CALCULUS II [Type the document MATH 2414 subtitle]

Turnell

2018

Review of Cal I

Trig : d d d [sinu ]  [cosu ]  [tanu ]  dx dx dx

d d d [secu ]  [cotu ]  [cscu ]  dx dx dx

dduddu1 ux() Exponential and Natural Log Functions: []eeuu and [lnu ]  or dx dx dx u dx u 2 f ()xe x x f ()xx ln(sin())

d : []xnn nx 1 dx f ()xx432 3 x  6 x  x

d : []uv uv vu dx f ()xx 3 cos x

du vuuv dT BTTB   :  OR  dx v v2 dx B B2 x3 1 fx() x2  2

Page 1 d n : unuu n1  dx

4 fx() 3 x2 5 x 1

f ()xx sin(5)3

fx() tan52 3 x 1

Page 2 Implicit Differentiation: dy Find , given xxyy3245439 dx

dy Find , given cos(xy ) y3 4 dx

Integrals:  f ()xdx= a set of Why a set? Review of Basic Forms u n1 kdu ku C udun   C n 1   n 1 sin(udu ) ______ csc(uudu )cot( ) ______ cos(udu ) ______sec(uudu )tan( ) ______sec2 (udu ) ______ csc2 (udu ) ______u  1 edu ______ du  ______  u

 3 3 1 23  x 3 xdx5 x(5xdx 4)  x 

Page 3 2  6x  cos x  35dx  3 dx  (4x  9)  sin x

 tanx sec2 xdx  xsin(6xdx2 )

 x x2  dx 2xedx  x2 1 

Page 4 I Review Worksheet

dy Find the ()yor for each function. dx 2 2/3 cos(3x ) 1. y  (x  4x  5) 6. fx() 2. y  tan(2 x) x  sin(3x ) 3. yx cos(5 ) 7. x3  2x2y2  5y3  10 4. y  sin2 (3x) 8. x2 y  3tany  4 3 5. y  x2 tan x 9. y  cos(sin(x )) 10. y  (2x  3)4 (3x  2)5 Find the indefinite or definite :  sin x 11.  dx 1.  x(4xdx24 3)  1cos x  dx 543 1 12. 2. x 2xx dx  4 x3  x  3 3. sec(5)tan(5)x xdx  ee42xx1 13.  dx  dx ex 4.    72x  14. sin6 x cos xdx  sin x  5.  dx 2  cos3 x 15.  tan(2x )sec (2xdx ) 6. sin 6xdx 16.  412x xdx2  dx 4 7.  17. sec (x )sec(xxdx ) tan( )  13 x  3 2 8.  xcos(xdx )  ex 18.  dx 1x2 x 9. xedx  e 1  1  x3 10.  dx  3 1 x4

Page 5 Calculus I Review Worksheet-Answers

2(2x  4) 22 1.  dy43 xy x y  1 7.  3(xx2  4 5) 3 dx 15yxy22 4

2 dy2 xy 2. yx  2sec(2)  8.  dx x22 3sec y 3. yx 5sin(5 ) 9. yxx 323 cos( )sin(sin( x 3 )) 4. yxx  6sin(3 )cos(3 ) yx 15(3 2)44 (2 x  3) 8(2 xx 3) 35 (3 2) 5. yx 22sec ( x ) 2 x tan( x ) 10. yx (3 2)43 (2 x  3) (54 x  29)

3sin(3x )(xx sin(3 )) cos(3 x )(1 3cos(3 x )) 6. y  (xx sin(3 ))2

(4x25 3) 2 1 3xx x 1.  C 7. 13x C 13. eeeC  40 3 3 1 7 8. 2 sin (x ) 4 sin(x )  C 14.  C xxx65233 1 2 7 2. 1 2  2 C 1x 1 65 42x 9. eC 15. (tan2 (2x ))  C 2 4 2 3 4 3 3 1 10. (1 x )  C 2 2 2 3. sec(5x )  C 8 16. 12 x  C 5 3  11. ln 1 cos x C sec5 (x ) 1 1 17.  C 4. ln 7x  2 C 12.  C 5 7 3 3(x 3) 3 1 3 e 1 5.  C 18. lnee 1 ln  1 ln 2cos2 x e 1 1 6. cos(6x ) C 6

Page 6 of Region Between Two

Review from Cal 1: The area under a .

Write the integral or to find the area of each region: 1. 2.

3.

Area Between 2 Curves:

4.

Page 7 5. With respect to x 6. With respect to y

7. 1

Page 8 Examples: Sketch a graph of the region bounded by the given equations and find the area of the region: y xx2 31  1. yx 1

y 2. fy() ,gy () 0, y 3 16  y2

(Use Desmos.com for graph)

Larson 11th ed: p. 450: 7,15,17,19,23,24,35,37

Page 9 Page 10 Finding

Example 1: yx4 2 . Find the from the y-axis and x-axis in terms of x and y. .

Example 2: x yy2 4 . Example 3: x2 Find the distance from the y-axis in terms of x and y. Find the distance from y 3 to the : 2 y 1 in terms of x and y.

x2 Example 4: Find the distance from y 3 Example 5: Find the distance from yx 2 2 2 to the line: y 1 in terms of x and y. to the line: x  2 terms of x and y.

Page 11 Example 6: Find the distance from yxx2,2 0 Example 7: Find the distance from yxx2,2 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: yx(2)3 2  xy(2),,6,1 2 yxyy   

Page 12 : The Disk Method

Review from : The volume of a

2 Vrh 

Determine the Volume of a : So, for the purposes of the derivation of the formula, let’s look at rotating the yfx () in the interval [,]ab 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: Vrx 2 . Approximating the volume of the solid by n disks of width x with rx(), produces: n 2 Volume of solid [(rxi )] x i1 This approximation appears to become better and better as x 0(n  ). Therefore, n b 22 Volume of solid lim [rx (i )] x [ rx ( )] dx x 0   i1 a

Page 13 As seen in animation, we can rotate functions around the y-axis:

The radius is now a function of y. d Volume of solid   [()]ry2 dy 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 about the x-axis. yx4 2 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 about the y-axis. yx4 2

Page 14 Example 3: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the y-axis. x yy2 4 (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 about the line x  2: yx 2 2 (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  ()R22rw Where R  the outer radius and r  inner radius If rotated around a horizontal axis, then b Volume of solid  ([()]R xrxdx22 [()]) a If rotated around a vertical axis, then d Volume of solid  ([()][()])R yrydy22 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: yx 2 , yx , about x-axis.

3 (answer is ) 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: yx 2 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: yx 2 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. yx 2 , yx , about the y-axis. 2. yx 2 , yx , about line x 1. 3. yx 2 , yx , about line y  3. 4. yx(1)12 , y 1, x  0 , x 1 about the x-axis. 5. yx(1)12 , y 1, x  0 , x 1 about line x 1. 6. yx(1)12 , y 1, x  0 , x 1 about line x  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 the graphs: yxx4 2 , 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()()  rx hxdx V 2()()  ry hydy a 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: yxx4 2 , y  0 , about y-axis.

128 Answer is 3

Example 2: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded by the graphs: xy(2)2 , yx , about the line y 1.

63 Answer is 2

Page 20 Example 3: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded by the graphs: yx2 , yx , y  0 about the x-axis. (p.470: 22)

16 Answer is 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: yx 2 , yx 3 , 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?

11th ed. P. 470: 3,5,9,11,13,16,17,23,25

Page 21 Page 22 Arc

Finding the length of a curve: Demo: http://demonstrations.wolfram.com/ArcLengthApproximation/

Examples: Find the length of the curve over the given interval. 3 1. yx23,[0,9]2

Page 23 x3 1 2. y ,[1,2] 62x

x5 1 3. y  ,[2,5] 10 6x3

Page 24 Area

Find the 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 yfx () in the interval [,]ab 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.

The approximation on each interval gives Now, rotate the approximations about the a distinct portion of the solid. Each of these portions is called frustums and the x-axis and we get the following solid. surface area a frustum is given by: SA 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 2 dy dx ds 1 or ds 1 dx dy

http://demonstrations.wolfram.com/SurfaceAreaOfASolidOfRevolution/

Review of algebra

Simplify the following

2 x 1 2 21x x 3  ______2  x  

1 x 1 x23

Page 26 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:

yx2,1 x 4 about the x-axis

Page 27 Find the surface area obtained by revolving the given curve about the indicated axis:

xy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:

5 x 21,yy  1 about the y-axis 8

Page 29

Homework problems:

P. 481: 7,9,11,39,43,45,47 (Hint: refer to notes on )and the following:

Find the arc length of the graph of the function over the indicated interval. (Exact values!)

x3 1 1. yx,1  4 12 x 3 x 2 1 2. yxx2 ,1  9 3 x4 1 3. yx,1  2 48x2

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!)

13 4. yxxx2,2  about the x-axis 22

y3 5. x ,0y 1 about the y-axis 3

15 6. xyy24 , 0  about the y-axis 4

7. yx1,012  x  about the y-axis

8. xy122 ,1  y  2 about the x-axis

1 3 9. xy(2),122 2  y 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 WFd .

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 WFxdx  () 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, F  kd where k is the spring constant.

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 C Fx() 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 —1 A line—2 A —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 01122mx mx  mxnn

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

M x  o , where mm  m  m is the total mass of the system. m 12 n

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 mm12, ,. . . , mn be located at (x11 ,yxy ),( 2 , 2 ), . . . ,( xynn , ) .

1. The moment about the y-axis is M ynnmx11  mx 2 2  mx .

2. The moment about the x-axis is M x my11  my 2 2  mynn.

3. The center of mass (,)x y (or center of gravity) is

M y M x  and y  x m m where mm12  m  mn is the total mass of the system. Example 3: Find the center of mass of the given system of point masses.

mi 3 4 2 1 6

(,xiiy ) (2,3) (5,5) (7,1) (0,0) (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 yfxgx(() ()) 2

Representative Thin Strip (rectangle): Full Region 1. Area of representative rectangle: 1. Area of region: b A (()fx  gx ())  x (Height*width) A (()fx gxdx ()) a 2. The moment about the y-axis is 2. The moment about the y-axis is b M xA xfx(()  gx ())  x M xfx(() gxdx ()) . y y  a 3. The moment about the x-axis is 3. The moment about the x-axis is

1 1 b M yA(() fx gx ())(() fx gx ()) x M ([fx ( )]22 [ gx ( )] ) dx. x 2 x 2  a 1 MyA   ([()]fx22 [()]) gx x x 2 4. The center of mass (,)x y (or center 4. The center of mass (,)x y (or center of gravity) of gravity) is 1 M y M x is x  x and yfxgx(() ()) x  and y  2 A A Example 5: Find the center of mass of a lamina of uniform density  covering the region bounded by the 2 y 4 x and y  0 . Let  1 (Hint: symmetry)

Page 35 Example 6: Find the center of mass of a lamina of uniform density  covering the region bounded by the 4 parabola y  x and yx . Let  1

Page 36 Example 7: Find the center of mass of a lamina of uniform density  covering the region bounded by the 2 parabola x y 2 and xy . Let  1

11th ed. P. 502: 7,9,13,17-27 odd, 31, 21

Page 37 Page 38 Review of Basic Integration Rules

Review Basic Inverse Trig Derivatives: du du du u (arcsinu )  (arctan )  2 (arcsecu )  dx 1 u2 dx 1 u dx uu2 1 du  du  du 

(arccosu )  (arccotu )  2 (arccscu )  dx 1 u2 dx 1 u dx uu2 1 Review Basic Inverse Trig Integration:  du  du  du    22    au22  au  uu22 a

Find the indefinite integral:  1  1  2 dx  dx  29 x  25 36x2

 1  dx  xx492 

Page 39 1  x  ecos x dx  dx  4  2  x 16  1 x

  1  sin x  dx  2 dx  xx(1 )  1cos x  2

Completing the Square x2 610x 2x  x2

Page 40  1  dx  86xx2

 6  dx  34xx 42

 25x   dx  xx2 22

Page 41 Improper Fractions: The 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.

 x4 1  dx  x2 1

 x2  dx  x2 1

2  47x   1  dx (Hint: Remember  du ln u C  23x   u

Separating Fractions: x44 xxx You CANNOT separate denominators. dx dx xx2211  28 x  dx  14 x2

Page 42 Review Basic Trig Integration: sin(udu ) cos( u )  C  csc(uuduuC )cot( )  csc( )  cos(udu ) sin( u ) C sec(uuduuC ) tan( ) sec( )  sec2 (udu ) tan( u ) C  csc2 (udu )  cot( u ) C What about  tan(udu )  cot(udu ) sec(udu )  csc(udu )

Multiplying by a Form of One: sec(udu )  csc(udu )

 dx  dx    1cos x  1csc x

Page 43 Trig Identities: sin 2  2sin cos cos22 sin 1 cos 2  cos22 sin 1tan22 sec 2cos2  1 cot22  1 csc  12sin2  (cscx  tanxdx )2

Review Other Bases Integration:

u x adu 8 dx

Log Rule: What about  ln(udu ) ?

 cot(x )[ln(sinxdx )]

"Math Magic Tricks"  1  dx  1 e x

Page 44 p.558 – Basic Integration Formulas (Thomas’ CALCULUS Media Upgrade 11th edition)

Basic Substitutions 1  16x dx  16x dx  dx 1. 3. 3sincosx xdx 5. 7.     2  81x2   82x   xx1 0  

t 9. cot(3 7x )dx 11. eedcsc( 1)  13. sec dt 15. csc(x  ) dx    3 

w ln 2 x2  2 17. 2xedx 19. evdvtanv sec 2 21. 3x1dx 23.  dw 0     2 w

2 1 3 6  dx  dx  2x dx 25. 27. 29. 30.  2    (tanx )[ln(cosxdx )]  49 x  19 x2  1 x4 0 0  3 e 2 6dx  dx  dx  ln x 31. 33. 35. 36.  dx   xx   xx252  1  ee  x cos(lnx )  x 1

Completing the Square 2  8dx  dx  dx 37. 39. 41.   2    xx22  xx2 43   (1)2x xx2 1

Trigonometric Identities 43.  (secx  cotxdx )2

Improper Fractions 3 32 x 3 2x 416x xx 47. dx 49. dx 51. dx  x 1  2 x2 1  x2  4

Separating Fractions  4  1 x  1sin x 53. dx 55.   2 dx  1 x2  cos x 0

Multiplying by a Form of 1 1 1 1 57. dx 59. d 61. dx  1sin x  sec  tan  1sec x

Trigonometric Powers 83.  cos3  d (Hint: cos22  1 sin  )

Page 45 Answers

2 [ln(cosx )]2 1. 28x  1 C 30.  C 3 2 3. 2(sinx ) 2  C 31. 6sec1 5x  C 5. ln10 ln 2 ln 5 1 x 7. 2ln(x  1) C 33. tan (eC )  9. 1 ln sin(3 7x ) C 35. ln(2 3) 7 36. (lnx )2  C or 11. ln csc(eeC 1) cot( 1) 37. 2 tt 1 13. 3ln sec tan C 39. sin (x  2)  C 33 41. sec1 x  1  C 15. ln csc(x  ) cot(xC ) 43. tanx  2ln cscxx cot cot xxC 17. eeln 2 0 211 tanv 47. x  lnxC 1 19. eC 3(1)x 49. 7ln8 21.  C 21 x 51. 22tanx  xC ln 3 2 w 2 12 23.  C 53. sinx  1xC ln 2 55. 2  25. 57. tanx  sec xC 18 59. ln 1 sin  C  27. 18 61. x  cotxxC csc 1 12 83. sin sin3  C 29. sin (x )  C 3

Page 46

If you are given a function yfxgx ()() and asked to take the derivative, you would use the product rule and get: d f ()()xgx f () xg'' () x gx () f () x dx If we let ufx () and vgx (), then du f' () x dx and dv g' () x dx

d We can write the product rule as: ()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

 5x  dx  e2x

Page 47  x5 ln(3xdx )

ln xdx

 4sec(2)x 2 xdx

 4arccosxdx

Page 48  x2 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  x33edxx

 x3 cos(2xdx )

***  x5 ln(3xdx ) Can you use the tabular method for this problem?

Page 49

Are we going in ? exdx4x cos(2 )

Thomas 11th ed. P. 529: 15-33 odd, 43, 45, 49, 53, 55

Page 50 Trigonometric Integrals

Pythagorean Identities: cos22 sin 1 sec22  1 tan  cos22  1 sin  tan22  sec 1 sin22  1 cos 

cos 2  2 cos2  1 cos 2  1 2sin 2  1cos2  1cos2  sin 2  2sin cos cos2   sin 2   2 2

Trig Review: d [sinx ] cos x cos(x )dx sin( x ) C dx  d [cosx ] sin x sin(x )dx cos( x )  C dx  d [tanx ] sec2 x sec2 (x )dx tan( x ) C dx  d [secx ] secxx tan sec(x ) tan(xdx ) sec( x ) C dx 

sec(x )dx ln sec( x ) tan( x ) C  tan(x )dx ln cos xC  or ln sec xC 

Powers of Sines and Cosines

sin3 (x )cos(xdx ) sin4 (x )cos(xdx )

sin7 (x )cos(xdx )  sin(x ) cos(xdx )

Conclusion:

Page 51  (1 sin35 (4x ) sin (4xxdx ))cos(4 )

 cos3 (x )sin(xdx )

Conclusion:

 (cos23 (2x ) cos (2xxdx ))sin(2 )

sin23 (x )cos (xdx )

sin54 (x )cos (xdx )

Page 52 Even Powers of Sines and Cosines

 cos2 (x )dx sin2 (4x )dx

sin22 (x )cos (xdx )

POWERS OF AND SECANT sec6 x (secxxdx tan ) sec23x tan xdx

Conclusion:

Page 53

 tan24 (x )sec (xdx )

 tan33 (x )sec (xdx )

 tan5 (x )dx sec3 (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: uuu sin tan sec aaa Or

auauausin  tan sec

 3 1.  dx  19 x2

2.  19 x2 dx

Page 55  y2  25 3.  3 dy  y

 4 4.  dx  xx2216 

Page 56  25x2  4 5.  4 dx  x

 1 6.  3 dx  (5)x2  2

11th ed. P. 547: 3, 5, 7, 9, 13, 19-29 odd, 35, 37, 41

Page 57

Page 58 Partial Fractions

g()x Suppose fx() where gx() and hx() are polynomials and the degree of hx()is hx() BIGGER than the degree of gx() (if this is not true, you must first do long division), we need to 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: hx()is a product of linear factors, none repeating, then the Partial Fraction Decomposition has the form: x 1 AB  (2)(211)2211xx x  x 

Watch this video: Partial Fractions-Part 1

Case II: hx() is a product of linear factors, some repeating, then the Partial Fraction Decomposition has the form: xABC1   (1)(2)xx22 x  1 x  2(2) x 

Watch this video: Partial Fractions-Part 2

Case III: hx()contains irreducible quadratic factors, none repeating, then the Partial Fraction Decomposition has the form: x 1 ABxC  (2)(1)2xx22 x  x  1

Case IV: hx()contains irreducible quadratic factors, some repeating, then the Partial Fraction Decomposition has the form: x 5 ABxCDxE   (2)(4)2xx22 x  x2  4( x 22  4)

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 1.  dx  28x3  x

 2x 2.  dx  275xx2 

Page 60  x3  3 3.  dx  xx2 2

 1 4.  2 dx  (1)(4)xx

Page 61  345xx2  5.  2 dx  (1)(1)xx

 882xx2  6.  dx  2 2  41x 

 1 11th ed. P. 557:3-17 odd, 21, 23, and  dx  2 2  xx1

Page 62 L'Hôpital's Rule

Review of Limits 23x  x2 16 45x2  x lim lim lim x2 x  4 x4 x  4 x 13 x2

Indeterminate Forms

0  0  1 00 0  0 

2cos x  2 23x  lim lim x2 x2  4 x2 x2  4

L'Hôpital's Rule

Suppose that we have one of the following cases: fx() 0 fx()  lim  OR lim  xa gx() 0 xa gx()  where a can be any real number, infinity or negative infinity.

In these cases we have: f ()xfx () lim lim xagx() xa g () x

Find the : ln x4 cos x x2 1. lim 2. lim 3. lim x x3 x01tan x x ex

Page 63 1 10 3 4. limx tan 5. lim  xx x0 x x2

x 1 1 6. lim 1 7. lim x x1 2  xx 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   c f ()x is continuous on [,a  ): f ()xdx lim f () xdx c aa

bb f ()x is continuous on (,] b : f ()xdx lim f () xdx c  c f ()x is continuous on (,)  : c fxdx() fxdx () fxdx ()   c

ca limf (xdx ) lim f ( xdx ) ba  bc

If the limit exists, then the integral converges . If the limit does not exist, then the integral diverges.

  1 1.  dx  x2 1

0  1 2.  dx  3  x 

Page 65   4 3.  dx  16  x2 

  (1 x ) 4.  dx  e x 1

Type 2 Vertical asymptotes within the given limits of integration. (Function is undefined) f ()x has an infinite discontinuity at xc , cab (,)

cb f ()xdx lim f () xdx bc  aa bb f ()xdx lim f () xdx ac  ca bcb f ()xdx f () xdx f () xdx aac

Page 66 2  1 5.  dx  x  2 0

5  1 6.  dx  2  x  9 3

1  1 1 7. 2  2 dx  x dx  x 1 1 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  ft() and the system of equations  is called a system of parametric equations. ygt ()

For the following, sketch the curve, indicate its orientation, and eliminate the parameter to get an equation involving just x and y. xt2 3 Ex. 1  yt21

Page 69 3 xt2 1 Ex. 2  yt

x  2cost Ex. 3  yt 3sin

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: x  ft() and ygt (),

then the of C at (,x y ) is dy dy dx  dt ,  0 dx dx dt dt d dy dy2 dx The second derivative is  dt dx2 dx dt Tangent Lines dy dx There is a horizontal tangent line if  0 and  0 dt dt dx dy There is a vertical tangent line if  0 and  0 dt dt Examples: dy d 2 y Find and . Find the slope and concavity (if possible) at the given value of the parameter. dx dx2 Find the (,x y ) coordinate at given value of t.

1. xtyt,31,1 t

Page 71 2. xtytt2sin(2 ), cos(2 ), 1 6

Find all points (if any) of horizontal and vertical tangency to the curve.

3. x t 432, y tt 4. x 8sinty , 4cos t

Page 72 5. Find the equation of the tangent line(s) to the given set of parametric equations at the given value of the parameter. x tt534, y t 2 when t 2

Arc Length

Definition: If a curve C is defined parametrically by x  ft() and ygt (), atb , where f  and g are continuous and not simultaneously zero on [,]ab , and C is traversed exactly once as t increases from ta to tb , then the length of C is the definite integral: b b 22 22  dx dy L [ft ( )] [ gt ( )] dt , L  dt   dt dt a   a

Find the arc length of the curve on the given interval:

6. x 6,t 23y 2,tt 1 4

Page 73 Area of a

If a smooth curve x  ft() and ygt (), atb , is traversed exactly once as t increases from ta to tb , then of the surfaces generated by revolving the curve about the coordinate axes are as follows:

1. Revolution about the x-axis (0)y  : 2. Revolution about the y-axis (0)x  :

b b 22 22  dx dy  dx dy Sy2 dt Sx2 dt      dt dt  dt dt a a

Find the area of the surface generated by revolving the curve about the given axis: t 2 7. xytt,2,05 about the x-axis 2

33  8. xycos , sin , 0 about the y-axis 2

11th ed. P. 715:9,11,13,15,33,35,37,39,49,55,63,65 and Worksheet (Parametric-)

Page 74 Parametric—Tangents

Find all vertical and horizontal points of tangency. x costt sin t 1. , 02t  ytttsin cos

x 1 t 2. ,  t   yt3 3 t

x tt2 2 3. ,  t   yt3 3 t

x 42cost 4. , 02t  yt1sin 

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(3ttyt ) 4sin(3 ),  3tan(6 ) when t  2

6. xt23211, t ytt  (4)3(4)7   tt2 2  when t 2

7. xt3 cos( t ), y  4 t  sin(2 t  6) when t 3

8. x tt2321,y ttt 72  8 when t 1

Page 75 Solutions:

1. Horizontal: t  (1,) 5. yx312

Vertical:  t  ,1 22

33 t ,1 22

2. Horizontal: t 1(2,2) 6. yx24  65

t 1(0,2)

Vertical: None 3. Horizontal: t 1(4,2) 252 7. yx t 1(2,2) 99

Vertical: 1311 t  , 248 4. Horizontal:  25 7 t  (4,0) 8. yx 2 42 3 t (4, 2) 2 Vertical: 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: ra ra2cos ra2sin

r  3 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 33cos , , r 34sin , r 53sin

Page 77 Limaçons rabcos rab sin r 13cos r 22sin r 32cos r 42sin

Limacon: https://youtu.be/MT1j6vdX2aI

Lemniscate (a figure 8) : ra22cos(2 ) ra22 sin(2 )

r 2  25cos(2 ) r 2 16sin(2 )

Rose Curves: racos( n ) ra sin( n ) Rose curve: https://youtu.be/tvPsjmcN2Yg r  3cos(3 ) r  3cos(2 ) r  3sin(5 ) r  3sin(2 )

Page 78 Slope and Tangent Lines dy Given that xr cos , yr sin and rf () , find . dx

dy 1. Find and the of the tangent lines at the given values of  . dx r 1sin   ,   0 and   

Page 79  2. Determine the equation of the tangent line to r 38sin at   6

Tangent Lines dy dx There is a horizontal tangent line if  0 and  0 d d dx dy There is a vertical tangent line if  0 and  0 d d

Page 80 3. Find all the points of vertical and horizontal tangency: r 1sin   https://www.youtube.com/watch?v=4SpQ9jOxd_E

4. Review quadratic formula. Solve: 4cos2   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  2cos3

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 1cos  B. Determine the equation of the tangent line to r  sin(4 )cos at   6  C. Determine the equation of the tangent line to r   sin(3 ) at   2  D. Determine the equation of the tangent line to r 12cos at   4 E. Find all the points of vertical and horizontal tangency: r 24cos

(you will need to use the quadratic formula)

Answers:

335  2  1. Horizontal: (,r  )  , , , 3. yx 23 2 3  2

12 14 11212 Vertical: (,r  )  2,0, , , , 4. yx(1 2 2 )    23 23 7 22

11 5. Bonus 2. yx 33 4

Page 83 Page 84 Area and Arc Length of Polar Curves

If we have a of radius r, and select a sector of angle  , where  , then the area of that sector is

2  Ar   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  .

1 Since  is infinitesimally small, the wedge shaped has an area of dA r2 d 2 If we partition the interval [,] into n equal subintervals:

01234     nn 1  . Then approximate the area of the region by the sum of the area of the sectors.

n 1 2 Then the approximate area is: Af [()]i i1 2

n 1 2 Aflim [ (i )] n 2 Taking the limit as n , i1 1   [()]fd2 2  1  Area=  rd2  2 

Page 85

Example #1: Find the area inside the curve: r 1cos

Example #2: Find area of one petal of rose curve: r  3sin2

Page 86

Example#3: Find the area shared by r 2(1 sin ) and r  2

Example#4: Area outside r 1cos 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 24cos .

Page 88

Arc Length in Polar Form

Consider a polar arc: rf(),  and f ' () is continuous on interval  . Remember in parametric form the b 22  dx dy Arc Length  dt  dt dt a Now, we are in polar coordinates. We change the variable t into  : b 22  dx dy Arc Length  d  dd a We can simplify this by remembering that: x  r cos and yr sin

Page 89

Example: Find the length of the curve over the given interval: r 1cos0  

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 2  6cos2 and outside r  3 Answer: 33  B. Area shared by r 22cos and r  2 Answer: 58  8 C. Inside r  4cos and to the right of r  sec Answer:  3 3 19 D. Find the length of the curve over the given interval: r 2 05 Answer: 3    3 E. Find the length of the curve over the given interval: r cos3 0  Answer:  34 88

Page 90 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: aa1234,,,,... aa an th an is called the n term of the sequence and is denoted by an

21n  st Example 1: Write out the 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 liman does not exist, then a diverges. x n

Theorem:

Let L be a real number. Let f be a function of a real variable such that limf (xL )  . x

If an is a sequence such that f ()na for every positive integer n, then limaLn  n n

In other words: limf (xa ) lim n Why is this good? nx 

Do the sequences converge or diverge? If the sequence converges, find its limit.

n 1 21n  Ex. 2: an (1)1  Ex. 3: an  n 13 n

Page 91 n2 Ex. 4: a  n 21n 

Theorem:

If liman  0 , then liman  0 n n

n 1 1n1 Ex. 5: an  (look at graph) Ex. 6: an  (look at graph) 2 n

Monotonic Sequence: A sequence that is non-decreasing or non-increasing. aaaa1234...  an ... or aaaa1234...  an ...

Theorem:

If a sequence an is bounded and monotonic, then it converges. 1 Look at the graph of: a  n n

Page 92 n Look at the graph of: an (1)

n2 Look at the graph of: a  How could we prove that this is monotonic? n n 1

Useful Limits to Know: ln n lim 0 limn n  1 n n n

1 limx n  1, for x  0 limxn  0 , for x 1 n n

n n x x x lim 1e lim 0 n n n n!

Page 93 : nn! 1234     and for the special case of 0!, we define 0! 1

Simplify the following expressions: (1)!n  (2n  1)! Ex. 6: Ex. 7: (1)!n  (2n )!

Do the sequences converge or diverge? If the sequence converges, find its limit.

11 Ex. 8: ann   cos( n ) Ex. 9: an 23nn  22

(2)!n  5n en Ex. 10: a  Ex. 11: a  Ex. 12: a  n n! n 3n n  n

n 11th ed. P. 596: 9, 17-37 odd AND show that a  is monotonic n 41n2  Page 94

Infinite Series:

Definition: If an is an infinite sequence, then   aaakn123   a   a  is an infinite series. k1

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. Sa11  n Saa212 aakk lim Saaa3123 kk11n  When this limit exists, we say that the series  converges; otherwise, it diverges.  Saaann123     a

 1111 Ex. 2: Ex. 1: 1   n 248 11111 1 n0 2        k1 kk(1)261220nn (1) Sequence of Partial Sums: Sequence of Partial Sums:

Page 95 Conclusions: For a convergent series, its sum is defined as the limit of its partial sums: n SSlimnk lim  a nn  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 Sn does not exists), then the series an diverges also. n n1 **Problem: Sn is often difficult if not impossible to calculate.

Special Series :  234nn 1 1 aarararar     ar  ar , a  0 n1 where a is the first term and r is the common ratio.  5 15  45  135   ______n1

11 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 ?

234n 1 Sn  a ar ar ar ar ar

2345 n rSn  ar  ar  ar  ar  ar  ar

Conclusion: The series converges if ______

Determine if the series converges or diverges. If it converges, find its sum.   n n1 2 Ex. 1: 53 Ex. 2: 3 n1 n1 5

Page 97   2n n 1 1 3 Ex. 3: (1)  Ex. 4: (1) n1  n  31n n1 3 n1 2

Telescoping Series: ()()()()bb12 b 23 b bb 34 b 45 b Determine if the series converges or diverges. If it converges, find its sum.

 11 Ex. 5:  n1nn1

 n 1 Ex. 6: ln  n1 n

Page 98  6 Ex. 7:  n1 (2nn 1)(2 1)

Nth   an  the asn ' 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 liman  0 , this tells us NOTHING an ! n th If liman  0 , this tells us the series diverges. The N Term Test. n

Compare the sequence to the series: n 1  n 1 an   21n  n1 21n 

Page 99 Determine if the series converges or diverges. If it converges, find its sum (if possible).

 n  n1 Ex. 8:  2 Ex. 9: 1 n n0 n1

 cos(n )  23nn Ex. 10: Ex. 11:  n  n n1 5 n1 4

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 afnn  (), where f is continuous, positive, and decreasing on [1, ) , then   a and f ()xdx  n 1 n1 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).  1 Ex. 1:  n1 n

 1 Ex. 2:  2 n1 n 1

**Note about sum: Page 101  n Ex. 3:  2 n1 n 1

Power Series (P‐Series) p 1111 These are series in the form: 1 ,where p  0  ppp nn11n n 23 When p  1, we get the harmonic series as seen in the first example:

 1111 1    , and the related improper integral diverged:  p n1 n 234 a  1 lim dx lim (ln( a ) ln1) aa 1 x 

But what if p  1? 1 Let fx() . This is a continuous, positive, decreasing function for x  1 x p a  1 a  p lim dx lim x dx p 1 aa1 x 

Page 102 Using the power rule to integrate, we get: p 1 a a  p x limxdx lim  aa1 p 1 1 Now substituting in our values of a and 1: a xappp111 1   lim lim   aappp111    1  

What happens when p 1 ? In other words, what happens when p is negative? a xappp111 1   lim lim ppp111    aa   1 a pp111 lim pp11 a And the integral diverges.

But what happens when p 1 ? In other words, what happens when p is positive? a xappp1111   lim lim  aa(1)ppp (1)(1)    1  a11pp1 lim  a(1pp ) (1 ) 0 111 Since (1-p) will now be a lim negative value, we can move a p1 ap(1 ) (1p ) 1 p it to the denominator.

And the integral converges.

Conclusions:  1  1 converges if p 1 diverges if p 1  p  p n1 n n1 n We call this the P-series test.

Page 103 Determine if the series converges or diverges.  1  3 Ex. 4: Ex. 5:  21n   5 n1 n1 n 3

 1  n! Ex. 6: Ex. 7:    5 n1 n n1 n

11th ed. P. 613:3, 5, 7, 17, 21, 25, 27, 33, 35, 37, 69-79 odd

Page 104 Comparisons of Series Let 0 abnn for all n .   1. If bn converges, then  an converges. n1 n1   2. If  an diverges, then bn diverges. n1 n1

Note: As stated, the Direct Comparison Test requires that 0 abnn 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 abnn for all n greater than some integer N.

Limit Comparison Test an If an  0 ,bn  0 and lim  L n bn   1. Where L is finite and positive, then  an and bn either both converge or both diverge. n1 n1   2. If the L  0 and bn converges, then  an converges. n1 n1   3. If the L  and 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 could be modified to require only that an and bn be positive for all n greater than some integer N.

Determine the convergence or of the following series. Clearly state your reasons.  n 1 Ex. 1: by DCT and by the LCT  4 n1 n  2

Page 105  1 Ex. 2:  n1 2  n

 n 1  n 1 Ex. 3: ,by DCT Ex. 4: ,by LCT  2  2 n1 n  2 n1 n  2

Page 106  1cos() n  53nn3  Ex. 5: Ex. 6:  2  22 n1 n n1 nn(2)(5) n

 1  ln(n ) Ex. 7: Ex. 8:  n  n1 23 n1 n 1

Page 107  1  45n  Ex. 9: Ex. 10: sin  n n1 n n1 742

11th ed. p. 620: 5-23 odd (use whatever test is best), 27-34 all

Page 108 Test

  (1) n a (1) n1a Given a series:  n or  n where an  0 (all terms are positive), n1 n1 then the series converges if the following conditions are met:

1. liman  0 (If the limit ≠ 0, then series will diverge by the nth-term test) n

2. a n eventually decreases. (aann1  ) 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1 1.  2.  n1 n n1 21n 

 n12  (1) n n 1 (1)ln1 3.  2 4.   n1 n 1 n1 n

Page 109 Absolute Convergence Test   If the series  an converges, then the series an converges absolutely. n1 n1

(The Ratio and Root Tests are tests for absolute convergence but not the ONLY tests.)

   If an converges BUT  an diverges, then an converges conditionally. n1 n1 n1

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

Go back to No Converges ORIGINAL CONVERGE Yes ABSOLUTELY series

No Yes Converges DIVERGE CONVERGE CONDITIONALLY

Page 110 Determine the convergence (absolute or conditional) or divergence of the following series. Clearly state your reasons.

 (0.1)n  (1) n1 Ex. 1:  (1) n1 Ex. 2:  n1 n n1 n

 (3 n )  sin(n ) Ex. 3: n1 Ex. 4: n  (1) (1) 2 n1 (5 n ) n1 n

Page 111  (1) n Ex. 5:  n1 ln(n  1)

11th ed. P. 629: 9-25 odd, 41-55 odd

Page 112 Ratio and

Ratio Test  Let  an be a series with nonzero terms . n1  an1 1. The series  an converges absolutely when lim 1. n1 n an  an1 an1 2. The series  an diverges when lim 1 or lim . n1 n an n an a 3. The is inconclusive when limn1  1 n an Note: Always use if you have factorials. Also, very useful if you have powers of n.

Root Test  n 1. The series an converges absolutely when liman  1.  n n1   n n 2. The series an diverges when liman  1 or lim an .  n n n1   n 3. The Root Test is inconclusive when liman  1 n  n212n  e2n Ex. 1: Ex. 2:  n  n n1 3 n1 n

Page 113  3n Ex. 3:  n n1 (1)n 

 n n 3 x x Ex. 4: 1 Review: lim 1e  n  n1 n n

 nn  (!)n 2 Ex. 5.  Ex. 6.  n1 n! n1 (3n )!

Recall limn n  1 11th ed P. 637:17-35 odd (omit 33), 39-51 odd n Page 114 Taylor and Maclaurin Series

 2 n Given f ()xaaxcaxc01 ()()...()  2   axcn  is differentiable (and therefore n0 continuous) on the interval (,)cRcR then the following is true:

(0) 2 3 4 f() x a01 axc ()()  axc 2 axc 3 ()  axc 4 ()  (1) 2 3 fxaaxcaxc()12  2 (  )  3 3 (  )  4 axc 4 (  ) 

(2) 2 fx() 2 a23  3!( axc  )  43  axc 4 (  )  (3) fx() 3! a34  4!( axc  )     ()n 2 f() x na !nn  ( n  1)! a12 ( xc  )  ( n  2)! a n ( xc  ) 

Evaluating each of these derivatives at x  c yields

(0) f ()ca 0!0 (1) f ()ca 1!1

(2) f ()ca 2!2 (3) f ()ca 3!3 ()n and, in general, f ()cna !n . By solving for an , you can find the coefficients of the power f n ()c series representation of f ()x are: a  . n n! fc() f n ()c f ()ca , f ()ca ,  a , . . . , a  0 1 2! 2 n n!

Definitions of nth Taylor Polynomial and nth Maclaurin Polynomial

If f has n derivatives at c, then the polynomial ()n fc() 23 fc() f() c n Px()()()()() fc fcxc  xc  () xc  ()xc  n 2! 3! n! is called the nth Taylor polynomial for f at c . If c  0 , then ()n ff(0)23  (0) f(0) n Px() f (0) f (0) x  x  x  x n 2! 3! n! is called the nth Maclaurin polynomial for f .

Page 115 Derivation of series for ex, cosx, sinx , and lnx f ()xe x f ()xx cos() and f ()xx sin()

f ()xx ln()

Page 116 Maclaurin Series for a Composite Function:

Find the Maclaurin series. 1. f ()xx sin 2 2. f ()xx cos 3. f ()xe 3x 4. f ()xx ln(12 )

Find nth degree Taylor or Maclaurin series

1. fx() e x , n 5, c 0 2. f ()xxnc cos , 4, 0

Page 117 3. f ()xxnc3 , 3, 8

11th ed. P. 648: 17,18,19, 27, 29, 31 and p. 677: 27-35 odd Page 118

Definition If x is a variable, then an infinite series of the form  nn23 axnn a01  ax  ax 2  ax 3  ax  n0 is called a power series. More generally, an infinite series of the form  nn2 axcnn()  a01  axc ()()   axc 2   axc ()   n0 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  n fx() an ( x c ) n0 where ()xc 0 is defined to  1when xc 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 2 n fc() a01  ac ()()  c  a 2 c  c  an () c  c 

a0 000

 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 xc and R = the radius of convergence, exactly one of the following is true: 1. The series converges at xc . The domain is only xc . R = 0, and the interval of convergence is [c]. 2. The series converges absolutely for xc R and diverges for xc R If xc 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.  xn Ex. 1:  n1 n

 nx2 n Ex. 2:  n n1 2!n

Page 120  Ex. 3:  nx!( 4)n n1

 (1) nnx 1 Ex. 4:  n0 3  n

Page 121  (3x  2)n Ex. 5:  n1 n

11th ed. p. 658: 15-33 odd, and Worksheet

Page 122 Power Series-Worksheet

Find the interval of convergence for each series.

  (1) nnx 1.  xn 6.  n0 n0 n!   4nnx2 2.  (1)(4nnx 1) 7.  n0 n1 n  (2)x  n  xn 3. 8.  n  2 n0 10 n0 n  3  nxn  nx(3) n 4. 9.  (2)n   n n0 n0 5  xn  nxn 5. 10.  n  n n1 nn3 n0 3

Answers:

1. (1,1) 6. (,)  1 11 2.  ,0 7.  , 2 22 3. (8,12) 8. [1,1) 4. (1,1) 9. (8,2) 5. [3,3] 10. (3,3)

Page 123 Representation of Functions as Power Series

Geometric Power Series

1 Consider the function fx() . The form of this function closely resembles the sum 1 x  a of a geometric series: arn ,0 r 1. In other words, when a 1 and rx , a n0 1 r 1 1  power series representation for centered at 0 is  ar n 1 x 1 x n0    xn n0 1...,1xx 23  x  x 

Finding a Geometric Power Series centered at 0

1 5 Ex. 1: fx() Ex. 2: fx() 1 x2 23x 

Page 124 x3 Ex. 3: fx() x  2

3 Ex. 4: fx() xx2 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.

 n If the power series axn () c has a radius of convergence R  0, then the function n0  2 n defined by f ()xaaxcaxc01 ()()...()  2   axcn  is differentiable (and therefore n0 continuous) on the interval (,)cRcR and the following is true:

  21n 1. f ()xa12 2 axcaxc (  ) 3 3 (  )...  anxcn (  ) n1

()x cxc23 () () xc n1 2. fxdxC() a ( x  c ) a  a ...  C a  01 2  n 23n0 n  1

Express the following as a power series: Ex 1: f ()xx ln(1 ) Ex 2: f (xx ) arctan

Page 126 1 Ex 3: fx() 1 x 2

What if we multiplied problems 1-3 by x? 1  1 Ex 4:  7 dx 0 1 x

Assignment: Worksheet

Page 127 Power Series Representation

Find a power series representation for the function:

1 4 1. fx() 10. fx() 1 x 4  x2

2 3x2 2. fx() 11. fx() 3  x 9  x 3 3. fx() 12. fx() ln( x 4) 21x  13. f (xx ) arctan(2 x ) 1 4. fx() 25x  1 14. fx() x  3 2 4  5. fx() 32x  1 15. fx() 2 3x 12 x 6. fx() x2 x 2 1  1 47x  16.  5 dx 7. fx() 2  1 x 232x x 0

x 8. fx() 2 9  x x 9. fx() 21x2 

Page 128 Power Series Representation—Answers

1  x  1. f ()xx (1)nn 9. f ()xx(2)nn21  2  1 x n0 21x  n0

  n 2 xn 41 2. fx() 2 10. f ()xx2n 33 x  n1 2  n0 44 x n0 

 22 n 3 n 3x x 3. f ()xx 3 (2) 11. fx() 21x    21n n0 93 x n0

 1(2)x n  xn 4. fx() 12. fx( ) ln( x 4) ( 1)n ln 4 25x   5n1  n1 n0 n0 4(n  1)

 4(3x)n  x22n 5. fx()(1)n 13. fx( ) x arctan(2 x ) ( 1)nn 221 32x   2n1  n0 n0 21n 

321x  n1 6.fx ( )  11 n1 2 14.f (xn ) 2  x x xxx221   x  3 3 n n1 nnx  n2 (1) x 1 n 2 OR  (1)n x nn00  3 n0  47x  2 3 7.fx ( )   2 1 232(21)2xx x  x  15.f (xn ) 2nn11x 2  n 12 x n1 nn1 nx  (2)x 3 ( 1) n1 2 nn nn00 OR 2( n 1) x n0  21n xxn 8. fx()(1) n 99 x21 n 1 1  1 n0 16.   5 dx   0 151x n0 n

Page 129