CALCULUS II
MATH 2414
Turnell
2018
[Type the
document
subtitle]
Review of Cal I
Trig Derivatives:
d
[sin u ] 
dx
d
[cos u ] 
dx
d
[tan u ] 
dx
d
[sec u ] 
dx
d
[cot u ] 
dx
d
[csc u ] 
dx
Exponential and Natural Log Functions:
f ( x)  e x
d u
du
[e ]  e u
dx
dx
2x
and
f ( x)  ln(sin( x))
d n
[ x ]  nx n 1
dx
4
3
f ( x)  x  3 x  6 x 2  x
Power Rule:
Product Rule:
d
[uv]  uv  vu
dx
f ( x)  x3 cos x
Quotient Rule:
d
1 du
u ( x )
or
[ln u ] 
dx
u dx
u
d  u  vu  uv
d  T  BT   TB


OR
2
dx  v 
v
dx  B 
B2
x3  1
f ( x)  2
x 2
Page 1
d
u n  nu n1u
dx
Chain Rule:
f ( x)   3 x 2  5 x  1
4
f ( x)  sin 3 (5 x)
f ( x)  tan 5

3x 2  1

Page 2
Implicit Differentiation:
dy
, given x 3  4 x 2 y 4  3 y 5  9
Find
dx
dy
, given cos( xy )  y 3  4
dx
Find
Integrals:
 f ( x)dx = a set of antiderivatives
Why a set?
Review of Basic Forms
u n1
n
 u du  n  1  C n  1
 k du  k u  C
 sin(u ) du  ___________________
 csc(u ) cot(u ) du  ___________________
 cos(u ) du  ___________________
 sec (u ) du  ___________________
 sec(u ) tan(u ) du  ___________________
 csc (u ) du  ___________________
u
 e du  ___________________
 1 du  ___________________

u
2
2
1
 3
3
  x  3 x  5 dx
x 

 x(5x
Page 3
2
 4)3 dx
 6x2
dx

3
5
 (4 x  9)
 cos x dx
 3
 sin x

 x sin(6 x )dx
tan x sec2 xdx
2
 x dx
 2
 x 1
x
 2 xe dx
2
Page 4
Calculus I Review Worksheet
Find the derivative ( y or
dy
) for each function.
dx
1. y  (x 2  4x  5)2/3
2. y  tan(2 x)
3. y  cos(5 x)
6. f ( x) 
cos(3x)
x  sin(3x)
4. y  sin 2 (3x)
7. x 3  2x 2 y 2  5y 3  10
8. x 2 y  3tan y  4
5. y  x 2 tan x
9. y  cos(sin(x 3 ))
10. y  (2x  3)4 (3x  2)5
Find the indefinite or definite integral:
1.
 x(4 x
2
11. 

sin x
dx
 1  cos x
 dx
12. 
4
  x  3
 3) dx
4
1 
 5
4 3



x
x
x
2

 dx

x3 
3.  sec(5 x ) tan(5 x ) dx
2.
 e4 x  e2 x  1
dx
x
e

13. 
4. 

dx
 7x  2
 sin x dx
5. 
 cos3 x
6.  sin 6xdx
14.
15.
16.
 dx
 1  3x
2
8.  x cos( x ) dx
7. 
9.
17.
 ex
18. 
dx
x
 e 1
2

x
1
3
 1 x
4
6
3
1 x
 xe dx
10. 
3
 sin x cos xdx
2
 tan(2 x)sec (2 x)dx
2
 4 x 1  2 x dx
4
 sec ( x)sec( x) tan( x)dx
dx
Page 5
Calculus I Review Worksheet-Answers
2(2 x  4)
1. y 
3( x 2  4 x  5)
1
dy 4 xy 2  3 x 2
7.

dx 15 y 2  4 x 2 y
3
2. y  2 sec (2 x )
2
8.
3. y  5sin(5 x )
9. y  3 x 2 cos( x 3 )sin(sin( x 3 ))
4. y  6sin(3 x ) cos(3 x )
10.
5. y   x sec ( x )  2 x tan( x )
2
6. y 
2
y  (3 x  2) 4 (2 x  3)3 (54 x  29)
( x  sin(3 x )) 2
5
4
2
1  3x  C
3
1
8.
sin( x 2 )  C
2
1 1 x 2
9.  e
C
2
2
3
4
10. (1  x ) 3  C
8
11.  ln 1  cos x  C
1
C
12.
3( x  3)3
7. 
3
2.
x
2x
3x
1


 2 C
6
5
4
2x
3.
1
sec(5 x )  C
5
1
ln 7 x  2  C
7
1
5.
C
2cos 2 x
1
6.  cos(6 x)  C
6
4.
y  15(3 x  2) 4 (2 x  3) 4  8(2 x  3)3 (3 x  2)5
3sin(3 x )( x  sin(3 x ))  cos(3 x )(1  3cos(3 x ))
(4 x 2  3)5
1.
C
40
6
dy
2 xy
 2
dx x  3sec2 y
13.
14.
15.
16.
17.
1 3x
e  e x  e x  C
3
sin 7 ( x)
C
7
1
(tan 2 (2 x))  C
4
3
2
2 2
C
1  2x
3
sec5 ( x)
C
5


e3  1
18. ln e  1  ln e  1  ln
e 1
3
Page 6
Area of Region Between Two Curves
Review from Cal 1: The area under a curve.
1.
Write the integral or integrals to find the area of each region:
2.
3.
Area Between 2 Curves:
4.
Page 7
5. With respect to x
7.
6. With respect to y
1
Page 8
Examples:
Sketch a graph of the region bounded by the given equations and find the area of the region:
y   x 2  3x  1
1.
y  x 1
2.
f ( y) 
y
16  y 2
, g ( y )  0,
y3
(Use Desmos.com for graph)
Larson 11th ed: p. 450: 7,15,17,19,23,24,35,37
Page 9
Page 10
Finding Distances
Example 1: y  4  x . Find the distance from the y-axis and x-axis in terms of x and y.
.
2
Example 2: x   y  4 y .
2
Example 3:
x2
to the line:
Find the distance from y  3 
2
y  1 in terms of x and y.
Find the distance from the y-axis in terms of x and y.
Example 4: Find the distance from y  3 
to the line: y  1 in terms of x and y.
x2
2
Example 5: Find the distance from y  2 x
to the line: x  2 terms of x and y.
Page 11
2
Example 6: Find the distance from y  2 x , x  0
Example 7: Find the distance from y  2 x , x  0
to the line: x  3 terms of x and y.
to the line: y  8 terms of x and y.
Example 7: Find the indicated distances:
Example 8: Find the indicated distances:
2
2
2
x  ( y  2) 2 , y  x, y  6, y  1
y  ( x  2)  3
Page 12
Volume: The Disk Method
Review from Geometry: The volume of a cylinder
V   r 2h
Determine the Volume of a Solid of Revolution:
So, for the purposes of the derivation of the formula, let’s look at rotating the continuous function
y  f ( x) in the interval [a, b] about the x-axis. Below is a sketch of a function and the solid of
revolution we get by rotating the function about the x-axis.
Short animation: https://youtu.be/i4L5XoUBD_Q
The volume of the disk (a cylinder) is given by: V   r x . Approximating the volume of the solid by n
disks of width x with radius r ( x) , produces:
2
Volume of solid 
n
  [r ( x )] x
2
i
i 1
This approximation appears to become better and better as x  0 (n  ) . Therefore,
Volume of solid  lim 
x 0
b
n
[r ( x )] x    [r ( x)] dx
i 1
Page 13
2
2
i
a
As seen in animation, we can rotate functions around the y-axis:
The radius is now a function of y.
d

Volume of solid   [ r ( y )] dy
2
c
Note: the radius is ALWAYS perpendicular to axis of rotation.
Example 1: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region
2
about the x-axis. y  4  x
Go to http://www.shodor.org/interactivate/activities/FunctionRevolution/ to see revolution.
Example 2: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region
2
about the y-axis. y  4  x
Page 14
Example 3: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region
2
about the y-axis. x   y  4 y (p.461: 12)
Revolving about a line that is NOT the x or y axis.
Example 4: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region
x2
about the line y  1 : y  3 
2
Example 5: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region
2
about the line x  2 : y  2 x (p.461: 14d)
Page 15
Page 16
Volume: The Washer Method
The disk method can be extended to cover solids of revolution with holes by replacing the
representative disk with a representative washer.
Volume of the washer   ( R  r ) w
2
2
Where R  the outer radius and r  inner radius
If rotated around a horizontal axis, then
b

Volume of solid   ([ R ( x)]  [r ( x)] )dx
2
2
a
If rotated around a vertical axis, then
d

Volume of solid   ([ R ( y )]  [r ( y )] )dy
2
2
c
Example 1: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region
bounded by the graphs: y  x 2 , y 
(answer is
x , about x-axis.
3
)
10
Page 17
Example 2: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region
bounded by the graphs: y  2 x 2 , y  0 , x  2 about the line y  8 . (p.461: 14d)
Example 3: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region
bounded by the graphs: y  2 x 2 , y  0 , x  2 about the y-axis. (p.461: 14c)
For the following problems, (a) Find the outer radius, R, and the inner radius, r, (b) find the limits of
integration, and (c) set up the integral that gives the volume of the solid bounded by the given functions.
For problems 1-3, also find the volume.
Homework
1. y  x 2 , y  x , about the y-axis.
x  1.
y  x , y  x , about line y  3 .
y  ( x  1) 2  1 , y  1 , x  0 , x  1 about the x-axis.
y  ( x  1) 2  1 , y  1 , x  0 , x  1 about line x  1.
y  ( x  1) 2  1 , y  1 , x  0 , x  1 about line x  2 .
2. y  x 2 , y  x , about line
3.
4.
5.
6.
2
7.2 p. 461: 5-15 odd, 19, 21, 29, 31
Page 18
Volume: The Shell Method
Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded by
2
the graphs: y  4 x  x , y  0 , about y-axis.
Perhaps, we need an alternate method. The shell method will help us out of this messy algebra.
In the shell method, a strip that is parallel to the axis of rotation is taken so that when it is
rotated, it forms a cylindrical shell. Animation: https://youtu.be/JrRniVSW9tg
Volume of the shell  2 rh * thickness . Where r  the distance from the axis of rotation.
Note: The representative rectangle will ALWAYS be parallel to the axis of rotation.
Page 19
There are two general formulas for finding the volume by the shell method.
Vertical Axis
Horizontal axis
b
d


V  2 r ( x )  h( x ) dx
V  2 r ( y )  h( y ) dy
a
c
Example 1: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region
2
bounded by the graphs: y  4 x  x , y  0 , about y-axis.
Answer is
128
3
Example 2: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region
2
bounded by the graphs: x  ( y  2) , y  x , about the line y  1 .
Answer is
Page 20
63
2
Example 3: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region
bounded by the graphs: y  x  2 , y  x , y  0 about the x-axis. (p.470: 22)
Answer is
16
3
Example 4: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region
bounded by the graphs: y  x , y  x , y  0 about the y-axis. Use the disk method and then use the shell
method. Which method is easier? Did you get the same result?
2
3
11th ed. P. 470: 3,5,9,11,13,16,17,23,25
Page 21
Page 22
Arc Length
Finding the length of a curve:
Demo: http://demonstrations.wolfram.com/ArcLengthApproximation/
Examples: Find the length of the curve over the given interval.
1. y  2 x
3
2
 3, [0,9]
Page 23
x3 1
 , [1,2]
2. y 
6 2x
x5
1
 3 , [2,5]
3. y 
10 6 x
Page 24
Surface Area
Find the surface area obtained by revolving a given curve about an indicated axis.
Recall from geometry how to find the lateral surface area of a cylinder.
So, for the purposes of the derivation of the formula, let’s look
at rotating the continuous function y  f ( x ) in the interval [a, b] about the x-axis. Below is a
sketch of a function and the solid of revolution we get by rotating the function about the x-axis.
Now, rotate the approximations about the
x-axis and we get the following solid.
The approximation on each interval gives
a distinct portion of the solid. Each of
these portions is called frustums and the
surface area a frustum is given by:
SA  Circumference  ArcLength
If we make the distances between the
points small, then we can say the surface
area is
SA  (Circumference)s .
Page 25
As s  0 , then we have the differential: dSA  (Circumference) ds
ds
SA   (circumference)( ArcLength)
r
SA   2 rds
2
 dx 
 dy 
ds  1    or ds  1   
 dx 
 dy 
http://demonstrations.wolfram.com/SurfaceAreaOfASolidOfRevolution/
Review of algebra
Simplify the following
x2
2  x 1
2 x
x 1
 x   _______
1
2
3
1
x2 3
Page 26
2
Examples
http://www.shodor.org/interactivate/activities/FunctionRevolution/
Find the surface area obtained by revolving the given curve about the indicated axis
TWO ways—one using dx and one using dy:
y  2 x , 1  x  4 about the x-axis
Page 27
Find the surface area obtained by revolving the given curve about the indicated axis:
x  y 2  1, 1  y  4 about the x-axis
Page 28
Find the surface area obtained by revolving the given curve about the indicated axis
TWO ways—one using dx and one using dy:
x  2 y  1,
5
 y  1 about the y-axis
8
Page 29
Homework problems:
P. 481: 7,9,11,39,43,45,47 (Hint: refer to notes on arc length)and the following:
Find the arc length of the graph of the function over the indicated interval. (Exact values!)
x3 1
1. y 
 , 1 x  4
12 x
3
1
x 2
2. y 
 x 2, 1 x  9
3
1
x4
3. y 
 2 , 1 x  2
4 8x
Set up the simplified definite integral that would calculate the area of the surface generated by
revolving the curve about the given axis. Evaluate the integral on #4 and #5 (Exact values!)
4. y 
2 x  x2 ,
1
3
 x  about the x-axis
2
2
y3
5. x 
, 0  y  1 about the y-axis
3
6. x  2 4  y ,
7. y  1  x ,
2
0 y
15
about the y-axis
4
0  x  1 about the y-axis
8. x  1  2 y , 1  y  2 about the x-axis
2
9. x 
3
1 2
( y  2) 2 , 1  y  2 about the x-axis
3
Page 30
Physics: Work Done by a Force
Definition:
If an object is moved a distance, d, in the direction of an applied constant force, F,
then the work, W, done by the force is defined as W  Fd .
However, most forces are not constant (varies-compressing a spring) and will depend upon
where exactly the force is acting. If the force is given by F ( x) , then the work done by the force
along the x-axis from x  a to x  b is given by,
b
W   F ( x)dx
a
Hooke's Law:
The force F required to compress or stretch a spring is proportional to the distance d that
the spring is compressed or stretched form its original length. That is,
where k is the spring constant.
F  kd
Body Weight:
The weight of a body varies inversely as the square of its distance from the center of the
Earth. The force F ( x) exerted by gravity is
F ( x) 
C
x2
where the radius of the Earth is approximately 4000 miles
Example 1: A force of 250 newtons stretches a spring 30 centimeters. How much work is done in stretching
the spring from 20 centimeters to 50 centimeters?
Page 31
Example 2: Seven and one-half foot-pounds of work is required to compress a spring 2 inches from its natural
length. Find the work required to compress the spring an additional one-half inch.
Example 3: A lunar module weighs 12 tons on the surface of Earth. How much work is done in propelling the
module from the surface of the moon to a height of 50 miles? Consider the radius of the moon to be 1100 miles
and its force of gravity to be one-sixth that of the Earth.
Larson 11th ed. P. 491: 9-17 odd
Page 32
Moments and Center of Mass
Archimedes used "moving power" to describe the effect of a lever in moving a mass on the other
end. He is famous for the quote " Give me a lever long enough and a fulcrum on which to place it, and I
shall move the world." We will define this idea as "moment"
We will look at moments about
A point—1 dimension
A line—2 dimensions
A plane—3 dimensions
Moment = (mass) * (distance)
I. One-Dimensional System
The measure of the tendency of this system to rotate about the origin is the moment
about the origin. M 0  m1 x1  m2 x2     mn xn
M 0 = moment about the origin. When M 0 = 0, the system is said to be in equilibrium.
Center of Mass: point where the system is in equilibrium, denoted by x
x
Mo
, where m  m1  m2      mn is the total mass of the system.
m
Example 1: Find the moment about the origin and center of mass of a system of four objects with
masses 10 g, 45 g, 32 g, and 24 g that are located at the points −4, 1, 3, and 8, respectively, on
the x-axis.
Example 2: If a 20 pound child and a 60 pound child sit on the ends of a seesaw that is 5 feet long, where must
the fulcrum be located so that the beam will balance?
Page 33
II. Two-Dimensional System
Let the point masses m1 , m2 ,. . . , mn be located at ( x1 , y1 ),( x2 , y2 ), . . . ,( xn , yn ) .
1. The moment about the y-axis is M y  m1 x1  m2 x2      mn xn .
2. The moment about the x-axis is M x  m1 y1  m2 y2      mn yn .
3. The center of mass ( x , y ) (or center of gravity) is
x
My
and y 
m
where m  m1  m2      mn
Mx
m
is the total mass of the system.
Example 3: Find the center of mass of the given system of point masses.
mi
( xi , yi )
3
(2, 3)
4
(5,5)
2
(7,1)
1
(0,0)
6
(3,0)
III. Three-Dimensional System-Using geometry
Mass = (density)*(Area) or Mass = (ρ)*(Area).
A planar lamina (a thin plate) with uniform density ρ. If we
take the density to equal 1, then the mass is numerically equal
to the area.
Example 4: Find the center of mass of the region shown.
Page 34
III. Three-Dimensional System-Irregular Laminas
1
x  x and y  ( f ( x)  g ( x))
2
Representative Thin Strip (rectangle):
1. Area of representative rectangle:
Full Region
1. Area of region:
b
A   ( f ( x)  g ( x))dx
A  ( f ( x)  g ( x))x (Height*width)
a
2. The moment about the y-axis is
2. The moment about the y-axis is
b
M y   x ( f ( x )  g ( x ))dx .
M y  x A  x( f ( x)  g ( x)) x
a
3. The moment about the x-axis is
3. The moment about the x-axis is
1
M x  y A  ( f ( x)  g ( x ))( f ( x)  g ( x)) x
2
1
M x  y A  ([ f ( x )]2  [ g ( x )]2 )x
2
b
Mx 
1
([ f ( x )]2  [ g ( x )]2 )dx .

2a
4. The center of mass ( x , y ) (or center
of gravity) is
4. The center of mass ( x , y ) (or center of gravity)
is x  x and y 
1
( f ( x)  g ( x))
2
x
Example 5: Find the center of mass of a lamina of uniform density
parabola
y  4  x 2 and y  0 . Let   1 (Hint: symmetry)
Page 35
My
A
and y 
Mx
A
 covering the region bounded by the
Example 6: Find the center of mass of a lamina of uniform density
parabola
y  x 4 and y  x . Let   1
Page 36
 covering the region bounded by the
Example 7: Find the center of mass of a lamina of uniform density
parabola x  y  2 and
x  y 2 . Let   1
11th ed. P. 502: 7,9,13,17-27 odd, 31, 21
Page 37
 covering the region bounded by the
Page 38
Review of Basic Integration Rules
Review Basic Inverse Trig Derivatives:
d
u
d
u
(arcsin u ) 
(arctan u ) 
dx
dx
1  u2
1  u2
d
u
(arccos u ) 
dx
1  u2
d
(arcsec u ) 
dx
u
d
(arccsc u ) 
dx
u
u
d
(arccot u ) 
dx
1  u2
Review Basic Inverse Trig Integration:
 du

 2
 a  u2
du



 u u2  a2
 du 
 2
 a  u2
Find the indefinite integral:
1

dx

2
 2  9x




1
dx

 x 4x2  9
Page 39
1
25  36 x 2
dx
u
u2 1
u
u2  1
1
x

dx
 4
 x  16
 e cos x
dx


2
 1 x



 sin x
dx

2
 1  cos x

1
dx
x (1  x)

Completing the Square
x 2  6 x  10
2
2x  x 2
Page 40






1
8  6x  x2
dx
6
3  4x  4x2
dx
 2x  5
dx
 2
 x  2x  2
Page 41
Improper Fractions:
The degree of the numerator is greater than or equal to the degree of the denominator. You must
DIVIDE to change to a "mixed" fraction before you integrate.
 x 1
dx
 2
 x 1
4
 x
dx
 2
 x 1
2
 4x  7
dx

 2x  3
2
(Hint: Remember 
 du  ln u  C
1
u
Separating Fractions:
x
x x
x
dx   2  dx
 2
1
 x 1
x
4
You CANNOT separate denominators.
 2  8x
dx

2
 1  4x
Page 42
4
Review Basic Trig Integration:
 sin(u ) du   cos(u )  C
 cos(u ) du  sin(u )  C
 sec (u ) du  tan(u )  C
2
 csc(u ) cot(u ) du   csc(u )  C
 sec(u ) tan(u ) du  sec(u )  C
 csc (u ) du   cot(u )  C
2
What about
 tan(u ) du
 cot(u ) du
 sec(u ) du
Multiplying by a Form of One:
 sec(u ) du
 csc(u ) du
dx


 1  cos x
dx


 1  csc x
Page 43
 csc(u ) du
Trig Identities:
cos 2   sin 2   1
1  tan 2   sec 2 
sin 2  2 sin  cos 
cos 2  cos 2   sin 2 
cot   1  csc 
2
2
 2 cos 2   1
 1  2 sin 2 
 (csc x  tan x) dx
2
Review Other Bases Integration:
a
u
8
du 
Log Rule:
What about
 ln(u ) du ?
 cot( x )[ln(sin x)] dx
"Math Magic Tricks"
 1 dx

 1  ex
Page 44
x
dx
p.558 – Basic Integration Formulas (Thomas’ CALCULUS Media Upgrade 11th edition)
Basic Substitutions
1
 16 x dx
1. 

 8x2  1
9.
17.
3.
0
 3 sin x cos x dx
2
2 x e x dx

e
19.
tan v
2
 e sec v dv
1
3
dx


2
 4  9x
0
csc(e  1) d
11.
2
25.
21.
x 1
 3 dx
23. 

0
6 dx
x 25 x 2  1
 2
29. 

30.
 1  x4
1  9x2
dx

33.  x
x
 e e
w
dw
 (tan x)[ln(cos x)]dx
 dx
3
 ln x
36. 

x


dx
35. 
 x cos(ln x)
1
2
Completing the Square
2

37. 

8 dx
x  2x  2
2

39. 


1

dx
x  4x  3
2
dx
41. 

 ( x  1) x  2 x
2
Trigonometric Identities
43.  (sec x  cot x) 2 dx
Improper Fractions
47.
x
 x  1 dx
49.
3
2
2 x3
dx
x2 1
51.
4 x3  x 2  16 x
dx

x2  4
61.
 1  sec x dx
Separating Fractions

 1 x
dx
53. 

 1  x2
4
 1  sin x
55. 
dx
2
 cos x
0
Multiplying by a Form of 1
57.
1
 1  sin x dx
59.
1
 sec   tan  d
Trigonometric Powers
83.  cos 3  d (Hint: cos 2   1  sin 2  )
Page 45
1

x 1
2 w
 2 x dx
dx

dx
 csc( x   ) dx
15.
6

27. 


t
x
 sec 3 dt




13.
e
31.
7. 
0
 cot(3  7 x)dx
ln 2

 16 x dx
5.  2
 8x  2
Answers
[ln(cos x)]2
C
2
31. 6sec 1 5x  C
1. 2 8 x 2  1  C
30.
3
3. 2(sin x ) 2  C
5. ln10  ln 2  ln 5
1 x
33. tan (e )  C
7. 2 ln( x  1)  C
35. ln(2  3)
1
7
9.  ln sin(3  7 x )  C

2
36. (ln x)  C or

11.  ln csc(e  1)  cot(e  1)  C
13. 3ln
t
sec
3
 tan
t
3
37. 2
1
39. sin ( x  2)  C
C
41. sec 1 x  1  C
15.  ln csc( x   )  cot( x   )  C
43. tan x  2 ln csc x  cot x  cot x  x  C
17. eln 2  e0  2  1  1
47. x  ln x  1  C
19. e tan v  C
21.
49. 7  ln8
3( x1)
C
ln 3
 x
2
2
1
51. 2 x  x  2 tan    C
2 w
C
ln 2

25.
18

27.
18
1 2
29. sin ( x )  C
53. sin 1 x  1  x 2  C
23.
55. 2
57. tan x  sec x  C
59. ln 1  sin   C
61. x  cot x  csc x  C
83. sin   sin 3   C
1
3
Page 46
Integration by Parts
If you are given a function
product rule and get:
y  f ( x) g ( x) and asked to take the derivative, you would use the
d
 f ( x) g ( x)   f ( x) g ' ( x)  g ( x) f ' ( x)
dx
'
'
If we let u  f ( x) and v  g ( x) , then du  f ( x)dx and dv  g ( x ) dx
We can write the product rule as:
d
(uv)  udv  vdu
dx
In terms of indefinite integrals, this equation becomes:
 d  uv dx  udv  vdu 


 dx
We can separate the right hand side:
 d  uv dx  udv  vdu



 dx
We can rearrange and get:
 d  uv dx  vdu  udv



 dx
We can replace and get:
uv   vdu   udv
Integration by Parts Formula:
 udv  uv   vdu
 5 x dx
 2x
e
Page 47
x
5
ln(3 x ) dx
 ln xdx
 4 x sec (2 x)dx
2
 4arccos xdx
Page 48
x
2
sin xdx
Tabular Method
Use this method if one part can be "derived" down to zero and other part can be integrated
repeatedly easily. Useful for: powers of x multiplied to e's, sines or cosines
x e
3 3x
x
3
dx
cos(2 x)dx

5
*** x ln(3 x ) dx Can you use the tabular method for this problem?
Page 49
Are we going in circles?
4x
 e cos(2 x)dx
Thomas 11th ed. P. 529: 15-33 odd, 43, 45, 49, 53, 55
Page 50
Trigonometric Integrals
Pythagorean Identities:
cos 2   sin 2   1
sec2   1  tan 2 
cos   1  sin 
2
2
tan 2   sec 2   1
sin 2   1  cos 2 
cos 2  2 cos 2   1
1  cos 2
cos 2  
2
cos 2  1  2 sin 2 
1  cos 2
sin 2  
2
Trig Review:
d
[sin x ]  cos x
dx
d
[cos x ]   sin x
dx
d
[tan x ]  sec 2 x
dx
d
[sec x ]  sec x tan x
dx
 sec( x) dx  ln sec( x )  tan( x)  C
sin 2  2sin  cos
 cos( x) dx  sin( x)  C
 sin( x) dx   cos( x)  C
 sec ( x) dx  tan( x)  C
2
 sec( x) tan( x) dx  sec( x)  C
 tan( x) dx   ln cos x  C
Powers of Sines and Cosines
 sin ( x) cos( x)dx
 sin
 sin

3
7
( x ) cos( x )dx
Conclusion:
Page 51
4
( x ) cos( x )dx
sin( x ) cos( x )dx
or ln sec x  C
 (1  sin (4 x)  sin
3
5
(4 x )) cos(4 x )dx
 cos ( x)sin( x)dx
3
Conclusion:
 (cos
2
(2 x )  cos 3 (2 x ))sin(2 x )dx
 sin
2
( x ) cos3 ( x )dx
 sin
5
( x ) cos 4 ( x )dx
Page 52
Even Powers of Sines and Cosines
 cos
 sin
2
2
 sin
( x )dx
2
(4 x )dx
( x ) cos 2 ( x )dx
POWERS OF TANGENT AND SECANT
6
 sec x(sec x tan x)dx
 sec
Conclusion:
Page 53
2
x tan 3 xdx
 tan
2
( x )sec 4 ( x )dx
 tan ( x)sec ( x)dx
3
 tan
5
3
 sec ( x)dx
3
( x )dx
11th ed. P. 538: 3, 4, 7, 11, 13, 23, 27, 29, 31, 37, 59
Page 54
Trig Substitution
Trig substitutions occur when we replace the variable of integration by a trig function. The most
common substitutions are:
sin  
u
u
u
tan  
sec 
a
a
a
Or
a sin   u a tan   u a sec  u
3

dx
1. 
 1  9 x2
2.

1  9x 2 dx
Page 55

3. 

y 2  25
dy
y3
4

dx
4. 
 x2 16  x 2
Page 56
 25 x 2  4
5. 
dx
x4


1
6. 
dx
3
2
 ( x  5) 2
11th ed. P. 547: 3, 5, 7, 9, 13, 19-29 odd, 35, 37, 41
Page 57
Page 58
Partial Fractions
g ( x)
where g ( x ) and h( x) are polynomials and the degree of h( x) is
h( x )
BIGGER than the degree of g ( x ) (if this is not true, you must first do long division), we need to
Suppose f ( x) 
decompose the rational function into simpler rational functions.
In order to integrate a partial fraction problem, you must first find the Partial Fraction
Decomposition:
Case I: h( x) is a product of linear factors, none repeating, then the Partial Fraction
Decomposition has the form:
x 1
A
B


( x  2)(2 x  11) x  2 2 x  11
Watch this video: Partial Fractions-Part 1
Case II: h( x) is a product of linear factors, some repeating, then the Partial Fraction
Decomposition has the form:
x 1
A
B
C



2
( x  1)( x  2)
x  1 x  2 ( x  2)2
Watch this video: Partial Fractions-Part 2
Case III: h( x) contains irreducible quadratic factors, none repeating, then the Partial
Fraction Decomposition has the form:
x 1
A
Bx  C


( x  2)( x 2  1) x  2 x 2  1
Case IV: h( x) contains irreducible quadratic factors, some repeating, then the Partial
Fraction Decomposition has the form:
x5
A
Bx  C Dx  E

 2

2
2
( x  2)( x  4)
x  2 x  4 ( x 2  4)2
Watch this video for Case III and IV: Partial Fractions-Part 3
Once you find A, B, etc., then you integrate the result.
Page 59
Integrate the following:
 x3
dx
1.  3
 2 x  8x

2.  2
 2x
2x
dx
 7x  5
Page 60
 x 3
dx
3.  2
 x  x2
3

1
4. 
dx
2
 ( x  1) ( x  4)
Page 61
 3x 2  4 x  5
5. 
dx
2
 ( x  1)( x  1)
 8x2  8x  2
dx
6. 
 4x2  1 2




1
11th ed. P. 557:3-17 odd, 21, 23, and 
2 dx

2
 x x 1

Page 62

L'Hôpital's Rule
Review of Limits
2x  3
lim
x2 x  4
x 2  16
lim
x4 x  4
lim
x 
4 x2  5x
1  3x 2
Indeterminate Forms


0
0
lim
0
1
0
00
2cos  x  2
x2
lim
2

2x  3
x2 x2
x 4
4
L'Hôpital's Rule
Suppose that we have one of the following cases:
f ( x) 0

x a g ( x)
0
lim
OR
f ( x) 

x a g ( x)

lim
where a can be any real number, infinity or negative infinity.
In these cases we have:
f ( x)
f ( x)
 lim
x  a g ( x ) x  a g ( x)
lim
Find the limit:
1. lim
x 
ln x 4
x3
cos x
2. lim
x 0 1  tan x
Page 63
3. lim
x2
x  e x

 1 

 x 
4. lim  x tan 
x  
1 

6. lim 1  2 
x  
x 
5.
 10 3 
lim   2 
x 0  x
x 
x
1
7. lim x x 1
x 1
11th ed. P. 369: 15, 19, 21, 23, 25, 29, 33, 43, 45, 49, 50, 51, 59
Page 64
Improper Integrals
Type 1
Limits of integration that involve
f ( x)
f ( x)
f ( x)
is continuous on
is continuous on
is continuous on


c
a
a
[ a ,  ) :  f ( x)dx  lim
 f ( x)dx
c 
b
b

c
(  , b ] :  f ( x)dx  lim
 f ( x)dx
c 
(  ,  ) :

c



 f ( x)dx   f ( x)dx   f ( x)dx
c
c
a
b
c
 lim
 f ( x)dx  lim
 f ( x)dx
b 
a 
If the limit exists, then the integral converges .
If the limit does not exist, then the integral diverges.


1. 

1
0

2. 



1
dx
2
x
1
dx
3 x
Page 65


4
3. 
 16 

x2
dx

 (1  x )
dx
4. 
x
 e
1
Type 2
f ( x)
Vertical asymptotes within the given limits of integration. (Function is undefined)
has an infinite discontinuity at x  c , c  ( a , b )
c
b
 f ( x)dx  lim  f ( x)dx
b c 
a
b
a
b
 f ( x)dx  lim  f ( x)dx
a c 
c
a
b
c
b
a
a
c
 f ( x)dx   f ( x)dx   f ( x)dx
Page 66
2

5. 

0
5

6. 


3
1
dx
x2
1
dx
x2  9
1
1
1
dx   x 2 dx
2
1
x
1

7. 

1
1

 (1  1)  2
x
1
Why is this incorrect?
11th ed. P 579: 17,19, 21,23,33-41 odd,45
Page 67
Page 68
Parametric Equations
In this section, we present a new concept which allows us to use functions to study
curves which, when plotted in the xy-plane, neither represent y as a function of x
nor x as a function of y.
We can define the x-coordinate of P as a function of t and the y-coordinate of P as a
(usually, but not necessarily) different function of t. (Traditionally, f(t) is used for x
and g(t) is used for y.) The independent variable t in this case is called a parameter
 x  f (t )
is called a system of parametric equations.
and the system of equations 
y

g
(
t
)

For the following, sketch the curve, indicate its orientation, and eliminate the
parameter to get an equation involving just x and y.
x  t2  3
Ex. 1 
 y  2t  1
Page 69
 x  t 3 2  1
Ex. 2 
 y  t
 x  2cos t
Ex. 3 
 y  3sin t
11th ed. P. 707: 5-15 odd, 23-29 odd
Page 70
Parametric Equations and Calculus
Parametric Form of the Derivative
If a smooth curve C is given by the equations:
then the slope of C at
( x, y )
x  f (t )
and
y  g (t ) ,
is
dy
dx
dy
0
 dt ,
dt
dx dx
dt
d  dy 
d y dt  dx 

2
dx
dx
dt
2
The second derivative is
Tangent Lines
There is a horizontal tangent line if
There is a vertical tangent line if
dy
0
dt
dx
0
dt
and
and
dx
0
dt
dy
0
dt
Examples:
Find
dy
dx
and
d2y
.
dx 2
Find the slope and concavity (if possible) at the given value of the parameter.
Find the ( x, y ) coordinate at given value of t.
1.
x  t,
y  3t  1, t  1
Page 71
2.
x  2sin(2 t ),
y  cos(2 t ), t   1
6
Find all points (if any) of horizontal and vertical tangency to the curve.
3.
x  t 4  2,
y  t3  t
4.
Page 72
x  8sin t ,
y  4 cos t
5. Find the equation of the tangent line(s) to the given set of parametric equations at the
5
3
2
given value of the parameter. x  t  4t , y  t when t  2
Arc Length
Definition: If a curve C is defined parametrically by x  f (t ) and y  g (t ) ,
a  t  b , where f  and g  are continuous and not simultaneously zero on [a, b] ,
and C is traversed exactly once as t increases from t  a to t  b , then the length
of C is the definite integral:
b
b
L   [ f (t )]  [ g (t )] dt ,
2
a
2
  dx 2  dy 2
L        dt
  dt   dt 

a
Find the arc length of the curve on the given interval:
6.
x  6t 2 ,
y  2t 3 , 1  t  4
Page 73
Area of a Surface of Revolution
If a smooth curve x  f (t ) and y  g (t ) , a  t  b , is traversed exactly once as t
increases from t  a to t  b , then areas of the surfaces generated by revolving the
curve about the coordinate axes are as follows:
1. Revolution about the x-axis ( y  0) :
b
2
2. Revolution about the y-axis ( x  0) :
b
2
2
2

 dx   dy 
S   2 x      dt

 dt   dt 


 dx   dy 
S   2 y      dt

 dt   dt 

a
a
Find the area of the surface generated by revolving the curve about the given axis:
7.
t2
x ,
2
8.
x  cos3  ,
y  2t , 0  t  5
about the x-axis
y  sin 3  , 0   

2
about the y-axis
11th ed. P. 715:9,11,13,15,33,35,37,39,49,55,63,65 and Worksheet (Parametric-Tangents)
Page 74
Parametric—Tangents
Find all vertical and horizontal points of tangency.
1.
2.
3.
4.
x  cos t  t sin t
, 0  t  2
y  sin t  t cos t
x 1 t
y  t 3  3t
,
x  t2  t  2
y  t 3  3t
  t  
,   t  
x  4  2cos t
, 0  t  2
y  1  sin t
Find the equation of the tangent line(s) to the given set of parametric equations at the
given value of the parameter.
5. x  2cos(3t )  4sin(3t ),
6.
x  t 2  2t  11,
7.
x  t 3  cos( t ),
8.
x  t 2  2t  1,
y  3tan(6t ) when t 

2
y  t (t  4)3  3t 2 (t  4) 2  7
when t  2
y  4t  sin(2t  6) when t  3
y  t 3  7t 2  8t
when t  1
Page 75
Solutions:
1. Horizontal:
Vertical:
t 
( 1,  )

 
 ,1
2 
t
t
2. Horizontal:
2
3
2
5. y  3 x  12
 3

  , 1 
 2

6. y  24 x  65
t  1 (2,2)
t  1 (0, 2)
Vertical: None
3. Horizontal:
Vertical:
4. Horizontal:
t  1 (4,2)
t  1 (2, 2)
t
t
2
52
x
9
9
8. y 
25
7
x
4
2
1  3 11 
 ,

2 4 8 

2
3
t
2
Vertical:
7. y 
(4,0)
(4, 2)
t  0 (6, 1)
t   (2, 1)
Page 76
Polar Graphs
How to plot polar coordinates: (r, )
The origin is now called the pole.
r
a directed distance from O to P.
  directed angle, counterclockwise from polar axis to OP
Special Polar Graphs
Circles: r  a
r 3
r  2a cos
r  2a sin 
r  2cos
r  6sin 
Use Graphing Calculator
For TI-83: MODE is Pol, Deg. or Rad,
WINDOW must correspond with Deg. or Rad.
Graph the following with calculator: r  3  3cos , , r  3  4sin  , r  5  3sin 
Page 77
Limaçons r  a  b cos
r  1  3cos
r  a  b sin 
r  2  2sin 
r  3  2cos
r  4  2sin 
Limacon: https://youtu.be/MT1j6vdX2aI
Lemniscate (a figure 8) : r 2  a 2 cos(2 )
r 2  a 2 sin(2 )
r 2  25cos(2 )
r 2  16sin(2 )
Rose Curves: r  a cos(n )
r  3cos(3 )
r  a sin(n )
Rose curve: https://youtu.be/tvPsjmcN2Yg
r  3cos(2 )
r  3sin(5 )
Page 78
r  3sin(2 )
Slope and Tangent Lines
Given that
1. Find
dy
dx
x  r cos  , y  r sin and r  f ( ) , find
dy
.
dx
and the slopes of the tangent lines at the given values of
r  1  sin  ,   0 and   
Page 79
.
2. Determine the equation of the tangent line to r  3  8sin  at
Tangent Lines
There is a horizontal tangent line if
There is a vertical tangent line if
dy
0
d
dx
0
d
and
and
Page 80
dx
0
d
dy
0
d


6
3. Find all the points of vertical and horizontal tangency: r  1  sin 
https://www.youtube.com/watch?v=4SpQ9jOxd_E
4. Review quadratic formula. Solve: 4 cos
2
  cos   2  0 within interval [0, 2 )
Page 81
Tangents at the Pole
There is a tangent line at the pole if
f ( )  0 and f ( )  0 , the line is:   
4. Find all the tangents at the pole: r  2 cos3
Page 82
10.4 11th ed. P. 727: 63, 64, 65, 67, 71, 73, 75, 77 AND
A. Find all the points of vertical and horizontal tangency: r  1  cos 
B. Determine the equation of the tangent line to r  sin(4 ) cos  at
C. Determine the equation of the tangent line to r   sin(3 ) at

D. Determine the equation of the tangent line to r  1  2cos  at


6

2


4
E. Find all the points of vertical and horizontal tangency: r  2  4 cos 
(you will need to use the quadratic formula)
Answers:
1. Horizontal:
Vertical:
2.
y
 3    3 5 
( r, )   ,  ,  , 
2 3 2 3 
 1 2
( r, )   2,0  ,  ,
2 3
1
3 3
x
  1 4 
, ,

 2 3 
1
4
2
y
4.

1
1 2  1 2
y  (1  2 2)  x 

7
2
2



5. Bonus
Page 83
x

3.
2
Page 84
Area and Arc Length of Polar Curves
If we have a circle of radius r, and select a sector of angle  ,
where      , then the area of that sector is
  
A   r 2  

 2 
Now suppose that we have a curve in polar coordintes. As we
did with rectangular coordinates, we take small increments of
 , which we will denote as  .
Since  is infinitesimally small, the wedge shaped has an area of dA 
1 2
r d
2
If we partition the interval [ ,  ] into n equal subintervals:
  0  1   2  3   4    n 1   n   .
Then approximate the area of the region by the sum of the area of the sectors.
n
1
Then the approximate area is: A    [ f (i )]2
i 1 2
1 n
[ f (i )]2
n  2 i 1
A  lim
Taking the limit as n   ,

1 
[ f ( )]2 d
2 

1
Area=  r 2 d
2
Page 85
Example #1: Find the area inside the curve: r  1  cos 
Example #2: Find area of one petal of rose curve: r  3sin 2
Page 86
Example#3: Find the area shared by r  2(1  sin  ) and r  2
Example#4: Area outside r  1  cos  and inside r  1
Page 87
Example#5: Set-up the integral to find the area shared by r  2(1  sin  and r  1
Example#6: Find the area of the inner loop of r  2  4 cos  .
Page 88
Arc Length in Polar Form
Consider a polar arc: r  f ( ),
     and
f ( ) is continuous on interval      .
'
Remember in parametric form the
b
  dx 2  dy 2
Arc Length   
    dt
dt

  dt 

a
Now, we are in polar coordinates.
We change the variable t into  :
b
  dx 2  dy 2
Arc Length   
 
 d
d
d







a
We can simplify this by remembering that: x  r cos  and y  r sin 
Page 89
Example: Find the length of the curve over the given interval: r  1  cos
0  
10.5 11th ed. p.735: 7,9,15,19, 21, 23, 39,41, 43, and Find the area of the following regions.
A. Inside r  6cos 2 and outside r  3
B. Area shared by r  2  2 cos  and r  2
Answer: 3 3  
Answer: 5  8
C. Inside r  4 cos  and to the right of r  sec 
Answer:
2
D. Find the length of the curve over the given interval: r  
Page 90
0    5 Answer:
2
E. Find the length of the curve over the given interval: r  cos
8
 3
3
3

3
0  

4
Answer:
19
3

8

3
8
Sequences
Sequence:
Definition: A sequence is a function whose domain (n) is the set of positive integers . In
other words, it is a list where order matters.
Terms: a1 , a2 , a3 , a4 , .. . an
an is called the nth term of the sequence and is denoted by an 
Example 1: Write out the
1st
 2n  1 
four terms of the sequence: an    n 
 2 
Find lim an 
n 
If a sequence an  agrees with a function f at every positive integer, and if f ( x)
approaches a limit L as x   , then the sequence must converge to the same limit L .
If lim an does not exist, then an diverges.
x 
Theorem:
Let L be a real number. Let f be a function of a real variable such that lim f ( x )  L .
x 
If an  is a sequence such that f ( n)  an for every positive integer n, then lim an  L
n
In other words: lim f ( x )  lim an Why is this good?
n 
x 
Do the sequences converge or diverge? If the sequence converges, find its limit.


Ex. 2: an  ( 1)  1 
n
1

n
Ex. 3: an 
Page 91
2n  1
1 3 n
Ex. 4: an 
n2
2n  1
Theorem:
If lim an  0 , then lim an  0
n 
n
 1
Ex. 5: an    
 2
1n1
Ex. 6: an 
n
n
(look at graph)
(look at graph)
Monotonic Sequence:
A sequence that is non-decreasing or non-increasing.
a1  a2  a3  a4 . . .  an  . . . or a1  a2  a3  a4 . . .  an  . . .
Theorem:
If a sequence an  is bounded and monotonic, then it converges.
Look at the graph of: an 
1
n
Page 92
Look at the graph of: an  (1)
n
n2
Look at the graph of: an 
How could we prove that this is monotonic?
n 1
Useful Limits to Know:
ln n
0
lim
n  n
lim x
n 
1
n
lim n n  1
n 
lim x n  0 , for x  1
 1 , for x  0
n
n
 x
lim 1    e x
n
 n
xn
lim  0
n n!
Page 93
Factorials:
n!  1  2  3  4    n and for the special case of 0! , we define 0!  1
Simplify the following expressions:
Ex. 6:
(n  1)!
(n  1)!
Ex. 7:
(2n  1)!
(2n)!
Do the sequences converge or diverge? If the sequence converges, find its limit.


Ex. 8: an  n cos( n )
(n  2)!
Ex. 10: an 
n!
Ex. 9: an   2 
5n
Ex. 11: an  n
3
1 
1
3 n 
n 
2 
2 
Ex. 12: an 
11th ed. P. 596: 9, 17-37 odd AND show that an 
Page 94
n
4n  1
2
en
n
is monotonic
Series
Infinite Series:
Definition: If an  is an infinite sequence, then

 ak a1  a2  a3    
k 1
 an     is an infinite series.
To find the sum of an infinite series, consider the sequence of partial sums. If the sequence of
partial sums converges, then the series converges.
S1  a1
S 2  a1  a2
S3  a1  a2  a3



n
k 1
k 1
 ak  nlim
 ak

When this limit exists, we say that the series
converges; otherwise, it diverges.

Sn  a1  a2  a3      an

Ex. 1:
1
1 1 1
 2n  1  2  4  8    
n 0
Ex. 2:

1
1
1
1
k 1
Sequence of Partial Sums:
1
1
 k (k  1)  2  6  12  20      n(n  1)    
Sequence of Partial Sums:
Page 95
Conclusions:
For a convergent series, its sum is defined as the limit of its partial sums:
S  lim Sn  lim
n 
n
 ak
n  n 1
In other words, 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.
n 1
If {Sn } diverges ( lim S n does not exists), then the series
n 

 an diverges also.
n 1
**Problem: Sn is often difficult if not impossible to calculate.
Special Series
Geometric Series:
2
3
4
a  ar  ar  ar  ar      ar
n 1

  ar n 1 , a  0
where a is the first term and r is the common ratio.
n 1

5  15  45  135     ________
n 1

1 1 1 1
 
    ________
3 9 27 81
n 1
When will this series converge or diverge?
r 1
r  1
Page 96
What about other values of r ?
S n  a  ar  ar 2  ar 3  ar 4      ar n 1
rS n  ar  ar 2  ar 3  ar 4  ar 5     ar n
Conclusion:
The series converges if _________________
Determine if the series converges or diverges. If it converges, find its sum.

Ex. 1:

2
Ex. 2:  3  
n 1  5 
53
n 1
n 1
Page 97
n

Ex. 3:
 (1)
n 1

n 1 
1 
 n
3 
Ex. 4:
 (1)
n 1
n 1 
32 n 
 3n 1 
2

Telescoping Series:
(b1  b2 )  (b2  b3 )  (b3  b4 )  (b4  b5 )   
Determine if the series converges or diverges. If it converges, find its sum.

Ex. 5:
  n  n  1 
1
1
n 1

Ex. 6:
 ln 
n 1
n 1

n 
Page 98

Ex. 7:
6
 (2n  1)(2n  1)
n 1
Nth Term Test

 an  the an ' s must be getting smaller and smaller if the series is going to converge.
n 1

However, if they are getting smaller it does NOT mean
 an will converge.
n 1
Also, if the terms are not getting smaller it DOES mean it diverges.
If lim an  0 , this tells us NOTHING an !
n 
If lim an  0 , this tells us the series diverges. The Nth Term Test.
n 
Compare the sequence to the series:
n 1
an 
2n  1

n 1
 2n  1
n 1
Page 99
Determine if the series converges or diverges. If it converges, find its sum (if possible).

Ex. 8:
  2
n

Ex. 9:
Ex. 10:
cos(n )
n 1
5n

n 1
n
n 1
n 0

  1
Ex. 11:

2n  3n
n 1
4n

11th ed. P. 605:11-23 odd, 29, 31, 35, 45, 47, 49, 55
Page 100
Integral Test and P-series
Integral Test:
Let an  be an infinite sequence of positive terms. If an  f ( n ) , where
f is continuous,
positive, and decreasing on [1,  ) , then

 an
n 1
and

1
f ( x)dx
either both converge or both diverge.
Use the integral test to see if the given series converges or diverges. (In order to use the
integral test, you need to state that the function is continuous, positive, and decreasing).

Ex. 1:
1
n
n 1

Ex. 2:
1
 n2  1
n 1
**Note about sum:
Page 101

Ex. 3:
n
 n2  1
n 1
Power Series (P‐Series)

p

1
1
1
1
These are series in the form:     



    ,where p  0
1
p
2 p 3p
n 1  n 
n 1 n
When p  1 , we get the harmonic series as seen in the first example:

1
1
1
1
 n p  1  2  3  4    , and the related improper integral diverged:
n 1
a
1

lim  dx  lim (ln(a)  ln1)  
a  1 x
a 
But what if p  1 ?
Let f ( x) 
1
xp
. This is a continuous, positive, decreasing function for x  1
a
a
1

lim  p dx  lim  x  p dx
a  1 x
a  1
Page 102
Using the power rule to integrate, we get:
a
a
lim x  p dx
a  1

 x  p 1 
 lim   p 1 

a  

1
Now substituting in our values of a and 1:
a
 x  p 1 
 a  p 1 1 p 1 
lim 
  lim   p 1   p 1 
a    p 1 

1 a  

What happens when p  1 ? In other words, what happens when p is negative?
a
 x  p 1 
 a  p 1 1 p 1 
lim 
  lim   p 1   p 1 
a    p 1 

1 a  

 a p 1 1 p 1 
 lim  p 1  p 1   

a  


And the integral diverges.
But what happens when p  1 ? In other words, what happens when p is positive?
a
 x  p 1 
 a  p 1
1 p 1 
lim 
lim





a   ( p  1) 

1 a   ( p  1) ( p  1) 
 a1 p
11 p 
 lim 


a   (1  p ) (1  p ) 


0

1
1
1 
 lim  p 1


a   a
(1  p ) (1  p )  1  p
Since (1-p) will now be a
negative value, we can move
it to the denominator.
And the integral converges.
Conclusions:

1
 np

1
 np
converges if p  1
n 1
diverges if p  1
n 1
We call this the P-series test.
Page 103
Determine if the series converges or diverges.


Ex. 4:
1
 2n  1
n 1

Ex. 6:

n 1
Ex. 5:
n 1

1
Ex. 7:
n

n 1
11th ed. P. 613:3, 5, 7, 17, 21, 25, 27, 33, 35, 37, 69-79 odd
Page 104
3

5
n
3
n!
n5
Comparisons of Series
Direct Comparison Test
Let 0  an  bn for all n .

n 1

n 1

 bn converges, then  an converges.
1. If
2. If

 an
diverges, then
n 1
 bn diverges.
n 1
Note: As stated, the Direct Comparison Test requires that 0  an  bn for all n. Because
the convergence of a series is not dependent on its first several terms, you could modify the test
to require only that 0  an  bn for all n greater than some integer N.
Limit Comparison Test
a
If an  0 , bn  0 and lim n  L

1. Where L is finite and positive, then
2. If the
L  0 and
3. If the L   and


n  bn
 an and  bn either both converge or both diverge.
n 1

n 1
 bn converges, then  an converges.
n 1

n 1

 bn diverges, then  an diverges.
n 1
n 1
b
a
The first case is the important one and it will work even if you accidentally write n instead of n .
an
bn
The second and third case require that you get the fraction "right side up".
Note: As with the Direct Comparison Test, the Limit comparison Test could be modified to
require only that an and bn be positive for all n greater than some integer N.
Determine the convergence or divergence of the following series. Clearly state your reasons.

Ex. 1:
n 1
 n4  2
by DCT and by the LCT
n 1
Page 105

Ex. 2:
n 1

Ex. 3:
1
2

n 1
n
n 1
n2  2

Ex. 4:
,by DCT

n 1
Page 106
n 1
n2  2
,by LCT

Ex. 5:

n 1

Ex. 7:

1  cos(n)
Ex. 6:
n2
5n3  3n
 n2 (n  2)(n2  5)
n 1

1
 2  3n
Ex. 8:
n 1
Page 107
ln(n)
n 1 n  1



1
Ex. 9:  sin  
n
n 1
Ex. 10:

4n  5
n 1 7
n
 42
11th ed. p. 620: 5-23 odd (use whatever test is best), 27-34 all
Page 108
Alternating Series Test

Given a series:
(1) a
n
n1
n

or
 (1)
n1
n1
an where an  0
(all terms are positive),
then the series converges if the following conditions are met:
1. lim a n  0 (If the limit ≠ 0, then series will diverge by the nth-term test)
n 
2. a n eventually decreases. ( an 1  an ) To determine if this is true, you can take the derivative
of the corresponding function. If the derivative is always negative, then the function is
decreasing.
NOTE: This test does not tell anything about divergence.
(1)n1
1. 
n
n1
(1)n1
2. 
n1 2n  1



(1)n1 n2
3. 
n2  1
n1

4.
n
n1
Page 109

1
(1) ln 1  n 
Absolute Convergence Test
If the series

a
n 1
n
converges, then the series

a
n
n 1
converges absolutely.
(The Ratio and Root Tests are tests for absolute convergence but not the ONLY tests.)
If

a
n 1
n
converges BUT

a
n 1
n
diverges, then

a
n 1
n
converges conditionally.
Steps to determine absolute convergence:
1. Take the absolute value of the given series. Use any necessary test to determine
convergence/divergence.
2. If absolute value of given series converges, then the original series converges ABSOLUTELY.
3. If absolute value of given series diverges, then you must go back to the original series and test
it to determine convergence/divergence.
4. If the original series diverges, then the series diverges.
5. If the original series converges, then the series converges CONDITIONALLY.
ORIGINAL
SERIES
Absolute Value of
ORIGINAL SERIES
DIVERGE
No
Go back to
ORIGINAL
series
No
CONVERGE
CONVERGE
Yes
Converges
CONDITIONALLY
Page 110
Yes
Converges
ABSOLUTELY
Determine the convergence (absolute or conditional) or divergence of the following series.
Clearly state your reasons.
Ex. 1:

 (1)
n 1 (0.1)
n 1

Ex. 3:
 (1)
n 1
n
Ex. 2:
n
(1)n 1
 n
n 1


n 1 (3 
n)
(5  n )
Ex. 4:
 (1)n
n 1
Page 111
sin(n)
n2
(1)n
Ex. 5: 
n1 ln(n  1)

11th ed. P. 629: 9-25 odd, 41-55 odd
Page 112
Ratio and Root Test
Ratio Test

Let
 an be a series with nonzero terms .
n 1

1. The series
n 1

2. The series
an 1
 1.
an
 an converges absolutely when nlim

 an diverges when nlim

n 1
an 1
a
 1 or lim n 1   .
n  an
an
3. The Ratio Test is inconclusive when lim
n 
an 1
1
an
Note: Always use if you have factorials. Also, very useful if you have powers of n.
Root Test

1. The series
n a
 an converges absolutely when nlim
n

n 1

2. The series
n a
 an diverges when nlim
n

 1 or lim
 1.
n
n 
n 1
3. The Root Test is inconclusive when lim n an  1
n 
n 2 2n1

3n
n 1


Ex. 1:
Ex. 2:
Page 113
e2 n

n
n 1 n
an   .

Ex. 3:
3n

n
n 1 ( n  1)

 3
Ex. 4:   1  
n
n 1 
n
n
 x
x
Review: lim  1    e
n
 n


(n!) 2
Ex. 6. 
n 1 (3n)!
nn
Ex. 5. 
n 1 n !
Recall lim n n  1 11th ed P. 637:17-35 odd (omit 33), 39-51 odd
n 
Page 114
Taylor and Maclaurin Series
Given f ( x)  a0  a1 ( x  c)  a2 ( x  c)  ... 
2

 a ( x  c)
n 0
n
n
is differentiable (and therefore
continuous) on the interval (c  R, c  R) then the following is true:
f (0) ( x)  a0  a1 ( x  c)  a2 ( x  c) 2  a3 ( x  c)3  a4 ( x  c) 4    
f (1) ( x)  a1  2a2 ( x  c)  3a3 ( x  c) 2  4a4 ( x  c)3    
f (2) ( x)  2a2  3!a3 ( x  c)  4  3a4 ( x  c) 2    
f (3) ( x)  3!a3  4!a4 ( x  c)    



(n)
f ( x)  n !an  (n  1)!an 1 ( x  c)  (n  2)!an  2 ( x  c)2    
Evaluating each of these derivatives at x  c yields
f (0) (c)  0!a0
f (1) (c)  1!a1
f (2) (c)  2!a2
f (3) (c)  3!a3
and, in general, f
(n)
(c)  n !an . By solving for an , you can find the coefficients of the power
f n (c )
series representation of f ( x) are: an 
.
n!
f (c)
f n (c )
 a2 , . . . , an 
f (c)  a0 , f (c)  a1 ,
n!
2!
Definitions of nth Taylor Polynomial and nth Maclaurin Polynomial
If f has n derivatives at c, then the polynomial
f (c)
f (c)
f ( n ) (c )
( x  c) 2 
( x  c )3     
( x  c)n
2!
3!
n!
is called the nth Taylor polynomial for f at c . If c  0 , then
Pn ( x)  f (c)  f (c)( x  c) 
f (0) 2 f (0) 3
f ( n) (0) n
x 
x  
x
2!
3!
n!
is called the nth Maclaurin polynomial for f .
Pn ( x)  f (0)  f (0) x 
Page 115
Derivation of series for ex, cosx, sinx , and lnx
f ( x)  cos( x) and f ( x)  sin( x)
f ( x)  e x
f ( x)  ln( x)
Page 116
Maclaurin Series for a Composite Function:
Find the Maclaurin series.
2
1. f ( x )  sin x
3. f ( x )  e
2. f ( x )  cos x
3 x
4. f ( x )  ln(1  x )
2
Find nth degree Taylor or Maclaurin series
x
1. f ( x )  e ,
2. f ( x)  cos  x,
n  5, c  0
Page 117
n  4, c  0
3. f ( x ) 
3
x , n  3, c  8
11th ed. P. 648: 17,18,19, 27, 29, 31 and p. 677: 27-35 odd
Page 118
Power Series
Definition
If x is a variable, then an infinite series of the form

a x
n0
n
n
 a0  a1 x  a2 x 2  a3 x3      an x n    
is called a power series. More generally, an infinite series of the form

 a ( x  c)
n 0
n
n
 a0  a1 ( x  c)  a2 ( x  c)2      an ( x  c)n    
is called a power series centered at c, where c is a constant.
A power series in x can be viewed as a function of x

f ( x)   an ( x  c) n
n 0
where ( x  c ) is defined to  1 when x  c
The domain of f is the set of all x for which the power series converges. Determination of the
domain of a power series is the primary concern of this topic. Of course, every power series
converges at its center c because
0
f (c )  a0  a1 (c  c )  a2 (c  c ) 2      an (c  c ) n     
 a0  0  0  0     
 a0
Thus c, always lies in the domain of f. The domain of a power series can take three basic forms:
R
R
The convergence of the series will depend upon the value of x that we put into the series.
Theorem:
For a power series at x  c and R = the radius of convergence,
exactly one of the following is true:
1. The series converges at x  c . The domain is only x  c . R = 0, and the interval of
convergence is [c].
2. The series converges absolutely for x  c  R and diverges for x  c  R
If x  c  R , then the interval of convergence can be
(c-R, c +R) or (c -R, c +R] or [c -R, c +R) or [c -R, c +R]
You must check the endpoints.
3. The series converges for all x. If R = ∞, the interval of convergence is (  ,  )
Page 119
Find the interval of convergence for each series.

Ex. 1:
xn

n 1 n

Ex. 2:

n2 x n
n 1 2
n
n!
Page 120

Ex. 3:
 n!( x  4)n
n 1
(1) n x n 1

n 0 3  n

Ex. 4:
Page 121

(3x  2)n
Ex. 5: 
n
n 1
11th ed. p. 658: 15-33 odd, and Worksheet
Page 122
Power Series-Worksheet
Find the interval of convergence for each series.

1.
n 0

2.

(4 x  1)
10n
nx n

n 0 ( n  2)

5.
n
( x  2) n
n 0

4.
6.
 (1)
n 0

3.

 xn

n 1 n
xn
n 3n
n
(1)n x n
 n!
n 0

4n x 2 n
7. 
n 1 n
8.
9.

xn
n 0

n2  3
n( x  3) n


n 0

10.

n 0
5n
nx n
3n
Answers:
( 1,1)
 1 
2.   ,0 
 2 
3. ( 8,12)
4. ( 1,1)
5. [ 3,3]
1.
(, )
 1 1
7.   , 
 2 2
8. [ 1,1)
9. ( 8, 2)
10. ( 3,3)
6.
Page 123
Representation of Functions as Power Series
Geometric Power Series
1
. The form of this function closely resembles the sum
1 x

a
n
, 0  r  1 . In other words, when a  1 and r  x , a
of a geometric series:  ar 
r
1

n 0

1
1
power series representation for
centered at 0 is
  ar n
1 x
1  x n 0
Consider the function f ( x ) 

  xn
n 0
 1  x  x 2  x3  ..., x  1
Finding a Geometric Power Series centered at 0
Ex. 1: f ( x) 
1
1  x2
Ex. 2: f ( x) 
Page 124
5
2x  3
x3
Ex. 3: f ( x) 
x2
Ex. 4: f ( x) 
3
x x2
2
Page 125
Differentiation and Integration of Power Series
We can do this we can do so by differentiating or integrating each individual term in the
series, just as we would for a polynomial. This is called term-by-term differentiation and
integration.

If the power series
 a ( x  c)
n 0
n
n
has a radius of convergence R  0 , then the function
defined by f ( x)  a0  a1 ( x  c)  a2 ( x  c)  ... 
2

 a ( x  c)
n 0
n
n
is differentiable (and therefore
continuous) on the interval (c  R, c  R) and the following is true:

2
n 1
1. f ( x)  a1  2a2 ( x  c )  3a3 ( x  c) ...   an n( x  c)
n 1
2.


( x  c) 2
( x  c )3
( x  c ) n 1
f ( x)dx  C  a0 ( x  c)  a1
 a2
...  C   an
2
3
n 1
n 0
Express the following as a power series:
Ex 1: f ( x )  ln(1  x)
Ex 2: f ( x)  arctan x
Page 126
Ex 3: f ( x) 
1
1  x 
2
What if we multiplied problems 1-3 by x?
Ex 4:
1
 1 dx

0 1  x 7
Assignment: Worksheet
Page 127
Power Series Representation
Find a power series representation for the function:
1. f ( x) 
1
1 x
10. f ( x) 
4
4  x2
2. f ( x) 
2
3 x
11. f ( x) 
3x 2
9 x
3. f ( x) 
3
2x 1
12. f ( x )  ln( x  4)
13. f ( x )  x arctan(2 x )
1
4. f ( x ) 
2x  5
5. f ( x) 
14. f ( x) 
4
3x  2
15. f ( x) 
3x
6. f ( x)  2
x  x2
7. f ( x) 
4x  7
2 x  3x  2
8. f ( x) 
x
9  x2
9. f ( x) 
1
 x  3
1
1  2 x 
1
1

dx
16. 
0 1  x5
2
x
2x 1
2
Page 128
2
2
Power Series Representation—Answers
1. f ( x) 
1

1 x


9. f ( x) 
( 1) n x n
n 0


2
xn
2. f ( x) 
2
3 x
3n 1
n 0
2x 1
2
 (2) x

n

3
(2 x) n
 3
2x 1
n 0
3x 2
11. f ( x) 

9 x


n0


n 0


4

3x  2

(1) n
n 0
12. f ( x )  ln( x  4) 


n0
n
 x
(1) n   
2

( 1) n
xn
 ln 4
4 n 1 ( n  1)

(3 x) n
2n 1
13. f ( x )  x arctan(2 x ) 

( 1) n 2 2 n 1
n 0
14. f ( x) 

x
xn2
32 n 1
n 0
3x
2
1


2
x  x  2 x  2 x 1
6. f ( x) 


n
 1  2n
  x
 4

1
(2 x ) n
4. f ( x) 

2x  5
5n 1
n0
5. f ( x) 
2 n 1
n0
4
10. f ( x) 

4  x2

3. f ( x) 

x

n
OR
n 0

1
 x  3
  3 
1
2


n 1
1
 
3
n 1
nx n 1
n2
(n  1) x n
n0
7. f ( x) 
4x  7
2
3


2 x  3 x  2 (2 x  1) x  2
2





(2) n 1 x n  3
n0
8. f ( x) 
x

9  x2
( 1) n
n0
n 0
xn
2 n 1
1  2 x 

2
2
n 1
nx n 1
n 1

OR
 2 (n  1) x
n
n
n 0

 (1)
15. f ( x) 

1
2 n 1
n
x
9n 1
1

1
 1

16. 
dx


0 1  x 5
n  0 5n  1
n
Page 129
x 2n2
2n  1