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# CALC ANALYT GEOM II MATH 125

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This 13 page Class Notes was uploaded by Addison Beer on Wednesday September 9, 2015. The Class Notes belongs to MATH 125 at University of Washington taught by Staff in Fall. Since its upload, it has received 5 views. For similar materials see /class/192084/math-125-university-of-washington in Mathematics (M) at University of Washington.

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Date Created: 09/09/15

71 75 Review Integration Techniques My reviews and review sheets are not meant to be your only form of studying It is vital to your success on the exams that you carefully go through and understand ALL the homework problems7 worksheets and lecture material Hopefully this review sheet will remind you of some of the key ideas of these sections 1 Integration by Parts 0 Integration by Parts Formula fudv uv 7 fvdu and f udv 11le 7 f vdu 0 Understand how to perform integration by parts and how to choose your u Remember u should be easy7 to differentiate and d1 should be easy7 to antidifferentiate To help you decide we discussed LIPET Logs7 Inverse trig7 Powers of X7 Exponentials7 Trig let u the part of the integrand that occurs earlier in LIPET 0 There is no one rule that tells you when to use which methods But here are some instances when integration by parts tends to work 7 Products speci cally powers of 1 times sinz7 cosz7 or 6 It also works with 6 times sinz or cosz but you have to pay attention to when the process repeats 7 Logs integrals involving logs take u lnz 7 Inverse Trig integrals involving inverse trigonometric functions take u inv trig7 2 Trigonometric Integrals 0 There is one general theme Rewrite the integral using identities and use usubstitution if needed 0 The identities that are most useful are cos2z sin2z 1 7 which we may use as cos2z 1 7 sin2z or sin2 1 7 cos2z Dividing these by cos2z gives more useful identities 1tan2z sec2z 7 which we may use as 1 sec2z 7tan2 or tan2z sec2z 71 We will also need the half angle identities 1 1 1 1 1 sin2 E 7 i cos21 7 cos2z E i cos21 7 and sinz cosz i sin21 Page 484 and 486 discuss strategies for when to use which identities and why For me7 it is most useful to attempt to set up a usubstitution7 this gives three possibilities for sinz and cosz and three possibilities for sec andtan z o Integrals with sinz and cosz a Using u sinz and du coszdz7 look for an odd power of cosz and pull out one actor Then use the identites to turn the rest of the problem into sinz7s b Using u cosz and du 7 sinzdz7 look for an odd power of sinz and pull out one factor Then use the identites to turn the rest of the problem into cosz7s c If both powers are even7 the above two methods don7t work Use the half angle identities to simplify the problem 0 Integrals with secz and tanz a Using u tanz and du sec2zdz7 look for an even power of secz and pull out two factors Then use the identites to turn the rest of the problem into tanz7s b Using u secz and du secz tanzdz7 look for an odd power of tanz and pull out one factor of tanz and one factor of secz Then use the identites to turn the rest of the problem into secz7s c If neither of the above cases work you may need to use identites or integration by parts Also it may be useful to have the following integrals tanz dz ln 1 seczl C and sec dz lnl secz tanzl C 3 Trigometric Substitution o This is a different type of substitution We let I asin19 z atan19 or z asec19 These methods often work well for the speci c integrals that have xaQ 7 I2 V a2 12 or V12 7 a2 somewhere in the problem 0 The basic idea Replace z with a trigonometric function do the substitution use a trig identity and hope for the best the result is a trig integral which we studied in 72 These steps will ensure that you get rid of the square root 0 Page 490 of the text has a table reminding you of which trig function to use along with the corresponding identity 0 At the end of the problem you will have a solution in terms of 19 To get back to I you need to know the triangle trick7 as is discussed in the text and we will discuss it in class 4 Partial Fractions 0 The basic idea Break up rational functions into sums of factors and integrate each sepa rately This is successfully because each individual part can be integrated easily If we have a quadratic that factors then we immediately factor it from the beginning For example 1 7 1 z2i6z5 7 171175i Then we will have a method for breaking this up and integration each part separately by using the rst set of integrals below If we have a quadratic that doesn7t factor then we will need to complete the square and use what we know about the second integrals below For example 1 7 1 7 1 z2z1 z2z14141 z12254 Then we will use usubstitution along with second and third integrals below 0 Thus we are trying to simplify into integrals in an easier form We will use all the following 2 facts 9 1 dzlnlziblC 17b 1 1 7zib2 dz7zibC 1 1 1 2 de 0 And so on b 1 1 1 u 7u2a2 duigtan ltggtC 1 12a2dz lnlz2a21c 0 Here s how to do it a If the highest power on top is bigger than or equal to the highest power on the bottom DIVIDE b Factor the denominator where possible c Complete the square for quadratics that don7t factor d Write out the general partial fraction expansion For example BLAH 7 A B C Dzi2E 111732zi227 mm 35732 17227 e Multiply both sides by the common denominator and solve for A B C D and E either y expanding and comparing coef cients or plugging in special values for I f lntegrate each part separately using the integrals above Here are a couple extra examples 7 Simple linear factors 12 2 d 7 122 d 7 122 d 133I22z I fxx3x2 I xac1x2 I 7 A B c 7f3x1x2 d1 Now we multiply the denominator out to get 12 2 Az 1z 2 Bzz 2 Czz 1 We plug in strategic values for z to nd A B and C z0gives22AsoA1 z 71 gives 3 7B so B 73 172 gives62CsoC3 Now we nish integrating 1 73 3 11mdzlnz73lnz13lnz2C 7 A quadratice factor I 1 md1 fmihdr w lmdr A Bx12C x ac12254 Now we multiply the denominator out to get I 1 Az 122 54 Bz 12 Cz We plug in strategic values for z to nd A B and C I 0 gives 1 64A so A 23 I 712 gives 12 A54 C712 so 12 2354 C712 so C 23 Now there is not an obvious choice of I but we can choose some other value or equate coef cients to nd B I 12 gives 32 2394 B 2312 so 0 12B 13 so B 723 Now we nish integrating LET u I 12 I i f7 i ti dz 1nltzgt f ilt gt3 dz Inltzgt f 3 dz g 1111 i 23 f Mrs4 dx 23 f Mia4 dz 1nz 7 231nu2 54 23 tan 1ltL C 1nz 723lnz 1225423 tan 1 C 5 Choosing and Combining Methods 0 The best way to learn which methods to use is to read pages 505 520 and PRACTICE LOTS OF PROBLEMS Problems are readily available in your homework textbook and on the weekly assignment 83 91 93 and 94 Review This review sheet discusses in a very basic way the key concepts from these sections This review is not meant to be all inclusive but hopefully it reminds you of some of the basics Please notify me if you nd any typos in this review 1 Section 83 Center of MassCentroid Be able to nd the centroid of a flat uniformly distributed plate bounded by given functions in the plane o If it is easier to determine the area in terms of I use ffz r dz and g Liam dz 1 7 I la 101 41 la 1 d1 o If it is easier to determine the area in terms of y use d d L WW dd 7 l0 l wl dy d and z 7 d fc y do a u do 0 For the centroid of a region bounded by two curves gt 91 use the following this is written in terms 0 z 7 szmzgt 791 dz and If mz e gltzgt2gt dz 7 b b lalf1 9Il d1 falf1 9Il dz 2 Section 91 Intro to Differential Equations Understand the basic terminology for differential equatlons In particular you should know the definition of all of the following these are loose de nitions f y El 0 A di erential equation is an equation involving a function and its derivatives An ordinary di erential equation ODE involves only two variables independent and dependent and a partial di erential equation PDE involves more than two variables We only talk about ODE s in this course An initial condition is a value of the function that accompanies a differential equation and allows us to nd the undetermined constant in our solution An initialvalue problem is a differential equation with an initial condition 0 The order of a differential equation is the highest derivative that appears in the equation In this course we talk mostly about rstorder differential equations A constant solution or equilibrium solution is a solution of the form c for some constant 5 notice that this causes the derivative to be 0 for all We solve for the equilibrium solution by setting our derivatives equal to zero 0 A solution to a differential equation is a function that satis es the equation There are typically infinitely many solutions which can be written in a general form by including an undetermined constant To check if you have a solution you must plug your function into the left and righthand sides of the differential equation and see if they are equal Differential equations can be used to model many natural phenomena and is a great analytic tool for testing hypotheses about natural phenomena Although this is not a course in mathematical modeling you should know how to interpret the physical signi cance of different parts of a differential equations Perhaps the most useful thing to remember when modeling is that d dig the instantaneous rate of change of y with respect to t7 So for example the differential equation kP says that the rate of change of P is proportional to the size of P 3 Section 93 Separation of Variables This is the only technique that you will learn in this course for solving differential equations A differential equations is a separable equation if you can write it in the form d f gltzgtfltygt I that is if you can separate the variables If you can separate the variables then you can use integrals to solve the differential equation as follows a SEPARATE VARIABLES you may need to simply or factor to start then use only mul tiplication or division to get all the 17s to one side and all the y7s to the other b INTEGRATE integrate each side separately put the undetermined constants all on one side c SOLVE FOR IF POSSIBLE to get a nice clean looking solution we often write our solutions as a function We will then replace all undetermined constants with a new constant d USE INITIAL CONDITION this will allow you to solve for undetermined constants g Section 94 Exponential Growth and Decay Understand the basic exponential differential equation and how to use it to model different situations We will work with the following model d i ky where k the relative growth rate at Solving this differential equation you should be able to do this using separation of variables we get W yoek where yo y0A If k gt 0 then we have exponential growth If k lt 0 then we have exponential decay You should understand our analysis of this basic differential equation 64 and 65 Review My reviews and review sheets are not meant to be your only form of studying It is vital to your success on the exams that you carefully go through and understand ALL the homework problems worksheets and lecture materiali Hopefully this review sheet will remind you of some of the key ideas of these sections 1 Work 0 Understand the basics of how to use integrals to study work o If z represents a distance and represents the force on an object at a given distance then b the work required to move the object from z a to z b dz a That is I Work Force dz 0 We also used Newton s Second Law F mar Pay attention to units I discuss the differences in class 0 We looked at three major work examples The rst springs t the above description nicely The other two required a little more work to derive a Springs if z distance beyond natural lengthi ii Force kz where k is the spring constant you must nd iiil Work to strecth from z a to z 1 beyond natural length77 f kz dz Lifting Cables if z the distance from the top of the building or cliff ii k the pounds per foot for the cable or Newtons per meter iiil Work to lift a slice m kAzdistance lifted iv Work to lift a rope of length z b all the way to the top77 f kz dz v Note Here force is represented by kdz as we discussed in class A 0quot A n V Pumping Liquid if z the distance below the pumps spout distance from the top ii Work to lift pump out a slice mweight of a slicedistance lifted weight per volume volume density gtlt gravity area of fact of a slice gtlt Az iiil Work densitygravityfArea ofa Horizi Slicez dz z a is the distance from the spout to the top of the liquid and z b is the distance from the spout to the lowest part that this being pumped out iv density for water 1000 kgmgi v gravity on Earth 98 ms2l vi For standard units density gravity 625 lbsftgl viii It is often useful to draw the trough7 or container7 on its side Then nd the equations for the curves This will help you nd a formula for the area of a horizontal slicer NOTE For all of these it is good to have an understanding of how we use integral calculusl That is we break up a problem into to small slices and we approximate each slice We approximate in such a way that more slices lead to better approximationsl Finally with the formula for each slice we let the number of slices go to in nity In this manner we can represent the exact answer to a problem in terms of integralsl A CL V 2 Average Value 0 Understand how to compute the average value of function on a given integral and how this value relates to areas and the mean value theoremi o If is a function7 then we de ne the average value of from I a to z b by b favebiaa 0 We can interpret fave as the height we should use to get a perfect approximation of the area under the curve with only one rectanglei o If the function is continuous from I a to z 127 then it must cross the horizontal line which represents the average values This is the Mean Value Theorem for lntegrals which is stated more techniquely as If is continuous from I a to z 127 then there is a value 5 between I a and z b where f dz 64 and 65 Review My reviews and review sheets are not meant to be your only form of studying It is vital to your success on the exams that you carefully go through and understand ALL the homework problems worksheets and lecture materiali Hopefully this review sheet will remind you of some of the key ideas of these sections 1 Work 0 Understand the basics of how to use integrals to study work o If z represents a distance and represents the force on an object at a given distance then b the work required to move the object from z a to z b dz a That is I Work Force dz 0 We also used Newton s Second Law F mar Pay attention to units I discuss the differences in class 0 We looked at three major work examples The rst springs t the above description nicely The other two required a little more work to derive a A 0quot V Springs if z distance beyond natural lengthi ii Force kz where k is the spring constant you must nd iiil Work f kz dz Lifting Cables if z the distance from the top of the building or cliff ii k the pounds per foot for the cable or Newtons per meter iiil Work f kz dz iv Note Here force is represented by kdz as we discussed in class Pumping Liquid if z the distance below the pumps spoutl ii Work densitygravityfArea of a Horizl Slicez dz iiil density for water 1000 kgmgi iv gravity on Earth 98 ms2l v For standard units density gravity 625 lbsftgl vi It is often useful to draw the trough7 or container7 on its side Then nd the equations for the curves This will help you nd a formula for the area of a horizontal slicer NOTE For all of these it is good to have an understanding of how we use integral calculusl That is we break up a problem into to small slices and we approximate each slice We approximate in such a way that more slices lead to better approximationsl Finally with the formula for each slice we let the number of slices go to in nity In this manner we can represent the exact answer to a problem in terms of integralsl 2 Average Value 0 Understand how to compute the average value of function on a given integral and how this value relates to areas and the mean value theoreml o If is a function7 then we de ne the average value of from I a to z b by b fave dz 0 We can interpret fave as the height we should use to get a perfect approximation of the area under the curve with only one rectangle o If the function is continuous from I a to z 127 then it must cross the horizontal line which represents the average values This is the Mean Value Theorem for lntegrals which is stated rnore techniquely as If is continuous from I a to z 127 then there is a value 5 between I a and z b where fa dz 71 75 Review Integration Techniques My reviews and review sheets are not meant to be your only form of studying It is vital to your success on the exams that you carefully go through and understand ALL the homework problems7 worksheets and lecture materiali Hopefully this review sheet will remind you of some of the key ideas of these sections 1 Integration by Parts 0 Integration by Parts Formula fudv uv 7 fvdu and f udv 11le 7 f vdu 0 Understand how to perform integration by parts and how to choose your ui Remember u should be easy7 to differentiate and d1 should be easy7 to antidifferentiatei To help you decide we discussed LIPET Logs7 Inverse trig7 Powers of X7 Exponentials7 Trig let u be the part of the integrand that occurs earlier in LIPETI 0 There is no one rule that tells you when to use which methodsi But here are some instances when integration by parts tends to wor 7 Products speci cally powers of 1 times sinz7 cosz7 or emf It also works with 6 times sinz or cosz but you have to pay attention to when the process repeats 7 Logs integrals involving logs take u lnz 7 Inverse Trig integrals involving inverse trigonometric functions take u invi trig7 2i Trigonometric Integrals 0 There is one general theme Rewrite the integral using identities and use usubstitution if needed 0 The identities that are most useful are cos2z sin2z 1 7 which we may use as cos2z 1 7 sin2z or sin2 1 7 cos2z Dividing these by cos2z gives more useful identities 1tan2z sec2z 7 which we may use as 1 sec2z 7tan2 or tan2z sec2z 71 We will also need the half angle identities 1 1 1 1 1 sin2 7 7 7 cos2z 7 cos2z 7 7 cos2z 7 and sinz cosz 7 sin2z 2 2 2 2 2 o Page 484 and 486 discuss strategies for when to use which identities and why For me7 it is most useful to attempt to set up a usubstitution o Integrals with sinz and cosz a Using u sinz and du coszdz7 look for an ODD POWER OF cosz and pull out one factor Then use the identites to turn the rest of the problem into sinz7si b Using u cosz and du 7 sinzdz7 look for an ODD POWER OF sinz and pull out one factor Then use the identites to turn the rest of the problem into cosz7si c If BOTH POWERS ARE EVEN7 the above two methods don7t worki Use the half angle identities to simplify the problemi o Integrals with secz and tanz a Using u tanz and du sec2 zdz7 look for an EVEN POWER OF secz and pull out two factors Then use the identites to turn the rest of the problem into tanz7si b Using u secz and du secz tanzdz7 look for an ODD POWER OF tanz and pull out one factor of tanz and one factor of seczi Then use the identites to turn the rest of the problem into secz7si c If neither of the above cases work you may need to use identites or integration by parts Also it may be useful to have the following integrals tanz dz ln l seczl C and sec dz lnl secz tanzl C 3 Trigometric Substitution o This is a different type of substitution We let I asint9 z atant9 or z asect9 for integrals involving the three cases a2 7 12 V a2 12 or V12 7 a2 respectively 0 The basic idea Replace z with a trigonometric function do the substitution use a trig identity and hope for the best the result is a trig integral which we studied in 72 These steps will ensure that you get rid of the square root Sometimes you may have to complete the square to set expressions like a2 7 I b2 you need to know how to do this 0 Page 490 of the text has a table reminding you of which trig function to use along with the corresponding identity 0 At the end of the problem you will have a solution in terms of 9 To get back to I you need to know the triangle trick7 as is discussed in the text and we will discuss it in class 4 Partial Fractions 0 The basic idea Break up rational functions into sums of factors and integrate each sepa rately This is successful because each individual part can be integrated easily For example A A A d Al 712 C d 7d 77 C Iib z nlz l an Iib2 z zib 0 Here s how to do it a If the highest power on top is bigger than or equal to the highest power on the bottom DIVIDE b Factor the bottom and write the general partial fraction decomposition here is an example BLAH 7 A B C 1azb2 m Ib2 c Multiply both sides by the common denominator and solve for A B and C either by expanding and comparing coef cients or plugging in special values for I d lntegrate each part separately 5 Choosing and Combining Methods 0 The best way to learn which methods to use is to read pages 505 520 and PRACTICE LOTS OF PROBLEMS Problems are readily available in your homework textbook and on the weekly assignment Chapter 6 Review As always my reviews and review sheets are not meant to be your only form of studying It is vital to your success on the exams that you carefully go through and understand ALL the homework problems worksheets and lecture material Hopefully this review sheet will remind you of some of the key ideas of these sections Please inform me if you nd any typos on this sheet WARNING Chapter 6 involves techniques for the application of integrals to areas and volumes Al though the techniques are not difficult after a little practice it is difficult to explain and illustrate all situations using only a few sentences on a review sheet I have tried to use common language to describe the techniques that we explore in class and in the homework However without carefully reviewing the lecture notes and working through homework problems this review may be difficult to follow 1 Areas Between Curves 0 Given a region bounded by two curves you should know how to nd the area If you are not given the picture you need to know techniques for guring out 1 where the curves intersect 2 whether to use I or y and 3 which function is on top7 or to the right 7 If the functions are easily expressed as y an expression only involving 1 then it is probably easiest to integrate with respect to z 1 Area between from I a to z b TOP FUNC 7 BOTTOM FUNC dz 7 If the functions are easily expressed as I an expression only involving y then it is probably easiest to integrate with respect to y 1 Area between from y c to y J7 RIGHT FUNC 7 LEFT FUNC dy 7 There are two choices using 17s or using y7s If you are unsure here is how you can try each a Draw two pictures 1 Draw a typical vertical rectangle7 with small width on the side closest to the zaxis and 2 Draw a typical horizontal rectangle7 with small width on the side closest to the yaxis b Choose the picture that always has the height of the rectangles bounded by the same two curves 96 If you choose 1 use dz and write the equations and endpoints in terms of I Write the integral as top function7 bottom function7 If you choose 2 use dy and write the equations and endpoints in terms of y Write the integral as right function7 left functionl 2 Volumes by the Slicing Method 0 Be able to effectively nd volumes using the method of slicing Usually we apply this methods to volumes of revolution a Draw a picture and imagine that you slice the volume across the axis of revolution If you slice across the zaxis then the width corresponds to dz If you slice across the y axis then the width corresponds to dy This means use I or y respectively throughout the rest of the problem b Visualize the crosssection and write a formula for the area of the cross section 7 If the crosssection is a circle Area of Slice7 7rradius function2 7 If the crosssection is a washer Area of Slice7 7router radius function2 7 7rinner radius function2 c Evaluate the integral using dz or dy depending on the your rst step 1 Volume Area of Slice7 a 3 Volumes by the Shells Method 0 Be able to effectively nd volumes using the method of cylindrical shells a Draw a picture and imagine that you draw a pop can7 with thin outer shell around the axis of revolution If the width of the outer shell corresponds to z use dz this also is when you rotate around the yaxis If the width of the outer shell corresponds to y use dy this also is when you rotate around the zaxis b Visualize the cylindrical shell and write a formula for the height and the circumference 7 The height function is simply the top function7 bottom function or right func tion7 7left function depending on whether you are using z or y 7 The circumference is 27rradius formula The radius formula is the formula for the distance from the axis of rotation to the edge of the shell This is often of the form z 7 h or y 7 k where z h or y k is the axis of rotation In fact in many problems the radius is just z or y because we rotate around an axis c Evaluate the integral using dz or dy depending on the your rst step 1 Volume 27f radius formula7 height formula7 4 Miscellaneous 0 These sections can seem a little overwhelming at rst But truly there are only a few simply principles at work Although it may seem easier initially to memorize each part of each techniques separately it is far better for now and for later in this class that you understand how to reason through a problem by using typical approximating rectangles7 typical approximating slicesl and typical approximating shells7 If you can comfortable raw and visualize all of these then you should be able to easily read off all the information you need including whether to use dz or dy what the arearadiusheight formulas are and which endpoints to use 7 For instance when I need to evaluate a volume I don7t have all the techniques memo rized and l don7t like to waste my time trying to look through textbooks for an example to guide me What I typically do is to draw a picture and rederive the basic ideas For example if I wanted to use shells I would write 1 Volume circumference77height7 Then I would let my picture guide me to the end of the problem 0 Even so I nd that some students struggle with these ideas initially As a check on your work you can remember the following this tables only tells you which variable to use if you are using the designated method and using either the z or yaxis to rotate Axis of Rotation Axis of Rotation Method zaxis y axis b d Slicing Volume Azdz Volume Aydy b d Shells Volume 27ryfydy Volume 27rzfzdz

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