CALCULUS IIOVERVIEWCalculus is the study of nicesmoothly changingfunctions. ! Differential calculus studies how quickly a function is j

changing at a particular point. For more on differential calculus, j se e the Calculus /SparkChart.

Integral calculus studies areas enclosed by curves and is used i to compute a continuous (as opposed to discrete) ! summation. Integration is used in geometry to find the length j of an arc, the area of a surface, and the volume of a solid; in ] physics, to compute the total work done by a varying force or j the location of the center of mass of an irregular object; in j statistics, to work with varying probabilities.

Differential equations (diff-eqs) express a relationship j between a function and its derivatives. Diff-eqs come up j when modeling natural phenomena.

Infinite series are special types of functions. Familiar j

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.

R E V IE W O F T E R M S A function is a rule that assigns to each value of the domain a

unique value of the range. Function f(x ) is continuous on some interval if whenever x i

is close to #2, f ( x i ) is close to / (x 2). Function f ( x ) is increasing on some interval if whenever

x i < x 2 > f { x 1 ) < f ( x 2) (so f ' ( x ) is positive). It is decreasing if f ( x 1 ) > /(X2) (so f ' ( x ) is negative). Afunction that either never increases or never decreases is called monotonic.

Function f(x ) 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(x ) is concave up on some interval if its second derivative f " ( x ) is positive there; its graph cups up. It is concave down if / " (x ) is negative; its graph cups down. The line x = a is a vertical asymptote for f(x ) if f(x ) blows up to (positive or negative) infinity as x gets closer and closer to a (from the left side, the right side, or both). Formally, limx_ a- f(x ) = 0 0 orlimx_ a+ f(x ) = 0 0 (or both).The line y b is a horizontal asymptote for f(x ) if the value of f(x ) gets close to b as \x\ becomes very large (when x is positive, negative, or both). Formally, limx_>+00 f(x ) = b or l im a ; - , - ^ f(x ) b (or both).

AREA UNDER A CURVE AND THE DEFINITE INTEGRALB A S IC P R O B L E M O F IN T E G R A L C A L C U L U SGiven a function y f(x ) on the interval [a, 6], what is the area enclosed by this curve, the x - axis, and the two vertical lines x = a and x = b?

NOTE: We alw ays sp eak of "signed' area: a curve above the x -ax is is said to enclose positive area, while a curve below the x -ax is is said to enclose negative area. The concept of "negative a re a ' m ay seem ridiculous, but signed area is more versatile and simpler to keep track of.

A P P R O X IM A T IO N S T O T H E A R E A Left-hand rectangle approximation: We can approximate

this area by a series of n rectangles. Divide the interval into n equal subintervals of width A x = and obtain n + 1 points on the x-axis at xq = a, x\ = a + A x , . . . , xn = a + nA x = b. These are the bottom corners of n rectangles, which well always number 0 to n 1. The height of each rectangle is the value of f ( x ) at the left x-axis corner. The A;th rectangle has height f ( x k ) and area A x f (x k ) . The total area of the n rectangles, then, is n_ 1

L n = A x ( / ( x 0) + f ( x 1) 4------+ / ( x _ 1) ) = A x f ( x k) k=0

Larger n will give more accurate approximation to the area.

0 a b

Ex: We approximate the area under the curve y = x 2 on the interval [0,1] with 4 rectangles of

width A x = | and heights 0, (| )2, ( | ) 25 ( f ) 2> f r a total area of

U = \((0 )2 + ( | ) 2 + ( i ) 2 + ( f ) 2) = * = 0.21875.

Right-hand rectangle approximation: Instead of taking the height of each rectangle to be the value of / ( x ) at the left x - axis corner, we can take the the value of / ( x ) at the right corner. The height of the A:th rectangle is now f ( x k + i ) for a total area of

y = f (x )

Rn = Ax(/(xi) + /(x2) + * + /(in)) = Ax ^ f{xk) fc=l

Right-hand and left-hand approximations are related by_________________________ Rn = Ln + A x ( / ( 6 ) - / ( a ) ) .___________

Ex: For f i x ) = x2 on the interval [0,1],

*4 = | ( ( i ) ? + ( I ) 2 + ( I ) 2 + ( I ) 2) = i = 0.46875.

Midpoint Rule: The height of each rectangle can be taken to be f(x) evaluated at the midpoint of each rectangle; the height of the kth rectangle is now

/ ( a + A * ( * + i ) ) = / ( s ^ )

for a total area of

y = f ( x )

UTrapezoidal Rule: We can approximate the area under the curve using trapezoids with the same two vertical sides and x - axis side as the rectangles. The area of a trapezoid is (average length of two parallel sides) x (distance between them). The area of the fcth trapezoid, then, for a total area of

A t- - ^~ { f ( xo) + 2/ (x i) + 2f ( x f ) + + 2/ (* _ !) + / (* ) ) .

Simpson's Rule: This time, we suppose n to be even and approximate the area with ~ parabola pieces with the kth parabola defined by points on the curve at x2k - 2 , x 2k - 1 , and x2k The total area is given by

Sn ^ ( / ( * o ) + 4/(xi) + 2/(x2) + 4/(x3) -I 1- 2/(xn_ 2) + 4/(xn_ i ) + f(x n)).

C O M P A R IN G A P P R O X IM A T IO N S T O T H E A R E AIf f(x ) is increasing on the interval [a, b], then L n 0 fc=0 rb

[a, b]. The limit represents the area under the curve and is denoted / f ( x ) dx.

In this notation, f is the integral sign, f(x ) is the integrand, and a and 6 are the lower and upper limits of integration, respectively.

The marker dx keeps track of the variable of integration and evokes a very small Ax;intuitively, the integral from a to b of f(x ) d x is a sum of heights (function values) times tiny widths dx (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 00 (equivalently, has a vertical asymptote).

Properties of the definite integral: Let f (x ) and g(x) be functions integrable on the interval [a, 6], and p be a point inside the interval.

rb rb rb

1. Sums and differences: / (f (x ) g(x)) dx= f(x) dx g(x) dx.

2 .Scalar multiples: J cf(x) dx = c J f(x ) dx. Here, c is any real number.

3. Reversing the limits: J f(x ) dx = J f(x) dx.

4. Concatenation: f f (x ) dx-\- f f(x) dx = f f(x) dx. J a J p J a

5. Betweenness: If f(x ) < g(x) on the interval [a, 6], then J f(x ) dx < J g(x) dx.In particular:

If f(x ) > 0 on [a, 6], then f ( x ) d x > 0.

If M is the maximum value of f ( x ) on [a, b] and m is the minimum value, then

m(b a) < f(x ) dx < M (b a).

ANTIDERIVATIVES AND THE INDEFINITE INTEGRALAntidifferentiation is the reverse of differentiation: an antiderivative of f(x ) is any function F (x ) whose derivative is equal to the original function: F '( x ) = f ( x ) in a pre-established region. Functions that differ by constants have the same derivative; therefore, we look for a family of antiderivatives F(x) + C , where C is any real constant.

The family of the antiderivatives of f ( x ) is denoted by the indefinite integral:

J f ( x ) dx = F (x ) + C if and only if F '{x ) = f(x ) .

The indefinite integral represents a family of functions differing by constants.

THE FUNDAMENTAL THEOREM OF CALCULUSThe Fundamental Theorem of Calculus (FTC) brings together differential and integral calculus.

MAIN POINT: Differentiation and integration are inverse processes. Finding antiderivatives is a lot like calculating areas under curves.

S T A T E M E N T O F T H E T H E O R E M Part 1: Let / (x ) be a function continuous on the interval [a, b]. Then the area function

F (x ) = J xf ( t ) dt is continuous on [a, b] and differentiable on [a, b] and F ' ( x ) = f { x ) . Part 2: If / (x ) is a function continuous on the interval [a, b] and F ( x ) is an antiderivative of

f { x ) , then f f ( x ) dx = F(b) F(a ) . The total change in the antiderivative function over an interval is the same as the area under the curve.

W H Y IS T H E F T C T R U E ?There are two ways of thinking about it:1. The change in the area function (function whose value at a is the area under / ( x ) up to a ) is chronicled by f(x ) itself: the area function changes quickly if / ( x ) is large, slowly if / ( x ) is small; it is increasing whenever f(x ) is positive and decreasing if / ( x ) is negative. So f(x ) behaves like the derivative of the function for the area under / ( x ) .2.The function for the area under / ' ( x ) behaves like f(x ) itself.

If / ( x ) is increasing (or decreasing), then f '( x ) is positive (or negative) and the area under / ' ( x ) is increasing (or decreasing).

If f(x ) is changing quickly, then |f'{x)\ is largeand the area under f ' ( x ) is changing quickly as well.

If the growth rate of / ( x ) is positive but slowing down to 0 (i.e., f(x ) is concave down approaching a local maximum), then f '( x ) is crossing the x-axis from the positive halfplane to the negative half-plane; at the same time, the area under / ' ( x ) , which has been growing while f '( x ) > 0, is stopping its growth and will start to decrease: like / ( x ) , the area under f '{x ) is nearing a local maximum.

U S IN G T H E F T CThe 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 = x 2 over the interval [0,1] is given by Jq X2 dx. Since F (x ) = | x 3

is an antiderivative of x2 (check that F ' ( x ) = x2), f ^ x 2 dx = F ( 1) F ( 0) = L 4 < 5 < i?4, and Z/4 is a slightly better approximation to the area, as expected since / (x ) = x2

is increasing and concave up on the interval.

1 C O M M O N IN T E G R A L S 1

J kdx = kx-\-C J o d x = C

f xn+1/ xn dx = ----------h C if n 1

J n -1-1J ^ dx = ln |x| + C

J ex dx = ex + C f ax dx = -----h CJ lna

J sin x dx cos x + C J cos x dx = sin x + C

j tan xdx ln | sec x| -1- C J cot x dx = ln | sin x| + C

J sec2 xdx tan x + C J sec x tan xdx = sec x + C

[ ---------dx = tan_ ^ x + C f 1/ r!t cin ^t -1- DJ x2 + 1 J y / l - x 2

TECHNIQUES OF INTEGRATIONUnlike differentiation, integration is hardthere are easy-to-write functions that dont 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.

S U B S T IT U T IO N R U L EIf u = g(x) is continuously differentiable on some interval and / ( x ) is integrable on the range of g(x), then

J f(9 (x ))g \x ) dx = J f(u ) du.

This is the analog of the Chain Rule for integrals (see the Calculus I SparkChart). It is useful for composite functions and products.

x :J x W .x3 + 8 dx.

x 2 is a lot like d^ xd^ Let u = x 3 + 8. Then du 3x 2 dx, so x 2 dx = ^ .

Substituting, we transform the original integral into J du, or

J ^u? du = + C = ? ( x 3 + 8 ) 5 + C.

When evaluating definite integrals using substitution, you have a choice about how to deal with the limits of integration. Lets say that youre integrating with respect to x .

1. Less thinking: Choose a useful u, substitute for x , integrate in terms of u, substitute x back, evaluate the integral with original limits.

t l 2 = ? 16 - ? 9 x: f x2s fifi + id x = ? ( z 3 + 8) f l = ? 1 6 i - ? 9J i 9 J i 9 9 9 '

2. Less work: Choose a useful u = g(x) substitute, integrate in terms of u, then evaluate the integral using, for limits, values o iu = g{x) evaluated at the original limits of integrationall without substituting x back in. Formally, if u = g{x), then

rsW

In the example above,

rb rg (b )

/ f {g (^ ) )g \ x ) d x = u du. Ja Jg {a )

!, f x 2y[:J x=1

/ x 3 + 8 dx -f u=9{2) i _ 2 3 1 1D 74/ - J u du = -it2 | = -.

J u = g ( 1) 3 9 J9 9

IN T E G R A T IO N B Y P A R T SSome 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 = f ' g + fg '.

Indefinite integrals: j f ( x ) g ' ( x ) d x = f ( x ) g ( x ) - j f ' ( x )g ( x )d x , or

J udv = uv J v du.

Definite integrals: Slap on limits: J f ( x ) g ' ( x ) dx = f ( x ) g ( x ) ] ba J f ' ( x ) g ( x ) dx.

A polynomial multiplied by a trigonometric, exponential, or logarithmic function is frequentlybest integrated by parts. Let the polynomial be u.

Ex: J (2x -I- l )e ~ x d x .

Let u = 2x -|- 1 (so du = 2 dx), and dv = e~xdx (so v = e~x). The integral becomes

(2x + 1) (e-x) - J 2e~x d x ,which simplifies to (2x + l ) e -x 2e~x -I- C, or ( - 2x - 3)e~x + C.

T R IG O N O M E T R IC S U B S T IT U T IO N S

Square roots of quadratics, such as \Ja? x2, cannot be integrated with the substitution u = a2 x 2 because the factor of du is missing. Enter trig substitutions, applying the substitution rule backwards. Trig substitutions are often necessary when calculating areas bounded by conic sections.

fc/\/4 a

dx. We can set x = 2 sin 0 with ^ < 6 < ^ .

Then dx = 2 cos 6 dO,and \/4 x2 = \/4 4 sin2 6 = V 4 cos2 0 = 2 cos 0 (since cos# > 0 on the interval). The integral becomesr 2 c o s 0 r f s in # = f

I 4 C 2 0 ^ C S ^^ = / t2 ^^ = / (csc2 0 ~ ^ )d 0 = co t 0 0 + C.

If we want to convert back from 0 to x, we note that cot 0 =

\ f 4 ; thus

/ V T - - dx = - ta - d j + o .T A B L E O F T R IG O N O M E T R IC S U B S T IT U T IO N S

Expression Trig substitution Expression becomes Range of 6 Pythagorean identity used

y/a2 a

x -f- a

x = a sin 0 dx = a cos 0 d0

yja2 x 2 = a cos 0 1 sin2 0 = cos2 0(a < x < a)

y/x2 a2 x = a sec # \/x2 a2 a tan 0 0 < 0 < | (when x > 0) sec2 0 1 = tan2 6dx = a sec 0 tan 0 d0 n < 0 < (when x < 0)

x = a tan 0 dx = a sec2 6

\/a2 + x 2 = as 1 -I- tan2 0 = sec2 0

Pay careful attention to the limits o f integration. The intervals fo r 0 correspond to the ranges o f the inverse trigonometric functions.

Expressions o f the form y/ x 2 + bx + c can be integrated by

completing the square and converting to the form

y/ (x + h)2 a2, where h = | and a = yj\c -^|. Then choose the appropriate trig substitution depending on the signs.

CALCULUS II

T E C H N I Q U E S O F I N T E G R A T I O N ( C O N T I N U E D )

P A R T IA L F R A C T IO N S p o l y n o m i a l s i n t r i g o n o m e t r i c f u n c t i o n s IIntegrating rational functionsratios of polynomialscan be tricky. However, after factoring thedenominator into linears and quadratics (which can always be done, though the coefficients maynot be rational numbers), a rational function can be expressed as a sum of simpler "partial"fractions. These come in four easy-to-integrate types:

m f du . . . - [ A dx A , . ,, _1. / = ln u -|- C. So / -------- = ln ox + b + C.

J u J ax + b a

For powers of trigonometric functions, regular u-substitution may not work. Use the following substitutions instead. Pythagorean identities:

sin2 6 + cos2 0 = 1 1 + tan2 0 = sec2 0

Square sine/cosine substitutions (from the half-angle formulas):

sin2 0 = ^ (1 - cos 20) cos2 0 = ^ (1 + cos 20) ^ f du un+1 _ .. . , f A dx A , . 11 _J u n + 1 * J (.ax + b) a(n + 1 ) ^ Odd powers of sine or cosine:

To compute J sinn 0 dO when n is odd, keep one sine factor and replace the rest with cosines

using sin2 0 = 1 cos2 0 to obtain

J ( l cos2 0) 2 sin 0 dO.

Integrate using the substitution u = cos#.For odd powers of cosine, co all the functions above (co cosine = sine).

3 J ~~ = ln lwl + C. So

[ UaX + B d x - A [ 2ax + b d x , [ B ~ % -d x J ox2 + bx + c 2a J ax2 + bx + c J ax2 + bx + cA E D

= ln lax2 + bx + cl + / 7---- --------dx, where D = B -2a 1 1 J ax2 -1- bx + c 2a

M f du 1 , u f D dx . , ,4. / ----- 7 = - tan l-C. So / 7---- -------- where the denominator has no real

J u2 -\-a2 a a J ax2 + b x + c

roots (62 4ac < 0) can be evaluated by completing the square in the denominator, which becomes

a(x -1- h)2 + k2 where h = ^ and k = yjc .

Even powers of sine or cosine: Use the square formulas

sin2 0 = - ( 1 cos 20) or cos2 0 = ^ ( 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 J sin710 cosm 0 dO, 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 0 or cos 0 (or both) each, which can be integrated using a a-substitution.

The step-by-step process for integrating / (x ) = |||| 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 / (x ) = s (x ) +

2. Factor the denominator, reducing it to linear factors in the form (ax -1- b)n and irreducible quadratic factors in the form (cx2 + dx + e )m where d2 4ce < 0.

3. Decompose into a sum of partial fractions: If q(x) has no repeated factors, express

r (x ) _ A i ( A e ( C ix + D i C kx + D k q(x ) a\x + 61 agx + be c\x2 -I- d\x + e\ ckx2 + dkx + ek

Solve for all the As, Cs, and Ds by multiplying the equation by q(x ) and equating coefficients.

TIP: If q(x ) factors a s a product of two linears ( x a )(x b), then w e can solve for A and B in

( x - M - b ) = A + ^~b bY takin9 A = and B =

Products of sines and cosines of different angles: Use the following identities to get rid ofproducts: \

sin A cos B = - (sin(A B ) + sin(A + B ) )

sin A sin B = ^ (cos(A B ) cos (A + B ))

cos A cos B = ^ (cos(A - B ) + cos(A + B ) )

MNEMONIC: Products of like terms use cosines; unlike terms use sines.

Powers of secant or tangent: Even powers of sec 0: Convert all but two secants to tangents using Pythagorean identity

1 -1- tan2 0 = sec2 #; use substitution u = tan#. Odd powers of tan # : Convert all but one tangent to secants, pull out a factor of sec # from

the polynomial in secants and use the substitution u = sec #.

If q(x) has repeated factors, then for each factor in the form (ax 4- b)n expect fractions A i ( A 2 ( t A n

ax + b A (ax -I- b)2 * * (ax -1- b)n C + D C + D Each (cx -j- dx -j- e) factor will give fractions 0 I 1 . n . .

cx2 + d x + e (cx2 + dx + e )m 4. Integrate s(x) and each partial fraction individually, using the four types of integrals above. Ex: J tan5 0 dO = J ( l + sec2 # )2 tan#d#

= J g + 2 sec # + sec3 #^ tan # sec 0 dO = J - + 2u + u3 du.4x3 17x2 28 A B C x + D ' ( x - 2 ) ( x + 2)(3x2 + 4 ) x - 2 1 x + 2 1 3x2 + 4

Cross-multiplying, simplifying, and equating coefficients gives the four equations 2(A B ) + D = 7, 6(A - B ) + D = -17, 3(A + B ) + C = 4, (A + B ) - C = 0. Solving this system of four linear 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 B and D independently), we get A = 1, B 2, and Cx + D = x + I. Now we can integrate.

Other powers: Combine tricks and use J tan 0 dO = ln | sec #| + C andJ sec 0 dO = ln | sec # + tan #| + C.

IMPROPER INTEGRALSImproper integrals come in two types:1. Definite integrals over an infinite interval. Ex: e~x dx.2. Definite integrals over an interval in which the function blows up to infinity (has a vertical

asymptote). Ex: f Q ^ dx.Not all improper integrals convergerepresent 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.

IN T E G R A L S O V E R A N IN F IN IT E IN T E R V A LImproper integrals of this type should be rewritten as one of three limit forms: y

1. Interval infiniteroc rt

nite to the right: / f ( x ) dx = l^im / f ( x ) dx.

/ y = f(x)2. Interval infinite to the left: J f(x ) dx = ^lim f(x ) dx.3. Integrals over the whole real line:

J f(x ) dx = J f(x ) dx + J f(x ) dx fo r any a. Improper integral J f(x ) dx

The original integral converges only if both integrals over half-intervals converge.

dx , / - converges (and equals 77) if and only if r > 1 .

'-'1 XT rOO

The integral / f(x ) dx will converge only if lim f(x ) = 0 (y = 0 is a horizontal

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: lim f(x ) = 0 alone does not imply convergence of an improper integral.

f dx f 1 dx ( i t \The classic exam ple is / = lim / = lim I ln lx l I = lim \nt = 0 0 ._. . , . . . /1 X >oo L x t* 00 V J 1 / t* 00The integral diverges. ' '

i n t e g r a l s o v e r a n i n f i n i t e d i s c o n t i n u i t y If bThe integral / f ( x ) dx is improper if at any point c in 3

the closed interval [a, 6], the function blows up. If c = a is the left endpoint, then the integral is

improper if* lim ztoo1

The integral is interpreted as lim / f(x ) dx. : i V \Jt 0

If c = b is the right endpoint then the integral isimproper if lim f(x ) = 0 0 f ( x ) Jp0'

x-*b~ . rf so / f\

and is reinterpreted as lim / f ( x ) dx. Ja Ja

If c (a, 6), then the original integral is understood to be tl integrals

J f(x ) dx + J f(x ) dx.

The original integral converges only if both endpoint-ir independently.

a c b ^

; a vertical asymptote at x = c, (x) dx is an improper integral.

he sum of the two improper

nproper integrals converge

f dx . converges (and is equal to y1 ^ ) if and only if r < 1 .

Jo xT

dx dx f ^ dx - diverges (and does not evaluate to 0) b ecau se both / and / diverqe,

1 * J _ x x J0 x ythough the two a re a s seem to be "equal' and opposite in sign.

/1 dx dx f 1 dx7= = 0 b ecau se both half-integrals / = and / = converge and1 io

their values are equal in m agnitude and opposite in sign.

GEOMETRY OF CURVESA R E A S B O U N D E D B Y C U R V E SSuppose that f ( x ) > g (x ) on the interval [a, b] and both functions are continuous. Then the area bounded by the two curves y = / (x ), y = g (x ) and the two vertical lines x = a and x = b

J ( f ( x ) - g (x ) ) dx.

Shaded area is g (x ) ) dx.

' In general, if the (continuous) curves cross each other on the interval, then the positive area defined by the curves between a and b is rb

J I f ( x ) - g ( x ) \ d x .

It is most easily evalutated by considering subintervals whose endpoints are all points c such that /(c ) = g(c).

1 If the area is bounded by horizontal lines, it may be easier to rewrite the curves in the form x = f ~ 1(y) and integrate the difference between them with respect to y.

V O L U M E S : S O L ID S O F R E V O L U T IO NSuppose 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 x-axis) is given by the function A(x ) . The volume of a slice of thickness A x is A (x ) Ax, and the volume of the solid bounded by the planes x = a andx = b is

A rea A(x) = n f( x )2 Volume 7r/(x)2A x

) dx.1 A (x )

olume of t as it revol x = b is gi

J Tr(radius)2 dx ottt J* (/ (x ) ) 2 dx.

Disk method: The volume of the solid swept out by the curve y = / (x ) as it revolves around the x-axis between x = a and x = b is given by

Washer method: If / (x ) > y (x ) between a and b, then the volume of the solid swept out between the , two curves y = f ( x ) and y = g (x ) as they revolve / around the x-axis between x = a and x = b is /

J 7r(outer radius)2 7r(inner radius)2 or

* J ( / ( * ) ) 2 - ( g ( x ) ) 2 dx.

Disk method

A rea A(x) = 7r(/(x)2 - g(x)2)y = f(x)

W asher method

PARAMETRIC CURVESA parametric curve defines both the x- and the y- coordinates in terms of a third variable, often t (as in time). Parametric curves dont necessarily represent functions and dont have to pass the vertical line test.Sometimes the domain of t is restricted to an interval a < t 0 and 0 < 9 < 27T, although

(r, 9) = (r, 9 7r) and (r, 9) = (r, 9 + 2mr) for integer n.

C O N V E R T IN G B E T W E E N C A R T E S IA N A N D P O LA R C O O R D IN A T E S From Cartesian to polar: r yjx2 + y2; 9 = tan- 1 | From polar to Cartesian: x = rcos0; y = rs in 9

F U N C T IO N SFunctions in polar coordinates usually define r in terms of 9. They need not (and almost never will) pass the vertical line test.Circles: The graph of r = a is a circle of radius \a\ centered at the origin. The graphs of r = a sin 9

and r = a cos 9 are circles of radius 1 1 centered at (0, ) and ( f , 0), respectively.Roses: The graphs of r = sin n9 and r = cos n9 are roses centered at the origin with n petals if

n is odd, 2n petals if n is even.Lima^ons and cardiods: The graphs o f a 6sin0 and a bcos9 are lima^ons. If | | > 1, the

limaQon has an inner loop; if | | = 1 , the limagon is heart-shaped and is called a cardiod

1 Shell method: A solid is obtained by revolving the region under the curve y = f ( x ) between x = a and x = b (the area of this region is / f ( x )d x ) 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 x and - thickness Ax. The surface area of a cylinder is (circumference) x (height) or 27rx/(x); thus, the volume of the solid is

V

y = /(x)

Shell method

27rx/(x) dx.

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.

A R C L E N G T H

If / (x ) has a continuous derivative on the interval (a, b), then the length of the curve fromx = a to x = b is

lengths* A + (f (.x ))2iJ V1 +(/'(*)) 2 dx

i:Fdx-In Leibniz notation this becomes LWhy? If we break up the interval into n subintervals each ofwidth A x = - with x k a sample point in the kth interval, then the length of the curve on the kth interval is approximately

the length of the vector (A x , A x / '(x ) ) , or Axy/l + (f' (x*k) f . Riemann sum to get the formula.

y = f (x )

Take the limit of the

S U R F A C E A R E A : S O L ID S O F R E V O L U T IO NThe surface area of a surface swept out by revolving the function y = f ( x ) about the x-axis between x = a and x = b is

> = f 2tr f {x )y f : Ja

1 + (/ '(x )) dx. perimeter width = A + (/'(x))2Ax= 2 7[/(*) / s ---------y = /(*)

The formula is obtained by approximating the surface area by cylindrical bands of radius / (x ) and 0 width equal to the tiny arc length on the tiny interval.

I fix)Ax

Concavity:

d ( dy \dry d ( dy\ dt \dxJdx2 dx \ d x ) ^

Area defined by curve: The area between the x-axis and the curve traced out from t = a tot = b is

A = J y(t)x' ( t )d t.

NOTE: The area is counted a s negative for the regions where the curve is moving "backwards" i.e., < 0 and positive when the curve is moving "forwards."

HAPPY CONSEQUENCE: The area enclosed by a loop wholly above the x-axis, traversed exactly once from t = a to t b, can be computed directly. i b

A = / y {t )x '(t )d t .The integral is positive for loops traced out clockwise, negative for those traced counterclockwise. Alternatively, break up [a, b] into subintervals depending on the sign of integrate separately.

Arc length: The length of a parametric curve traced out from t = a to t = b is

L =

This formula also works works for loops.

Surface area of revolved solid: If the same curve always stays above the x-axis (y (t) > 0), then the surface area swept out when it is revolved around the x-axis is

S =

For more on polar cooixlinates, see the Pre-calculus SparkChart.

G E O M E T R Y O F T H E C U R V E Slope of tangent: Convert the curve r = / (9) into parametric equations in Cartesian

coordinates: x = r cos 9 = f (9 ) cos 9, y = f (9 ) sin 9. The slope of the tangent to the curve at (x (0), y (0) ) is ^ ^

dy _ dx

sin 9% + r cos 9

cos r sin b

Area: The area enclosed by rays at 9 a and 9 -/ 3 bounded by the curve r = /(0) isr0 1I ^ r2 d9.

Ja ^Why? The area of a circle of radius r is 7rr2 (angle sweep of 27r). The area of a sector of a circle of radius r and angle measure 0 is thus \r29. We approximate the area by a Riemann sum of sliver-sectors with radius r and angle measure A0.

Arc length: The length of an arc r = f (9 ) from 0 = a to 0 = /3 is

L =7. ' \R W *The formula is derived by converting the curve to Cartesian parametric equations with 0 as the parameter.

INTEGRATION: APPLICATIONS TO PHYSICSNewton's Second Law, F = ma, states that the force F on an object is proportional to the objects 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 = Fx. If the force F (x ) is variable and depends on the distance x, then the work done by F (x ) in moving an object from x = a to x = b is

W - f w dx.

1 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 x units away from its natural position is given by F (x ) kx; here, A: is a constant that depends on the tightness of the spring.

C E N T E R OF M A S SThe center of mass (CM) of any system is the point on which (if connected) it could balance on a fulcrum.

Moment: The farther away something is from the fulcrum, the heavier its mass counts. Each mass of weight m a distance x from some point contributes mx worth of moment (or torque) with respect to that point. If the point is the CM, then all the moments of the system have to balance.The system b ehaves a s though all of its m ass w ere

concentrated at the CM.

~ L

mass m-i m o

fu lcru m

Torques: m i(x x\) counterclockwise m2 (x 2 x) clockwise

M asses balance, so x = ^ i ? i.+ m2x 27711 -+-7712

2 masses, 1 axis: If a massless rod with objects of masses m\ and m2 at each end balances on a fulcrum at its CM, then the distances d\ and d2 from the masses to the fulcrum must satisfyTfl\d\ = 77l2 d 2 .

If such a rod has length d, then distance d\ from the m\ mass to the fulcrum satisfies m\d\ = m2{d d\). Solving, d\ = We can view this as positioning the rod along

where

x\ and x 2 are the distances of the masses from the origin. Here, x\ = 0 and x2 = d, the length of the rod.

Discrete masses, 1 axis: In general, a system of n objects of masses m i,m 2, ... ,m n positioned at points x i ,x 2, ... , x n along the x-axis (respectively) has CM at the point

+ mnxnmi + m2 H 1- mn

This is the moment of the system about the x-axis.The moment of the system about the y-axis can be computed independently.

1 Discrete masses, 2 axes: The CM of a system of objects of masses at points(x i, j/i) , . . . , (x , yn) on a coordinate system is at the point

( m ix i -I------ h mnx n miy i -\------ 1- mnyn \

'm J

, yn ) uu a lui

(x ,y ) =m j + mi +

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 = /(x ) from x = a to x = b. As usual, we approximate the object by thin rectangular strips of width Ax, height /(x ) , area / (x ) Ax, mass p f ( x ) Ax, and CM at the point (x, | / ( x ) ) .

x-coordinate of the CM: For each strip, the moment is given by(mass) x (x-coordinate of strip CM ) = p x f ( x )A x.

The x-coordinate of the CM (equivalently, the moment about the y-axis) is therefore

[ x f ( x ) dx = \ [ x/ (x ) dx. m J a A Ja

y-coordinate of the CM: The moment of each strip is

(mass) x (y-coordinate of strip CM ) = (p f ( x )A x ) (^ / (x )).1 f b 2

The y-coordinate of the CM is therefore I (/ (x )) dx.

The density p doesnt appear in the final result; all that matters is that the density is uniform.

INTEGRATION: APPLICATIONS TO PROBABILITY & STATISTICSA V E R A G E V AL UEFor a discrete set of values, their average multiplied by their number gives their sum. The analog of an average for a continuous function f ( x ) on the interval [a, 6] is the average value /, which has the property that the rectangle of height / and width b a has the same area as is enclosed under the curve y = / (x ). Thus

b - a j af ( x ) dx.

f = / (c) is the average value of / (x ) on the interval [a, 6].

The two shaded regions have

equal area.

The Mean Value Theorem for Integrals states that a continuous function attains its average value. Like the MVT for derivatives (see the Calculus I SparkChart), this is a completely intuitive statement.

G E N E R A L P R O B A B I L I T Y D E N S I T YA probability density function describes how likely it is that the outcome of some trial is x. The probability that the outcome is any specific point a is negligible; instead, we talk about the chances that the outcome falls in some range and think of areas under the probability density curve as representing actual probabilities. The probability that the outcome is between a and b is

r.f ( x ) dx.

> The probability that the outcome is something is 1 ; therefore, / / (x ) dx = 1J OO

The mean of a probability density function is the long-run average outcome; it can be seen as

the x-coordinate of the CM of the region on a graph and is given by p = J x/(x) dx.

' The median is the point m such that the probability that x < m is equal to the probability that x > m. (Again, the probability that x = m is negligible.) Solve for m in the equation

fm dx=k fZo f (x) dx=\-

THE N O R M A L D I S TR I BU TI O NThe normal distribution, or bell curve, is a probability density that often arises from repeated random events.

1The probability density N ( x ) =

oV27r The mean is p. The variable a is the standard deviation, a measure of

how clustered the outcomes are around the mean. Theprobability that the outcome is within a of the mean isabout 68% : [ + a

/ N ( x ) dx 0 .68 . Normal distribution with m ean p andJp-tr standard deviation a. The blue region

The probability that the outcome is within 2

SEQUENCES AND SERIESWhy is this a Calculus topic? Complicated functions can often be approximated with polynomialsor 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.

S E Q U E N C E S

A sequence is an ordered list of real numbers, called terms. An infinite sequence has infinitely many terms. Shorthand: { a f c } ^ represents the sequence ai, a%, as, A sequence is defined explicitly if each of its terms can be found independently of the other

terms. Ex: an = n2 is the sequence 1,4,9,16,.... A sequence is defined recursively if the nth term is found using the preceding term(s). Ex: ai = 1; an = an_ i + (2n 1) is again the sequence 1,4,9,

Limit of a sequence:

The limit of an infinite sequence, denoted lim a , if it exists, is the value that the sequence approaches. If the limit exists and is finite, then the sequence is called convergent. I f not, the

sequence is divergent. Formally, the limit exists and is equal to a if for all e > 0 there exists an N so that whenever

n > N, we have |an a| < e.For a divergent sequence whose terms tend toward infinity, we can say that lim an = oo if for all integers A there exists an N so that if n > N, then an > A.

A sequence {a n} is called increasing if a* < a*;+i for all k and decreasing if a* > 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.

S E R I E S : D E F IN I T I O NS A N D B A S I C T Y P E S

A series is a summed sequence: ai + a2 + 03 + . An infinite series has infinitely many terms.OO

An infinite series is often den o ted ^ a/t or just ^ a*;.fc=i

A partial sum of a series is a cut-off series sum sn = a\ + 02 H h an = X )*= i afe- 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.

A geometric series has the form a + ar + ar2 + ar3 -|-------- Yl'kLo ark where 0. It is convergent if and only if |rj < 1, in which case its sum is y^p-

We can compute the partial sum sn = ark a ( l rn+1)

if r 7^ 1 .

A p-series has the form converges if and only if p > 1.

The special divergent series *s ca^ed the harmonic series.

G E N E R AL T E S T S FOR C O N V E R G E N C E Divergence test: I f limn_MX) an ^ 0 (or if the limit does not exist), then the series

ai + a2 + 03 -I diverges.

Comparison tests: Suppose J2 an and bn are series with positive terms.

Convergence: If J2 bn converges, and an < bn for all n, then ^ an converges.

Divergence: If ^ bn diverges and a > bn for all n, then an diverges.

Limit comparison test: If limn_oo ^ exists and is positive, then either both series converge or both diverge.

Integral test: If {a n} is a monotonically decreasing positive sequence and / (x ) is a continuous function with the property thatOn = / (n)> then the series anconverges if and only if the improper integral f ( x ) dx converges.

Integral test: The sum of the infinite series (gray region) is strictly smaller than the area under f ( x ) (blue and gray region).If / j00 f {x )d x converges, then so does the series.

ABSOLUTE C O N V E R G E N C E A N D A L T E R N A T I N G S E R I E S

A series ^ an converges absolutely i f the series of absolute values lan| converges. If the series

of absolute values does not converge, but the original series does, then it converges conditionally. Absolute convergence test: If a series conveges absolutely, then it is convergent.

Ratio test: Suppose lim ^ n+1 ^ = L exists and is finite.n-*oo |an |

If L < 1, then the series converges absolutely. I f L > 1 (or if the limit is infinite) then the series

diverges. If L = 1 then the test is inconclusive. Root test: Suppose lim \/|an | = L exists and is finite.

I fL < 1, then the series converges absolutely. I f L > 1 (or if the limit is infinite) then the series

diverges. If L = 1 then the test is inconclusive.

TIP: If the ratio test is inconclusive on a particular series, then so is the root test. Try something else.

Alternating series test: An alternating series has terms with alternating signs.

If an are all positive, then the alternating series ^ ( l ) nan will always converge if both

fln+i < On for all n, and 2. limn^oo an = 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.

G E N E R A L P O W E R S E R I E SA power series is a formal function in the form of an infinite polynomial:

00

an(x a)n = a0 + ai (x a) 4- a2 (x - a)2 -|-----n = 0

Here, x is the variable and the an 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:

1. Only at x = a;2. For all real x;3. In an interval of radius R around a (i.e., a R < x < a + R) . R is called the radius of

convergence. NOTE: The endpoints a R and a + R have to be tested for each function.

When functions are represented as power series, they can be integrated or differentiated term by term in the usual way: if f { x ) = 0 an{x ~ o )n, then

f ' ( x ) = Y ^ = i na>n{x a ) -1 , and

J f ( x ) dx = J2n=o T rM * ) n+1 + c -The radii of convergence of f ' ( x ) and J f ( x )d x are the same as that of f ( x ) (but check endpoints individually).

Ex: The classic example is the power series for tan- 1 x :

Start with the power series xU = with radius of convergence 1. Substituting x > x 2, we get the power series for yq^j = 1 x 2 + x4 x6 -\----- , also with

radius of convergence 1 .1 3 5 7

Integrating, we get tan 1 x = C + x yy + yj \ H Finally, since tan- 1 0 = 0, we know that C = 0. Check convergence at x = 1 to find that the

power series converges when |x| < 1 .

T AYLOR A N D M A C L A U R I N S E R I E SIf f ( x ) can be represented by a power series around a, then the coefficients an are given by

an = ^ J ) .Here, is the nth derivative and n\ = 1 2 3 n with 0! =f 1. The Taylor series for f ( x ) centered at a has the form

r-It converges at a or in some interval around a.

The M adaurin series for f ( x ) is the Taylor series centered at 0, s o f ( x ) -n=0 ' t;

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 ( x ): The Taylor series for f ( x ) about a can be used toapproximate f ( x ) by a polynomial of any degree for x-values near a. Ignore the higher-orderterms. The linear polynomial is the tangent line to the curve at x a.

Applications to limits:

tan- 1 x * - ( * T + T - ' - )

< M o )

Ex: lim : - = lim -x >0

. . f i x 2 \ 1 = H( 3-y+-j = 3- 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 ( x ) converges at h, and |/(n+1)(:r)| < M for all

h < x < h, then the error in evaluating f ( h ) by the Maclaurin series truncated after the nth degree term is less than ^ ^ 1

IMPORTANT MACLAURIN SERIES

Function Series Domain of convergence

1

1 - X^ x n = 1 + x + x 2 + x 3 H-----n0

|x| < 1

1

( 1 - z ) 2y ^ (n -1- l ) x n = \ + 2x-\- 3x2 -\-----n=0

|x| < 1

ln (l x ) x n X2 X3

2 ^ ^ = ~ x ~ - 1 < X < 1

ex^ x n X2 X3

E ^ = 1 + * + 2 ! + 3T + "all real x

sin re ^ X2 +1 X3 X5 X7^ ) (2n + 1)! - X 3! + 5! 7! +

all real x

cos a; 1 \n x2n X2 X4 x&(2n )! _ 2 f + 4 f _ 6 f + '

all real x

tan-1 x lN n x 2n+1 X 3 x 5 x 7^ 2 + i = x _ y + y ~ T + 1*1 0, then the series converges at 1 ; if 1 < r < 0, only at x = 1 ; otherwise at neither endpoint.