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Calculus Notes BC PG. Topic 3 − 6 CH. 1 Limits / Continuity
1-2 Find Limits Graphically and Numerically
1-3 Evaluating Limits Analytically
1-4 Continuity and One-sided Limits
1-5 Infinite Limits/ 3-5 Limits that Approach Infinity
7 − 13 CH. 2 Derivatives
2-1 Derivative and the Tangent Line Problem
2-2 Basic Derivatives and Notation
2-3 Product and Quotient Rule
2-4 Chain Rule
2-5 IImplicit Differentiation
14 − 15 2-6 Related Rates
16 − 24 CH. 3 Applications of Derivatives
3-1 Extrema on an Interval
3-2 Rolle's Theorem and Mean Value Theorem
3-3 Increasing and Decreasing Functions and 1st Derivative Test
3-4 Concavity and the 2nd Derivative Test
3-6 A Summary of Curve Sketching
3-7 Optimization Problems
3-8 Newton's Method
3-9 Differentials
25 − 32 CH. 4 Integration
4-1 Antiderivatives and Indefinite Integration
4-3 Riemann Sum and Definite Integrals
4-4 Fundamental Theorem of Calculus
4-5 Integration by Substitution
4-6 Numerical Integration Trap. Rule, Left, Right, and Midpoint( )33 − 39 CH. 5 Transcendental Functions
5-1 Natural Logarithm Function and Differentiation
5-2 Natural Logarithm Function and Integration
5-4 Exponential Functions: Differentiation and Integration
5-5 Bases other than e and Applications
5-6 Inverse Trig. Functions: Differentiation
5-7 Inverse Trig. Functions: Integration
40 − 43 CH.6 Differential Equations
6-1 Slope Fields
PG. Topic 6-2 Growth, Double-Life, and Half-Life formulas
6-3 Separation of Variables
44 − 50 CH.7 Applications of Integration
7-1 Area
7-2 Volume rotated about x-axis
7-3 Volume rotated about y-axis
Volume rotated about other lines
Volumes of known cross sections
7-4 Arclengths and Surface of Revolution
51 − 58 CH. 8 Integration Techniques
8-1 Basic Integration Rules
8-2 Integration by Parts
8-5 Partial Fractions
8-7 Indeterminate Forms and L'Hopital's Rule
8-8 Improper Integrals
59 − 75 CH. 9 Infinite Series
9-1 Sequences/ 9-2 Series and Convergence
9-3 The Integral Test and p-series
9-4 Comparison of Series
9-5 Alternating Series
9-6 The Ratio and Root Tests
9-7 Taylor Polynomials and Approximations
9-8 Power Series
9-9 Representation of Functions by Power Series
9-10 Taylor and Maclaurin Series
76 − 80 CH. 10 Parametric and Polar Coordinates
10-2 Plane Curves and Parametric Equations
10-3 Parametric Equations and Calculus
10-4 Polar Coordinates and Polar Graphs
10-5 Area and Arc Length in Polar Coordinates
81 Hooke's Law
82 Finding Area Using Limits
83 Lagrange Error Bound
84 − 89 AP Review
90 − 91 8-3 Trigonometric Integrals
92 − 94 8-4 Trigonometric Substitution
2
Calculator Procedures Calculator SettingsYour calculator should be in Radian mode when performing Calculus problems. (Those of you in Physics need to be in Degree mode while in Physics class. You will be switching modes daily).
Graphing WindowsPush ZOOM for pre-set windows1) ZDecimal is the best for tracing (− 4.7 to 4.7). (−6.3 to 6.3 for TI-89)2) ZStandard shows most of the graph (−10 to 10).3) ZoomFit will help you find the graph if the graph is not in your current window.
Intersection TI - 83 / 84 TI - 89Push CALC Intersection Push CALC Intersection1) Move cursor near point of intersection. 1) Move cursor near point of intersection.2) Push enter for first equation. 2) Push enter for first equation.3) Push enter for second equation. 3) Push enter for second equation.4) Push enter for guess. 4) Move cursor to left of intersection and press enter. 5) Move cursor to right of intersection and press enter.
ZeroesPush CALC Zero1) Move cursor to the left of intercept (lower bound) and push enter.2) Move cursor to the right of intercept (lower bound) and push enter.3) Push enter for guess Derivatives Derivative at all points 2nd Derivative Derivative at a point
TI-83: NA NA nDeriv equation, x, point( )
TI-84: NA NA ddx
equation( ) x = point
TI-89: d equation, x( ) d equation, x,2( ) d equation, x( ) x = point
Integrals Indefinite Integral Definite Integral
TI-83: NA fnInt equation, x, lower bound, upper bound( )TI-84: NA equation( )
lower bound
upper bound
∫ dx
TI-89: equation, x( ) equation, x, lower bound, upper bound( )∫∫
3
CH.1 LIMITS** When evaluating limits, we are checking around the point that we are approaching, NOT at the point.**Every time we find a limit, we need to check from the left and the right hand side (Only if there is a BREAK at that point).
1-2 Finding Limits Graphically and NumericallyNo breaking point Hole in the graph piece-wise function radicals
limx→a
f x( ) =limx→a+
f x( ) =limx→a−
f x( ) =
limx→a
f x( ) =limx→a+
f x( ) =limx→a−
f x( ) =
limx→a
f x( ) =limx→a+
f x( ) =limx→a−
f x( ) =
limx→a
f x( ) =limx→a+
f x( ) =limx→a−
f x( ) =
Asymptotes Asymptotes Asymptotes
limx→a+
f x( ) = limx→a+
f x( ) = limx→a+
f x( ) =limx→a−
f x( ) = limx→a−
f x( ) = limx→a−
f x( ) =limx→a
f x( ) = limx→a
f x( ) = limx→a
f x( ) =
**If left and right hand limits DISAGREE, then the limit Does Not Exist (DNE) at that point. **If left and right hand limits AGREE, then the limit exists at that point as that value.**Even if you can plug in the value, the limit might not exist at that point. It might not exist from the left or right side or the two sides will not agree.
limx→a+
f x( ) = right hand limit limx→a−
f x( ) = left hand limit
ac
b
aa
bb
a
aaa
4
**Breaking Points are points on the graph that are undefined or where the graph is split into pieces. Breaking Points :
1) Holes (when the numerator and denominator equals 0)2) Radicals (when the radical equals 0)3) Asymptotes (when the denominator equals 0)4) Piece-wise functions (the # where the graph is split)
Note : In general when doing limits, #0= ∞ − #
0= −∞ #
∞= 0
**In limits, if the two sides of a graph don’t agree, then the limit does not exist.
1-3 Analyzing Limits AnalyticallyLIMITS AT NON - BREAKING POINTS Very easy. Just plug in the #( )
EX#1: limx→1
x2 = EX#2 : limx→5
x + 4 = EX#3 : limx→1
x −1x +1
=
HOLES IN THE GRAPH 00( ) Factor and cancel or multiply by the conjugate and cancel, then plug in #( )
EX#1: limx→3
x2 + 3x −18x − 3
= EX#3 : limx→−4
x + 5 −1x + 4
=
EX#2 : limx→4
x − 4x − 4
= EX#4 : limx→25
x −636 − x
=
TRIG. FUNCTIONSTrig. Identities to know : sin2x = 2sin xcos x cos2x = cos2 x −sin2 x
FACTS : limx→0
sin xx
= 1 limx→0
1− cos xx
= 0 limx→0
tan xx
= 1
limx→0
sinaxbx
= ab
limx→0
1− cosaxbx
= 0 limx→0
tanaxbx
= ab
EX#1: limx→0
sin2x5x
= EX#4 : limx→0
sin x tan xx2 =
EX#2 : limx→0
2 tan5x9x
= EX#5 : limx→0
8sin xcos x3x
=
EX#3 : limx→π
2 4sin 3x
x= EX#6 : lim
x→0 7sin2 x tan2 x
x4 =
5
1-4 Continuity and One-Sided LimitsRADICALS You must first check that the limit exists on the side(s) you are checking( )If a # makes a radical negative, the limit will not exist at that #.When we check at the breaking point (the # that makes the radical zero) there are two possible answers:1) 0 if the limit works from the side that you are checking.2) DNE if the limit does not work from the side that you are checking.
EX #1: limx→5 +
x −5 = EX #2 : limx→5 −
x −5 = EX #3 : limx→5
x −5 =
EX #4 : limx→2
16 − x2 = EX #5 : limx→5
16 − x2 = EX #6 : limx→4
16 − x2 =
PIECE - WISE FUNCTIONS
f x( ) =3− x x < − 3
2x +12 − 3≤ x < 49 x ≥ 4
⎧
⎨⎪
⎩⎪
The breaking points are − 3 and 4.
EX #1: limx→−3+
f x( ) = EX #2 : limx→4+
f x( ) = EX #3 : limx→−3 −
f x( ) =
EX #4 : limx→4−
f x( ) = EX #5 : limx→−3
f x( ) = EX #6 : limx→4
f x( ) =EX #7 : lim
x→7+f x( ) = EX #8 : lim
x→−5f x( ) = EX #9 : lim
x→2f x( ) =
CONTINUITYContinuous functions have no breaks in them. Discontinuous functions have breaks in them (Asymptotes or Holes / Open Circles).
** To check for continuity at “a”, you must check left hand limits limx→−a−
f x( ) and right hand limits limx→−a+
f x( ) as well as the value of the function at that point f a( ). If all three are equal then the function is continuous at a.
If f a( ) = limx→−a−
f x( ) = limx→−a+
f x( ) then the function is continuous at a.
If f a( ) is not equal to either one - sided limit, then the function is not continuous (discontinuous) at a.
Discontinuous at aContinuous at aaaaa
6
1-5 Infinite LimitsASYMPTOTES #
0( ) (Since the point DNE we have to check a point that is close on the side we are approaching)
There are three possible answers when checking near the breaking point (the # that makes bottom = zero)1) ∞ → If we get a positive answer the limit approaches ∞2) −∞ → If we get a negative answer the limit approaches − ∞3) DNE → If we get a positive answer on one side and a negative answer on the other side, then the limit DNE
EX #1: limx→7+
1x − 7
= EX #2 : limx→7−
1x − 7
= EX #3 : limx→7
1x − 7
=
EX #4 : limx→3
2x − 3( )2 = EX #5 : lim
x→8 −6x −8( )2 = EX #6 : lim
x→−5+ 4xx + 5
=
EX #7 : limx→0−
cos xx
= EX #8 : limx→0
cos xx
= EX #9 : limx→π
2+tan x =
3-5 Limits at Infinity LIMITS THAT APPROACH INFINITYCheck the powers of the numerator and denominator.1) If the denominator (bottom) is a bigger power the limit = 0.2) If the numerator (top) is a bigger power the limit = ∞ or -∞.
3) If powers are the same the limit = coefficient of the highest power of numeratorcoefficient of the highest power of denominator
EX#1: limx→∞
2x2 + 35x2 − 7
= EX#2 : limx→∞
6x3x2 +1
= EX#3 : limx→−∞
x2
3x −1= EX#4 : lim
x→−∞
4x2
8 − 3x2 =
EX#5 : limx→−∞
7x + 28x7 −1
= EX#6 : limx→∞
5 ⋅7x −203⋅7x + 4
= EX#7 : limx→−∞
5 ⋅7x −203⋅7x + 4
=
FINDING VERTICAL ASYMPTOTES AND HOLESA vertical asymptote is the # that makes only the denominator = 0.A hole occurs at the points that make the numerator and denominator = 0 at the same time.EX#1:
a) f x( ) = x −2x + 3
b) f x( ) = 8xx2 + 9
c) f x( ) = x + 5( ) x − 7( )x − 7( ) d) f x( ) = x +1( ) x + 4( )
x + 4( ) x − 6( )vert.asym. hole vert.asym. hole vert.asym. hole vert.asym. holes
7
CH.2 DIFFERENTIATION 2-1 The Derivative by Definition and the Tangent Line ProblemDerivative at all points Derivative at the point a, f a( )( )
′f x( ) = limh→0
f x + h( )− f x( )h
′f a( ) = limh→0
f a + h( )− f a( )h
slope of secant line l =f x + h( )− f x( )
x + h − x lim
h→0
f x + h( )− f x( )h
means that the distance h is approaching 0 and
the points get closer to each other and the two points become the same point and line l is now a tangent line.
The derivative of a function finds the slope of the tangent line!
EX #1: f x( ) = x2 − 3x + 2 Find ′f x( ) and ′f 4( ). Use ′f x( ) = limh→0
f x + h( )− f x( )h
′f x( ) =
EX #2 : f x( ) = x2 − 3x + 2 Find ′f 4( ) Use ′f a( ) = limh→0
f a + h( )− f a( )h
′f 4( ) =
Line l is a tangent lineLine l is a secant lineLine l is a secant line
ll
l
f(x+h)=f(x)
xx x+h
h
f(x)f(x+h)
f(x+h)
f(x)
h
x+hx
8
normal line to curve
tangent line to curve
The derivative finds the slope of the tangent line.The normal line is perpendicular to the tangent line.
EX #3 : f x( ) = 1
x Find equation of the tangent line and normal line at x = 3.
Equation of the tangent line : Equation of the normal line :
9
2-2 Basic Differentiation Rules, Notation and Rates of Change Properties of DerivativesDerivative is a rate of change; it finds the change in y over the change in x, dy
dx , which is slope.
The Derivative finds the slope of the tangent line to a curve!1st derivative ⇒ max. and min., increasing and decreasing, slope of the tangent line to the curve, and velocity.2nd derivative ⇒ inflection points, concavity, and acceleration.
The 2nd Derivative finds the rate of the slopes of the tangent line to a curve.
*Power Rule *Constant Rulef (x) = xn f (x) = c′f x( ) = nxn−1 ′f x( ) = 0
EX : y = 4x3 EX : y = 8 ′y = 12x2 ′y = 0
EX#1: y = 15x4 Find dydx
and dydx x=2
.
Derivative notation Lagrange Leibniz Newton
y = x3 y = x3 y = x3
′y = 3x2 dydx
= 3x2 !y = 3x2
′′y = 6x d2ydx2 = 6x !!y = 6x
′′′y = 6 d3ydx3 = 6 !!!y = !!!y = 6
y 4( ) = 0 d4ydx4 = 0 !y
4= 0
EX#2 : Find ′f x( ) for each.
a) f (x) = 18x8 −6x3 + 29x + 3 b) f (x) = x
9+ 5x2 c) f (x) = x −11
x
Slope of the tangent line to the curveEX : Given f x( ) = 3x2 −10x Find equation of the tangent line and normal line at x = 4.
f x( ) = 3x2 −10x ′f x( ) = 6x −10 f 4( ) = 8 ′f 4( ) = 14
Equation of a Line (point - slope form) : y − y1 = m x − x1( )Equation of the tangent line : y − 8 = 14 x − 4( )
Equation of the normal line : y − 8 = −1 14
x − 4( )
10
EX #3 : Find the slope, write the equation of the tangent line and the equation of the normal line at x = 3. f x( ) = 5x2 − 7 Equation of the tangent line Equation of the normal line
*Trig. Functions
2 STEPS : Derivative of the trig. function ⋅ Derivative of the angle
Function Derivativesin x cos xcos x −sin x
Function Derivative
tan x sec2 xcot x −csc2 x
Function Derivative
sec x sec x tan xcsc x −csc xcot x
EX #4 : Find each derivative
a) y = sin 8x( ) b) y = cos x4( ) c) y = tan 7x3( )
d) y = sec 6x2 + 5( ) e) y = csc 2x5( ) f ) y = cot 9x −1( )
g) y = cos sin x( ) h) y = sec tan 3x( )
Use s t( ) = −16t2 + v0t + s0. s t( ) = ending height s0 = initial height v0 = initial velocity t = time
EX #5 : To estimate the height of a building, a weight is dropped from the top of the building into a pool at ground level. How high is the building if the splash is seen 2 seconds after the weight is dropped?
EX #6 : A red ball is thrown upward from a building 100 feet above the ground with an initial velocity of 10ft/s. At the same time, a blue ball is thrown downward from a height of 150 feet with an initial velocity of 10ft/s. Which ball hits the ground first? How much faster?
11
2-3 Product and Quotient Rule*Product Rule 4 STEPS : Derivative of First equation ⋅ Second equation + Derivative of Second equation ⋅ First equation
y = f (x) ⋅g(x) dydx
= ′f (x)g(x)+ ′g (x) f (x)
*Quotient Rule
5 STEPS : Derivative of Top equation ⋅ Bottom equation − Derivative of Bottom equation ⋅ Top equationBottom equation( )2
y = f (x)g(x)
dydx
= ′f (x)g(x)− ′g (x) f (x)g(x)( )2
2-4 Chain Rule
3 STEPS : 1) Power in front 2) Lower power by 1 3) Multiply by derivative of insideOR Derivative of outside function ⋅Derivative of inside function
y = f (x)( )n OR y = f g x( )( ) OR y = f 4x( ) OR y = f x2( ) ′y = n f (x)( )n−1 ⋅ ′f (x) ′y = ′f g x( )( ) ⋅ ′g x( ) ′y = ′f 4x( ) ⋅4 ′y = ′f x2( ) ⋅2x
EX #1: f x( ) = x2 tan x EX #2 : y = x3 − 5( )6
EX #3 : f x( ) = sin xcos x
EX #4 : y = 2x4 x2 −5
12
2-5 Implicit DifferentiationThe differentiable functions we have encountered so far can be described by equations in which "y" is expressed in terms of "x". We can also find the derivative of the equation expressed in terms of x and y.*Implicit Differentiation: function in terms of x’s and y’s
must write dydx
everytime you take a deriv. of y⎛⎝⎜
⎞⎠⎟
EX : x2 − xy + 3y2 = 7
2x − y(1)+ dydxx⎡
⎣⎢⎤⎦⎥+ 6y dy
dx= 0
2x − y − x dydx
+ 6y dydx
= 0
− x dydx
+ 6y dydx
= −2x + y
dydx
(−x + 6y) = −2x + y
dydx
= −2x + y−x + 6y
EX#1: x2 + y2 = 16 (3, 7 ) Find the equation of the tangent line at the given point.
EX #2 : Find first two derivatives of x3y2 = 5
EX#3 : Find dydx
. 5x2 − xy2 + 3y = x EX#4 : Find dydx
. 5x2 − xy2 + 3y = x
Use shortcut( )
13
Graphs and ChartsDerivative of a graph′f 1( ) = ′f 3( ) = ′f 4( ) =
′f 6( ) = ′f 7( ) = ′f 8( ) =
Equation of the tangent line at x = 1 Equation of the tangent line at x = 4
Derivative of a chart
x 0 20 40 60 80 100 120 140 160
f x( ) 58 63 72 61 62 69 61 74 67
EX : ′f 40( ) = 61− 6360 − 20
= −2 40
= −1 20
EX : ′f 130( ) = 74 − 61140 −120
= 1320
EX#1: ′f 140( ) = EX#2 : ′f 110( ) =
How to read Derivative by Definition ProblemsThis problem means take a derivative of cos x. This problem means take a derivative of 10x3 at x = 2.
EX : limh→0
cos x + h( )− cos x
h= −sin x EX : lim
h→0 10 2 + h( )3 − 80
h= 120
Solve each
EX#1: limh→0
5 x + h( )4 − 5x4
h= EX#2 : lim
h→0
3+ h( )4 − 81h
=
EX#3 : limh→0
sec x + h( )− sec x
h= EX#4 : lim
h→0 sin π
6 + h( )− 12
h=
2
5 10
-1
3
2
1
109876543210
14
2.6 Related RatesWe take derivatives with respect to t which allows us to find velocity. Here is how you take a derivative with respect to t:
derivative of x is dxdt
, derivative of y2 is 2y dydt
, derivative of r3 is 3r2 drdt
, derivative of t 2 is 2t dtdt
= 2t
V means volume ; dVdt
means rate of change of volume (how fast the volume is changing)
r means radius ; drdt
means rate of change of radius (how fast the radius is changing)
dxdt
is how fast x is changing; dydt
is how fast y is changing
Volume of a sphere Surface Area of a sphere Area of a circle Circumference of a circle
V = 43πr3 A = 4πr2 A = πr2 C = 2πr
dVdt
= 4πr2 drdt
dAdt
= 8πr drdt
dAdt
= 2πr drdt
dCdt
= 2π drdt
Volume of a cylinder Volume of a cone
V = πr2h V = 13πr2h use r
h=
r is not a variable in a cylinder Due to similar triangles, the ratio of the radiusbecause its' value is always the same to the height is always the same. Replace r or h depending on what you are looking for.
!!
EX: In!the!cylinder!below!the!radius!will!!!!!!!!!!!always!be!4!so!we!can!replace!r !with!!!!!!!!!!!4!BEFORE!we!take!the!derivative.!!!!!!!!!!!!!V =πr2h!!!becomes!!V =16πh.
!!
EX: In!the!cone!to!the!right!the!ratio!of!!!!!!!!!!!!the!radius!and!height!will!always!be!!
!!!!!!!!!!!! rh= 59 !,!!so!!!r !=
59h!!!!or!!!!h= !
95r.
!!!!!!!!!!!!We!replace!the!appropriate!one!
!!!!!!!!!!!!V = 13πr2h!!!becomes:!
!!!!!!!!!!!!V = 13π59h
⎛⎝⎜
⎞⎠⎟
2
h!⇒ !V = 25243πh
3
!!!!!!!!!!!!OR!!!V = 13πr2 95r
⎛⎝⎜
⎞⎠⎟!⇒ !V = 35πr
3
4
16
9
5
15
EX #1: Given 8x2 −5y3 = 112 find dydt
,
when x = 3 and dxdt
= 5.
EX #2 : Water is spilling on the ground. The radius is changing at 4 in/s. How fast is the volume changing when the radius is 10 in. ? (A = πr2 )
EX #3 : Suppose a spherical balloon is inflated at the rate of 14 in3 /min. How fast is the radius of the
balloon changing when the radius is 7 inches? V = 43πr3⎛
⎝⎜⎞⎠⎟
EX #4 : Water is poured into a cylinder with dimensions below at the rate of 11 in3 /s. How fast is the
height of the water changing when the height is 4 inches? V = πr2h( )
5
12
16
EX #5 : Water is leaking out of a cone with diameter 10 inches and height 9 inches at the rate of 5 in3 /s. How fast is the radius of the water changing when the radius is 4 in?
EX #6 : A 17 foot ladder is leaning against the wall of a house. The base of the ladder is pulled away at 3 ft. per second.
a) How fast is the ladder sliding down the wall when the base of the ladder is 15 ft. from the wall?
b) How fast is the area of the triangle formed changing at this time?
c) How fast is the angle between the bottom of the ladder and the floor changing at this time?
EX #7 : A person 6 ft. tall walks directly away from a streetlight that is 13 feet above the ground. The person is walking away from the light at a constant rate of 4 feet per second.a) At what rate, in feet per second, is the length of the shadow changing?
b) At what rate, in feet per second, is the tip of the shadow changing?
10
9
θ
y
x
17
17
CH.3 APPLICATIONS OF DIFFERENTIATION 3-1 Extrema on an IntervalLet f be continuous on a closed bounded interval [a,b]. Then f has an absolute maximum and absolute minimum on the interval [a,b].Procedure for finding absolute max. and absolute min. :
Compute the values of f at all critical points (when ′f x( ) = 0) in a,b( ) and at the endpoints a and b.The largest of these values is the absolute maximum value of f on [a,b].The smallest of these values is the absolute minimum value of f on [a,b].
Find the extreme values (max. and min.) of f on given interval and determine at which #'s in they occur.
EX#1: Let f x( ) = 12x − x3 ; [0,3] EX#2 : Let f x( ) = 3xx −5
; [6,21]
There are critical points that show up as undefined in the derivative, but work in the original equation. I call these special points Hardpoints. The textbooks don't have a name for these points so I gave them one.
Find the extreme values (max. and min.) of f on given interval and determine at which #'s in they occur.
EX#3 : f x( ) = x23 −2 ; [ −8, 27] EX#4 : f x( ) = x − 4 ; [1, 10]
18
3-2 Mean Value Theorem and Rolle's Theorem *Mean-Value Theorem (Only applies if the function is continuous and differentiable)
Slope of tangent line = slope of line between two points
′f (c) = f (b)− f (a)b − a
According to the Mean Value Theorem, there must be a number c between a and b that the slope of the tangent line at c is the same as the slope between points (a, f (a)) and (b, f (b)).
The slope of secant line from a and b is thesame as slope of tangent line through c.
Use the Mean Value Theorem to find all values of c in the open interval a, b( ) such that ′f (c) = f (b)− f (a)b − a
EX #1: f x( ) = sin x 0,π[ ] EX #2 : f x( ) = − x2 + 7 −1, 2[ ]
EX #3 : f x( ) = x − 3 0, 5[ ]
4
2
5
f(b)
f(a)
ba
4
2
5c
f(b)
f(a)
ba
19
3-3 Increasing and Decreasing Functions and the First Derivative Test Properties of First Derivativeincreasing : slopes of tangent lines are positive (derivative is positive.) ′f x( ) > 0
decreasing : slopes of tangent lines are negative (derivative is negative). ′f x( ) < 0
maximum point : Slopes switch from positive to negative at maximum point. found by setting ′f x( ) = 0( )minimum point : Slopes switch from negative to positive at minimum point. found by setting ′f x( ) = 0( )
***1st Derivative Testa) If ′f changes from positive to negative at the critical point c, then f has a relative maximum value at c.
b) If ′f changes from negative to positive at the critical point c, then f has a relative minimum value at c.
3-4 Concavity and the Second Derivative Test Properties of Second Derivativeconcave up : slopes of tangent lines are increasing (2nd Derivative is positive). ′′f x( ) > 0
concave down : slopes of tangent lines are decreasing (2nd Derivative is negative). ′′f x( ) < 0
inflection points : points where the graph switches concavity. found by setting ′′f x( ) = 0( ) slopes of tangent line switch from increasing to decreasing or vice versa.
***2nd Derivative Test (Alternate method for finding rel. max. and rel. min.)Set ′f c( ) = 0 to find your critical points c. Plug your critical point(s) into ′′f x( ).a) If ′′f c( ) < 0, then c is a rel. max.b) If ′′f c( ) > 0, then c is a rel. min.c) If ′′f c( ) = 0, then from this test alone we can't draw any conclusions about a relative extreme value at c
M is a Maximum; m is a minimum;I is an inflection point
Im, I
M
mm is a Minimum; slopes switchfrom negative to positive
I is an inflection point; slopes switchfrom decreasing to increasing
M is a Maximum; slopes switchfrom positive to negative
Im
M
decreasing/concave downslopes are negative and decreasing
decreasing/concave upslopes are negative and increasing
increasing/concave upslopes are positive and increasing
increasing/concave downslopes are positive and decreasing
20
EX #1: From 0,7[ ] tell me about the function. (Use graph above)List the x - coordinates for each : Find each : On which interval(s) is the graph:Inflection points Abs. max. value increasing/concave up Relative maximum Abs. min. value increasing/concave down Relative minimum Abs. max. value occurs at decreasing/concave up Hard points Abs. min. value occurs at decreasing/concave down
EX #2 : f x( ) = x2
3x − 2
Find relative extreme values and determine the intervals on which f x( ) is increasing and decreasing.Find all inflection points and the intervals on which f x( )is concave up and concave down.
EX #3 : f x( ) = − x −1( )3 + 27x −27
Find relative extreme values and determine the intervals on which f x( ) is increasing and decreasing.Find all inflection points and the intervals on which f x( )is concave up and concave down.
2
5
3
2
1
76543210
21
3.6 A Summary of Curve Sketching GRAPHING TRIG. REVIEWx - intercepts : where a graph crosses the x-axis. The # that makes y = 0. To find the x-intercept, set the numerator = 0 (plug in zero for y).y - intercepts : where a graph crosses the y-axis. The # found when x = 0. To find the y-intercept, plug in zero for x.Holes : the # that makes both the numerator and denominator = 0. To find the hole in the graph, you plug the # into the remaining function after canceling out the like factors. Find the hole first before finding the vertical asymptote.vertical asymptotes : An undefined point on the graph. A graph will never cross the vertical asymptote. To find the vertical asymptote, set the denominator = 0. The vertical asymptote is x = the # that makes only the denominator = 0.horizontal asymptotes : The graph will approach the horizontal asymptote as x approaches ∞ and − ∞. To find the horizontal asymptote, check the highest powers of the numerator and the denominator.1) If the denominator (bottom) is a bigger power the horizontal asymptote is y = 0.2) If the numerator (top) is a bigger power there is no horizontal asymptote (there is a different kind of asymptote).
3) If powers are the same the horizontal asymptote is y = coefficient of the highest power of numeratorcoefficient of the highest power of denominator
slant asymptotes : Occur when the numerator is one power higher than the denominator. To find the slant asymptote, you must use long division to divide the denominator into the numerator. The quotient is your slant asymptote.quadratic asymptotes : Occur when the numerator is two powers higher than the denominator. To find the quadratic asymptote, you must use long division to divide the denominator into the numerator. The quotient is your quadratic asymptote.
Sketch the graph Label the maximum, minimum and inflection points( )EX#1: y = x3 − 3x2 ′y = 3x2 −6x ′′y = 6x −6x−int y−int v.asym. h.asym. rel.max. rel.min. inc. dec. inf .pts. conc.up conc.down0, 0( ) 0, 0( ) NONE NONE 0, 0( ) 2, − 4( ) −∞, 0( ) 0, 2( ) 1, −2( ) 1, ∞( ) −∞, 1( )3, 0( ) 2, ∞( )
x3 − 3x2 = 0 3x2 −6x = 0 6x −6 = 0 x2 x − 3( ) = 0 3x x −2( ) = 0 6 x −1( ) = 0 x = 0, 3 x = 0, 2 x = 1
f '(x)00 +-+20
f "(x)0 +-1
2
-2
- 4
5
I
M
m-5-4-3-2
-2-1
-1 0 321 4
321
22
Note all relevant properties of f and sketch the graph Label the maximum, minimum and inflection points( )
EX#2 : y = 3x x −2( )3
x − int y − int v.asym. h.asym. rel.max. rel.min. inc. dec. inf .pts. conc.up conc.down
Note all relevant properties of f and sketch the graph Label the maximum, minimum and inflection points( )
EX#3 : y = 12xx +1( )2
x − int y − int v.asym. h.asym. rel.max. rel.min. inc. dec. inf .pts. conc.up conc.down
-5
-5 -4
-4-3
-3
-2
-2-1
-1 0
54
321 54
321
-5
-5 -4
-4-3
-3
-2
-2-1
-1 0
54
321 54
321
23
3-7 Optimization Problems1) Draw and label picture.2) Write equation based on fact given and write equation for what you need to maximize or minimize.3) Plug in fact equation into the equation you want to maximize or minimize.4) Take derivative and set equal to zero. 5) Find remaining information.
EX#1: An open box of maximum volume is to be made from a square piece of material, 12 inches on a side, by cutting equal squares from the corners and turning up the sides. How much should you cut off from the corners? What is the maximum volume of your box? EX#2 : A farmer plans to fence a rectangular pasture adjacent to a river. The farmer has 56 feet of fence in which to enclose the pasture. What dimensions should be used so that the enclosed area will be a maximum? What is the maximum area?
12-2X
12-2X
X
X
X
XX
X
X
XX
X
X
X X
X
X
X
12
1212
12
Y Y
X
RIVER
24
EX#3 : A crate, open at the top, has vertical sides, a square bottom and a volume of 864 ft3. What dimensions give us minimum surface area? What is the surface area?
EX#4 : A rectangle is bounded by the x-axis and the semicircle y = 16 − x2 . What length and width should the rectangle have so that its area is a maximum?
Y
XX
4
2
-2
- 5 5
Y= 16-X2
Y
XX
25
3-8 Newton's Method
Newton's method is used to approximate a zero of a function.
*Newton’s Method
c −f c( )′f c( ) where c is the current approximation for the zero.
To perform Newton's Method you will need the equation, its' derivative and a first approximation.
EX#1: If Newton’s method is used to approximate the real root of x3 + x −1= 0, then a first approximation x1 = 1 would lead to a third approximation of x3 =
f (x) = x3 + x −1
EX#2 : Given f (x) = x2 − 22 and first approximation x1 = 5. Find third approximation x3 = .
26
3-9 Differentials∗Differentials are tangent lines. Tangent lines hug closely along a graph near the point of tangency.∗Sometimes it's easier to use the differential (tangent lines) to approximate a value of a function as opposed to using the function itself.
EX#1: f x( ) = ln x
a) Find the equation of the tangent line (differential) at e,1( ).
b) Use the differential to approx. f 3( ).
c) Actual value of f 3( ) =
EX#2 : xy2 + 3y − x = 17a) Find the equation of the tangent line at 1,3( ).
b) Use the differential to approx. f 1.2( ).
c) Actual value of f 1.2( ) =
6
4
2
5
(e, 1)
g x( ) = 1e · x – e( ) + 1
f x( ) = ln x( )
27
CH.4 INTEGRATION 4-1 Antiderivatives and Indefinite Integration
** Integrals : They find the antiderivatives, sum/total, Area and Volume.
Integration Formulasf x( ) + g x( )( )∫ dx = f x( )∫ dx + g x( )∫ dx
c ⋅ f x( )∫ dx = c ⋅ f x( )∫ dx where c is a constant. Constants can move in and out of an integral.
f x( )∫ dx = F x( ) + C (notation)
*Integral of a constanta∫ dx = ax +C EX : 5 dx = 5x + C∫ EX : π dx = π x∫ + C
*Polynomials
xa∫ dx = xa+1
a +1 + C EX : x2 dx =x3
3+ C∫ EX : 4x6 dx =
4x7
7∫ + C
*Fractions Bring up denominator, then take integral( )1xa∫ dx⇒ x−a∫ dx = x −a+1
−a +1 + C EX : 1x4∫ dx ⇒ x− 4∫ dx =
x− 3
− 3+ C =
−1 3x3 + C
*Trig Functions Always divide by derivative of the angle( )
sin x dx = − cos x + C∫ EX : sin 7x dx = − cos 7x7∫ + C
cos x dx∫ = sin x + C EX : cos 3x dx = sin 3x3∫ + C
28
EX#1: a) x8∫ dx = b) 12x3∫ dx = c) x3∫ dx =
d) 9∫ dx = e) 7x2∫ dx = f ) 1
x47∫ dx =
EX#2 : a) x3 − 7x + 4( )∫ dx = b) x2 + 8xx∫ dx =
EX#3 : a) cos12x dx =∫ b) sin 7x dx =∫
Preview of Rectilinear MotionSymbols used
position : x t( ), y t( ), s t( )velocity : v t( )acceleration : a t( )
EX#4 : v t( ) = t 2 −6t −16 x 0( ) = 3 t > 0a) Find position at any time t. b) Find acceleration at any time t.
c) Find velocity at t = 7. d) Find acceleration at t = 7.
e) Is the speed inc. or dec. at t = 7? f ) When is the particle at rest?
IntegralDerivative
Rectilinear Motion position velocity acceleration
29
4-3 Riemann Sums and Definite Integrals
f x( )a
b
∫ dx + f x( )b
c
∫ dx = f x( )a
c
∫ dx EX : f x( )2
5
∫ dx + f x( )5
10
∫ dx = f x( )2
10
∫ dx
f x( )a
b
∫ dx = − f x( )b
a
∫ dx EX : f x( )3
8
∫ dx = − f x( )8
3
∫ dx
EX #1: Given g x( )1
3
∫ dx = 8 g x( )3
6
∫ dx = 27 Find the following:
a) g x( )1
6
∫ dx = b) g x( )6
3
∫ dx = c) g x( )3
3
∫ dx = d) 5 g x( )1
3
∫ dx =
EX #2 : Given f x( )−2
3
∫ dx = 40 f x( )3
7
∫ dx = −17 f x( )12
7
∫ dx = −10 Find the following:
a) f x( )−2
12
∫ dx = b) f x( )12
3
∫ dx = c) f x( )−2
12
∫ dx = d) 10 ⋅ f x( )−2
7
∫ dx =
EX #3 : Use picture to the right to find each.
a) f x( )0
10
∫ dx = b) f x( )10
7
∫ dx =
c) f x( )4
10
∫ dx = d) f x( ) +10( )0
4
∫ dx =
EX #4 : Use picture to the right to find each.
a) f x( )0
4
∫ dx = d) f x( )−4
6
∫ dx =
b) f x( )−2
2
∫ dx = e) f x( )−4
6
∫ dx =
c) f x( )6
0
∫ dx = f ) f x( ) + 3( )−4
6
∫ dx =
2
-2
- 5 5
-2
2
-4 -2 6420
2
-2
5 109
y = f(x)
107402
35
8
30
4-4 The Fundamental Theorem of Calculus∗∗∗ 1st Fundamental Theorem of Calculus After you take the integral, just plug in the top # minus the bottom #.
f x( ) dxa
b
∫ = F x( ) ab = F b( )− F a( )
EX #1: Evaluate 3x2 − x +5( )0
3
∫ dx = EX #2 : Evaluate 1x1
16
∫ dx =
***** Area= top equation −bottom equation( )a
b
∫ dx
EX #3 : Find Area of each shaded regiona) b) c)
EX #4 : Find Area of the region between y = sin x and the x-axis from 0,2π[ ]
2
5
1
2
3
43210
2
3
2
1
3210
4
2
y=x 2
3
4
3
2
1
210
2
-2
5
-1
1
3π2
π2
2ππ0
y=sinx
31
4-4 The Fundamental Theorem of Calculus (Cont.)∗∗∗∗∗Average Value (use this when you are asked to find the average of anything)
If f is integrable on the closed interval a,b[ ], then the average value of f on the interval is
Average value = 1b − a
⋅ f (x) dxa
b
∫EX#1: Find the average value of f (x) on the closed interval. EX #2 : Find average acceleration from 0, 2[ ]a) f x( ) = x2 0,3[ ] b) f x( ) = x −2 3, 11[ ] v(t) = t 3 + 2t2 −5
∗∗∗2nd Fundamental Theorem of Calculus (When taking the derivative of an integral)Plug in the variable on top times its derivative minus plug in the variable on bottom times its derivative.
ddx
f t( ) dt0
x
∫ = f x( ) ddx
f t( ) dt0
x
∫ = f x( ) ⋅1− f 0( ) ⋅0 = f x( )
EX #3 : Evaluate each.
a) ddx
t 3 dt0
x
∫ = b) ddx
t 3 dtx
0
∫ =
c) F x( ) = t 7 +1( ) dt1
x2
∫ = d) F x( ) = 1+ t2 dtx
sin x
∫ =
′F x( ) = ′F x( ) =
32
4-5 Integration by Substitution*Substitution When integrating we usually let u = the part in the parenthesis, the part under the radical, the denominator, the exponent, or the angle of the trig. function.
f g x( )( )∫ ⋅ ′g x( )dx = f u( )∫ ⋅du = F u( ) +C = F g x( )( ) +C Let u = g x( ) du = ′g x( )dx
EX#1: 3x2 x3 + 5( )20
∫ dx = EX#2 : cos x( )4 sin x∫ dx =
EX#3 : x4 x5 −12( )7
∫ dx = EX#4 : xsin x2 dx∫ =
EX#5 : x2 x3 + 2( )3
0
2
∫ dx = EX#6 : 4x6
x7 + 8( )5∫ dx =
EX#7 : x x + 9∫ dx = EX#8 : x3 x2 −6∫ dx =
33
4-6 Numerical Integration (Approximating Area)We approximate Area using rectangles (left, right, and midpoint) and trapezoids.*Riemann Sumsa) Left edge Rectangles f (x) = x2 +1 from [ 0, 2] using 4 subdivisions (Find area of each rectangle and add together)
A = b −an
⎛⎝⎜
⎞⎠⎟ left height of each rectangle ( ) A = 2 −0
4⎛⎝⎜
⎞⎠⎟ 1+ 5
4+ 2 + 13
4 ⎛
⎝⎜⎞⎠⎟
Total Area= 308
3.750
b) Right edge Rectangles f (x) = x2 +1 from [ 0, 2] using 4 subdivisions (Find area of each rectangle and add together)
A = b −an
⎛⎝⎜
⎞⎠⎟ right height of each rectangle ( ) A = 2 −0
4⎛⎝⎜
⎞⎠⎟
54+ 2 + 13
4+ 5⎛
⎝⎜⎞⎠⎟
Total Area = 468
5.750
c) Midpoint Rectangles f (x) = x2 +1 from [ 0, 2] using 4 subdivisions (Find area of each rectangle and add together)
A = b −an
⎛⎝⎜
⎞⎠⎟ midpt. height of each rectangle ( ) A = 2 −0
4⎛⎝⎜
⎞⎠⎟ 17
16+ 25
16+ 41
16+ 65
16 ⎛
⎝⎜⎞⎠⎟
Total Area = 14832
4.625
*Trapezoidal Rule (used to approximate area under a curve, using trapezoids).
Area ≈ b − a2n
f (x0 )+ 2 f (x1)+ 2 f (x2 )+ 2 f (x3).....2 f (xn−1)+ f (xn )[ ]where n is the number of subdivisions.
d) f (x) = x2 +1 Approximate the area under the curve from [ 0, 2 ] using the trapezoidal rule with 4 subdivisions.
A = 2 − 02 4( ) f (0)+ 2 f 1
2⎛⎝⎜
⎞⎠⎟ + 2 f (1)+ 2 f 3
2⎛⎝⎜
⎞⎠⎟ + f (2)⎡
⎣⎢⎤⎦⎥
= 14
1+ 2 54
⎛⎝⎜
⎞⎠⎟ + 2(2)+ 2 13
4⎛⎝⎜
⎞⎠⎟ + 5⎡
⎣⎢⎤⎦⎥
= 14
764
⎡⎣⎢
⎤⎦⎥= 76
16= 4 3
4= 4.750
All you are doing is finding the area of the 4 trapezoids and adding them together!
e) Actual Area = x2 +1( ) dx0
2
∫ = x3
3+ x 0
2 = 83+ 2⎛
⎝⎜⎞⎠⎟ − 0 + 0( ) = 14
3= 4.667
5
4
3
2
1
-1
2
5
4
3
2
1
232
112
0
5
4
3
2
1
-1
2
5
4
3
2
1
232
112
0
5
4
3
2
1
-1
2
5
4
3
2
1
232
112
0
5
4
3
2
1
-1
2
5
4
3
2
1
232
112
0
34
*Approximating Area when given data only no equation given( )To estimate the area of a plot of land, a surveyor takes several measurements. The measurements are taken every 15 feet for the 120 ft. long plot of land, where y represents the distance across the land at each 15 ft. increment.
x 0 15 30 45 60 75 90 105 120
y 58 63 72 60 62 69 61 74 67
a) Estimate using Trapezoidal Rule b) Estimate using 4 Midpoint subdivisions
A 120 −02 8( ) f 0( ) + 2 f 15( ) + .....+ 2 f 105( ) + f 120( )⎡⎣ ⎤⎦ A 120 −0
4f 15( ) + f 45( ) + f 75( ) + f 105( )⎡⎣ ⎤⎦
A 152
58 +126 +144 +120 +124 +138 +122 +148 + 67[ ] A 30 63+ 60 + 69 + 74[ ]A 7852.5 A 7980c) Estimate Avg. value using Trapezoidal Rule d) What are you finding in part c?
Avg.Value 1120
7852.5( ) 65.4375 The average distance across the land.
e) Estimate using Left Endpoint f) Estimate using Right Endpoint
A 120 − 08
f 0( ) + f 15( ) + f 30( ) + .....+ f 105( )⎡⎣ ⎤⎦ A 120 − 08
f 15( ) + f 30( ) + f 45( ).....+ f 120( )⎡⎣ ⎤⎦
A 15 58 + 63+ 72 + 60 + 62 + 69 + 61+ 74[ ] A 15 63+ 72 + 60 + 62 + 69 + 61+ 74 + 67[ ]A 7785 A 7920
*Approximating Area when given data only no equation given( )Unequal subdivisions: You must find each Area separately.
x 0 2 5 10 y 10 13 11 15
a) Estimate using Trapezoids A = 12b1 + b2( )h⎛
⎝⎜⎞⎠⎟
A ! 12⋅ 10 +13( ) ⋅2 + 1
2⋅ 13+11( ) ⋅3 + 1
2⋅ 11+15( ) ⋅5 ! 124
b) Estimate using Left Endpoint A = width ⋅ left height( )A ! 2 10( )+ 3 13( )+5 11( ) ! 114 c) Estimate using Right Endpoint A = width ⋅right height( ) A ! 2 13( )+ 3 11( )+5 15( ) ! 134 Trapezoids shown
35
EX #1: To estimate the surface area of his pool, a man takes several measurements. The measurements are taken every 5 feet for the 50 ft. long pool, where y represents the distance across the pool at each 5 ft. increment.
Estimate each Area using 10 subdivisionsa) Use Trapezoidal Rule b) Estimate Avg. value using Trapezoidal Rule
c) Use Right Endpoint d) Use Left Endpoint
e) Use 5 Midpoint subdivisions
EX#2 : Estimate each Area using 3 unequal subdivisionsa) Use Trapezoids b) Estimate Avg. value using Trapezoids
c) Use Right Endpoint d) Use Left Endpoint
3015
35 40 45 5020 18 15 11172219171210
2520151050yx
32402825
10520yx
20
10
20 404020 504535302515105
5
0
10
15
20
25
8
6
4
2
5 10
40
35
30
25
20
15
10
5
0 1052
36
CH. 5 Logarithmic,Exponential, and other Transcendental Functions 5-1 The Natural Logarithm Function and Differentiation
Properties of Logarithms
limn→∞
1+ 1n
⎛⎝⎜
⎞⎠⎟n
= e
e = natural numberlogarithmic form ⇔ exponential form
y = ln x ⇔ ey = xLog Laws :
y = ln xy ⇔ y = y ln x ln 8 = ln23 = 3ln2ln x + ln y = ln xy ln2 + ln5 = ln10
ln x − ln y = ln xy
⎛⎝⎜
⎞⎠⎟
ln7 − ln2 = ln 72
⎛⎝⎜
⎞⎠⎟
Change of Base Law :
y = loga x ⇒ y = ln xlna
or log xloga
Memorize : lne = 1 ln1= 0
Fact : You can’t take a ln/log of a negative # or zero.We use logarithms to solve any problem that has a variable in the exponent.
EX : e5x = 24 ⇒ lne5x = ln24 ⇒ 5x lne = ln24 ⇒ x = ln245
EX : ln x = 3 ⇒ eln x = e3 ⇒ x = e3
*Derivative of Natural Log 2 STEPS: 1 divided by function i Derivative of function( ) y = ln f (x)( ) ′y = 1
f (x)⋅ ′f (x)
Find derivative of each
EX#1: ddxln(x) = EX#2 : d
dxln(x10 ) = EX#3 : d
dxln(x5 −1) =
EX#4 : ddxlog x = EX#5 : f x( ) = ln x x2 +1( )2
2x3 −1=
4
2
-2
5
(1,e)
(0,1)(e,1)
(1,0)
y=lnx
y=ex
Memorize these graphs.They are inverses of each other so they are symmetric about the line y=x.
37
5-2 The Natural Logarithmic Function: Integration
∗∗∗′f x( )f x( )∫ dx = ln f x( ) +C ⇒ top is the derivative of the bottom
∗∗∗ 1x∫ dx = ln x +C
Integrals of other four trig. functions
tan x dx =∫sin xcos x∫ dx = − −sin x
cos x∫ dx = − ln cos x +C
cot x dx = cos xsin x∫ dx∫ = ln sin x +C
sec x dx =∫sec x(sec x + tan x)
sec x + tan x∫ dx = sec2 x + sec x tan xsec x + tan x∫ dx = ln sec x + tan x +C
csc x dx = csc x(csc x + cot x)csc x + cot x∫ dx = csc2 x + csc xcot x
csc x + cot x∫ dx∫ = − − (csc2 x + csc xcot x)csc x + cot x∫ dx
= − ln csc x + cot x +C
EX#1: 3x2
x3 − 5∫ dx = EX#2 : x4
x5 −1∫ dx = EX#3 : 19x + 4∫ dx =
EX#4 : sec2 xtan x∫ dx = EX#5 : 1
x−6
−2
∫ dx = EX#6 : 2xx2 +1( )7∫ dx =
EX#7 : x2 + x +1x2 +1∫ dx =
EX : tan5x dx = − 15
ln cos5x∫ +C EX : sec9∫ x dx = 19
ln sec9x + tan9x +C
EX : csc8x dx = − 18
ln csc8x + cot 8x +C∫ EX : cot13x dx = 113
ln sin13x +C∫
38
5-3 Inverse FunctionsINVERSES : To find an inverse, f −1 x( ), you switch the x 's and y 's and solve for y.
EX : f x( ) = 2x + 3 Find f −1 x( ). EX : f x( ) = ex Find f −1 x( ).Inverse : x = 2y + 3 Inverse : x = ey ⇒ lnx = lney
x − 32
= y so f −1 x( ) = x − 32
ln x = y so f −1 x( ) = ln x
Facts about inverses1) When you plug one inverse into the other you always get the answer x and vice versa.
( f f −1) x( ) = f f −1 x( )( ) = x and ( f −1 f ) x( ) = f −1 f x( )( ) = xEX : f x( ) = ex f −1 x( ) = ln x
( f f −1) x( ) = eln x = x ( f −1 f ) x( ) = lnex = x
2) Graphs that are inverses are symmetric about the line y = x.
∗∗∗dydx
finds the slope of the tangent line.
dxdy
= 1dydx
∗∗∗dxdy
finds derivative the slope of the tangent line( ) of the inverse.
Formula for finding derivative of the inverse at a point : ′g y( ) = 1′f x( ) (where g y( ) is the inverse of f x( ))
EX #1: Let y = x3 + x. If h is the inverse function of f , then ′h 2( ) =
A) 113
B) 14
C) 1 D) 4 E) 13
EX #2 : Let f be a differentiable function such that f 4( ) = 20, f 8( ) = 4, ′f 4( ) = − 7, and ′f 8( ) = −5.
The function g is differentiable and g x( ) = f −1 x( ) for all x. What is the value of ′g 4( )?
A) −1 5
B) −1 7
C) 18
D) 1 4
E) The value of ′g 4( ) cannot be determined from the information given.
4
2
-2
5
(1,e)
(0,1)(e,1)
(1,0)
y=lnx
y=ex
39
5-4 Exponential Functions:Differentiation and IntegrationTrig. Revieweln x = eln 3 = e2 ln 5 =
Exponential Differentiation ConstantVariable( )ddxex = ex ⋅1⋅ lne = ex 3 steps : itself, multiplied by deriv. of exponent, multiplied by ln of base( )
ddxe2x =
ddxe5x3
=
Exponential Integration ConstantVariable( )ex∫ dx = ex
1⋅ lne+C = ex +C 3 steps : itself, divided by deriv. of exponent, divided by ln of base( )
e2x∫ dx =
e5x3dx =∫
EX#1: Find derivative of each
a) y = e7x2 b) y = ex sin x
EX#2 : Evaluate each integral
a) ex
ex +1∫ dx = b) e3x
0
ln2
∫ dx =
EX#3 : Evaluate each integral
a) x2 ⋅e5x3dx =∫ b) x ⋅e4x2
dx =∫
40
5-5 Bases other than e and Applications*Derivative of ConstantVariable y = a f (x ) ′y = a f (x ) ⋅ ′f (x) ⋅ lna3 steps : itself, multiplied by derivative of exponent, multiplied by ln of base( )
EX#1: ddx
7x = EX#2 : ddx
3x2
= EX#3 : f t( ) = t 2 ⋅5t
′f t( ) =
*Integral of ConstantVariable ax∫ dx = ax
1⋅ lna+C
3 steps : itself, divided by deriv. of exponent, divided by ln of base( )EX#4 : 9x dx∫ = EX#5 : 54 x dx∫ =
EX#6 : 8x2
dx∫ = EX#7 : x ⋅8x2
dx∫ =
*Derivative of VariableVariable y = f (x)g(x )
take ln of both sides then take derivative of both sides( )
ln y = g x( )ln f x( ) 1ydydx
= ′g x( )ln f x( ) + ′f x( )f x( ) g x( ) dy
dx= f x( )g x( ) ′g x( )ln f x( ) + ′f x( )
f x( ) g x( )⎛⎝⎜
⎞⎠⎟
EX#8 : y = xsin x EX#9 : y = 2x( )5x
*Derivative of VariableVariable y = f (x)g(x ) alternate way( )Need to change y = f x( )g x( ) to y = eln f (x)g(x)
⇒ y = eg(x)ln f (x) then take derivative.
EX : y = xsin x ⇒ y = esin x ln x
′y = esin x ln x ⋅ cos x ⋅lnx + 1x
sin x⎡⎣⎢
⎤⎦⎥= xsin x cos x ⋅lnx + sin x
x⎡⎣⎢
⎤⎦⎥
*Derivative of VariableVariable (Shortcut) 2 steps : itself, multiplied by product rule of exponent and ln of base( )
41
5-6 Inverse Trigonometric Functions: Differentiation
sinθ = 12
means give me all the answers. θ = π
6+ 2nπ
= 5π6
+ 2nπ ; n∈Z
⎧
⎨⎪⎪
⎩⎪⎪
⎫
⎬⎪⎪
⎭⎪⎪
sinθ = 12
; 0, 2π[ ] means give me only the answers from 0, 2π[ ]. θ = π6
, 5π6
⎧⎨⎩
⎫⎬⎭
arcsin 12
means give me the principal (first) answer only between −π 2
≤ x ≤ π2
. θ = π6
⎧⎨⎩
⎫⎬⎭
EX #1: Trig. Review EX #2 : Trig. Review EX #3 : Trig. Review
a) arcsin 12= b) arcsin − 1
2⎛⎝⎜
⎞⎠⎟ = a) tan arccos 2
3⎡⎣⎢
⎤⎦⎥= a) sin arccos x
3⎡⎣⎢
⎤⎦⎥=
c) arcsin 12
⎛⎝⎜
⎞⎠⎟= d) arcsin − 1
2⎛⎝⎜
⎞⎠⎟=
e) arctan 1= f ) arctan −1( ) = b) cos arccsc178
⎡⎣⎢
⎤⎦⎥= b) sec arctan x[ ] =
g) arcsin 1= h) arcsin −1( ) =i) arctan 0 = j) arctan 3 =
*Inverse Trig. Functions
y = arcsin f x( ) ′y = 1
1− f x( )( )2⋅ ′f x( ) y = arccos f x( ) ′y = −1
1− f x( )( )2⋅ ′f x( )
y = arctan f x( ) ′y = 11+ f x( )( )2 ⋅ ′f x( ) y = arccot f x( ) ′y = −1
1+ f x( )( )2 ⋅ ′f x( )
y = arcsec f x( ) ′y = 1
f x( ) f x( )( )2−1
⋅ ′f x( ) y = arccsc f x( ) ′y = −1
f x( ) f x( )( )2−1
⋅ ′f x( )
Take derivative of eachEX#1: y = arcsin 3x4 EX#2 : y = arctan 3x4 EX#3 : y = arcsec6x
EX#4 : y = arcsinex EX#5 : y = arctan ln x( ) EX#6 : y = arcsec sin x( )
42
5-7 Inverse Trigonometric Functions: Integration*Inverse Trig Functions
1a2 − x2∫ dx = arcsin x
a+C 1
a2 + x2∫ dx = 1a
arctan xa+C 1
x x2 − a2∫ dx = 1a
arcsec xa+C
OR OR OR
= −arccos xa+C = − 1
aarccot x
a+C = − 1
aarccsc x
a+C
Find variable v and constant a. The top MUST be the derivative of the variable v.
EX#1: 125 − x2
dx∫ = EX#2 : 1x2 + 9∫ dx = EX#3 : 1
x x2 −16∫ dx =
EX#4 : 1025 − 4x2∫ dx = EX#5 : 7
9x2 + 36∫ dx =
EX#6 : 12x 16x2 − 49∫ dx = EX#7 : 11
x 81x4 − 4∫ dx =
EX#8 : xx2 +100∫ dx =
43
6-1 Slope Fields and Euler's MethodDraw the slope field for each
EX #1: dydx
= x − y EX #2 : dydx
= −2x y
Here are the slope fields for the given differential equations. Sketch the solution for the given point.
EX #3 : dydx
= y8
6 − y( ) EX #4 : dydx
= x + y
Euler's Method
New y = Old y + dx ⋅ dydx
dx = change in x (the step), dydx
= derivative(slope) at the point
EX #5 : ′y = x − y passing thru 0, 1( ), step of h = 0.1 EX #6 : dydx
= xy2
passing thru 1, 4( ), step of h = 0. 5
Find f (0.3) = Find f (2) =
2
–2-2-1
2
1
10-1
2
–2-2-1
2
1
10-1
44
6-2 Differential Equations: Growth and Decay
*Growth Formula ( Can be used at any time)
y =C⋅ekt Comes from ′y = ky ( )*Compound Continuous Formula (Money equation)
A = P ⋅ert
*Double - Life Formula (Use only when doubling is mentioned)
y =C ⋅(2)td
*Half - Life Formula (Use only when half - life is mentioned)
y =C ⋅ 12
⎛⎝⎜
⎞⎠⎟th
A, y = ending amountC, P = initial amountt = timed = double-life timeh = half-life timek = growth constant
EX #1: A certain kind of algae doubles every 6 days. If the beginning population of the algae is 400, what will the population be after 2 weeks?
EX #2 : If I invest $20,000 compounded continuously for 25 years and it grows to $150,000. At what rate was the money invested?
EX #3 : SHHS population in 1980 was 1200 people. SHHS population in 2000 was 1700 people. What will the population be in 2015 at the same growth rate? (Round answer to nearest whole #)
EX #4 : Carbon has a half-life of 5730 years. We measure the amount of carbon in a tree and it has 40% less carbon than when it was planted. How old is the tree? (Round answer to nearest whole #)
45
6-3 Separation of Variables and the Logistic EquationDIFFERENTIAL EQUATIONS Separating Variables( ) (used when you are given the derivative and you need to find the original equation. We separate the x's and y's and take the integral).
EX #1: Find the general solution given
dydx
= x2 + 43y2
EX #2 : Find the particular solution of each:
a) xy dydx
− ln x = 0 y(1) = 8 b) ′y −12x3y = 0 y(0) = 11
EX #3 : AP Test 2000 BC#6 dydx
= x y −1( )2
a) Find the particular solution y = f x( ) given f 0( ) = −1 b) Draw slope field at 11 points indicated.
If the rate of growth of something is proportional to itself ′y = ky( ), then it is the growth formula y = C1e
kt( ) .Proof : ′y = ky ⇒ dy
dt= ky ⇒ dy
y= k dt ⇒ dy
y∫ = k dt∫ ⇒ ln y = kt +C ⇒
eln y = ekt+C ⇒ y = ekt ⋅eC ⇒ y = C1ekt
2
-2
-2 -1
-2
-1
1
2
210
46
Logistical Growth
dydt
= ky 1− yL
⎛⎝⎜
⎞⎠⎟ ⇒ y = L
1+ be− k t; b = L −Y0
Y0
k = constant of proportionalityL = carrying capacityY0 = initial amount
dPdt
= kP 1− PL
⎛⎝⎜
⎞⎠⎟ ⇒ P t( ) = L
1+ be− k t; b = L − P0
P0
k = constant of proportionalityL = carrying capacityP0 = initial population
EX #1: dPdt
= P5
1− P12
⎛⎝⎜
⎞⎠⎟
a) If P 0( ) = 3, what is the limt→∞
P t( )?
If P 0( ) = 20, what is the limt→∞
P t( )?
b) If P 0( ) = 3, for what value of P t( ) is the population growing the fastest?
c) Find original equation if P 0( ) = 3.
EX #2 : dPdt
= 2P − P2
5000 P 0( ) = 3000
a) limt→∞
P t( )?
b) Find P t( ).
c) At what time is the population growing the fastest?
47
CH.7 Application of Integration 7-1 Area of a Region Between Two Curves
*Area A= [top equation-bottom equation] dxa
b
∫
Area from a to c Area from a to c
A = [ f (x)− g(x)] dx + [g(x)− f (x)] dxb
c
∫a
b
∫ A = [ f (x)− h(x)] dx + [g(x)− h(x)] dxb
c
∫a
b
∫
Steps to find Area :1) Find out where the equations meet.2) Find out which equation is on top.Find the Area of each enclosed region.
EX #1: f x( ) = x2 + 2 g x( ) = − x −1, 2[ ] .
EX #2 : f x( ) = x g x( ) = 2 − x2
EX#3 : f x( ) = ex g x( ) = x + 3
g(x)f(x)
cbab ca
h(x)
g(x)f(x)
4
2
-2
- 4
- 5 5
-5
-5 -4
-4-3
-3
-2
-2-1
-1 0
54
321 54
321
2
-2
-3
-3
-2
-2-1
-1 0 321
321
4
2
g x( ) = x + 3
f x( ) = ex
48
We can find Area using vertical (top to bottom) or horizontal (right to left) cross sections.
EX#4 : Find the Area of the shaded region bounded by the x-axis and the graphs of f x( ) and g x( ).
EX #5 : . f x( ) = x −2 ; 2 , 11[ ] Find the area of the shaded region.
EX #6 : Find the Area of the region bounded by the graphs of x = 3− y2 and x = y +1.
2
5 10
2
951
h x( ) = 9 – xf x( ) = x – 1
2
5 10 11
3
2
f x( ) = x – 2
4
2
-2
- 4
- 6
- 5 5
-5
-5 -4
-4-3
-3
-2
-2-1
-1 0
54
321 54
321
49
7-2 Volume: The Disc method∗∗∗∗Volume V = A x( )∫ dx where A x( ) is the Area of the cross section.
This is the formula for all volume problems.( ) Volume if cross section rotated is a circle A = π r2( )V = π top function( )2 − bottom function( )2⎡
⎣⎤⎦
a
b
∫ dx
radius
EX #1: f x( ) = x 0, 9[ ] Find the volume of the enclosed region between f x( ) and the x-axis rotated about the x-axis.
EX #2 : f x( ) = 5x − x2 g x( ) = xFind the volume of the enclosed region rotated about the x-axis
6
4
2
-2
51 2 3
456
7-1-1
123
4 5 6
7
0
4
2
5 101 2 3
4
7 8 9-1-1
10
123
4 5 60
50
7-3 Volume: The Shell methodVolume if cross section rotated is a cylinder A = 2πrh( )
V = 2π x top function − bottom function[ ]a
b
∫ dx
radius height
EX #1: f x( ) = x −1 1, 5[ ]Find the volume of the enclosed region between f x( ) and the x-axis rotated about the y-axis
EX #2 : f x( ) = 3− x2 f x( ) = 3x2 −1 0, 1[ ]Find the volume of the enclosed region revolved about the y-axis
4
2
-2
- 4
- 6
- 5 5
-5
-5 -4
-4-3
-3
-2
-2-1
-1 0
54
321 54
321
51
Volumes: Rotations about other lines *Volume rotated about : Vertical cross section( ) the x - axis the y - axis
π f (x)( )2 − g(x)( )2⎡⎣
⎤⎦
a
b
∫ dx 2π x f (x)− g(x)[ ]a
b
∫ dx
the line y = −k the line x = − c
π f (x)+ k( )2 − g(x)+ k( )2⎡⎣
⎤⎦
a
b
∫ dx 2π x + c( ) f (x)− g(x)[ ]a
b
∫ dx
the line y = m the line x = d
π m − g(x)( )2 − m − f (x)( )2⎡⎣
⎤⎦
a
b
∫ dx 2π d − x( ) f (x)− g(x)[ ]a
b
∫ dx
EX #1 : f x( ) = 4x − x2 g x( ) = x2 0, 2[ ]Find the volume of the solid formed when rotating the enclosed region about the given lines.the x - axis the line y = − 3 the line y = 8
the y - axis the line x = −1 the line x = 4
EX #2 : f x( ) = x 0, 4[ ]Find the volume of the solid formed when rotating the enclosed region about the given lines.the x - axis the line y = − 5 the line y = 6
the y - axis the line x = −3 the line x = 4
y=-k
y=m
x=dx=-c
ba
g(x)
f(x)
52
ba
f(x)
g(x)
Volumes of known cross sections*Volume (Region is not rotated) V = A x( )∫ dx where A x( ) is the Area of the cross section.
- Sometimes we will find the volume of regions that have different cross sections (not a circle or a cylinder).- These regions are not rotated but come out at us. - We must first find the Area of the cross section, then take it's integral.
A = multiplier ⋅ top equation −bottom equation( )2
a
b
∫ dx
EX#1 : Let R be the region in the first quadrant below f x( ) and above g x( ) from x = a to x = b.Find the volume of the solid whose base is the region R and whose cross sections cut by planes perpendicular to the x-axis are : Squares A = s2( ) f x( ) − g x( ) V = f x( ) − g x( )( )2 dx
a
b
∫
f x( ) − g x( )
Equilateral Δ's A = s2 34
⎛⎝⎜
⎞⎠⎟
V = f x( )− g x( )( )2 34
⎛⎝⎜
⎞⎠⎟dx
a
b
∫ V = 34
f x( )− g x( )( )2
a
b
∫ dx
f x( )− g x( )
Semicircle A = πr2
2⎛⎝⎜
⎞⎠⎟
V =π2
f x( )−g x( )2( )2
dxa
b
∫ V = π8
f x( )− g x( )( )2
a
b
∫ dx
f x( )− g x( ) = diameter f x( )−g x( )
2= radius
Rectangle with h = 5 ⋅b A = 5bh( )
5 ⋅ f x( ) − g x( )( ) V = 5 f x( ) − g x( )( ) f x( ) − g x( )( )dxa
b
∫ V = 5 f x( ) − g x( )( )2
a
b
∫ dx
f x( ) − g x( ) Rectangle with h = 12 − x A = bh = b ⋅ 12 − x( )( )
12 − x V = f x( )− g x( )( ) 12 − x( )dxa
b
∫ f x( )− g x( )
53
Isosceles Right Triangle A = 12b ⋅h⎛
⎝⎜⎞⎠⎟
V =12⋅ f x( )− g x( )( ) ⋅ f x( )− g x( )( )dx
a
b
∫ V = 12
f x( )− g x( )( )2
a
b
∫ dx Leg is base( )
V =12⋅f x( )− g x( )
2⎛⎝⎜
⎞⎠⎟⋅f x( )− g x( )
2⎛⎝⎜
⎞⎠⎟dx
a
b
∫ V = 14
f x( )− g x( )( )2
a
b
∫ dx Hyp. is base( )
Multipliers for other figures : 30-60-90 SL( ): 32
30-60-90 LL( ): 12 3
30-60-90 HYP( ) : 38
Regular Hexagon : 3 32
Regular Polygon : # of sides4
⋅ tan half of the interior angle( )
EX#1: Let R be the region in the first quadrant under the graph of y = x − 3 for 3≤ x ≤ 7.Find the volume of the solid whose base is the region R and whose cross sections cut by planes perpendicular to the x-axis (vertical cross sections) are :
a) Squares b) Equilateral Δ's c) Semicircle
d) Rectangle with h = 5 ⋅b e) Rectangle with h = 6 − x f ) Isos.Rt. △ Leg is base( )
Find the volume of the solid whose base is the region R and whose cross sections cut by planes perpendicular to the y-axis (horizontal cross sections) are :g) Squares h) Isos.Rt.△ (Hypotenuse is base)
f(x)-g(x)
f(x)-g(x)
2
5
2
0
R73
f x( ) = x – 3
54
7-4 Arclengths and Surface of RevolutionThis formula can be used to find the length of an arc or the distance a particle travels along an arc.
∗∗∗ Let the function y = f x( ) represent the smooth curve on the interval a, b[ ]. The arclength of f between a and b is:
s = 1+ ′f x( )( )2
a
b
∫ dx
Also, for a smooth curve given x = g y( ), the arclength between c and d is:
s = 1+ ′g y( )( )2
c
d
∫ dy
EX #1: Find the arclength of the graph of y = ln cos x from x = 0 to x = π4
.
EX #2 : Set up and find the arclength of the graph of y −1( )3 = x2 from 0, 8[ ].
EX #3 : Set up and find perimeter of the shaded region.
d
c
ba
4
2
552
4
0
f x( ) = x2
4
2
5
1
5
80
55
Preview for CH.8 Procedures for fitting integrands to Basic Rules
Technique Example
Expand (numerator) 1+ ex( )2dx∫ = 1+ 2ex + e2x( )∫ dx = x + 2ex + e
2x
2+C
Separate numerator 1+ xx2 +1∫ dx = 1
x2 +1+ xx2 +1
⎛⎝⎜
⎞⎠⎟∫ dx = arctan x + 1
2ln x2 +1( ) +C
Complete the square 3x2 +8x + 41∫ dx = 3⋅ 1
x + 4( )2 + 25∫ dx = 35
arctan x + 45
+C
12x − x2∫ dx = 1
1− x −1( )2∫ dx = arcsin x −1( ) +C
Divide improper rational function x2
x2 +1∫ dx = 1− 1x2 +1
⎛⎝⎜
⎞⎠⎟∫ dx = x − arctan x +C
Add and subtract terms in the numerator 2xx2 + 2x +1∫ dx = 2x + 2
x2 + 2x +1− 2x2 + 2x +1
⎛⎝⎜
⎞⎠⎟∫ dx =
2x + 2x2 + 2x +1
− 2x +1( )2
⎛
⎝⎜⎞
⎠⎟∫ dx = ln x2 + 2x +1 + 2x +1
+C
Use trigonometric Identities tan2 x = sec2 x −1 sin2 x = 1− cos2 x cot2 x = csc2 x −1
Multiply and Divide by Pythagorean conjugate 11+ sin x∫ dx = 1
1+ sin x⎛⎝⎜
⎞⎠⎟
1− sin x1− sin x
⎛⎝⎜
⎞⎠⎟∫ dx = 1− sin x
1− sin2 x∫ dx
= 1− sin xcos2 x
dx∫ = 1cos2 x
− sin xcos2 x
⎛⎝⎜
⎞⎠⎟∫ dx
= sec2 x − sec x tan x( ) dx∫ = tan x −sec x +C
Derivatives and Integrals of Trig. Functions
Derivative:
Integral:
Trig. Function:
-ln cscx+cotx ln secx+tanx ln sinx-ln cosx
-cscx·cotx secx·tanx -csc2xsec2x-sinx
sinx
cosx
-cosx
cotxsecxcscxtanxcosxsinx
56
8-1 Basic Integration Rules*Integral top is higher or same power than bottom( ) *Two problems in One
Must divide bottom equation into top equation( ) Split top binomial into two integrals( )
EX#1: x2 + 2x + 3x2 + 3∫ dx = EX#2 : 1+ x
x2 +1∫ dx =
*Complete the square *Add and Subtract to numerator Turn problem into arctangent( ) Turn problem into natural logarithm( )
EX#3 : 2x2 −6x +13∫ dx = EX#4 : 2x
x2 −6x + 9∫ dx =
57
8-2 Integration by Parts*Integration by Parts
(used when taking an integral of a product and the products have nothing to do with each other)Always pick the function whose derivative goes away to be u. There are two special cases. Case 1: When ln x, arcsin x or arctan x is in the problem they must be u. Case 2: When neither equation goes away, either equation can be u (the equation we pick as u must be u both times) and we perform int. by parts twice and add to other side.
f x( ) ′g x( )dx =∫ f x( )g x( )− g x( ) ′f x( ) dx∫ more simply u dv =∫ uv − v du∫
EX : xex∫ dx = x ⋅ex − ex∫ dx *Special case 1 : EX : x2 ln x∫ dx = x3
3⋅ ln x − x2
3∫ dx
= xex − ex +C = x3
3⋅ ln x − x
3
9+C
u = x dv = exdx u = ln x dv = x2dx
du = dx v = ex du = 1x
v = x3
3
*Tabular method
EX : x2∫ cos x dx = x2 sin x + 2xcos x − 2sin x +C
Deriv. Integral
x2 + cos x2x − sin x 2 + − cos x 0 − sin x
EX : x∫ ⋅3x dx = x ⋅3x
ln 3− 3x
ln 3( )2 +C
Deriv. Integral
x + 3x
1 − 3x
ln 3
0 3x
ln 3( )2
*Special case 2
neither function's derivative goes away so we use integration by parts twice and add integral to the other side( )
1st timeu = ex dv = sin xdu = exdx v = −cos x2nd timeu = ex dv = cos xdu = exdx v = sin x
EX : ex∫ sin x dx = −ex + ex∫ cos x dx
= −ex cos x + ex sin x − ex∫ sin x dx
2 ex∫ sin x dx = −ex cos x + ex sin x
ex∫ sin x dx = −ex cos x + ex sin x2
+C
58
EX#1: xcos x∫ dx =
EX#2 : ln x∫ dx =
EX#3 : arctan x∫ dx =
EX#4 : x3 sin x∫ dx =
EX#5 : e2x sin 3x∫ dx =
59
8-5 Partial FractionsPartial fractions are used when you can factor the denominator.Simple Partial fractions
EquationLinear Factor 1( ) Linear Factor 2( )∫ dx = A
Linear Factor 1( )∫ + BLinear Factor 2( ) dx
(multiply by common denominator)
EX : 2x + 3x − 4( ) x +1( )∫ dx = A
x − 4( )∫ + Bx +1( ) dx 2x + 3= A x +1( ) + B x − 4( )
=11
5x − 4( )∫ +
−15
x +1( ) dx Let x = −1 1= −5B B = −1 5
= 115
ln(x − 4)− 15
ln(x +1)+C Let x = 4 11= 5A A = 115
Special Cases : (Not on AP Test)Equation
Linear Factor( )2∫ dx = ALinear Factor( )∫ + B
Linear Factor( )2 dx
(multiply by common denominator)
EX : x − 7x + 5( )3∫ dx = A
x + 5( )∫ + Bx + 5( )2 +
Cx + 5( )3 dx x − 7 = A x + 5( )2 + B x + 5( ) +C
= 0x + 5( )∫ + 1
x + 5( )2 +−12x + 5( )3 dx Let x = −5 −12 = C
= x + 5( )−2 −12 x + 5( )−3 dx∫ Let x = 0 − 7 = 25A + 5B +C⇒ 5 = 25A + 5B
= −1x + 5
+ 6x + 5( )2 +C Let x = 1 −6 = 36A + 6B +C⇒ 6 = 36A + 6B
By sub./elim. A = 0 B = 1
Equationx2 + #( ) Linear Factor( )∫ dx = Ax + B
x2 + #( )∫ + CLinear Factor( ) dx
(multiply by common denominator)
EX : 2x2 + x − 3x2 + 9( ) x −2( )∫ dx = Ax + B
x2 + 9( )∫ + Cx −2( ) dx 2x2 + x − 3 = Ax + B( ) x −2( ) +C x2 + 9( )
=
1913x + 51
13x2 + 9( )∫ +
713
x −2( ) dx Let x = 2 7 = 13C C = 713
= 1926
ln x2 + 9( ) + 1713
arctan x3
+ 713
ln x −2( ) +C Let x = 0 − 3 =−2B + 9C B = 5113
Let x = 1 0 =−A −B +10C A = 1913
60
EX#1: 1x − 3( ) x −2( )∫ dx
EX#2 : x + 2x2 + 4x + 3∫ dx
EX#3 : 5x2 + 20x + 6x3 + 2x2 + x∫ dx
EX#4 : 2x3 − 4x −8x x −1( ) x2 + 4( )∫ dx
61
8-7 Indeterminate Forms and L'Hopital's Rule Normal Forms Indeterminate Forms
00
and ∞∞
1∞ , 00, 0 ⋅∞, ∞−∞, ∞0
***If a limit is in indeterminate form, we convert it to normal form then use L'Hopital's Rule.
L'Hopital's Rule: If limx→a
f x( )g x( ) =
00
or ∞∞
, then limx→a
f x( )g x( ) = lim
x→a
′f x( )′g x( )
EX#1: limx→2
x2 + 3x −10x −2
= EX#2 : limx→0
sin xx
=
EX#3 : limx→∞
ex
x3 =
EX#4 : limx→∞
x2
ex −1=
EX#5 : limx→0
3ex − 3x − 3x2 =
EX#6 : limx→0+
x ln x =
EX#7 : limx→∞
x1x =
EX#8 : limx→3+
18x2 −9
− xx − 3
=
62
8-8 Improper Integrals
f x( ) dx =a
∞
∫b→∞lim f x( ) dx
a
b
∫ =b→∞limF x( )
a
b=
b→∞lim F b( )−F a( )
f x( ) dx = f x( ) dx + f x( )c
∞
∫−∞
c
∫−∞
∞
∫ dx where c is undefined ⇒ f x( ) dx =b→c−lim f x( ) dx +
a→c+lim f x( )
c
∞
∫−∞
c
∫−∞
∞
∫ dx
If f x( )a
b
∫ dx = finite # then the function converges. If f x( )a
b
∫ dx = ∞ then the function diverges.
EX#1: 1x1
∞
∫ dx =
EX#2 : 1x2
1
∞
∫ dx =
EX#3 : e− x0
∞
∫ dx =
EX#4 : 1x2 +10
∞
∫ dx =
EX#5 : 1x0
1
∫ dx =
63
CH.9 INFINITE SERIES 9-1 SequencesSequence : A sequence is a list of numbers, called terms, in a definite order. Sequences of objects are most commonly denoted using braces. If the limit of a sequence exists, then we say the sequence converges. Otherwise the sequence diverges.
nth term of an arithmetic sequence :an = a1 + (n −1)dRecursively defined sequence : Each subsequent term depends on the previous term.EX : ak+1 = 2(ak −1) ; a1 = 4 EX#1: ak+1 = 5ak +10 ; a1 = −2a2 = 6 a3 = 10 a4 = 18 a5 = 34 a2 = a3 = a4 = a5 =
Most famous recursively defined sequence is the Fibonacci Sequenceak+2 = ak + ak+1 ; a1 = 1, a2 = 11,1, 2, 3, 5, 8,13, 21, , , , .......
Fibonacci Petals3 petals lily, iris5 petals buttercup, wild rose, larkspur, columbine8 petals delphiniums13 petals ragwort, corn marigold, cineraria21 petals aster, black-eyed susan, chicory34 petals plantain, pytethrum55, 89 petals michelmas daisies, the asteraceae familyHumans exhibit Fibonacci characteristics. The Golden Ratio is seen in the proportions in the sections of a finger.
EX#2 : Find limn→∞
an .
a) an =3n2 + 52n2 − 7
b) an =7n −28n2 − 4
c) an =6n + 89n2 +1
d) an = 5 − 1n
EX#3 : Simplify each
a) 20!19!
b) 7!10!
c) n +1( )!n!
d) n +1( )!n + 2( )!
e) n + 3( )!n!
f ) 6n
6n+1 g) x + 2( )n+1
x + 2( )n h) xn+1 ⋅5n
xn ⋅5n+1
64
9-2 Series and ConvergenceInfinite series :
ann=1
∞
∑ = a1 + a2 + a3 + a4 + ...........+ an + .......
Definition of Convergent and Divergent series :
For the infinite series an∑ , the nth partial sum is given by Sn = a1 + a2 + a3 + a4 + ...........+ an .
If the sequence of partial sums {Sn} converges to S, then the series an∑ converges. The limit S is called the sum of the series. S = a1 + a2 + a3 + a4 + ...........+ an + .....If {Sn} diverges, then the series diverges.
Geometric series :
arnn=0
∞
∑ = a + ar + ar2 + ar3 + ..........+ arn + ....... a ≠ 0
***Geometric series Test :
A geometric series a ⋅rnn=m
∞
∑ converges iff r <1. A geometric series a ⋅rnn=m
∞
∑ diverges iff r ≥1.
If a geometric series converges, it converges to the Sum : S = a1
1− r
∗∗∗ nth term test (Used to show immediate divergence)If lim
n→∞an = 0 then it may converge. (bottom power is greater and we must proceed and use a different test,
because this test cannot prove convergence).If lim
n→∞an ≠ 0 then it diverges (top power is the same or greater than the bottom power).
EX #1: Test for Convergence
a) 32n
n=0
∞
∑ b) 32
⎛⎝⎜
⎞⎠⎟n
n=0
∞
∑ c) nn +1n=0
∞
∑
EX #2 : A ball is dropped from a height of 6 feet and begins bouncing. The height of each bounce is 34
the height of the previous bounce. Find the total vertical distance travelled by the ball.
65
***Telescoping Series are convergent Limit = 0 and the terms get smaller as we approach ∞( )Find the sum of the following telescoping series.
EX #3 : 1nn=1
∞
∑ − 1n +1
EX #4 : 1n− 1n + 2n=1
∞
∑
EX #5 : 4n(n +1)n=2
∞
∑
Repeating decimals0.222222.... 0.333333....
0.242424.... 0.833333....
66
9-3 Integral Test / p-series***Integral Test : If f is positive, continuous and decreasing for x ≥1 and an = f n( ), then
ann=1
∞
∑ and f x( )1
∞
∫ dx either both converge or both diverge.
EX #1: Use Integral Test to test for convergence
a) nn2 +1n=1
∞
∑ b) 1n2 +1n=1
∞
∑
***p -series : 1np
n=1
∞
∑1) Converges if p >12) Diverges if 0 < p ≤1
If p = 1, it is called the harmonic series. 1nn=1
∞
∑ is the divergent harmonic series
EX #2 : Use p-series to test for convergence
a) 1n2
n=1
∞
∑ b) 3nn=1
∞
∑ c) 7n
32n=1
∞
∑
EX #3 : Review (Test each series for convergence)
a) 7 910
⎛⎝⎜
⎞⎠⎟n=1
∞
∑n
b) n2 − 7n +5n=1
∞
∑
c) 6n −1n=2
∞
∑ − 6n +1
d) 117
⎛⎝⎜
⎞⎠⎟n=1
∞
∑n
67
9-4 Comparison of Series***Direct Comparison TestLet 0 ≤ an ≤ bn for all n.
1) If bnn=1
∞
∑ converges, then ann=1
∞
∑ also converges.
2) If ann=1
∞
∑ diverges, then bnn=1
∞
∑ also diverges. If ann=1
∞
∑ converges and bnn=1
∞
∑ diverges.
***Limit Comparison Test
Suppose then an > 0, bn > 0, and limx→∞
anbn
= L, where L is finite and positive. Then the two series ∑an and ∑bn
either both converge or both diverge.
Assume limn→∞
anbn
> 0 If bnn=1
∞
∑ converges, then ann=1
∞
∑ converges. If bnn=1
∞
∑ diverges, then ann=1
∞
∑ diverges
*** Limit test works well when comparing "messy" algebraic series with a p-series. In choosing an appropriate p-series, you must choose one with an nth term of the same magnitude as the nth term of the given series.
Determine Convergence and Divergence for each
EX #1: 1n −8n=1
∞
∑ EX #2 : 1n2 +14n=1
∞
∑ EX #3 : 2n
5 + 3nn=1
∞
∑
EX #4 : 12 + nn=1
∞
∑ EX #5 : n2n2 + 5n=1
∞
∑ EX #6 : n1
2
n2 +1n=1
∞
∑
Can't tell
Converges
Diverges
b na na n
b n
68
9-5 Alternating Series***Alternating Series Test
Let an > 0. The alternating series −1( )n ann=1
∞
∑ and −1( )n+1ann=1
∞
∑ converge
if the following two conditions are met.
1) limn→∞
an = 0 (the bottom is bigger than the top)
2) an+1 ≤ an for all n. (each succeeding term is getting smaller than the preceding term)
This test does not prove divergence. If the condtions are not met, usually bottom not smaller than the top, then the series diverges by nth term test.
EX #1: (−1)n+1
n=1
∞
∑ 1n
EX #2 : (−1)n+1
n=1
∞
∑ n +1n
*Alternating Series RemainderIf a convergent alternating series satisfies the condition an+1 ≤ an , then the absolute value of the remainder Rn involved in approximating the sum S by Sn is less than or equal to the first neglected term. That is,S −Sn = Rn ≤ an+1
EX #3 : Approximate the sum of the series by its first 6 terms (S6 ) (−1)n+1
n!n=1
∞
∑
EX#4 : Find the remainder (error).
69
Absolute Convergence (Alternating Series)
If the series an∑ converges, then the series an∑ also converges.
Definition ofAbsolute and Conditional Convergence
1) an∑ is absolutely convergent if an∑ converges.
2) an∑ is conditionally convergent if an∑ converges, but an∑ diverges.
Determine if each series is for absolutely convergent, conditionally convergent, or divergent
EX #1: −1( )n 1nn=1
∞
∑ EX #2 : −1( )n 1n2
n=1
∞
∑
EX #3 : −1( )n 2n +13n +1n=1
∞
∑ EX #4 : −1( )n 89
⎛⎝⎜
⎞⎠⎟n=1
∞
∑n
EX #5 : −1( )n 2n=1
∞
∑ EX #6 : −1( )n nn2 +1n=1
∞
∑
70
9-6 The Ratio and Root Test***Ratio Test : Let an∑ be a series with nonzero terms
1) an∑ converges absolutely if limn→∞
an+1
an<1
2) an∑ diverges if limn→∞
an+1
an>1 or lim
n→∞
an+1
an= ∞
3) The ratio test is inconclusive if limn→∞
an+1
an= 1
EX : −1( )n n5
2nn=1
∞
∑ limn→∞
n +1( )5
2n+1
n5
2n
= limn→∞
n +1( )5
2n+1 ⋅ 2n
n5 = limn→∞
n +1( )5
2n5 = 12
so series Converges
***Root Test : Let an∑ be a series with nonzero terms
1) an∑ converges absolutely if limx→∞
ann <1
2) an∑ diverges if limx→∞
ann >1 or = ∞
3) The root test is inconclusive if limx→∞
ann = 1
EX #1: −1( )n 2n
n!n=0
∞
∑ EX #2 : n2 ⋅2n+1
3nn=0
∞
∑
EX#3 : e2n
nnn=1
∞
∑ EX #4 : 2nn +1
⎛⎝⎜
⎞⎠⎟n
n=1
∞
∑
71
Summary of tests for Series
Test Series Converges Diverges Comment
nth-Term ann=1
∞
∑ limn→∞
an ≠ 0 This test cannot be used to show convergence.
Geometric Series arnn=0
∞
∑ r <1 r ≥1 Sum: S = a1− r
Telescoping bn − bn+1( )n=1
∞
∑ limn→∞
bn = L Sum: S = b1 − L
p-Series 1np
n=1
∞
∑ p >1 p ≤1
Alternating Series −1( ) n−1 ann=1
∞
∑ 0 < an+1 ≤ anand lim
n→∞an = 0
Remainder:RN ≤ aN+1
Integral( f is continuous,positive, anddecreasing)
an
n=1
∞
∑ ,
an = f n( ) ≥ 0
if f x( )dx converges1
∞
∫
f x( )dx is1
∞
∫ finite if f x( )dx diverges
1
∞
∫
f x( )dx = ∞1
∞
∫
Remainder:
0 < RN < f x( )dxN
∞
∫
Root ann=1
∞
∑ limn→∞
ann <1 limn→∞
ann >1 Test is inconclusive if
limn→∞
ann = 1.
Ratio ann=1
∞
∑ limn→∞
an+1an
<1 limn→∞
an+1an
>1 Test is inconclusive if
limn→∞
an+1
an= 1.
Direct comparisonan ,bn > 0( ) an
n=1
∞
∑ 0 < an ≤ bn and
bnn = 1
∞∑ converges
0 < bn ≤ an and
bnn = 1
∞∑ diverges
Either show your series
is less than a convergent
series or greater than a
divergent series.
Limit Comparisonan ,bn > 0( ) an
n=1
∞
∑ limn→∞
anbn
= L > 0
and bnn=1
∞
∑ converges
limn→∞
anbn
= L > 0
and bnn=1
∞
∑ diverges
If you know your series
is convergent then compare
your series to a convergent
series and vice versa.
72
9-7 Taylor Polynomials and ApproximationsDefinition of nth Taylor Polynomial and MacLaurin Polynomial ( c = 0 )If f has n derivatives at c, then the polynomial:
Pn x( ) = f c( ) + ′f c( )1!
x − c( ) + ′′f c( )2!
x − c( )2 +′′′f c( )3!
x − c( )3 + ......+ f n (c)n!
(x − c)n
is the nth Taylor Polynomial for f at c.
If c = 0, then
Pn x( ) = f 0( ) + ′f 0( )1!
x − 0( ) + ′′f 0( )2!
x − 0( )2 +′′′f 0( )3!
x − 0( )3 + ......+ f n (0)n!
(x − 0)n
is called the nth MacLaurin polynomial for f .
Taylor series of f x( ) → this is a power series f n c( )n!n=0
∞
∑ x − c( )n
EX #1: Find the Taylor Polynomials P4 x( ) for f x( ) = ln x centered at 1.
EX #2 : Find the fifth degree MacLaurin Series centered at c = 0( ) for f x( ) = sin x.
73
EX #3 : Find the fourth degree MacLaurin Series for f x( ) = cos x.
EX #4 : Find the fourth degree MacLaurin Series for f x( ) = ex
EX #5 : Find the fourth degree Taylor Polynomial for f x( ) = sin x centered at c = π4
.
74
9-8 Power SeriesWhen finding the interval of convergence we use either Geometric Series Test, Root Test or Ratio Test.
Find the interval of convergence, radius of convergence and the center for each example below.
EX #1: 3 x −2( )nn=0
∞
∑ EX #2 : xn
nn=1
∞
∑
interval of convergence : interval of convergence :radius of convergence : radius of convergence :center : center :
EX #3 :−1( )n x +1( )n
2nn=0
∞
∑ EX #4 : x − 8( )nn ⋅15n
n=0
∞
∑
interval of convergence : interval of convergence :radius of convergence : radius of convergence :center : center :
75
Special Cases
When finding the interval of convergence :
If limn→∞
= 0 then the power series converges from −∞, ∞( ).If lim
n→∞= ∞ then the power series converges at the center only.
EX #5 : n! x + 9( )n3nn=0
∞
∑ EX #6 : x − 7( )nn!n=0
∞
∑
limn→∞
n +1( )! x + 9( )n+1
3n+1 ⋅ 3n
n! x + 9( )n lim
n→∞
x − 7( )n+1
n +1( )! ⋅ n!x − 7( )n
limn→∞
n +1( ) x + 9( )3
= ∞ limn→∞
x − 7( )n +1
= 0
Never less than 1. Always less than 1.Converges at center −9 only. Converges always.
interval of convergence : Converges at center −9 only interval of convergence : −∞ < x < ∞
radius of convergence : 0 radius of convergence : ∞center : −9 center : 7
76
9-9 Representation of Functions by Power Seriesa1r
n
n=0
∞
∑ = a1
1− r, r <1 We will work backward and convert a1
1− r to a1r
n
n=0
∞
∑ .
There are two techniques to convert a1
1− r to a power series.
1) Rewrite function to look like a1
1− r.
2) Use Long Division
Write the power series and find the interval of convergence for each example below.
EX #1: f x( ) = 12 − x
, c = 0 EX #2 : f x( ) = 12 − x
, c = 5
EX #3 : f x( ) = 12x −5
, c = 0 EX #4 : f x( ) = 12x −5
, c = 3
77
Power Series for Elementary FunctionsFunction Interval of Convergence
1x
= 1− x −1( ) + x −1( )2 − x −1( )3 + x −1( )4 −!+ −1( )n x −1( )n +! 0 < x < 2
11+ x
= 1− x + x2 − x3 + x4 − x5 +!+ −1( )n xn +! −1< x <1
11− x
= 1+ x + x2 + x3 + x4 + x5 +!+ xn +! −1< x <1
ln x = x −1( )− x −1( )2
2+x −1( )3
3−x −1( )4
4+!+
−1( )n−1 x −1( )nn
+! 0 < x ≤ 2
ex = 1+ x + x2
2!+ x
3
3!+ x
4
4!+ x
5
5!+!+ x
n
n!+! − ∞ < x < ∞
sin x = x − x3
3!+ x
5
5!− x
7
7!+ x
9
9!−!+
−1( )n x2n+1
2n +1( )! +! − ∞ < x < ∞
cos x = 1− x2
2!+ x
4
4!− x
6
6!+ x
8
8!−!+
−1( )n x2n
2n( )! +! − ∞ < x < ∞
arctan x = x − x3
3+ x
5
5− x
7
7+ x
9
9−!+
−1( )n x2n+1
2n +1+! −1≤ x ≤1
arcsin x = x + x3
2 ⋅3+ 1⋅3x5
2 ⋅4 ⋅5+ 1⋅3⋅5x7
2 ⋅4 ⋅6 ⋅7+!+
2n( )!x2n+1
2n n!( )22n +1( )
+! −1≤ x ≤1
1+ x( )k = 1+ kx +k k −1( )x2
2!+k k −1( ) k − 2( )x3
3!+k k −1( ) k − 2( ) k − 3( )x4
4!+! −1< x <1*
*The convergence at x = ±1 depends on the value of k.
78
9-10 Taylor and MacLaurin SeriesWe will find other series by adjusting known series. We can adjust by adding/subtracting, multiplying/dividing, replacing(substituting) or taking a derivative/integral.Taylor Series to memorize :
sin x = x − x3
3!+ x
5
5!− x
7
7!+ x
9
9!− ....
cos x = 1− x2
2!+ x
4
4!− x
6
6!+ x
8
8!− ....
ex = 1+ x + x2
2!+ x
3
3!+!
11− x
= 1+ x + x2 + x3 + x4!
EX #1: Given cos x = 1− x2
2!+ x
4
4!− x
6
6!+ ....
Find each:We can multiply: xcos x = x4 cos x =
We can substitute: cos2x = cos x3 =
We can differentiate/integrate: −sin x = sin x =
We can combo:sin 3x =
EX #2 : Given 11+ x
= 1− x + x2 − x3 +! Find arctan x Hint: ddx
arctan x = 11+ x2
⎛⎝⎜
⎞⎠⎟
79
EX #3 : Given ex = 1+ x + x2
2!+ x
3
3!+! Find e− x
2
0
1
∫ dx
EX #4 : Find sin2 x (sin2 x = 1− cos2x2
)
Formulas for sign changes
−1( )n n+1( )
2 = + − − + + − − + +
−1( )n+1( ) n+2( )
2 = − − + + − − + +
−1( )n+2( ) n+3( )
2 = − + + − − + + − −
−1( )n+3( ) n+4( )
2 = + + − − + + − −
80
10-2 Plane Curves and Parametric EquationsParametric equations : equations in terms of a third variable (usually t or θ )
EX #1: Graph x = t 2 − 4 y = t2
, −2 ≤ t ≤ 3 and show direction.
dxdt
is the change in x over time.
dydt
is the change in y over time.
dydx
is the change y over change in x. (slope)
EX #2 : Graph x = 3t 2 − 7t y = e4 t Find dxdt
and dydt
and dydx
at t = 2.
-5
-5 -4
-4-3
-3
-2
-2-1
-1 0
54
321 54
321
3210-1-2
yxt
81
10-3 Parametric EquationsIf a smooth curve c is given by the equation, x = f t( ) and y = g t( ), then the slope of c at x, y( ) is
Slope = dydx
=dydt
dxdt
, dxdt
≠ 0 Also, d2ydx2 = d
dxdydx
⎛⎝⎜
⎞⎠⎟ =
ddt
dydx
⎛⎝⎜
⎞⎠⎟
dxdt
d3ydx3 =
ddx
d 2ydx2
⎛⎝⎜
⎞⎠⎟=
ddt
d 2ydx2
⎛⎝⎜
⎞⎠⎟
dxdt
speed = dxdt
⎛⎝⎜
⎞⎠⎟
2
+ dydt
⎛⎝⎜
⎞⎠⎟
2
or ′x t( )( )2+ ′y t( )( )2
EX #1: x = t y = 14t 2 − 4( ) t ≥ 0 Find slope, concavity, and speed at 2, 3( ).
∗∗∗Arclength in parametric form :
If a smooth curve c is given by x = f t( ) and y = g t( ) such that c does not intersect on the interval a ≤ t ≤ b(except possibly at the endpoints) then the arclength of c over the interval is given by
s = dxdt
⎛⎝⎜
⎞⎠⎟
2
+ dydt
⎛⎝⎜
⎞⎠⎟
2
a
b
∫ dt or s = ′f t( )( )2+ ′g t( )( )2
a
b
∫ dt
∗∗∗Reminder : Arclength in function form :
s = 1+ ′f x( )( )2
a
b
∫ dx
EX #2 : Find arclength of x = 2sin t and y = 2cos t 0, 2π[ ]
82
Using Parametric Equations to solve Rectilinear Motion problemsEX #3 : 1997 #1 t = 0 to t = 6 x t( ) = 3cos πt( ) y t( ) = 5sin πt( )a) Find the position of the particle when t = 2.5. b) Sketch from t = 0 to t = 6.
c) The # of times it passes thru (0, 5) d) Find velocity vector
e) Distance travelled from t = 1.25 to t = 1.75.
EX #2 : 2000 #4 Position at t = 1 is (2, 6) Velocity vector at any time t > 0 is given by 1− 1t 2 , 2 + 1
t 2⎛⎝⎜
⎞⎠⎟
a) Find acceleration vector at t = 3. b) Find position at t = 3.
c) When is slope = 8 d) limt→∞
dydx
=
EX #3 : 2001#1 dxdt
= cos t 3( ), dydt = 3sin t 2( ), 0 ≤ t ≤ 3 At t = 2 position is (4, 5).
a) Equation of tangent line at (4, 5) b) Find speed at t = 2
c) Distance travelled from 0 ≤ t ≤1 d) position at t = 3.
83
120°= 2π3
0°,360°=0,2π
90°= π2 60°= π
345°= π
4
315°= 7π4
330°= 11π6
300°= 5π3
240°= 4π3
270°= 3π2
180°=π
225°= 5π4
210°= 7π6
150°= 5π6
135°= 3π4
30°= π6
10-4 Polar Coordinates and Polar GraphsEquations to convert rectangular to polar and polar to rectangular polar to rectangular rectangular to polar
Rectangular: (x, y) x = r cosθ x2 + y2 = r2
Polar: (r, θ ) y = r sinθ θ = tan−1 yx
⎛⎝⎜
⎞⎠⎟
EX #1: Convert each
a) 5, 6( ) R → P b) − 3, 7π6
⎛⎝⎜
⎞⎠⎟ P → R
EX #2 : Graph r = 1+ 2cosθ
!!
r θ!!!!!!!!!! 0!!!!!!!!!! π
6!!!!!!!!!! π
4!!!!!!!!!! π
3!!!!!!!!!! π
2!!!!!!!!!! 2π
3!!!!!!!!!! 3π
4!!!!!!!!!! 5π
6!!!!!!!!!! π!!!!!!!!!! 7π
6!!!!!!!!!! 5π
4!!!!!!!!!! 4π
3!!!!!!!!!! 3π
2!!!!!!!!!! 5π
3!!!!!!!!!! 7π
4!!!!!!!!!! 11π
6!!!!!!!!!! 2π
θ
ry
x
84
10-5 Area and Arclength in Polar Coordinates***Area in Polar Coordinates
If f is continuous and nonnegative on the interval α , β[ ], then the area of the region bounded by the
graphs of r = f θ( ) between the radial lines θ =α and θ = β is given by
A = 12
f θ( )⎡⎣ ⎤⎦2
α
β
∫ dθ
∗∗∗A = 12
r2
α
β
∫ dθ
EX #1: r = 3cos 3θ Find area of one leaf.
EX #2 : r = 1−2sinθ
Area of the Inner Loop : Area of the whole figure : Area of the Outer Loop :
EX #3 : r = 4 + 4 cosθ r = 2 Find area of the enclosed region.
Arclength of a polar curve (Not on AP Test)
s = f θ( )( )2+ ′f θ( )( )2
α
β
∫ dθ or s = r2 + drdθ
⎛⎝⎜
⎞⎠⎟
2
a
b
∫ dθ
2
2
2 2
6
4
2
2
4
6
5 10
g θ( ) = 4 + 4·cos θ( )
f θ( ) = 2
85
Hooke's LawHooke's Law: The Force required to compress or stretch a spring (within its elastic limits) isproportional to the distance d that the spring is compressed from its original length.
F = kd
W = kxa
b
∫ dx
EX#1: A force of 750 lbs compresses a spring 3 inches from its natural length of 15 inches. Find the work done in compressing the spring an additional 3 inches.
EX#2 : A force of 10 lbs is required to stretch a spring 4 inches beyond its natural length. Assuming1997 MC( ) Hooke's Law applies, how much work is done in stretching the spring from its natural length to 6 inches beyond its natural length.
86
Finding Area Using Limits Area = lim
n→∞ f a + b −a
ni⎛
⎝⎜⎞⎠⎟
height! "## $##i=1
n
∑ b −an
⎛⎝⎜
⎞⎠⎟
width!"# $#
i = interval, n = # of subdivisions
Area = b −an
⎛⎝⎜
⎞⎠⎟ f a + b −a
n⋅1⎛
⎝⎜⎞⎠⎟ + f a + b −a
n⋅2⎛
⎝⎜⎞⎠⎟ + f a + b −a
n⋅3⎛
⎝⎜⎞⎠⎟ + ......... f a + b −a
n⋅n⎛
⎝⎜⎞⎠⎟
⎡⎣⎢
⎤⎦⎥
Summation Formulas
1) ci=1
n
∑ = cn
2) ii=1
n
∑ =n n +1( )
2
3) i2
i=1
n
∑ =n n +1( ) 2n +1( )
6
4) i3
i=1
n
∑ =n2 n +1( )2
4Summation Properties
5) ai ± bi( )i=1
n
∑ = aii=1
n
∑ ± bii=1
n
∑
6) kaii=1
n
∑ = k aii=1
n
∑ , where k is a constant
EX#1: f x( ) = x3 0, 1[ ] n subdivisions EX#2 : f x( ) = x2 1, 3[ ] n subdivisions
EX#3 : 1988 MC #41
limn→∞
1n
1n+ 2
n+ ...... n
n⎡
⎣⎢
⎤
⎦⎥ =
A) 12
1x0
1
∫ dx B) x0
1
∫ dx C) x0
1
∫ dx D) x1
2
∫ dx E) 2 x x0
1
∫ dx
87
Lagrange Error Bound
Error = f x( )−Pn x( ) ≤ Rn x( ) where Rn x( ) ≤ f n+1 z( ) x −c( )n+1
n +1( )!f n+1 z( ) is the MAXIMUM of n +1( )th derivative of the function
EX : Find the 4th degree Maclaurin Series for cos x.f x( ) = cos x f (0) = 1′f x( ) = −sin x ′f (0) = 0
′′f x( ) = −cos x ′′f (0) = −1 cos x = 1− x2
2!+ x
4
4!.....
′′′f x( ) = sin x ′′′f (0) = 0
f 4 x( ) = cos x f (4 )(0) = 1
EX : Use Maclaurin Series to approximate value of cos 0.1( ).
P4 .1( ) ! 1− (0.1)2
2!+ (0.1)4
4!! 0.9950041667 Actual value of cos 0.1( ) = 0.9950041653
EX : Lagrange Error Bound at 0.1f 5 x( ) = −sin x The MAXIMUM of −sin x is 1( )
cos x −P4 x( ) ≤ 1( ) 0.1− 0( )5
5!= 0. 000000083 Actual gap: 0.0000000014
EX #1: f 1( ) = 2, ′f 1( ) = 5, ′′f 1( ) = 7, ′′′f 1( ) = 12a) Write a 2nd Degree Taylor Polynomial
b) Use Taylor Polynomial to approximate 1.1
c) Lagrange Error Bound at 1.1
88
Interpreting the graph of ′f (x)
Looking at the graph of ′f (x) is different than looking at f x( ). ′f (x) = 0 (x-int) is where the possible max. and min. are.
Remember a graph is :
increasing when ′f x( ) > 0decreasing when ′f x( ) < 0rel.max. occurs when ′f x( ) switches from + to − .rel.min. occurs when ′f x( ) switches from − to + .
The ′′f x( ) is the slope of the tangent line of ′f x( ).concave up when slope is positive ′′f x( ) > 0.( )concave down when slope is negative ′′f x( ) < 0( ).inf. pts. occur when slopes switch from + to − or − to + peaks and valleys of ′f x( )( ).
EX : Graph of ′f x( ) from − 6 ≤ x ≤ 6
f x( ) is increasing from − 4, 0( ) and 3, 6( ) because ′f x( ) > 0
f x( ) is decreasing from −6, − 4( ) and 0, 3( ) because ′f x( ) < 0
f x( ) has a rel.max. at x = 0 because ′f x( ) switches from + to − .f x( ) has a rel.min. at x = − 4 and x = 3 because ′f x( ) switches from − to + .
The ′′f x( ) is the slope of the tangent line of ′f x( ).f x( ) is concave up from −6, −2( ), 2, 4( ) and 5, 6( ) because ′f x( ) is increasing slope is positive( ).f x( ) is concave down from −2, 2( ) and 4, 5( ) because ′f x( ) is decreasing slope is negative( ).x = −2, 2, 4, 5 are inf. pts. because ′f x( ) switches from + to − and vice versa.
slopes switch from + to − or − to + ( ) peaks and valleys of ′f x( )( ).
6
4
2
-2
- 4
- 5 5
Graph of f '(x)
-6 -5
5
5 6
-4-3-2-1
-4 -3 -2 -1
432
1
43210
89
-2
-1
2
1
876543210-1-2-3-4
Graph of f'(x)
2π
42
1.5
Integral as an accumulatorA definite integral finds the change in the equation above it. The integral of velocity from a to b is the change in position (distance travelled) from a to b.The integral of acceleration from 0 to 3 is the change in velocity from time 0 to time 3.The integral of ′f (x) is the change in f (x).
EX#1: If f 0( ) = 5 then find f 1( ).
Since ′f (x) dx0
1
∫ = f 1( )− f 0( ) it finds the change in f from 0 to 1 .
Since the area under ′f from 0 to 1 = 2 this will help find f 1( ).
f 1( ) = f 0( ) + ′f (x) dx0
1
∫ = 5 + 2 = 7
EX#2 : Integrals going left are negative. Integrals going right are positive.
If f 0( ) = 10 then:
f − 4( ) = 10 + ′f x( ) dx0
-4
∫ = 10 + 2 −1.5 = 10.5 f −2( ) = 10 + ′f x( ) dx0
-2
∫ = 10 + 2 = 12
f 4( ) = 10 + ′f x( ) dx0
4
∫ = 10 − 4 = 6 f 8( ) = 10 + ′f x( ) dx0
8
∫ = 10 − 4 + 2π = 6 + 2π
EX#3 : Given v 2( ) = 8 and a t( ) = sin t2 +1( ) find v 5( ).v 2( ) = 8
v 5( ) = 8 + a t( )2
5
∫ dt = 8 + −0.02336 = 7.97663 (The integral of acceleration finds the change in velocity)
4
2
f '
2
4
3
2
1
0 1
90
4
2
- 2
5 10
Graph of f(x)
1 2 3
4
7 8 9
-2-3
-110
123
4 5 60
10
8
6
4
2
- 2
5 10
velocity of runner in meters per second 1 2 3
456
7 8 9
10
10
123
4 5 6
789
0
10
8
6
4
2
- 2
5 10
7015
velocity of runner in meters per second 1 2 3
456
7 8 9
10
10
123
4 5 6
789
0
Finding Derivatives and Integrals given a graph of f(x)
The derivative is the slope of each line.
′f 1( ) = 32 ′f 7( ) = DNE
′f 2( ) = DNE ′f 8( ) = 2′f 3( ) = 0 ′f 9( ) = DNE′f 5( ) = −2 ′f 9.5( ) = 0
The integral finds the total area between f(x) and the x - axis.
f x( )0
10
∫ dx = 17 − 2 + 3= 18 f x( )10
8
∫ dx = − 3
f x( )0
10
∫ dx = 17 + 2 + 3= 22 f x( )8
2
∫ dx = 2 −12 = −10
f x( ) + 5( )0
10
∫ dx = 68 f x( )6
10
∫ dx = 1
EX :Find the velocity of the runner at t = 2 and t = 7 seconds.
v 2( ) = 103⋅2 = 20
3 v 7( ) = 10
Find the acceleration of the runner at t = 2 and t = 7 secondsSince ′v t( ) = a t( ), you find acceleration by finding the derivative (slope) of velocity.
a 2( ) = 103
a 7( ) = 0
Find the distance travelled by the runner from t = 0 and t = 10 seconds
Distance travelled = v t( )0
10
∫ dt v t( )0
10
∫ dt = 85
EX : Given x(0) = 45 find x(3) and x(10).
x(3) = 45 + v(t)0
3
∫ dt =
45 +15 = 60
x(10) = 45 + v(t)0
10
∫ dt =
45 + 85 = 130
4
2
-2
5 10
Graph of f(x)
3
2
4
8
2
3
1 2 3
4
7 8 9-1
-2-3
-110
123
4 5 60
91
Approximations /Charts
EX#1 : The rate of snow fall, in cubic feet per hour, recorded over a 48 hour period is given by a twice differentiable function S of time t. The table of selected values of S(t) for the time interval 0 ≤ t ≤ 48 hours, are shown above.a) Use data from the table to find an approximation for ′S (21). Show the computations that lead to your answer. Indicate units of measure.
b) Approximate the value of S(t)0
48
∫ dt using a midpoint Riemann sum with the four subintervals indicated by the data
in the table. Using correct units explain the meaning of S(t)0
48
∫ dt .
c) Approximate the value of 124
S(t)0
24
∫ dt using the trapezoidal rule with the four subintervals indicated by the data
in the table. Using correct units explain the meaning of 124
S(t)0
24
∫ dt .
d) Find ′S (t)0
48
∫ dt . Indicate units of measure. e) For 0 ≤ t ≤ 48 what is the fewest number of times that ′S (t) = 0.
Give a reason for your answer.
23 31 36 34 22 30 37 38 29
36 42 483024181260
S t( )ft3/hr( )
thours( )
92
Rectilinear Motion (Position, Velocity, Acceleration Problems) ∗We designate position as x t( ) (moving along the x-axis), y t( ) (moving along the y-axis), or s t( ).∗The derivative of position, ′x t( ) = v t( )⇒ velocity.∗The derivative of velocity, ′v t( ) = a t( )⇒ acceleration.
∗A particle is at rest or is changing direction when v t( ) = 0.
∗A particle is moving to the right or up when v t( ) > 0 and to the left or down when v t( ) < 0.
∗The integral of velocity finds the position equation; v t( )dt∫ = x t( ) +C
∗The integral of velocity from a to b , v t( )dta
b
∫ , finds the change in position from a to b (displacement).
∗Distance travelled is v t( )a
b
∫ dt
∗The integral of acceleration finds the velocity equation; a t( )dt∫ = v t( ) +C
∗The integral of acceleration from a to b , a t( )a
b
∫ dt , finds the change in velocity from a to b.
∗Average velocity of a particle ⇒ 1b − a
v(t)dta
b
∫ or Average acceleration of a particle ⇒ 1b − a
a(t)dta
b
∫
∗To find the maximum or minimum acceleration of a particle set ′a t( ) = 0.
∗To find the maximum or minimum velocity of a particle set ′v t( ) = 0 a t( ) = 0( ).∗To find absolute maximum or minimum velocity on an interval we must check the endpoints and the critical pts.
∗To find when velocity increases find ′v t( ) > 0 a t( ) > 0( ) or when velocity decreases find ′v t( ) < 0 a t( ) < 0( ).
∗Speed is the absolute value of velocity, v t( ) .
If v t( ) and a t( ) agree (both positive or both negative) v t( ) a t( )+ +− −
⎫
⎬⎪
⎭⎪
then speed is increasing.
If v t( ) and a t( ) disagree (one positive or both negative)v t( ) a t( )+ −− +
⎫
⎬⎪
⎭⎪
then speed is decreasing.
Derivatives Integrals
A
V
P
93
Continuity / Differentiability ProblemWe must show a function is continuous before we discuss its differentiability.If a function is not continuous, then it cannot be differentiable.To show continuity, we must show that lim
x→a−f x( ) = lim
x→a+f x( ) = f a( ).
To show differentiability, we must show that limx→a−
f x( ) = limx→a+
f x( ) = f a( ) and that ′f a( ) exists.
EX : f x( ) = x2 , x < 36x − 9 , x ≥ 3
⎧⎨⎪
⎩⎪ ′f x( ) = 2x , x < 3
6 , x ≥ 3⎧⎨⎪
⎩⎪
At 3 f 3( ) = 9 lim
x→3−f x( ) = 9 lim
x→3−′f x( ) = 6
limx→3+
f x( ) = 9 limx→3−
′f x( ) = 6
Therefore f (x) is continuous at x = 3. Therefore f (x) is differentiable at x = 3.
f (x) is continuous iff f 3( ) = limx→3−
f x( ) = limx→3+
f x( )f (x) is differentiable iff lim
x→3−′f x( ) = lim
x→3−′f x( ) and the function is continuous.
Determine the continuity and differentiability of each.EX#1: EX#2 :
f x( ) = 2x3 , x < −1
− 3x2 +1 , x ≥ −1
⎧⎨⎪
⎩⎪ f x( ) = x3 −1 , x ≥ 2
27 −5x2 , x < 2
⎧⎨⎪
⎩⎪
EX#3 :
To be continuous a graph can't have any holes or asymptotes.To be differentiable a graph can't have any holes or asymptotes or hard points.The graph is continuous at the points a and d.The graph is differentiable at the point a only. (d is a hard point)
dcba
94
8-3 Trigonometric Integralssinm x ⋅cosn x dx∫
If n is odd, set aside cos x and change everything to sine. Substitute u = sin x, use cos2 x = 1−sin2 x.If m is odd, set aside sin x and change everything to cosine. Substitute u = cos x, use sin2 x = 1−cos2 x.If m and n are both even, reduce so that either m or n are odd using trig. double angle identities.
secm x ⋅ tann x dx∫If m is even, set aside sec2 x and change everything to tangent.If m is odd, set aside sec x tan x and change everything to secant.
Identities we should know :Double angle identities Reduction Formulas Pythagorean Identities
sin2x = 2sin xcos x sin2 x = 1−cos2x2
sin2 x = 1−cos2 x / cos2 x = 1−sin2 x
cos2x = cos2 x −sin2 x cos2 x = 1+cos2x2
1+ tan2 x = sec2 x / sec2 x −1= tan2 x
1+ cot2 x = csc2 x / csc2 x −1= cot2 x
EX#1: sin3 x ⋅cos4 x dx∫
EX#2 : sin6 x ⋅cos3 x dx∫
95
EX#3 : cos2 5x dx∫
EX#4 : sec4 x ⋅ tan2 x dx∫
EX#5 : sec3 x ⋅ tan3 x dx∫
EX#6 : sec4 x ⋅ tan5 x dx∫
96
Useful trigonometric integrals
From trigonometry half angles: sin x2
⎛⎝⎜
⎞⎠⎟ =
1−cos x2
cos x2
⎛⎝⎜
⎞⎠⎟ =
1+cos x2
sin x = 1−cos2x2
cos x = 1+cos2x2
sin2 x = 1−cos2x2
cos2 x = 1+cos2x2
cos x∫ dx = sin x +C cos2 x∫ dx = 1+cos2x2
dx =∫12x + sin2x
4+C
sin x∫ dx = −cos x +C sin2 x∫ dx = 1−cos2x2
dx =∫12x − sin2x
4+C
tan x∫ dx = sin xcos x
dx =∫ − ln cos x +C tan2 x∫ dx = sec2 x −1dx =∫ tan x − x +C
cot x∫ dx = cos xsin x
dx =∫ ln sin x +C cot2 x∫ dx = csc2 x −1dx =∫ −cot x − x +C
sec x∫ dx = ln sec x + tan x +C sec2 x∫ dx = tan x +C
csc x∫ dx = − ln csc x + cot x +C csc2 x∫ dx = −cot x +C
Useful trigonometric derivatives
Reminder : ddx
sin x = cos x ddx
cos x = −sin x ddx
tan x = sec2 x
ddx
csc x = −csc x ⋅cot x ddx
sec x = sec x ⋅ tan x ddx
cot x = −csc2 x
Useful trigonometric identities
Reminder : sin2 x + cos2 x = 1 ⇒ sin2 x = 1−cos2 x ⇒ cos2 x = 1−sin2 x
1+ tan2 x = sec2 x ⇒ sec2 x −1= tan2 x
1+ cot2 x = csc2 x ⇒ csc2 x −1=cot2 x
97
8-4 Trigonometric SubstitutionTrigonometric Substitutions
∗If the integrand contains a2 − x2 then substitute x = asinu dx = acosu du
EX : a2 − x2 dx∫ = a2 − a2 sin2 u( ) ⋅acosu du∫ = a2 1− sin2 u( ) ⋅acosu du cos2 x = 1−sin2 x( )∫= a2 cos2 u ⋅acosu du∫ = acosu ⋅acosu du∫ = a2 cos2 u ⋅ du∫ = a2 cos2 u du∫
∗If the integrand contains a2 + x2 then substitute x = a tanu dx = asec2 u du
EX : a2 + x2 dx∫ = a2 + a2 tan2 u( ) ⋅asec2 u du∫ = a2 1+ tan2 u( ) ⋅asec2 u du 1+ tan2θ = sec2θ( )∫= a2 sec2 u ⋅asec2 u du∫ = asecu ⋅asec2 u du∫ = a2 sec3u ⋅ du∫ = a2 sec3u du∫
∗If the integrand contains x2 − a2 then substitute x = asecu dx = asecu tanu du
EX : x2 − a2 dx∫ = a2 sec2 u − a2( ) ⋅asecu tanu du∫ = a2 sec2 u −1( ) ⋅asecu tanu du sec2θ −1= tan2θ( )∫= a2 tan2 u ⋅asecu tanu du∫ = a tanu ⋅asecu tanu du∫ = a2 secu tan2 u du∫ = a2 secu tan2 u du∫
EX#1: dxx2 −9∫ =
EX#2 : dx4x2 +1∫ =
98
EX#3 : x16 − x2∫ dx =
EX#4 : dx
x2 +1( )32∫ =
EX#5 :25 − x2
x∫ dx =
