Calculus2 - SPARKCHARTS CALCULUS II CHARTS OVERVIEW...

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Unformatted text preview: SPARKCHARTS CALCULUS II CHARTS OVERVIEW Calculus is the study of “nice"7smoothly changingifunctions. - Differential calculus studies how quickly a function is changing at a particular point. For more on differential calculus, see the Cu/cu/us /SparkChart Inteng calculus studies areas enclosed by curves and is used to compute a continuous (as opposed to discrete) summation. Integration is used in geometry to find the length of an arc, the area of a surface, and the volume ofa solid; in physics, to compute the total work done by a varying force or the location of the center of mass of an irregular object; in statistics, to work with varying probabilities. Differential equations (diff-eqs) express a relationship between a function and its derivatives. Diff-eqs come up functions can often be represented as infinite Taylor polynomials. Infinite series are used to differentiate and integrate difficult functions, as well as to approximate values of functions and their derivatives. REVIEW OF TERMS O A function is a rule that assigns to each value of the domain a unique value of the range. - Function f (1') is continuous on some interval if whenever an is close to 1:2, f(;rl) is close to f(.T2). - Function f (.r) is increasing on some interval if whenever or, < 1172, < [(172) (so f’(.'r:) is positive). It is decreasing if f(;r,) > f(;l72) (so f’(r) is negative). A function that either never increases or never decreases is Function is differentiable on some open interval if its derivative exists everywhere on that interval. A differentiable function must be continuous, and it cannot have vertical tangents on the interval. Function f (r) is concave up on some interval if its second derivative f”(;r) is positive there; its graph “cups up." It is concave down if f”(.r) is negative; its graph “cups down.” The line at : a is a vertical asymptote for if “blows up" to (positive or negative) infinity as J,‘ gets closer and closer to a (fmm the left side, the right side, or both). Formally, liniTga : :00 orlirn, out : ::oc (or both). The line ,7; : b is a horizontal asymptote for f (1') if the value of f(;lr) gets close to b as |ar| becomes very large (when x is when modeling natural phenomena. positive, negative, or both). Formally, liniIaJrm f(.r) : b or ' Infinite series are special types of functions. Familiar liniJg .m f(m) : b (or both). AREA UNDER A CURVE AND THE DEFINITE INTEGRAL BASIC PROBLEM OF INTEGRAL CALCULUS Given a function y = on the interval [11, b], what is the area enclosed by this curve, the ar- axis, and the two vertical lines .77 : a and :r, : b? called monotonic. - Simpson’s Rule: This time, we suppose it to be even and approximate the area with parabola pieces with the k“' parabola defined by points on the curve at [ligh- 2, must , and $21,. The total area is given by Sn = ¥Ui70l + 4f(-Tr1) + 2f(12) + 4f(1'3) t ' ' ‘ f 2fl-Tne2) + 4f(-'Iinei) + flirnll- COMPARING APPROXIMATIONS TO THE AREA - Iff(m) is increasing on the interval [a, b], then L,l <Area< R" for each n. Iff(:r) is decreasing 0n the whole interval, the inequalities are reversed. - If f (:17) is concave up on the whole interval, then L,, is a better approximation to the area than R". If f(z) is concave down on the whole interval, then R." is a better approximation than Ln. 0 The Midpoint Rule approximates area more accurately than the Trapezoidal Rule. Both are better than the left- or right-hand rectangle approximations. Simpson’s Rule is best of all. NOTE We always speak of "signed" area: a curve above the .reaxis is said to enclose positive area, while a curve below the .Izrctxis is said to enclose negative area The concept of "negative area" may seem ridiculous, but signed area is more versatile and simpler to keep track of. APPROXIMATIONS TO THE AREA - Left-hand rectangle approximation: We can approximate y y : fix) 0 this area by a series of n. rectangles. Divide the interval into 1L equal subintervals of width AI : h’" and obtain n + 1 points on the J—axis at 1:0 : a, ,r, : a + An, .. 1:” : a + nAm : I). These are the bottom corners of n rect- angles, which we’ll always number (J to n 7 1. The height of each rectangle is the value of f (I) at the left x-axis corner. The km rectangle has height f(.l?k;) and area A:L'f(1;,.). The total area of the n rectangles, then, is n, 1 L, = AI(f(In) + f(:1:1)+m+ new.» : Air. Z flank). k:0 - r The left— and right-hand rectangle approximations and the Midpoint Rule all use a prescribed point in the subinterval as the height of the rectangle, In general, we can pick any sample point at}; in the Icfl‘ subinterval. The area approximation, then, is Ar 22;; f (33;). This general area approximation is called a Riemann sum, and its limit as n increases will give the area of the region. E DEFINITE INTEGRAL 7171 If the limit liIn Am 2 f exists, then the function f(x) is called integrable on the interval new a b 0 Larger n will give more accurate approximation to the area. I Ex: We approximate the area under the curve 11 : 1'2 on the interval [0, 1] with 4 rectangles of width Ax : I; and heights 0, (it), (%)2, (fill, for a total area of h l 2 l 2 I 2 3 2 7 [a, b], The limit reprecszeiits the area under the curve and is denoted/ dc. L4 : Z + (Z) + (5) + (Z) ) : E : 0'21875' 0 In this notation, is the integral sign, f(x) is the integrand. and a and l) are the lower and ‘ upper limits of integration, respectively. - The marker dz keeps track of the variable of integration and evokes a very small Am; intuitively, the “integral from a to b off(z) d1)" is a sum of heights (function values) times tiny widths d1. (i.e., a sum of many minute areas). - Being “integrable” says nothing about how easy the symbolic integral is to write down. Often the integral is difficult to express. - All functions made up of a finite number of pieces of continuous functions are integrable. In practice, every function encountered in a Calculus class will be integrable except at points where it “blows up" towards ::oo (equivalently, has a vertical asymptote), - Right—hand rectangle approximation: Instead of taking the height of each rectangle to be the value of f at the left :1?- axis corner, we can take the the value of at the right corner. The height of the k“ rectangle is now f (n+1) for a total area of Ru : A1'(f(an)+ 1m) + - - - + f(wn)) : Ax: rm). kil - Right-hand and left~hand approximations are related by R. : L. + MW) 7 NO) 0 a b Ex: For f(ar) : $2 on the interval [0, 1], ‘< a II is e Properties of the definite integral: bet and g(1:) be functions integrable on the interval [a, b] R4 : g (a)? + (92 + (3)2 + (1)2) : = 0.46875. 3' y = fl-l‘) andp beapoint inside the interval. m, , .t W b t, b - Midpoint Rule: The height of each rectangle can be taken to 'l. Sums and differences: / (Hm) :: g(m)) d1? 2 / f(ar) d1: : / g(x) (11?. be f (2) evaluated at the midpoint of each rectangle; the a ' a a height of the lcLh rectangle is now I b ( )d b ( )d 2.5ca ar multiples: /r:f 1' z : c/ f I I. Here, c is any real number. fra+ax(k+;))=f(w) .. a x for a total area of 0 a b 3.Reversingthe limits: /bf(1n) (1x : — [fonds [frmmHAb/(x) d1 :/abf(1)dz. 2 M,,*Az(f(fl’+1’) l -- _f(x, 5+1.» _AI::1f(n+§rn)_ :II *- as W **** :- r '- 4.Concatenation: 0 'I'rapezoidal Rule: We can approximate the area under the r . v | I r I I I t v t r i u v r | I r r r u r 1 i r r x I I l r r I l I I . - r u r I r I | | r r I . r I \ I r I t a h r u r r u r t r r u . r H r u r r r u I x r r n x I u x r I r I I r u t I r n I r r r r r r r I u r I r r r r h | I t r r r a i r r V | v r r u l n I r y t t r r u I t I r r I r . u r I I I r u r | I r u r t r | I n r u r t r r v I r x r r l h - curve using trapezoids with the same two vertical sides and r- y y = fl” p, 5, axis side as the rectangles. The area ofa trapezoid is 5.Berweenness: If S on the interval [(1, b], then/ dlL‘ S (11'. (average length of two parallel sides) >< (distance between them). In pamcuhn “ " The area “the km trapewid’ then’ is % + f<zk+ll) ’ 0 If > 0 on [a b] then [b dz > (J. for a total area of — i 7 . a 7 ' If M is the maximum value of flat) on [(1, b] and m is the minimum value, then »b T. : %(f(arn)+2f(rt)+2f(1'2)+~~+2fi1nei)+f(zn))~ x milieu) S l. fwd]: S Wheat 0 a b ANTIDERIVATIVES AND THE INDEFINITE INTEGRAL Antidifferentiation is the reverse of differentiation: an antiderivative of f (I) is any function F (1) whose derivative is equal to the original function: F’(z) : f (an) in a pro—established region. Functions that differ by constants have the same derivative; therefore, we look for a family of antiderivatives F(:c) + C, where C is any real constant. The family of the antiderivatives of f is denoted by the indefinite integral: /f(z)dz : + C ifand onlyifF’(z) : f(I), The indefinite integral represents a fornin of functions differing by constants, “SO IF A MAN’S WIT BE WANDERING, LET HIM STUDY THE MATHEMATICS.” FRANCIS BACON THE FUNDAMENTAL THEOREM OF CALCULUS The Fundamental Theorem of Calculus (FI'C) brings together differential and integral calculus. ) MAIN POINT Differentiation and integration are inverse processes Finding ontiderivotives IS a lot like calculating areas under curves ’ STATEMENT OF THE THEOREM 0 Part 'I: Let f(.r) be a function continuous on the interval [(1.1)]. Then the area function m) *7 (flit) In is continuous on [11.1,] and differentiable on [in 2.] and F’(.r) : _/'(..-). fit). the“ (1.1- : 19(0) 7 F(u.). The total change in the antiderivative function over an interval is the same as the area under the curve. WHY IS THE FTC TRUE? There are two ways of thinking. about it: I. The change in the area function (function whose value at u is the area under ]'(.r) up to u) is chronicled by ./'(.r) itself: the area function changes quickly if f(.1') is large, slowly if ‘/'(.r) is small; it is increasing whenever is positive and decreasing if f(.r) is negative. So f(.r) behaves like the derivative of the function for the area under f( 2.'l‘he function for the area under ‘/"(I) behaves like itself. - Iff(.r) is increasing (or decreasing), then f’(.1t) is positive (or negative) and the area under j"(;r) is increasing (or decreasing), 0 If f(a') is changing quickly, then lf’(.r)l is large7and the area under f’(.1‘) is changing quickly as well. - If the growth rate of f(.l') is positive but slowing down to (J (i.e., [(1') is concave down approaching a local maximum), then f’(;r) is crossing the Juaxis from the positive half- plane to the negative half-plane; at the same time, the area under j"(:1t), which has been growing while f’(:r) > 0, is stopping its growth and will start to decrease: like f(;i'), the area under f’(1’) is nearing a local maximum. 1 l t z I v t r t . r s r . t I r O r I t V r l a r r r r l t x » r a r r 0 Part 2: 1f _f(.r) is a function continuous on the interval lo. bl and [3(1) is an antiderivative of 7 USING THE FTC The FTC justifies using the integral sign for both antiderivatives and areas under curves. And it gives us a simple way to calculate the area under many curves. Ex: The area under the curve y 4 1:2 over the interval [0, l] is given by r].r. Since F(.r) : is an antiderivative of .1'2 (check that F”(.l‘) : 12),. 61.1241 : F(1) 7 1"(0) : - L4 < g R4, and Ll is a slightly better approximation to the area, as expected since ./'(.I‘) : .r is increasing and concave up on the interval. / (MJ‘ : (7 COMMON INTEGRALS / (1.1- : lulrl + C / k' (1.1‘ : [or + (.' /u*d.1~: 3 l (' . lurz - (.ri+l 1‘” III : ~ 71 + l /eos J" (1.1' : sin.r + C' l 3. 2 + (' ifrl 71 /(‘J’ (hr 7 ("r + (' /si11.1‘ 11.1' : 7 cos .1' + (' tauJ'dJ' : In | scorl + C cot 2- (1.1- : In lsin .rl + (7 dJ':taii ].r+(7 fidr:sin 1.L'+(Y / see2 :1‘ (1.1- : lzui .I' + C /S(‘('.T (2111 .r LIJ‘ : sec .1: + (' TECHNIQUES OF INTEGRATION Unlike differentiation, integration is “hard”7there are easy-to—write functions that don’t have easy antiderivatives. The art of integration requires a bag of tricks. These are some of them. NOTE: All techniques work with both definite and indefinite integrals; pay special attention to the limits of integration. TIP: All functions that you will work with are integrable except at points where they blow up; all “smoothly” changing functions are differentiable. SUBSTITU N LE g(;r), then /f(.a(.r))y’(w> = / numu. - This is the analog of the Chain Rule for integrals (see the Calculus lSporkChort). It is useful for composite functions and products. If u : t/(r) is continuously differentiable on some interval and f(.r) is integrable on the range of INTEGRATION BY PART Some function products (or quotients) cannot be integrated by substitution alone. Integration by parts works when one piece of the product has a simpler derivative and the other piece is easy to integrate. - This is the integral analog of the Product Rule 51:1?) :.f’.<1+fy’- I Indefinite integrals: /-f(1')g’(1) (1.)" : f(a')g(1') 7 f/(.’I‘)g(£lf)d.’lf. or /udzI : 11117 / edit. .I: ' Definite integrals: Slap on limits: f(.r)g/(.'L‘) d.r : f(r)g(;r)]f; 7 /h f’(a:)g(.rt) (1er I A polynomial multiplied by a trigonometric, exponential, or logarithmic function is frequently best integrated by parts. Let the polynomial be u. Ex: /.I'2 \/ .r“ + 8 11.1: (1(1" i a) 2 . . .l isalotllke d1, . Let n : .1'3 + 8. Then do, : 3.1'2 1LT, so .172 ILI‘ : 1 Substituting, we transform the original integral into/ do, or i 1 t 2 g , 2 ‘ go- do: (71k +6 — i J 9(1“+8)%+C. 0 When evaluating definite integrals using substitution, you have a choice about how to deal with the limits of integration. Let’s say that you’re integrating with respect to 1'. I. less thin/(mg Choose a useful u, substitute for .l‘, integrate in terms of u, substitute 1: back, evaluate the integral with original limits. 2 . Exz/ :r2\/.r3+8dar: 5(13+8)%] l 2 74 Ms ML. 2 . o = 71 SJ 1 9 2 6 9 ' 2. less work Choose a useful u : g(:r:) substitute, integrate in terms of u, then evaluate the integral using, for limits, values of u : g(w) evaluated at the original limits of integration7all without substituting .1: back in. Formally, ifu : 9(1), then b gibl ' r , _ d, : [ff/(“)9”) I ./gm> r .u:y(2) 1 2 16 .r2 1'“ + 8 (LT : / 7 ’u. ((71. : r-ué] : 1t:g(l) 1’ 9 u do. 2;:2 In the example above, / .1:] Expression Trig substitution Expression becomes Range of f} x/a27m2 .r:asin9 x/a2—I2:acosfi 7ggagg (11; : ueosfirlt‘) 7a S S u.) x/ur27a2:atan0 0 .r : (L seed dm : a. sec 9 tan 9 d0 1‘2 7 :12 < (when a: > 0) < (when 7 < (l) \/172 + a2 at : otanf) d]: : a sec2 0 d0 \/a2+x2=asec9 7% sec26 71 :tan20 Ex: /(2:r + DPT” 'r Letu : 21' + 1 (so (In : 2d.)'), and do : 0 "’dm (so 1) : 70”). The integral becomes (21 + 1) (*FT'T) 7 / 720 "d1, which simplifies to 7(21+ he” 7 29’“ + C, or (723- 7 3y” + C. r omoernrc usrruroms Square roots of quadratics, such as i/a? :2, cannot be integrated with the substitution n 2 11.2 7 .12 because the factor of do. is missing. Enter trig substitutions, applying the substitution rule backwards. Trig substitutions are often necessary when calculating areas bounded by conic sections. . ,74 2 .'I‘ Then (1:17 : 2cos6' d6, 6 and v4 7 :r : V4 7 451112 : V4cos20 : 2cos9 (since (:05 9 Z 0 on the interval). The integral becomes /" 2 cos 0 4 sin2 0 d1: Wecansetar:2sinfi with 7g :6: .r sinH : 3 7 cotf) 7 6‘ + (I. 213059 (10 : / (tot2 0110 : /(cs(:2 9 7 1) 116 >149 7 V4772 If we want to convert back from 6 to 51:, we note that col. 0 : 7 I ; thus " v4 7 41:2 :- ‘d.r:7 I 7 7sin'l(%)+C. TABLE OF TRIGONOMETRIC SUBSTITUTIONS Pylhogoreanidenfilyused 17 sin20 : 00520 - Pay careful attention to the limits ofintegration. The intervalsfor 6' correspond to the ranges ofthe inverse h‘iyonometricfilncrions. - Expressions of the farm \/ _::1r2 +b1+c can be integrated by completing the square and converting to lheform :(a: + h)2 i (1.2. whereh : :3 undo : l0 7 Then choose the appropriate trig substitution depending on the :: signs. 2 1 + tan2 0 = see2 9 W 3, _ PARTIAL FRACTIONS Integrating rational functions7ratios of polynomials7can be tricky. However, after factoring the denominator into linears and quadratics (which can always be done, though the coefficients may not be rational numbers), a rational function can be expressed as a sum of simpler "partial" fractions. These come in four “easy”-to-integrate types: ' d A l.‘ A I./l:ln‘u‘+('. So/ (I :7ln‘a:1‘+bl+C. u I (LI + b a ' (in u.” l ' ‘ A IILI' A 2. 7 (I'fu 1. s ."ll”" C. u" 'n +1 + l n 0/ ((1:1' + b)" (L('n + I) (“I )) + l 3./ I I! : ln lul + (Y. So ’1] ‘A.-»B A‘Z-I ' 7:47“ . I i d.r / , m + I (Lr +/ . 2" ([J‘ I our) + 10.1" + t‘ 20 I (1,;1'2 + I).1' + r' "4‘1 + (II + (‘ A r ' I) : E In laur2 + hr + (‘I + mm, where D : B 7 1' l ‘ i I.) 1.’ 4. I ( u ,) m 7 tan’1 2 l 0 So , I I where the denominator has no real I 112 + a- u a, ang + lur + c roots (h2 7 4m < 0) can he evaluated by completing the square in the denominator, which becomes tr(.r + h)2 + Is'2 Where It : and Ir : The step-by-step process for integrating I/'(.r) : follows: 1. If necessary, use long division to get to the point where the degree of the numerator is less than the degree of the denominator. The function has the form f(.1:) : 5(1) + . 2. Factor the denominator, reducing it to linear factors in the form ((1.1' + b)“ and irreducible quadratic factors in the form (err2 + 41;]: + r)" where (12 7 4(‘6‘ < 0. 3.Decompose into a sum of partial fractions: 0 If (1(1) has no repeated factors, express 7’(.lf) Al A1 011+ D] (‘ka- + Dr. 7 + - « « + . r , . all + I); (Ir/.17 + I); ('11'1 + If. .1' + cl (141:2 + dk1'+ ck Solve for all the As, Cs, and Us by multiplying the equation by ([(J') and equating coefficients. TIP lf «)(J') factors as a product of two linears (.r 7 n)(.r 7 b), then we can solve for A and B in 072371 1.) j%.. + 33,, qU'Ck‘VrbVlefg A ?.-(.“%70nctB f [T72 77 I o If q(.1:) has repeated factors, then for each factor in the form (rut + b)" expect fractions A1 A2 + + All a1 + b ((1,at+ b)2 ((1.17 + 11)” E h( +4 + )mf t '11 iefat' C‘IH)‘ “MID” " ’ ac r cmn— .\. ac (T L I 0 WI gv r gar +d1+€ (612+dar+e)7" 4Jntegrate 5(1;) and each partial fraction individually, using the four types of integrals above. 41:3 7 17:13 7 28 A B I C]: + D (172)(r+2)(312+4) 1:72+1+2‘ 3x2+4‘ Cross~multiplying, simplifying, and equating coefficients gives the four equations 72(A—B)+D:7, 6(A7B)+D:717, 3(A+B)+C 4, (A+B)7C:0. Solving this system of fourlinear equations (in this case, it is easier to View this as two systems of two equations, and solve for A + B and C, and for A 7 B and D independently), we get A : 71., B : 2, and Cr + D : J? +1.Nowwecanintegrate. Ex: IMPROPER INTEGRALS Improper integrals come in two types: t . . t . n , 1 1. Definite integrals over an infinite interval, Ex: c” dzr. 2. Definite integrals over an interval in which the function blows up to infinity (has a vertical : asymptote). Ex: d1. Not all improper integrals converge—represent a finite area. To evaluate an improper integral, we interpret it as a limit. If a finite limit exists, the integral converges; otherwise the integral diverges. INTEGRALS OVER AN INFINITE INTERVAL Improper integrals of this type should be rewritten as one of three limit forms: 3' no I I. Interval infinite to the right/ f(37) d1 = Llllll / f(:c) dc. b [limos/t f(:c)ria:. b 2.1nterval infinite to the left:/ fix) dz 7 oo 3. Integrals over the whole real line: / rlm :/ d1: +/ (11 for any a. Improper inlegml/ f0") ‘IZI' r The original integral converges only if both integrals over half—intervals converge do (1', - / converges (and equals fi) if and only ifr > 1. 1 x O The integral / f(zr:) (1.1- will converge only if llIll f(.1:) :0 (y: 0 is a horizontal 37m asymptote). If the function does not tend to zero, the area underneath it will certainly not be finite, and the integral will diverge. Analogous statements are true for other infinite—interval : improper integrals. HOWEVER." lllll f(.r;) : (l alone does not imply convergence of an improper integral I Cold] > r: in) lnl : 0e 1 I7-x _ 0‘ (If I 1 (If _ The ClOSSIC example rs 7 7 hm 7 : Inn 1 .I' t7~mI l .I' I7'x The integral diverges, ‘ PO YNOMIALS IN TRIGONOMETRIC FUNCTIONS For powers of trigonometric functions, regular 'u-substitution may not work. Use the following substitutions instead. - Pythagorean identifies: sin2 (9 + cos2 9 : l 1 + tan2 0 : set-.21? 0 Square sine/cosine substitutions (from the half7angle formulas): . l 4 sin2 9 : §(1 7 (:05 20) cos‘) 0 : $(1+ cos 26) - Odd powers of sine or cosine: To compute/ sin" 0 (10 when 71 is odd, keep one sine factor and replace the rest with cosines using sin2 9 : I 7 0052 (7' to obtain. /(1 7 cos}: 0)% sinf) 119 . Integrate using the substitution u : cos 6. For odd powers of cosine, “co” all the functions above (“‘co” cosine : sine). - Even powers of sine or cosine: Use the square formulas . 1 . 1 sin) 9 : 7 (‘os2fl) or cos2 9 : §(1+ cos 20) to reduce the power by half. Expand the expression (using the Binomial Theorem) and use appropriate tricks for odd or even powers on each factor individually. - Mixed powers of sine and cosine: When computing f sin" 0 cos'” 6 d6. use combinations of substitutions outlined above. The end goal is always to reduce to a sum of terms with only one power of either sin 6 or cos 9 (or both) each, which can be integrated using a u—substitutzion. 0 Products of sines and cosines of different angles: Use the following identities to get rid of products: siiiAcosB : 7 (sin(A — B) + sin(A 7 3)) siuAsinB : ((‘os(A 7 B) 7 cos(A 7 B)) cos/leosB : 7 (cos(A 7 B) + cos(A + 3)) MHNlt-‘ml—t MNE/VION/C Products of like terms use cosines, unlike terms use sines - Powers of secont or tangent: - Even powers of sec 0: Convert all but two secants to tangents using Pythagorean identity 1 + tan2 {J : sec2 6; use substitution u, : tanfi. 0 Odd powers of tan 0 :Convert all but one tangent to secants, pull out a factor of see (I from the polynomial in secants and use the substitution u : sec 9. Exz/tans 0 :19 : /(I + $962 6)2 tanfil (if) no... 0 other powers: Combine tricks and use + 2see6’ + sec3 6’) taut} scc 0 d0 : / l + 2n +143 du. I u /tanr'}d6:lnisec6‘ +0 and /scc9d9=lnlscc9+tan€l +0. 1 l i r I l r s r r a x t 1 t r r l x x A r x r r r i l I r I I r r a . r r r x I I r r t r r r t r r u t p I r I r I r x h l r l l r r I r I I t I r t t r x 1 r r v r r I r r x x I r I x x r r r - u r x r I I | r r s i y I w o r r r r I r r r r r r u x - l r I r r - INTEGRALS OVER AN INFINITE OISCONTINUITY r The integral f (I) (is is improper if at any point c in y the closed in'telrval [u, b], the function blows up. 0 If c : u is the left endpoint, then the integral is [7 Hm) (1-7:. 0 If c : b is the right endpoint then the integral is improper if lim f(l‘) : :00 mg I and is reinterpreted as lim / f (it) [11. 17b a improper if link 7 :00, 17a The integral is interpreted as lim l7~a‘ f(.l') has a vertical asymptote at .r' : (3 h 50/ f(.1.‘)rf.’r is an improper integral, . a 0 If r' E (a, b), then the original integral is understood to be the sum of the two improper integrals [foo d$+/be(z) date. The original integral converges only if both endpoint—improper integrals converge independently. 1 0 / —I converges (and is equal to o d, fififandonlyif r<1. l o .1 1 ' 1.- l.‘ NOTE/ diverges land does not evaluate to 0) because both/ : and 571 diverge, even . 1 -' r l -" , o «1' though the two areas seem to be “equal” and opposite in Sign. -l U l (/J' 111‘ (1.1‘ On the otherhond,/ I : l) because both halHntegrals/ 7— and/ 77 converge—and I ,l W . 71 . 0 \fi their values are equal in magnitude and opposite in sign GEOMETRY OF CURVES I Shell method: A solid is obtained by revolving the region under the curve y : between : a and I, : b (the area of this region is f(1)dw) around the y-axis. Instead of cross—sectional slabs perpendicular to the axis of revolution we consider the volume of a small cylindrical shell of radius a: and thickness Am. The surface area of a cylinder is AREAS BOUNDED BY CURVES Suppose that f(:1') 2 9(1) on the interval [(1. b] and both functions are continuous. Then the area bounded by the two curves y : fix), y : 9(a) and the two vertical lines (I; : a and .I' = b ' y IS I) / (we) — gm) dz. I In general, if the (continuous) curves cross each other on the interval, then the positive area defined by the curves (circunifere cc) >< (height) or 27r:rf('I.); thus, the Surface between a and I) is b volume of the solid is She" memod Area / mar) —g(w)l dz. WW II I, V : / 27r:11f(J') (11'. It is most easily evalutated by considering shaded area is/ (“1.) 7 9(1)) (171.4 - a subintervals whose endpoints are all points I: such ~ (I that = 9(8). I If the area is bounded by horizontal lines, it may be easier to rewrite the curves in the form 2 f ” 1 and integrate the difference between them with respect to y. VOLUMES: SOLIDS OF REVOLUTION Suppose that a solid is oriented along the x»axis so that the area of a cross-section (the slice of solid intersecting with a plane perpendicular to the :Ir—axis) is given by the function The volume of a slice of thickness A1: is A(z)A;r, and the volume of the solid bounded by the planes at = a and Area Am : "we .1" : I) is " Volume II Wflxlex V : ‘ib A(I) (12;. The shell method is often used when it is hard to compute the inside or outside radius of the cross-sectional slabs perpendicular to the axis of revolution. ARC LE NGTH If f(.’L') has a continuous derivative on the interval (a, b), then the length of the curve from :1::rit017:bis b ‘ t‘ll' \: ll\ 2 ‘ L 7/ (1+ (f,(1‘))2 (11, y I gll V1+If . ii ,\.I 2 h d In Leibniz notation this becomesL : / 1 + a ‘- y : fm I Why? If we break up the interval into n subintervals each of width A1: : b3“ with 17; a sample point in the 19‘" interval, then the length ofthe curve on the k‘“ interval is approximately the length of the vector (A115,A.Itf, ). or A1: 1 + Take the limit of the Riemann sum to get the formula. SURFACE AREA: SOLIDS OF REVOLUTION The surface area of a surface swept out by revolving the function y : f(r) about the r-axis between .1' : a and :1: : b is b s :/ 27rf(1')I/l + (f’(;r))2 In. (I 0 I Disk method: The volume of the solid swept out by the curve 3/ : f('J:) as it revolves around the 1—axis between .I' = a and .t : b is given by . I, I _ / 7r(radius)2 (11' or7r / (ix. D'Sk melhod .a I Washer method: If f(.’I,') > g(1:) between a and b, I Amman: 7r(f(xlz-glxl2l then the volume of the solid swept out between the two curves 3.; : and y : g(x) as they revolve around the 1'»axis between .17 : a and 1' : b is V : for perimeter width = 1 + (f'lxllex y :ful I The formula is obtained by approximating the surface area by cylindrical bands of radius f (1.) and width equal to the tiny arc length on the tiny interval. /7r(outcr radius)2 7 7r(inner radius)2 or b I . I / (nor 7 (9(1))2 d1. Washer method I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I A parametric curve defines both the x— and the y- coordinates in terms of a third variable, often t (as in “time”). Parametric curves don’t necessarily represent functions and don't have to pass the vertical line test. Sometimes the domain of t is restricted to an interval (L S t S b. (II It via 2 i T . comm. d y : i : I Area defined by curve: The area between the r»axis and the curve traced out from t : a to 7 . h trb‘s A:/ y(t)m’(l)dt. Ex: .1: 2 cost and y : shit for 0 S t < 277 are para- metric equations that describe the unit circle. NOTE. The area is counted as negative for the regions where the curve is moving ”bockwords"Ai e, :17“; < U—ond positive when the curve is moving "forwards" I HAPPY CONSEQUENCE: The area enclosed by a loop wholly above the .I‘iflxls, traversed exactly once from t : (l to r : It. can be computed directly- A :. I To convert a curve described parametrically to Carte— sian coordinates, try to relate :r. and 3; directly, eliminating t. If possible, solve for t in one of the equations, and plug that expression into the other equation Again, this may not give a function for y in terms of .’L'. Ex: In the example above, solving and plugging in will give something like y : sin (cos’1 1'), which is not very useful. However, if you use the fact that cos2 6 + sin? 6' : 1 for all angles 0, you can relate :1: and y with the familiar x2 + 1/2 : l -l.r i/(tlr’ltldl a The integral is positive for loops traced out clockwise, negative for those traced counterclockwise. Alternatively, break up [(1. h] into subintervals depending on the sign of % , integrate separately I Arc length: The length ofa parametric curve traced out from t : a to I, : b is 2 equation for the unit circle. To define a function, you have to choose a piece of the curve; for example, y : I/ 1 7 .12, Doing so is equivalent to restricting t to the interval [1), 7T]. ’15 This formula also works works for loops. GEOMETRY OF THE CURVE The following formulas are obtained using the chain rule. I Surface area of revolved solid: If the same curve always stays above the I-aXIS (LI/(t) Z (l), l dy ,(1) l d then the surface area swept out when it is revolved around the .Ir-axis is "II_E7III (T7 1 y I I .\ I Slope oftangent. — 2:? 7 II“) If a 0 andfi # (l,then the tangent is vertical. b d 2 ‘2 s:/ 27ry(t) (it. P o R c o o R D I N E s For more on polar coordinates, see the Pro-calculus Spork(.‘hart, I Slope of tangent: Convert the curve 1' : f (9) into parametric equations in Cartesian coordinates: .I' : rcos 9 : f(0) ms 0; y : f(0) sin 9. The slope of the tangent to the curv at (1:09), y((9)) is S6 41.x % cos 9% e '7' sin 6 Polar coordinates describe a point P = (7', 6') on a plane in terms of its distance 1' from the pole (usually, the origin 0) and the (counterclockwise) angle 6 that the line W makes with a reference line (usually, the positive ar~axis). I To identify a point, it is standard to limit 1‘ Z (l and 0 S 9 < 27r. although (7r, 9) : (r, f) :: 7r) and (730) : (r, 6‘ + 2n7r) for integer n. CONVERTING BETWEEN CARTESIAN AND POLAR COORDINATES I From Cartesian to polar: r : I/Ii-2 + 112: 9 : tan" ' From polar to Cartesian: .r : rcos 9; y : T sinf) Functions in polar coordinates usually define ’l' in terms offi. They need not (and almost never will) pass the vertical line test. Circles: The graph of 1‘ : a. is a circle of radius lal centered at the origin. The graphs of 7' : a sin (I and 7' : ac0s9 are circles of radius centered at (U, and ($5.0), respectively. Roses: The graphs of7' : sin m9 and r : cos 11.9 are roses centered at the origin with n petals if n. is odd, 2n petals if 71. is even. Limacons and cardiods: The graphs of (L :: b sin 6 and II, i bum) are limacons. If > 1, the Iimacon has an inner loop; if : l, t e limacon is “heart—shaped“ and is called a cardiod. d . Ily 7 j 7 smdfl: +1“ I Area: The area enclosed by rays at f} : (r and 0 = l)’ bounded by the curve 1' = is B] . A:./u Err-1 (19. Why? The area of a circle of radius r is 7FI'2 (angle sweep of27r). The area ofa sector of a circle of radius 7' and angle measure 9 is thus £130. We approximate the area by a Riemann sum of slivervsectors with radius r and angle measure A6. [4, I Arc length: The length of an are 1' : from H : (I to (I : If is ,rt 1 2 .2 i I + ((10) ([0. The formula is derived by converting the curve to Cartesian parametric equations with f) as the parameter. Newton’s Second Law, F : ma, states that the force F on an object is proportional to the object’s mass m and its acceleration a : Work is the product of a force and the distance through which it acts. If the force is constant, then W’ : Fm. If the force F(:r.) is variable and depends on the distance 1:, then the work done , . . I : V : . b by F(r.)1n movmg an object from I a to .1, bis W : / F”) din I The classic situation of a force dependent on distance is the subject of Hooke’s Law: the force required to stretch or compress a spring m units away from its natural position is given by F (:11) 2 k1“: here, I: is a constant that depends on the tightness of the spring. The center of moss (CM) of any system is the point on which (if connected) it could balance on a fulcrum. I Moment: The farther away something is from the fulcrum, the “heavier” its mass counts. Each mass of weight in a distance x from some point contributes m1: worth of moment (or torque) with 361 E 952 respect to that point. If the point is the CM, then all x the moments of the system have to balance. mass m1 I mass m2 The system behaves as though all of its mass were fulcrum conceniroied Ol lhe CM Torques: ml(T 7 .n) counterclockwise I 2 masses, 1 axis: If a massless rod with objects of “"3”? T I) CIOCEwise Masses balance, 50 : masses m1 and "12 at each end balances on a fulcrum "ll +an at its CM, then the distances d; and (12 from the masses to the fulcrum must satisfy mull : 7712(12. I If such a rod has length d, then distance (11 from the in] mass to the fulcrum satisfies m. 111 : 7112(11 7 (11). Solving, 111 : We can View this as positioning the rod along the .1-«axis with the mi mass at the origin. The CM, then, is at the point , where INTEGRATION: APPLICATIONS TO PHYSICS _ I Discrete masses, 2 axes: The CM of a system of objects of masses m1, ,. 11 and 1‘2 are the distances of the masses from the origin. Here, .r. : 0 and 12 : d, the length of the rod. I Discrete masses, 1 axis: In general, a system of n objects of masses m1,rri2,...,m,, positioned at points 1‘1,.T2, . . . ,1:,, along the x-axis (respectively) has CM at the point mill + 771212 + "'+ mum-n ml+m2+~-+m,, ' This is the moment of the system about the x—oxis. The moment of the system about the y—axis can be computed independently. i: .,m,, at points (1:1 , y] ), . . . , (1:7,, y,,) on a coordinate system is at the point (1?) (min + - - - + mnruxmly1+m+ rn,,y,,>‘ ' ml +‘umn rri1+---m,, I Continuous mass, uniform density: The CM of a flat plate-like object of uniform density p is computed by taking the limit of a Riemann sum. If its area is A, the total mass is given by m : pA, Suppose that the shape of the object is given by the curve y = f(a:) from .7: : a to .1? : I). As usual, we approximate the object by thin rectangular strips of width Adi, height f(a:), area f(.r)A17, mass pf(ir)A;c, and CM at the point 5(1)). I x-coordinate of the CM: For each strip, the moment is given by (mass) X (a:—coordina.te of strip CM) : pu>f(at)Aar. The .r-coordinate of the CM (equivalently, the moment about the y-axis) is therefore b h :L'fil?) (1L5 : mf(;p) duh I y—coordinute of the CM: The moment of each strip is (mass) >< (y-coordinat,te of strip CM) : (pf(I)Aa7) 1 b r The y—coordinate of the CM is therefore Z / 2 d1. The density p doesn't appear in the final result; all that matters is that the density is uniform. AVERAGE VALUE For a discrete set of values, their average multiplied by their number gives their sum. The analog of an average for a continuous function on the interval [a,b] is the average value 7, which has the property that the rectangle of height 7 and width bio, has the same area as is enclosed under the curve 7 : f(.'1r). Thus h 1 bio a 7: 7 : f((’) is the average value of [(1) on the interval [(1. b], The two shaded regions have equal area, I The Mean Value Theorem for Integrals states that a continuous function attains its average value. Like the MVT for derivatives (see the Calculus / SparkChort), this is a completely intuitive statement. GENERAL PROBABILITY DENSITY A probabiiity density function describes how likely it is that the outcome of some “trial” is 1;. The probability that the outcome is any specific point a is negligible; instead, we talk about the curve as representing actual probabilities. The probability that the outcome is between a and b is I.» (it. DIFFERENTIAL EQUATIONS E An (ordinary) dilfermtial equation (diff-eq) involves the derivative(s) of a (single—variable) function. I The order of a diff-eq is the highest degree of a derivative involved in the equation. y" : y” + y’ + x is a second—order diff—eq. I A solution to a diff-eq is any curve y : f (1:) which satisfies the diff—eq. A general solution is the complete family of curves that satisfy the diff»eq. Ex: The general solution to the diff-eq y’ : 4si112.r is y : —2 cos 2.r + C. I An initial condition, often the value of y([)), isolates a particular solution from the family of general solutions. Ex: If y’ : 4 sin 21 and 31(0) : 3, then y : 7200s 23: + 5. EXPONENTIAL GROWTH AND DECAY: d /clt = k 2% : Icy is a common type of diff-eq. The general solution is y : AU“. 0 Solution: Separating and rewriting, we get L1] : I: (it. Integrating yields ln |y| : kt + C or in : ('kHF. Since 1" is a positive multiplicative . factor, we replace :3” by the constant A and rewrite ,1] : Ar)“. I If k > 0. the solution represents exponential growth; if k: < (l, exponential decay. I A is the initial value of the function at t : t). Several solutions to the differential equation '71,? : 2;) Word problems that often reduce to diff-eqs of this type: 0 Unlimited population growth: k is called the relative growth rate; 1% growth (per year) means k = 0.01 (if! is measured in years). I Radioactive decay: The function measures the mass remaining at time t. The constant Is: is negative; it is often conveyed in terms of the (constant) half—lite of the elementithc amount oftime it takes for half of the remaining mass ofthe element to decay. If II is the halfvlife, then 1,: _ e 1L! ’ h ‘ I Compounded interest: The final value P of an investment compounded n times a year with , in! initial value P0 and yearly interest 7' after t years is PU) : 1’” (l + L) compounded continuously (n A» ()0), then value is lint”, ,3, , If the interest is PU) : Fur”. The (continuously compounded) investment is changing at a rate proportional to its value. INTEGRATION: APPLICATIONS TO PROBABILITY & STATISTICS The probability densityN(m) : chances that the outcome falls in some range and think of areas under the probability density I To solve: Rewrite as Ir dt = P—l‘s‘fidf’ and integrate I The probability that the outcome is something is 1; therefore, / f (:17) d1 : 1, ego - The mean of a probability density function is the long-run average outcome; it can be seen as I, the uncoordinate of the CM of the region on a graph and is given by n : / arf(1) 0 The median is the point m such that the probability that .17 < m is equal to the probability that 1: > m, (Again, the probability that (l‘ : m is negligible.) Solve for m in the equation do" = or dun : THE NORMAL DISTRIBUTION The normal distribution, or “hell curve,” is a probability y density that often arises from repeated random events. 1 ,imriri‘l e 205" - The mean is p. r7 2” I The variable (7 is the standard deviation, a measure of how clustered the outcomes are around the mean. The 0 probability that the outcome is within a of the mean is about 68% : /“+" “*0” 11 11+” N(.t) d; z 0.68. Normal distribution with mean it and Iii” standard deviation n The blue region The probability that the outcome is within 2(70f the mean is 68% 0f lhe l0l0l Shaded Oreo. is about 95%. SEPARABLE DIFFERENTIAL EQUATIONS A diff~eq is called separable if it is a first—order equation that can be expressed in the form %% : f(r)g(y), where f and y depend only on one variable. The exponential growth diff—eq r : kg is separable. I To solve a separable diff-eq, we abuse Leibniz notation to rewrite it as i : f(.1;) dr- and d , ~ , gun integrate each Side separately. Only one constant C is necessary. I Ex: Mixing problems: A tank filled with a solution of one concentration is draining at one rate while a solution of different concentration is being pumped in at another rate. The rate of change ofthe concentration y at time I. is given by % : (rate in) ~ (rat 9 out). The rate out is proportional to the current concentration. I Ex: Logistic (limited) population growth: More accurately P Several solutions to :m A a). 1n represents population growth taking into account limited natural resources. A population I’(I,) with natural growth rate k and maxiumum carrying capacity 17,,lax will satisfy the logistic differential equation (1;; : AP (1 e Pf“) . Pu using partial fractions to obtain the general solution Pa) 2 ear, 3-, ., where A : 7P<0;;,{;-w condition. is the initial I If P(U) : l) or P(()) : I’m“, then the original diff»eq gives fl = 0; the population is stable. iii 0 If P(()) 9e 0, then limtaDo P(t) : Pmax, which makes sense. LINEAR EQUATIONS A linear differential equation is an inseparable equation of the form 3/ + f(.)')]/ = g(.r), with f(.r). g(.r) continuous functions. I To solve, multiply both sides by the “integrating factor" u(1t) = (J “Jim”: note that 3% : r-l “WI” [(1') : ll.(.l')v/I(.I'). Moreover, Md?) : ug’ + My : uy’ + ufy. This gives : My; now solve for the function my. 1 ' . I The general solution is g : fl u(.1:)y(.1t) 11.1: + C) . (Note that the exponent in "(.r) u 1: can be any of the family of functions f f(r)d.r: the constant will drop out.) SEDUENCES AND SERIES Why is this a Calculus topic? Complicated functions can often be approximated with polynomials—or with infinite polynomials called “power series." Polynomials, even infinite ones, are easy to differentiate and integrate. So we can find an approximate integral or derivative of a complicated function by representing it as a power series. A sequence is an ordered list of real numbers, called terms. An infinite sequence has infinitely many terms. - Shorthand: {tidal represents the sequence a1, 0.2, (13, . . .. - A sequence is defined explicitly if each of its terms can be found independently of the other terms. Ex: u,, = 7L2 is the sequence 1,4,9. 16, A sequence is defined recursively if the 71"“ term is found using the preceding term(s). Ex: a1 : 1; a." and + (271 7 1) is again the sequence 1, 4, 9. . . .. limit of a sequence: - The limit of an infinite sequence, denoted lirn on, if it exists, is the value that the sequence approaches. If the limit exists and is finitejltlizn the sequence is called convergent. If not, the sequence is divergent. 0 Formally, the limit exists and is equal to u if for all E > 0 there exists an N so that whenever it > N. wehave |a,, — a < E. For a divergent sequence whose terms tend toward infinity, we can say that lim on : 00 if for IHIIII all integers A there exists an N so that if n > N, then a" > A. I A sequence {o,,} is called increasing if (1k 3 In,“ for all k and decreasing if oh 2 III.“ for all k. A sequence is called monotonic if it is either increasing or decreasing. - A sequence is said to be bounded above if every term is smaller than some fixed constant and bounded below if every term is greater than some fixed constant. A sequence bounded both above and below is called simply bounded. ' Monotonic Sequence Theorem: All bounded, monotonic sequences are convergent. A bounded increasing sequence cannot increase too much; the terms must cluster around some limit. SERIES: DEFINITIONS AND BASIC TYPES A series is a summed sequence: :11 + (12 + (1.3 + I I -. An infinite series has infinitely many terms. as An infinite series is often denoted: (II. or justZ ak . k:l - Apartial sum ofa series is a cut—off series sum 3,, : a1 + a2 + I ~ - + (1,, : 22:1 (1],. 0 The sum of a series exists if the sequence of partial sums converges to a limit sum. If the limit of partial sums exists, the series is called convergent, otherwise it is divergent. I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I l I I I I I I I i I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I Ageometric series has the form a + or + ("'2 + or3 + I I - = 2:0 ar’“, where a 7£ 0. D It is convergent if and only if lr| < 1, in which case its sum is If ,. u a (1 n+1) 0 We can compute the partial sums,, : Z m"C : 16:0 ifr¢ 1. 17' 0 A p-series has the form 2 00 “:1 L M' I It converges if and only ifp > 1. The special divergent series 22:, % is called the harmonic series. GENERAL TESTS FOR CONVERGENCE 0 Divergence test: If lim,H00 an #0 (or if the limit does not exist), then the series 111 + a2 + a3 + - I I diverges. Comparison tests: Suppose 2a,, and 2b,, are series with positive terms. - Convergence: If 2 b,, converges, and a", S b" for all n, then 2 a" converges. 0 Divergence: If 2 b,, diverges and on 2 b", for all n. then 2 a" diverges. - Limit comparison test: If liinnew exists and is positive, then either both series converge or both diverge. 0 Integral test: If {a,,} is a monotonically de- creasing positive sequence and is a continuous function with the property that (1,, : f(n), then the series 22:10." converges if and only if the improper integral f (w) div converges. A "fi‘d‘fl" 1234567x Integral test: The sum of the infinite series (gray region) is strictly smaller than the area under f(.r) [blue and gray region} If if f (1:)IIJ- converges, then so does the series. series 2 (1,, converges absolutely if the series of absolute values 2 (ant converges. If the series of absolute values does not converge, but the original series does, then it converges conditionally. 0 Absolute convergence test: If a series conveges absolutely, then it is convergent. 1. 0 Ratio test: Suppose lim ‘1 "Hi "75x3 lanl IfL < 1. then the series converges absolutely. IfL > 1 (or if the limit is infinite) then the series diverges. If L = 1 then the test is inconclusive. 0 Roattest: Supposenlgi;c m : L exists and is finite. IfL < 1, then the series converges absolutely. IfL > 1 (or ifthe limit is infinite) then the series diverges. If L : 1 then the test is inconclusive. 2 L exists and is finite. T/P If the ratio test is inconclusive on 0 port icutar series, then so is the root test Try something else. 0 Alternating series test: An alternating series has terms with alternating :: signs. Do If (1,, are all positive, then the alternating seriesZ(71)"a" will always converge if both II:1 I. on“ g on for all n, and 2.1irn,,eDC a” Z 0. These conditions are sufficient but not necessary. For an alternating series that satisfies these conditions, the error from truncation is always smaller than the next term. I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I l I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I l I I I I I I I I I I I I I I I I I I I I I I I I I I I l I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I : I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I t I I I I I I GENERAL POWER SERIES A power series is a formal function in the form of an infinite polynomial: 00 E a,,(1r 7 u)” : (1.0+ (L1(.’T 7 a.) + a2(J' 7 a.)2 + I «- II:0 Here, :1: is the variable and the a" are coefficients; this series is “centered at a.” Many complex functions can be represented as power series; we need to know when these series converge. A power series about a can converge in one of three ways: I. Only at :17 : a; 2. For all real r; , 3.ln an interval of radius R around a (i.e., a 7 R < :1: < a + R). R is called the radius oft convergence. NOTE: The endpoints a 7 R and Ii + B have to be tested for each function ‘ When functions are represented as power series, they can be integrated or ditterentiated term by term in the usual way: if f(r) : a,,(.1' 7 u)”, then ' f’(J) = 2:11 Muir — a)"’1Iand ' d1: : 230:0 7 a’>"+1 ‘i’ C- The radii of convergence of f’(1:) and f(r)dm are the same as that of f(m) (but check endpoints individually). Ex: The classic example is the power series for tan' ‘ J: : Start with the power series 2:” .Ir" 7 1 171* with radius of convergence 1. - Substituting .1: 7 712, we get the power series for H1)”; : 1 7 r2 + .r" 7 .T6 + - I - , also with radius of convergence 1. / 0 Integrating, we get tan’1 ' : C + :1: — + $5 — + - - -. 0 Finally, since tan’1 0 = 0, we know that C = 0. Check convergence at .r = ::1 to find that the power series converges when g 1. TAYLOR AND MACLAURIN SERIES If f (r) can be represented by a power series around a, then the coefficients a" are given by a g .f‘”)(u) ’rl _ ' 7i! Here, fl”) is the nu‘ derivative and n! : 1 I 2 - 34 I In with 0! déf l. - The Taylor series tor f(w) centered at a has the form 0° f ("l (a) II M) : E; n, (I 7 a) . II: It converges at a or in some interval around o. oo f(,,) (0) 0 The Maclaurin series for f (:r) is the Taylor series centered at 0, so f(r) : ‘ r”. n. II:o Arithmetic with Taylor series: Functions written in Taylor series form can be added, subtracted, multiplied (painstakingly collecting like terms), and even divided if the constant term of the denominator is non—zero. Polynomial approximations to f(zlf): The Taylor series for f (1:) about II can be used to approximate f (.1:) by a polynomial of any degree for arr-Values near a. Ignore the higher~order terms. The linear polynomial is the tangent line to the curve at m : a. Applications to limits: 1 .r 2 I 1 3 5 ' 3' 1 l, h :1} i (I — 11m 170 L]: 0 Error bound: Rule of thumb: The error of a truncated Taylor series is less than something a lot like the next term after the cut—off. - Formally, if the Maclaurin series for f(a;) converges at h, and lf(”+’)(ur)l 3 1M for all 7h S 1: g h, then the error in evaluating f (h) by the Maclaurin series truncated after the n” I7tan’ Ex: lint 170 lirri 170 1.3 3 degree term is less than IMPORTANT MACLAURIN SERIES Function Series Domain of convergence 1 w . . 1 Xm":l+.rr+.r>2+rj+«~ lari<l 7 r ":0 1 9° . (171V Z(n+1)1'":1+2x+3;1:2+-II m<1 ’ II:u I2 , 3 lIl(l7.I') %7%7--~ 71S4r<1 I DO 11:" 1:2 1‘3 e Zm:l+a'+§+37!+-~ allrealm 71:0 ‘ Do u Jim-#1 1.3 1,5 1,7 sum 2f—1) mzrr7§+57fi+~ allrealar II: I 60 1,1“ 1,2 J 1 3 cos .1: ZED" (2n)! : r a I r all real .1: g ":0 ; 0° JI27l+I E tan’l a: 2(71)”271 +71 £1 E n:() BINOMIAL SERIES The binomial series is the Maclaurin series for functions in the form (1 + It is finite for positve integers 7', but works for all real numbers. (1+x)" (7)1" = 1+ rar+ LT 7 1) 2 n I I I Notation: “7' choose it” : . - Defined for all real r and non—negative integer n. 0 If r' is an integer and r' < n, then : 0. 0 If r is a non~negative integer, then is the number of ways t chosen from a set of 7' objects. 0 If r is negative, then (717‘) (71)"(7'4‘371). The infinite binomial series converge for tact < 1. Convergence at ::1 depends on r: if r 2 0, then the series converges at :1; if 71 < r < 0, only at a; = 1: otherwise at neither endpoint. +--I. iat a group of 71. objects can be ...
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Calculus2 - SPARKCHARTS CALCULUS II CHARTS OVERVIEW...

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