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Calculus II

by: Braeden Lind

Calculus II MA 241

Braeden Lind
GPA 3.93

Thomas Wears

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Thomas Wears
Class Notes
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This 16 page Class Notes was uploaded by Braeden Lind on Thursday October 15, 2015. The Class Notes belongs to MA 241 at North Carolina State University taught by Thomas Wears in Fall. Since its upload, it has received 38 views. For similar materials see /class/223698/ma-241-north-carolina-state-university in Mathematics (M) at North Carolina State University.

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Date Created: 10/15/15
0 4 Cannot had LH 1c JLW a 39 NM 6 397 if 03 bp 9 en drdi 5quot funk E D c E ro homeng EM A at Hn L ou6w n7 Con0xhvtf is mud Mr WM team 4 600 a we 571 3 HJL worK Veeruhmk Cr Jro bun HA wailk Mi of 4121 1M MA 241 Spring 2009 Review For Final Exam April 22 2009 Wears DISCLAIMER This is intended to be used as a study guide for the nal exam in MA 241 009 It is intended to put as much information in place as possible When crossed with your in class notes7 I suspect that this could be quite useful No effort has been made in ensuring that everything covered in the course appears in this study guide No effort has been made in trying to make this study guide 77All Inclusive7 for the nal exam It is probably impossible to learn any of the material from these notes7 but these notes should help you remember material already learned but possibly forgotten T Mistakestypos are quite possible7 so if something seems wrong7 please ask A list of topics That come to mind immediately Integration by parts U sub Trigonometric lntegration Partial Fraction Decomposition lmproper integrals Area bounded by curves Volumes of Solids of revolution ARC LENGTH Average Value Work and other applications to physics Differential Equations What does it mean for a function to be a solution of a Differential Equation Euler7s Method Unique solutions lnitial Value Problems Separable Differential Equations and How To Solve Them Exponential Growth and Decay Newton7s Law of Cooling 2nd order linear differential equations Homogeneous 2nd order and Non homogeneous 2nd order Particular solutions7 complementary solutions7 general solutions and how does every thing relate H 3 Applications to Physics Motion of a mass on a spring Sequences What does it mean for a sequence to be increasing or decreasing How do you show that a sequence is increasing or decreasing In nite Series Geometric Series P series Harmonic Series Integral Test Comparison Test Limit Comparison Test Alternating Series Alternating Series Test Absolute Convergence Ratio Test Power Series Radius of Convergence Interval of Convergence Representations of functions as power Series Taylor Series centered at X a for a function fX how to nd it Taylor Series centered at X 0 Maclaurin Series for sinx cosz em tan 1x Obtaining Taylor series centered at X 0 for functions related to the ve listed above Differentiation and Integration of power seriesTaylor Series representations of func tions Integration by Substitution USub and Integration by Parts 55 and a Both topics will be tested repeatedly throughout the eXam in the course of inte gration andor solving differential equations Trigonemtric Integration 5 and notes online also see notes from 42209 a sinmx cos x dz with m odd b sinmx cos x dz with 71 odd c tanm sec dz with 71 even 3 Integration by Trigonometric Substitution 57 a As wonderful as this is and it is wonderful7 I currently just don7t see a place for this on the nal exam 4 Partial Fraction Decomposition 57 and Appendix G a Given a proper rational function7 we can always integrate the function using P Partial Fraction Decomposition Assume f Q62 where Pz and Q P are polynomials with degPx lt degQx The we decompose fa QEx m which allows us to integrate i The denominator7 Qx7 always factors into a product of linear factors tn b7 andor irreducible quadratic factors a2 lnc with b2 7 4ac lt 0 ii For each factor in the denominator7 we 77build up77 to the multiplicity of the factor A Each linear factor of multiplicity k tn bk contributes L l i l l L tn b1 tn 1 tn bk PW to the decomposition of QW B Each lrreducible quadrate factor of multiplicity k th in ck con tributes A1B1 A2B2 tn in c1 tn in C2 tn in ck t t Pltzgt to the decomposition of QW PW b To solve dz one Qltzgt i Sets up the decomposition of E QW ii Solves for the unknown variables using the method of undetermined coef cients A Clear Denominators7 B Expand the RHS7 C Group and gather like terms on the RHS7 D Equate coef cients between the RHS and LHS iii Integrate the decomposed form of 3 2 C Example Ch 43822x5 i 2 2 1 2 2 1 PW QM 39 decomposes as 4x38x22x57A B CD 2 2 1 z 2 x21 A Clearing Denominators produces 4x3 8x2 2x 5 Azx2 1 B2 1 Ox D2 B Expanding the RHS and gathering like terms produces 43822x5ACz3BDz2AzB C Equating Coef cients yields7 ii Thus7 4z38z22z5 7 2 5 4AO 8BD A B iii Finally7 we integrate the Decomposed form 43822x5 dx i 7 23 221 ix x2 x2139 2 5 23 7 i d xx2z21 z 2 2 1 5 21n117 E lnz2 1 3tan 1z K DH 4 AAAAA U 0 VVVVV 7 5 Improper Integrals 510 a Type I In nite intervals ii if Suppose f is is de ned for all t 2 a ie7 all t on the interval acgto7 then de ne 0 f dx tlirgio f dz Suppose f is is de ned for all t S a ie7 all t on the interval 70041 then de ne a a fx d tlim f ds 700 H OO t 00 I The improper integrals f dx and f dx are convergent if the I 00 corresponding limit exists and divergent if the corresponding limit does not exist it When f dx is de ned for all 51t7 then de ne fltzgtdzfltzgtdz Ooofds noosofd gigOt 00 In this case7 f is convergent if and only if both of the 77split pieces 700 are convergent on their own Note the choice of splitting the integral at z 0 00 is arbitrary You are free to split f at any number you wish7 even if this number happens to be 612 The0 result is well de ned iv Example 00 x d 1 t x d d 8 lIIl 0 ltz22gt2 we 0 ltz22gt2 2 1 t2du m7 i taoo2 2 u2 l 1 t22 lim 7 7 taco 2 u 2 u sub with u 2 2 9 10 11 1 Thus7 L dx is convergent to 7 0 2 22 4 b Type ll Integrating overat a discontinuity i Suppose f is continuous on ab with a discontinuity at b7 then de ne abfx dx tlirlnatf dz ii Suppose f is continuous on 17 b with a discontinuity at 17 then de ne b b afx dztlir3tfzdz b In either case7 the improper integral the corresponding limit exists and divlozargent if the limit does not exist and 01 with a discontinuity at b Namely7 one splits f dx is said to be convergent if Similar considerations can be applied to when f is continuous on 10 b f dx into two improper integrals and then evaluates each of the 77split paieces77 individually That is b b l mmdmll mdzl mmdm Again7 both of the 77split pieces77 must converge individually in order for the entire integral to converge If either of the 77 split pieces77 gratuitous use of 77 7 is divergent7 then the entire integral is said to be divergent iv Example 10 10 1 l 7 d l 7 d A wihs 3E ains 6 du lim 7 ta4 74 U3 1 6 lim 7 H4 2u2 74 71 71 71 7 Willi 72 2t742 00 u sub with u t 7 4 10 1 Thus7 4 W dx is divergent c Recall that we have all ofthe Comparison Theorems for improper integrals These are all very similar to the comparison theorems for improper integrals 6 Area Between Curves 61 a When integrating with respect to 7 the general idea is that the area bounded by the curves z 17 z b7 y x7 and y g can be found using an integral Assuming that f 2 g on I 17 then the area is b A f79 dz i It is probably easiest to remember this as b Top 7 Bottom dz ii Recall that in class the idea was to think of 7 as representing the height of an approximating rectangle for the region in question7 with dx representing the in nitesimal base of the approximating rectangle b When integrating with the respect to y7 the general idea is that the area bounded by the curves y 07 y d7 z My7 and x py can be found using an integral Assuming hy 2 py on I 07d7 then 1 A hy7pydy i Geometrically7 when integrating with respect to y you should probably think d A Right 7 Left dy ii Again7 think of hy 7 py Right Left as giving the length of an ap proximating rectangle for the region in question7 with dy representing the in nitesimal height of the approximating rectangle c Example Find the area of the region bounded by the curves y 4x and y x2 by both integrating with respect to z and y Tlir auumiril negiun i With respect to a The intersection points ere a 0 end a 4 obteined by setting 49 at end solving for or The eree of the region is l A Albee dac2x77 n ii With respect to y First We must solve for a M We obtei d W7 The intersection points ere y 0 end y 16 obteined by either setting A W7 end so ving for y or evelueting one of the originel curves et a 0 end The eree is thus 64 32 32 e 7 7 0 754mm who 7 Volumes of Solids of Revolution 62 Y 7 0 isq omits 797256 32 7 3 3 e Trying to nd the Volume of the solid obteined from revolving e region R in the plene ebout some speci ed axis b WasherDisk Method it Generelly used when the epproximeting rectengles of the region to be re volved R ere drewn perpendiculerly to the axis of revolution ii If the region R is bounded by curves y e end y 909 end the axis revolution is e horizontel line y k7 the volume of the resulting solid is given b b V meter rams 77rmner rooms do 8 A Think of 7T0uter radius 7 Minnaquot radius as giving the area of a cross sectional slice of the resulting solid a washer or a disk UCJ For most regions ie7 the ones in this class7 one can decide which functionf or 9z determines the outer radius and and which func tion determines the inner radius by comparing the values lfc 7 kl and lgc 7 kl for any number 0 belonging to I cub Note Avoid the endpoints lf lfc 7 kl is larger7 then f determines the outer radius lf lfc 7 kl is smaller7 then f determines the inner radius 0 lf f determines the outer radius and g determines the inner radius7 the the formula for the resulting solid is given by b b V 7Tf96k27r996k2 196 W f96k2996k2 d96 a Example Determine the volume of the solid obtained by revolving the regionR7 bounded by y 4x and y 2 about the line y 0 See the gure above 4 V M495 7 my dab 17 0 4 7T 16x2 7 x4 dx 18 0 3 5 4 WCM 5 19 3 5 0 7r cubic um ts 20 21 For the same region7 R7 set up the integral that gives the volume of the solid obtained by revolving the region R about the line y 20 2 Note the functions determining the inner radius and the outer radius have changed 4 V 7T20 7 22 7 7T20 7 4amp2 dz 0 c One can apply the exact same techniques when the region R is bounded by func tions of y and the axis of revolution is vertical i Find the volume of the solid obtained by revolving the region R bounded by y 4x and y 2 about the line z 0 Note that the bounding curves should be viewed as x Z and x The limits of integration will now be y 0 and y 16 v we 7 ma dy lt22 16 yz 7 7 d 23 WO y 16 y 2 3 16 y y 7 7 7 24 7T 2 48 0 256 7T cub c um ts 25 26 I For the same region7 R7 set up the integral that gives the volume of the solid obtained by revolving the region about the line z 7 Note the functions determining the inner radius and the outer radius have changed 16 9 V 7 m 7 12 7m 7 ya dy 0 8 ARC LENGTH 63 a Three formulas for arc length that allow us to compute the arc length of a curve in the plane when the curve is given by i parametric equations7 ii y is given as a function of x or iii m is given as a function of y i When we have a smooth curve parametric equations have derivatives given by z ft and y gt7 with the curve being traversed exactly once as t ranges from a to b a g t g b then the length of the curve is 17 d2 d2 17 La CT 67 dta f 2g 2dt ii When the curve is given as the graph of a function y x7 then as X ranges from a to b for a S x S b7 the length of the curve is b 2 b L11ll7 dz 1f 2d iii When the curve is given as the graph of a function x fy7 then as y ranges from a to b for a S y S b7 the length of the curve is 17 d2 17 La 167 1dya xf 21dy 9 AVERAGE VALUE 64 a The average value of a function f over the interval I 17 is 1 17 2mg a fx dx 10 Applications to Physics 65 and many more examples online a This is tough to summarize in one place but I intend to work on it You can be sure that this section will have a fair representation on the nal exam we spent at least 1 week on it7 afterall TO BE CONTINUED H H Differential Equations The Basics 71 a What is a differential Equation b What does it mean for a function to be a solution of a differential equation c What is the difference between the general solution of a differential equation and the unique solution to an initial value problem d How do you determine the order of a differential equation 12 Euler s Method and Direction Fields 72 d a What is the signi cance of a rst order differential equation given by i fx7 y How do differential equations of this form relate to direction elds At a point d x07y0 what is the signi cance of g fx0y0 d 0410 13 Separable Differential Equations 737 747 75 a A separable differential equation is a rst order differential equation of the form dy 7 f969y b We solve separable differential equations by i Separating the variables7 dy 7 f z dz 99 ii and then integrate both sides7 f dx dy 632 sin2 A E l 7 7 0 0 Kampe dx y secx 7 y 7 2 sin2 ye y dy 7 dx separation of variables 27 secx 7 2 sin2x ye y dy dx lntegrate both sides 28 sec s 1 7 67y2 sin2z cosx dx LHS u sub with u 7y2 29 1 3 7E6 y2 M K RHS u sub with u sinz 30 1 3 0 7560 K Using lnitial Condition 31 1 77 K 32 2 1 3 1 726 3 2 Slnga 7 5 The Unique Solution irnplicit 33 c Applications of Separable Differential Equations i Orthogonal Trajectories 737 237 247 25 d A Find d7y for the original family of curves s B Solve the differential equation determined by the negative reciprocal ii Mixing Problems 73 dA A 7 Rate in Rate Out dt iii Exponential Growth and Decay 74 dP A E kP ltgt Pt Poe with P0 P0 iv Newton7s Law of Cooling 74 A Newton7s Law of Cooling states that the rate of cooling of an object is proportional to the temperature of the surrounding medium provided the difference is not to large lf Tt represents the temperature of the object at time t and TS is the temperature of the surrounding medium7 then dT dt 7 kT T9 B The Temperature of the object at time t is thus7 M 7 No 7 mar T 14 2nd Order Linear Differential Equations 777 78 a Solving Ag By Cy gz7 where A7 B7 0 are constants i When g 07 ie7 Ag By 7 Cy 07 the general solution depends en tirely on the roots of the characteristic equation The general solution of the differential will be one of three cases determined by the roots of the charac teristic equation A When the characteristic equation has two7 distinct7 real roots T1 and r2 B2 7 4A0 gt 07 the general solution is Ole1m 0267 EU When the characteristic equation has 1 real root of multiplicity two B2 7 4A0 07 the general solution is 016quot ngemm O When the characteristic equation has two complex roots7 r a i in B2 7 4A0 lt 07 the general solution is e 01 cosbz Cg sinbz 13 ii When gx 31 07 then nding the general solution of the Non homogeneous Differential Equation Ay By Cy gx is a two step process A First nd the solution of the corresponding homogeneous differential equation Ay By Cy 0 This is called the complementary solu tion7 yo B Second nd any particular solution yp of the differential equation Ay By Cy You need to know how to nd the particular solution using the method of undetermined coef cients Namely7 how do you de termine what form yp should have based on g7 C The general solution of Ay By Cy g is y yo l y 15 The motion of a mass on a spring 79 a Damped Vibrations i If t is the position relative to equilibrium of the mass at time t 7 the motion of the mass on the spring satis es mz t cz t kzt 07 Where7 m is the mass7 c is the damping constant and k is the spring constant A When there is no damping constant c 07 the motion of a mass on a spring satis es Newton7s Second Law mx t kzt 0 B When 0 07 this is Simple Harmonic Motion 16 Taylor Series Expansion 87 a The Taylor Series Expansion centered at z a for a function fx assuming it converges to the function on some interval 00 fna Z 77 f 96 x 7 a b When they Taylor Series is centered at z 07 it is often called the Maclaurin Series expansion for In this case it takes the form f Z 14 i Important meaning you should know them Taylor Series centered at z 0 ie7 Maclaurin Series Check your notes or with the book for the radius of convergence and the interval of convergence 00 A cosx 271 2 n0 0 x2n1 B 71 7 s1nx 23 2n1 m i 00 i C e 771 n0 D 1 fix 17 7 0 0 2nl E tanil z 71 n lt gt gt 2 1 ii Be able to use the known expansions to nd expansions for related functions iii Be able to derive the Taylor Series expansion of a function centered at z ausing the de nition iv Be able to use the known expansions to nd the sums of in nite series


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