, then ç gèxè dx 4
Averge vlue of f : formlize our old ide! Z vgèfè =, Men Vlue Theorem for integrls: If f is continuous in ë; ë, then Z there is c in ë; ë sotht fècè =, x7: The fundmentl theorem of clculus then Formlly, Z x depends on nd. Mke this explicit: fètè dt = F èxè is function of x. F èxè = the re under the grph of f, from to x. Fund. Thm. of Clc èè è: If f is continuous, then F 0 èxè =fèxè èf is n ntiderivtive off!è Since ny two ntiderivtives diæer y constnt, nd F èè = Ex: fètè dt, we get Fund. Thm. of Clc èè è: If f is continuous, nd F is n ntiderivtive of f, Z ç 0 Z x = F èè, F èè = F èxè j sin x dx = è, cos çè, è, cos 0è = Building ntiderivtives: p p F èxè= sin t dt is n ntiderivtive of fèxè = sin x Z 3 x p Gèxè = +t dt = F èx 3 è, F èx è, where x F 0 èxè = p +x,sog 0 èxè = F 0 èx 3 èè3x è, F 0 èx èèxè... x8: sustitution nd deænite integrls We cn use u-sustitution directly with deænite integrl, provided we rememer tht Ex: u =5; so relly mens Z Z x= x= nd we rememer to chnge ll of the x's to u's! Z xè + x è 6 dx; set u =+x, du =x dx. when x =,u =;when x =, xè + x è 6 dx = Z 5 u 6 du =... 5