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by: Jack Magann

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# Exam 3 Study Guide MTH 162

Jack Magann
UM
GPA 3.865
Calculus 2
Dr. Bibby

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This covers all topics for exam 3 Topics: improper integrals, arc length, sequences, series, and tests for convergence and divergence
COURSE
Calculus 2
PROF.
Dr. Bibby
TYPE
Study Guide
PAGES
11
WORDS
CONCEPTS
University of Miami calculus 2
KARMA
50 ?

## Popular in Mathematics (M)

This 11 page Study Guide was uploaded by Jack Magann on Friday March 27, 2015. The Study Guide belongs to MTH 162 at University of Miami taught by Dr. Bibby in Spring2015. Since its upload, it has received 206 views. For similar materials see Calculus 2 in Mathematics (M) at University of Miami.

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Date Created: 03/27/15
Calculus 2 Chapter 66A Improper Integrals Integrals that are over an in nite interval This means there is no nite area under the curve There are three cases for improper integrals l faoo f x dx Here the interval is a through all values in the 00 direction This can be solved by taking the limit of the integral For most functions the integral as the left most point of the region t approaches in nity is going to one set value that can be calculated by its limit This is the process to solve this is fa fxdx 11mm fa39fxdx limtooFt Fa Where F is the antiderivative of fX 2 If fxdx This can be thought of the same as the previous case but in the opposite direction This is solved by fb fxdx limtoo fbtxdx limtooFt Fa 3 If fxdx Here there is an endless interval in both directions f ofxdx ffoofxdx famfxdx SO f ofxdx limtooFt Fa limtooFt Fa If such an integral as these has a nite limit solvable limit the integral is said to converge at the value of the integral If such an integral as these has a limit that does not exist the integral is said to diverge Calculus 2 Chapter 66B Discontinuous lntegrands Limit approaches a vertical asymptote This means f x will be discontinuous at some point Recall A vertical asymptote is created Where the denominator equals 0 Case 1 Suppose x a is a vertical asymptote for f x Then ffxdx 11mm ft fxdx limtaFb Ft Note that if the limit exists then Ft exists This is integration from a vertical asymptote Case 2 Suppose x b is a vertical asymptote for f x Then f fxdx 11mm f fxdx limtbFt Fa This is integration to a vertical asymptote Case 3 Suppose x c is a vertical asymptote for f x Where a lt c lt b b b Then fa f xdx f xdx fc f xdx limtc fifoMx limtc ftb fxdx Ft Fa Fb Ftl in case three for the integral to converge both limits must exists Calculus 2 Chapter 74 Arc Length arc length s Def1nition The distance along line f x from a to b The arc length is considered smooth if f x is continuous and differentiable and f x is differentiable along a b Derivation of arc length formula Consider that P a lt x0 lt x1 lt xn lt b is a partition of a b Start by taking a line segment from xi1 f xi1 to xi f x0 Using the distance formula we get Li JCxi xi 12 fxi fxi 12 Multiply by xi xi 12 2 Li1W xcxrx xixi 1 Li1fxi fxi 12 Ax xixi 1 By the mean value theorem There exists wt in a b such that f w so i i 1 Li 1 f39W2 Ax Take the integral of this Adds up all the line segments on the interval a b to approximate s b s 1 f w2 dx a There are two cases when calculating arc length Case 1 When differentiating with respect to X yfx an Sb Case 2 When differentiating with respect to y 9901 6133 Sb 3de 1 2 dx Chapter 81 Infinite Sequences Sequence An ordered listing of numbers separated by commons Ex NIH wIN whlw U1IIgt three dots at the end are required since it is an infinite sequence Each of these values can be represented by the term an n is the position in the sequence WIN 1 3 so a1 a2 2613 andsoon The nth term might be represented by an nth term formula 39 n For th1s sequence an 2 n1 This formula describes sequence for every term Ex 1 Write the first 5 terms of sequence Who s nth term is given by n1 SOa21a3a42a a 63 quotn21 12 25 3105 417 5 26 13 3 2 5 3 Th1s 1s wr1tten as 1 3 E E E dontforget the three dots at the end Another way to found the terms in a sequence is through recursion An equation is given relating one term to the nest in the sequence Therefore the previous term is needed to nd the next 2 Write the rst 5 terms of the sequence de ned recursively by a12 an1an4 so a2a146 a36410 a41O414a5144 18 Answer 2 6 10 14 18 3 Write the rst 10 terms of the Fibonacci sequence de ned recursively by a1 1a2 1an an2 an1 a3112 a4123 c1525 c1628 a713 a821 a934 a1055 answer 11235813213455 Factorial Notation If n is apositive integer Then n factorial notated 11 1 x 2 x 3 x x n 1 x 11 It s important to know that 25 25X24gtlt23gtlt22 X21 or 25 25X24 4 Find the value oft quotI 10 x 9 x 8 720 2n2 2n2x2n2 1x2n2 2 2n2 x 2n1 x 2n 5 Slmpllfy 2n 2n 2n 2n2gtlt2n1 4n26n2 Common sequences Sequence Example nth term pos Integers 1234 an 2 11 pos even ints 2468 an 2 211 pos odd ints 1357 an 2 2n 1 pos powers of 2 24816 an 2 2quot pos powers of 2 w 124816 an 2 2quot 1 All int 01122 an 2 W If a1 a2 a3 a4 is an in nite sequence consider limnoo an 4 n E so an Ex consider n1 NIH wIN rhlw I I lim a lim n This means as the terms increase the values added up could approach a number Since polynomials of same degree are in numerator and denominator 1 11mnoo E 1 There is a horlzontal asymptote at y 1 n x Inth1s case can be treated as n1 x1 Basic limits 1 1 1 n 11mnoo Z 0 11mnoo Zn 1 0 11mnoo z n 11mnoo 2 0 Calculus 2 Chapter 82 In nite Series This is the notation for a series 2300 an It means that all the values of an from 0 to in nity will be added up So fo an a1 a2 a3 an1 an 1 1 1 1 1 Ex E127E23T6H 1 Consider the in nite series 291 an 2 a1 a2 a3 an1 an Rules Form sequence of partial sums SPS 1 52 S3 S4Sn Where Sn a1a2 an oo Zn 1 00 Zn 1 0 1 39EX 2111 2n 2111 2 1 2 1 2111 1 2 1 limnooSn limnoo 1 i 1 0 1 2n 1 If sequence of partial sums converges then the series 291 an also converges to S and is referred to the sum of the series If it converges nd the sum 2 If the SPS diverges then the series 291 an diverges 0 1 l l l l l l EXZn1 1234 SPS 51 152 12S3 123 3 11 25 hm oo Sn 1 E g E D1verges Test for Divergence and Convergence Nth test for Divergence Given the series 291 an If limnaoo an i 0 the limit DNE Then Eff an diverges Proof suppose 291 an converges Thus limnoo Sn 2 S is the sum of the series Sn n l an gt Sn Sn 1 an limnoo an 2 limngooSn Sn1 S S 0 The limits are the same since the two are the same other then one term Calculus 2 chapter 83 If the series converges there are 6 other tests that can be used to determine convergencedivergence l Integral Test apply only to series with all positive numbers or all negative numbers Compute the series as an in nite integral since If the limit converges then so does the series and same for divergence 0 1 oo 1 EX 1 Zn1 n24 compute f1 x24 2 lt2 lt2 lt2 2 2 o This integral converges so the series converges but not to the same number dx limtootan 1 2 221 flooidx 1imtoolnx 11m1nt 1n 1 oo This diverges so the series diverges Side note 2le an is the sum of the areas of the rectangles over the line while I100 f x dx is the area under a continuous function so2 1an gt f1 fltxgtdx means the sum of the series doesn t equal the sum of an 2 Basic Comparison test apply only to series with all numbers or all numbers Idea is to compare a series to another that you know converges or diverges Two types of series to compare to l Geometric series a Effzoarnaarar2m raquot n a rst term r common ratio Converges if 1 lt r lt 1 limnnoo rquot 0 Diverges if Irl gt 1 For this series Sn 2 a ar a r2 arquot So rSn arar2ar3arn n ThCI CfOI C7 Sn TSn a arn gt Sn n Now test the limit lim 00 HT L sum Of a n 1 7 1 1 71 EX 1 233204 371 612 r Since 1ltrlt1 Sum 22 2 ziz 5 2 Zai 2 5 2531 for both 1 lt r lt 1 5n 5 E E 3 2 13 Sum 5 5 1 1 2 3 6 2 Pseries 231 nip If p gt 1 the series converges If p S 1 the series diverges In the basic comparison test Compare to 2 bn which is known if it converges or diverges 1 If an lt bn and 2 bn converges then 2 an converges 2 If an gt bn and 2 bn diverges then 2 an diverges 3 If an lt bn and 2 bn diverges then no conclusion 4 If an gt bn and 2 bn converges then no conclusion 3 Limit Comparison Test apply only to series with all numbers or all numbers Suppose limnoo L gt If 0 lt L lt 00 then 1 If 2 bn converges then 2 an diverges 2 If 2 bn diverges then 2 an diverges 4 Ratio Test apply to both series with all numbers or all numbers Note on alternating series 1n1 Altemat1ngtest2an E 2 1 Z n 1 1 1 1 Ser1es ofabsolute values Elanl E 1 E E Z lf 2 an and Elanl converge 2 an is absolutely convergent If 2 an converges and Elanl diverges 2 an is conditionally convergent Ratio test Let 2 an be a series and let 1quot limnoo an 1 If 0 S r lt 1 then 2 an is absolutely convergent 2 If r gt 1 then 2 an is divergent 3 If r 1 then the test fails 5 Root Test apply to both series with all numbers or all numbers Let 2 an be a series and let L limnoo quotJ Ianl 1 If 0 S L lt 1 then 2 an is absolutely convergent 2 If L gt 1 then 2 an is divergent 3 If L 1 then the test fails 6 Alternating Series Test for convergence Tl An alternating series can be expressed as Z 1 Let 2 an be an alternating series If 1 an1 lt Ianl and 2 limnnmlanl 0 Then 2 an is convergent

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