CALCULUS II [Type the document MATH 2414 subtitle]
Turnell
2018
Review of Cal I
Trig Derivatives: 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 Power Rule: []xnn nx 1 dx f ()xx432 3 x 6 x x
d Product Rule: []uv uv vu dx f ()xx 3 cos x
du vuuv dT BTTB Quotient Rule: OR dx v v2 dx B B2 x3 1 fx() x2 2
Page 1 d n Chain Rule: unuu n1 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 xxyy3245439 dx
dy Find , given cos(xy ) y3 4 dx
Integrals: f ()xdx= a set of antiderivatives Why a set? Review of Basic Forms u n1 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 Calculus I Review Worksheet
dy Find the derivative ()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 integral: 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 ee42xx1 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 1x2 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 dy2 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 1x 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 Area of Region Between Two Curves
Review from Cal 1: The area under a curve.
Write the integral or integrals 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 Distances
Example 1: yx4 2 . Find the distance 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 line: 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 yxx2,2 0 Example 7: Find the distance from yxx2,2 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 Volume: The Disk Method
Review from Geometry: The volume of a cylinder
2 Vrh
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 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 radius rx(), produces: n 2 Volume of solid [(rxi )] x i1 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 i1 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. yx4 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. yx4 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: yxx4 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: yxx4 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: yx2 , 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 Length
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. yx23,[0,9]2
Page 23 x3 1 2. y ,[1,2] 62x
x5 1 3. y ,[2,5] 10 6x3
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 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 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 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 21x 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:
yx2,1 x 4 about the x-axis
Page 27 Find the surface area obtained by revolving the given curve about the indicated axis:
xy2 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 arc length)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. yxx2 ,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. yxxx2,2 about the x-axis 22
y3 5. x ,0y 1 about the y-axis 3
15 6. xyy24 , 0 about the y-axis 4
7. yx1,012 x about the y-axis
8. xy122 ,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 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 01122mx 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 ynnmx11 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 mm12 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 parabola 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 86xx2
6 dx 34xx 42
25x dx xx2 22
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.
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 xx2211 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 1tan22 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 xx1 0
t 9. cot(3 7x )dx 11. eedcsc( 1) 13. sec dt 15. csc(x ) dx 3
w ln 2 x2 2 17. 2xedx 19. evdvtanv sec 2 21. 3x1dx 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 xx22 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. 6sec1 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. sec1 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 1xC 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 Integration by Parts
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 circles? 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 TANGENT 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: xABC1 (1)(2)xx22 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)2xx22 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)2xx22 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 xx1
Page 62 L'Hôpital's Rule
Review of Limits 23x x2 16 45x2 x lim lim lim x2 x 4 x4 x 4 x 13 x2
Indeterminate Forms
0 0 1 00 0 0
2cos x 2 23x lim lim x2 x2 4 x2 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 xagx() xa g () x
Find the limit: ln x4 cos x x2 1. lim 2. lim 3. lim x x3 x01tan x x ex
Page 63 1 10 3 4. limx tan 5. lim xx x0 x x2
x 1 1 6. lim 1 7. lim x x1 2 xx x1
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. xt2 3 Ex. 1 yt21
Page 69 3 xt2 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 slope 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. xtytt2sin(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 Surface of Revolution
If a smooth curve x ft() and ygt (), atb , is traversed exactly once as t increases from ta to tb , then areas 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 Sy2 dt Sx2 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. xycos , 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-Tangents)
Page 74 Parametric—Tangents
Find all vertical and horizontal points of tangency. x costt sin t 1. , 02t ytttsin cos
x 1 t 2. , t yt3 3 t
x tt2 2 3. , t yt3 3 t
x 42cost 4. , 02t yt1sin
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. xt23211, t ytt (4)3(4)7 tt2 2 when t 2
7. xt3 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. yx312
Vertical: t ,1 22
33 t ,1 22
2. Horizontal: t 1(2,2) 6. yx24 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 ra2cos ra2sin
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 rabcos rab sin r 13cos r 22sin r 32cos r 42sin
Limacon: https://youtu.be/MT1j6vdX2aI
Lemniscate (a figure 8) : ra22cos(2 ) ra22 sin(2 )
r 2 25cos(2 ) r 2 16sin(2 )
Rose Curves: racos( 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 slopes 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 circle 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 i1 2
n 1 2 Aflim [ (i )] n 2 Taking the limit as n , i1 1 [()]fd2 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 1n1 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 1e lim 0 n n n n!
Page 93 Factorials: 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 Series
Infinite Series:
Definition: If an is an infinite sequence, then aaakn123 a a is an infinite series. k1
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 kk11n When this limit exists, we say that the series converges; otherwise, it diverges. Saaann123 a
1111 Ex. 2: Ex. 1: 1 n 248 11111 1 n0 2 k1 kk(1)261220nn (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 SSlimnk lim a nn n1 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. n1
If {}Sn diverges ( lim Sn does not exists), then the series an diverges also. n n1 **Problem: Sn is often difficult if not impossible to calculate.
Special Series Geometric Series: 234nn 1 1 aarararar ar ar , a 0 n1 where a is the first term and r is the common ratio. 5 15 45 135 ______n1
11 1 1 ______3 9 27 81 n1
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 n1 2 Ex. 1: 53 Ex. 2: 3 n1 n1 5
Page 97 2n n 1 1 3 Ex. 3: (1) Ex. 4: (1) n1 n 31n n1 3 n1 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: n1nn1
n 1 Ex. 6: ln n1 n
Page 98 6 Ex. 7: n1 (2nn 1)(2 1)
Nth Term Test an the asn ' must be getting smaller and smaller if the series is going to converge. n1 However, if they are getting smaller it does NOT mean an will converge. n1 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 n1 21n
Page 99 Determine if the series converges or diverges. If it converges, find its sum (if possible).
n n1 Ex. 8: 2 Ex. 9: 1 n n0 n1
cos(n ) 23nn Ex. 10: Ex. 11: n n n1 5 n1 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 anbe an infinite sequence of positive terms. If afnn (), where f is continuous, positive, and decreasing on [1, ) , then a and f ()xdx n 1 n1 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: n1 n
1 Ex. 2: 2 n1 n 1
**Note about sum: Page 101 n Ex. 3: 2 n1 n 1
Power Series (P‐Series) p 1111 These are series in the form: 1 ,where p 0 ppp nn11n 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 n1 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 aa1 x
Page 102 Using the power rule to integrate, we get: p 1 a a p x limxdx lim aa1 p 1 1 Now substituting in our values of a and 1: a xappp111 1 lim lim aappp111 1
What happens when p 1 ? In other words, what happens when p is negative? a xappp111 1 lim lim ppp111 aa 1 a pp111 lim pp11 a And the integral diverges.
But what happens when p 1 ? In other words, what happens when p is positive? a xappp1111 lim lim aa(1)ppp (1)(1) 1 a11pp1 lim a(1pp ) (1 ) 0 111 Since (1-p) will now be a lim negative value, we can move a p1 ap(1 ) (1p ) 1 p it to the denominator.
And the integral converges.
Conclusions: 1 1 converges if p 1 diverges if p 1 p p n1 n n1 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 n1 n1 n 3
1 n! Ex. 6: Ex. 7: 5 n1 n n1 n
11th ed. P. 613:3, 5, 7, 17, 21, 25, 27, 33, 35, 37, 69-79 odd
Page 104 Comparisons of Series Direct Comparison Test Let 0 abnn for all n . 1. If bn converges, then an converges. n1 n1 2. If an diverges, then bn diverges. n1 n1
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. n1 n1 2. If the L 0 and bn converges, then an converges. n1 n1 3. If the L and bn diverges, then an diverges. n1 n1 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. n 1 Ex. 1: by DCT and by the LCT 4 n1 n 2
Page 105 1 Ex. 2: n1 2 n
n 1 n 1 Ex. 3: ,by DCT Ex. 4: ,by LCT 2 2 n1 n 2 n1 n 2
Page 106 1cos() n 53nn3 Ex. 5: Ex. 6: 2 22 n1 n n1 nn(2)(5) n
1 ln(n ) Ex. 7: Ex. 8: n n1 23 n1 n 1
Page 107 1 45n Ex. 9: Ex. 10: sin n n1 n n1 742
11th ed. p. 620: 5-23 odd (use whatever test is best), 27-34 all
Page 108 Alternating Series Test
(1) n a (1) n1a Given a series: n or n where an 0 (all terms are positive), n1 n1 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. (aann1 ) 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) n1 (1) n1 1. 2. n1 n n1 21n
n12 (1) n n 1 (1)ln1 3. 2 4. n1 n 1 n1 n
Page 109 Absolute Convergence Test If the series an converges, then the series an converges absolutely. n1 n1
(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. n1 n1 n1
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) n1 Ex. 1: (1) n1 Ex. 2: n1 n n1 n
(3 n ) sin(n ) Ex. 3: n1 Ex. 4: n (1) (1) 2 n1 (5 n ) n1 n
Page 111 (1) n Ex. 5: n1 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 . n1 an1 1. The series an converges absolutely when lim 1. n1 n an an1 an1 2. The series an diverges when lim 1 or lim . n1 n an n an a 3. The Ratio Test is inconclusive when limn1 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 n1 n n 2. The series an diverges when liman 1 or lim an . n n n1 n 3. The Root Test is inconclusive when liman 1 n n212n e2n Ex. 1: Ex. 2: n n n1 3 n1 n
Page 113 3n Ex. 3: n n1 (1)n
n n 3 x x Ex. 4: 1 Review: lim 1e n n1 n n
nn (!)n 2 Ex. 5. Ex. 6. n1 n! n1 (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 ()xaaxcaxc01 ()()...() 2 axcn is differentiable (and therefore n0 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)! a12 ( 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 ()xxnc3 , 3, 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 nn23 axnn a01 ax ax 2 ax 3 ax n0 is called a power series. More generally, an infinite series of the form nn2 axcnn() a01 axc ()() axc 2 axc () n0 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 ) n0 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: n1 n
nx2 n Ex. 2: n n1 2!n
Page 120 Ex. 3: nx!( 4)n n1
(1) nnx 1 Ex. 4: n0 3 n
Page 121 (3x 2)n Ex. 5: n1 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. n0 n0 n! 4nnx2 2. (1)(4nnx 1) 7. n0 n1 n (2)x n xn 3. 8. n 2 n0 10 n0 n 3 nxn nx(3) n 4. 9. (2)n n n0 n0 5 xn nxn 5. 10. n n n1 nn3 n0 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 n0 1 r 1 1 power series representation for centered at 0 is ar n 1 x 1 x n0 xn n0 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 n0 2 n defined by f ()xaaxcaxc01 ()()...() 2 axcn is differentiable (and therefore n0 continuous) on the interval (,)cRcR and the following is true:
21n 1. f ()xa12 2 axcaxc ( ) 3 3 ( )... anxcn ( ) n1
()x cxc23 () () xc n1 2. fxdxC() a ( x c ) a a ... C a 01 2 n 23n0 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 n0 21x n0
n 2 xn 41 2. fx() 2 10. f ()xx2n 33 x n1 2 n0 44 x n0
22 n 3 n 3x x 3. f ()xx 3 (2) 11. fx() 21x 21n n0 93 x n0
1(2)x n xn 4. fx() 12. fx( ) ln( x 4) ( 1)n ln 4 25x 5n1 n1 n0 n0 4(n 1)
4(3x)n x22n 5. fx()(1)n 13. fx( ) x arctan(2 x ) ( 1)nn 221 32x 2n1 n0 n0 21n
321x n1 6.fx ( ) 11 n1 2 14.f (xn ) 2 x x xxx221 x 3 3 n n1 nnx n2 (1) x 1 n 2 OR (1)n x nn00 3 n0 47x 2 3 7.fx ( ) 2 1 232(21)2xx x x 15.f (xn ) 2nn11x 2 n 12 x n1 nn1 nx (2)x 3 ( 1) n1 2 nn nn00 OR 2( n 1) x n0 21n xxn 8. fx()(1) n 99 x21 n 1 1 1 n0 16. 5 dx 0 151x n0 n
Page 129