This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 (diffeqs) express a relationship
between a function and its derivatives. Diffeqs 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 2flTne2) + 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 righthand 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  Lefthand 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 xaxis 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 righthand 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  Righthand 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 = fll‘) 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 halfplane; 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 easyto—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. : rué] :
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”tointegrate 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‘ZI ' 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 stepbystep 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 I7x _ 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 'usubstitution 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) (17:. 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 yaxis. 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 crosssection (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
crosssectional 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 raxis
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(xlzglxl2l
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 IaXIS (LI/(t) Z (l), l dy ,(1) l d then the surface area swept out when it is revolved around the .Iraxis 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 Procalculus 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/Ii2 + 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 Err1 (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 xaxis (respectively) has CM at the point
mill + 771212 + "'+ mumn 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 platelike 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 xcoordinate of the CM: For each strip, the moment is given by
(mass) X (a:—coordina.te of strip CM) : pu>f(at)Aar.
The .rcoordinate of the CM (equivalently, the moment about the yaxis) is therefore b h :L'fil?) (1L5 : mf(;p) duh I y—coordinute of the CM: The moment of each strip is (mass) >< (ycoordinat,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 (diffeq) involves the derivative(s) of a (single—variable) function. I The order of a diffeq 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 diffeq 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 diffeq
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 diffeq. 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 diffeqs 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 longrun 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 diffeq, 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% : rl “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 pseries 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 arrValues 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 ...
View
Full Document
 Spring '10
 STUDYGUIDES
 The Land, break up, Riemann

Click to edit the document details