MTH 251

Calculus II

Miami University

Continuation of Calculus I. Plane analytic geometry, techniques of integration, parametric equations, polar coordinates, infinite series, approximations, applications. Credit not awarded for both MTH 249 and 251. CAS-E.
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| Konstantinos Beros
MTH 251- Final Exam Guide - Comprehensive Notes for the exam ( 32 pages long!)
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[MTH 251] Comprehensive spring guide including any lecture notes, textbook notes and exam guides.find more resources at oneclass.com MTH251-Lecture #1-Review of Syllabus and Integration and Introduction to Integration...

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MTH 251
Konstantinos Beros
MTH 251- Midterm Exam Guide - Comprehensive Notes for the exam ( 25 pages long!)
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[MTH 251] Comprehensive spring guide including any lecture notes, textbook notes and exam guides.find more resources at oneclass.com MTH251-Lecture #1-Review of Syllabus and Integration and Introduction to Integration...

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MTH 251
Konstantinos Beros

Exam Solutions for MTH 251

MTH 251 Lecture 30: MTH 251 Lecture 30
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Recall 1 1)2+1 tan = =0 2+1 , 1 Note: This holds for = 1 ( ) = 1 =1 + + 1 1 4 =0 2+1 3 5 7 9 Homework example: = tan ( ) 3 (1)(3)2+1 tan...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 36 - Cartesian Coordinate System
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1 = 0 36 + 36 2 1 = 1 + = 6 + 1 + 2 = 2 + 1 0 = 2 2 = 3 4 1 2 32 = 3 3 1 = 28 2 Section 10.3 Polar coordinates Consider a point in the plane. We call this point the pole (or the origin) and label it . We draw a ray extendi...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 35 - Telescoping Series
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Exam Review 2 telescoping series tan 1( + 1 tan1 )] =1 = tan (2) tan 1(1 + tan 1(3 tan 1(2 + tan 1(4 1 1 1 1 1 1 tan (3 + tan ( tan ( 1 + tan ( + 1 tan ( + 1 tan () 1 1 = tan 1 + tan ( + 1) 2 lim = lim[ + ( + 1)] 4 = + = 4...

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MTH 251
Konstantinos Beros

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All MTH 251 Materials | Spring 2018

MTH 251 Lecture 30: MTH 251 Lecture 30
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Recall 1 1)2+1 tan = =0 2+1 , 1 Note: This holds for = 1 ( ) = 1 =1 + + 1 1 4 =0 2+1 3 5 7 9 Homework example: = tan ( ) 3 (1)(3)2+1 tan...

Mathematics
MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 36 - Cartesian Coordinate System
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1 = 0 36 + 36 2 1 = 1 + = 6 + 1 + 2 = 2 + 1 0 = 2 2 = 3 4 1 2 32 = 3 3 1 = 28 2 Section 10.3 Polar coordinates Consider a point in the plane. We call this point the pole (or the origin) and label it . We draw a ray extendi...

Mathematics
MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 35 - Telescoping Series
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Exam Review 2 telescoping series tan 1( + 1 tan1 )] =1 = tan (2) tan 1(1 + tan 1(3 tan 1(2 + tan 1(4 1 1 1 1 1 1 tan (3 + tan ( tan ( 1 + tan ( + 1 tan ( + 1 tan () 1 1 = tan 1 + tan ( + 1) 2 lim = lim[ + ( + 1)] 4 = + = 4...

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MTH 251
Konstantinos Beros
MTH 251 Lecture 34: Exam Review
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Exam Review (2) Telescoping series [tan 1 + 1 tan ()] =1 = tan (2) tan 1(1 + tan1(3 tan1(2 tan 1(4 1 1 1 1 1 1 tan (3 + tan ( tan 1 + tan + 1 tan ( + 1 tan () = tan 1(1 + tan ( + 1) 2 1 = l[ 4 + tan ( + 1 ] = + = 4 2 4 (3)...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 33 - Parametric Equation
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Section 10.1 : Parametric equations Suppose an ant traces out a path on a flat surface Suppose the curve is something like: This curve is neither a function of nor But both and are functions of ; we can write = ; = () So, ...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 32 - Squeeze Theorem, Ratio Test
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Lecture Notes + Tnis the degree Taylor polynomial of at . Setting = () where () is called the remainder of the Taylor series If we can show that lim = 0, Then we have: lim( )( ) = lim =() Theorem: If = + ( ) Where is the n...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 31 - Convergent Series
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Section 11.10 Taylor and McLaren series Two big question: What functions have power series representations and how can we find them? Suppose has a power series of a with radius ; that is, = =0 , < 2 3 = +0 1 + 2( ) + 3 ) +...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 27 - Ratio Test
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MTH251Lecture 27Section 11.8: Power Series Definition: A power series is a series of the form = + 0 + 1 + 2 + 3 + =0 Where x is a variable and the are constant coefficients For a given x, the series may or may not converge...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 26 - Alternating Series Test, Conditional Convergence, Direct Comparison Test
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MTH 251Lecture 26The ratio and root tests Ratio Test Proof: lim +1 = < 1 < < 1 There is an N, so that if n> , < +1< For +1< < < 2 +2 +1 3 +3 < +2 < Generally, + < for 1 =1 This series is geometric and converges, s=1 + = +1...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 28 - Alternating Series, Ratio Test, Ibm System P
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MTH251Lecture 28Power Series ( 2) 2 =0 + 1 Ratio test converges for 1 < < 3 But the ratio test gives no information when 2 = 1,=> 1,3 We must test x=1,3 separately X=1 (1) , = 1 =0 +1 +1 (1) And, lim = 0 < 1 absolutely con...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 29 - Bessel Function
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MTH251-Lecture 29 Notes-Power Series Consider, 2 ( ) = + 16 We want to represent this as a p...

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MTH 251
Konstantinos Beros
MTH 251 Lecture 19: Series Continuation
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MTH 251Lecture 19Series Continuation 1= + + + =1 Converges t1if < 1, and diverges if 1 Example: 12 (5 ) =0 1 = 12( )5 =0 1 = 12( )1 5 =1 Geometric with a=12, r= , so, < 1, the series converge= 10 5 1( ) 5 Example: 12 +1 =0...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 20 - Convergent Series
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MTH251Lecture 20Series Review Theorem: Ifand are both convergent series then, i. , for any constant c, ii. ( + ) iii. ( are also convergent 1. =1 = =1 2. =1(+ )= =1 + =1 3. =1( ) ==1 =1 Note: A finite number of terms can n...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 18 - Telescoping Series
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MTH 251-Lecture 18 Notes-Series Consider 1 =12 1 1 1 =2 4 8 If we sum the first n terms, we get: 1 3 7 15 { , , , .} ...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 23 - Monotonic Function, Sequence, Alternating Series Test
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MTH251Lecture 2311.5: Alternating Series Definition: An alternating Series is a series whose terms alternate between positive and negative For example, (1 ) =1 This series is alternating harmonic 1 1 1 1 => 1 + + + 2 3 4 5...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 25 - Limit Comparison Test, Conditional Convergence, Ibm System P
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MTH 251Lecture 2511.6: Absolute Convergence and ratio and root tests For a given serieswe can consider , the series whose terms are the absolute value of . Defenition: A seriesis called absolutely convergent ifconverges. (...

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MTH 251
Konstantinos Beros
MTH 251 Lecture 16: Sequence
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MTH251Lecture 16 NotesSection 11.1 Section 11.1: Sequence A sequence is simply a list of numbers that are ordered: 1, 2 3 , 1is the first term, 2 is the second, and is the term in the sequence Note: We will only deal with ...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 24 - Ibm System P, Alternating Series, Alternating Series Test
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MTH 251Lecture 24Exam 2Quiz Review Quiz Review: 1. 1 =2 () Using the integral test for divergence, 1 2 () 1 = 2 ) Using usubstitution with u=ln(n), and du= () 1 = (2) = ( ( () ) ( (2))) = So, by the integral test for diver...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 21 - Ibm System P, Monotonic Function, Order Of Newfoundland And Labrador
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MTH 251Lecture 21The Integral Test Section 11.3: The Integral Test We have seen I the previous sections that it is sometimes possible to compute the exact sum of a series (geometric and telescoping series), however, in gen...

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MTH 251
Konstantinos Beros
MTH 251 Lecture 12: Arc Length
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MTH251Lecture 12Arc Length Example Find the length of = from (0,0) to (1,1) 1 2 = 0 1 + 2 dx 1 1 If 2 = , x= and, = sec 2 2 1 1 = 1 + tan sec 2 0 1 1 tan 2 = sec 2 0 1 1 tan 2 = sec 2 0 1 1 tan 2 = sec 2 0 2 Use integratio...

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MTH 251
Konstantinos Beros
MTH 251 Lecture 10: Type Two Integrals
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MTH251Lecture 10Type Two Integrals Definition of Improper Integrals Type 2(when the interval is not infinite) i. Suppose f is continuous on the interval , , and f is continuous at a, then we define, = lim+ ( ) provided tha...

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MTH 251
Konstantinos Beros
MTH 251 Lecture 15: Simpson’s Rule
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MTH 251Lecture 15 Notes Simpsons Rule 7.7 Approximation Of Integrals: Recall: If f is continuous on , then, ( ) ( ) = lim =1 Where = and = + The area under the curve can be estimated using right and left midpoints. The are...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 11 - Farad
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MTH251-Lecture 11 Notes- Comparison Test of integrals continued and 8.1 Example 1 + ...

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MTH 251
Konstantinos Beros
MTH 251 Lecture 13: Surface Area of Solids of Revolution
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MTH251-Lecture 13-Section 8.2 For any point on a curve, 2 | 1 + ( ) 1 ...

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MTH 251
Konstantinos Beros
MTH 251 Lecture 14: Exam Review Problems
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MTH251Lecture 14Exam Review Problems Example y = + 0 1, 2 = 2() 1 + ( ) = + = 21 + 1 2 = 2 + 1 + ( ) 0 21 + 1 ( ) 2 2 1 + 4 1 + + 0 4 1 + ) 4 1 + ) 1 2 4 + 4 + = 2 1 + 4 1 + ) 0 1 = 2+ 4 + 4 0 1 = ( + 2 )2 0 1 = + 2 0

Mathematics
MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 17 - Monotonic Function, Bounded Function
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MTH 251Lecture 17Section 11.1 For what values of r does the sequence converge? We know that > 1, then lim = and when 0< < 1 then, lim = So, if a=r, then diverges for r> 1, and converges for 0< < 1 Obviously, lim 0 = 0 lim ...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 22 - Geometric Progression, Direct Comparison Test, Ibm System P
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MTH 251Lecture 22Section 11.4 Recall: 1 The series=1 converges for P> 1, and diverges otherwise, this is called a pseries Section 11.4: Comparison Test Consider, 1 2 + 1 =1 The terms of this series are very similar to =12 ...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 9 - Improper Integral
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MTH 251Lecture 97.8 continuation Definition of improper Integrals of Type 1(The interval is infinite): i. () exists for every t , then we can dein = (), provided that this limit exists. ii. If the integal exists for every ...

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MTH 251
Konstantinos Beros
MTH 251- Final Exam Guide - Comprehensive Notes for the exam ( 32 pages long!)
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[MTH 251] Comprehensive spring guide including any lecture notes, textbook notes and exam guides.find more resources at oneclass.com MTH251-Lecture #1-Review of Syllabus and Integration and Introduction to Integration...

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MTH 251
Konstantinos Beros
MTH 251 Lecture 8: Longest Partial Fraction Integral Form
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MTH 251-Lecture 8- Longest Partial Fraction Integral Form and start 7.8 Recall From Last Class 1+ 1 = (1)(+1) where u is 1 + So, 1 = + 1 +1 ) 1 +1 ...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 7 - Partial Fraction Decomposition
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MTH 251Lecture 7Case 4 of Partial fraction integrals Recall Example from last class 2 4 + 4 3 + 4 Where A=1,B=1, and C=1, so using case III, 1 1 = + 2 + 4 1 = + + 4 + 4 2 For, +4 use u substitution where = + 4 and = 2 1 2 ...

Mathematics
MTH 251
Konstantinos Beros
MTH 251- Midterm Exam Guide - Comprehensive Notes for the exam ( 25 pages long!)
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[MTH 251] Comprehensive spring guide including any lecture notes, textbook notes and exam guides.find more resources at oneclass.com MTH251-Lecture #1-Review of Syllabus and Integration and Introduction to Integration...

Mathematics
MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 6 - Partial Fraction Decomposition
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MTH 251-Lecture 6-Methods if integrating with partial fractions Case I (Review) Q(x) is a product of distinct linear factors, Q(x)=( + + . 1 1 In thi...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 1 - Product Rule
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MTH251-Lecture #1-Review of Syllabus and Integration and Introduction to Integration By Parts Important Points from The Syllabus Review The textbook we will be using is Calculus:Early Transcendentals, 8 Edition b...

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MTH 251
Konstantinos Beros
MTH 251 Lecture Notes - Spring 2018, Lecture 5 - Partial Fraction Decomposition
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MTH 251-Lecture 5-Review and Section 7.4 Introduction Example 1 + 2 + 5 = 1 ...

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MTH 251
Konstantinos Beros
MTH 251 Lecture 3: Strategies for Solving Trig Functions
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MTH251-Lecture#3-Strategies For Integrating Trigonometric Functions Strategy For Integrating sin cos 1. If either m or n is odd, w...

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MTH 251
Konstantinos Beros
MTH 251 Lecture 2: Cont of Integration by Parts and Introduction to Trigonometric Integrals
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MTH251-Lecture#2-Integration By Parts Continuation and Introduction to trigonometric integrals Integration By Parts Continued Example ...

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MTH 251
Konstantinos Beros
MTH 251 Lecture 4: 7.3 - Trigonometric Substitution
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MTH251-Lecture #4-Trigonometric Substitution Example For Review 2 cos 5 cos 10 ) 0 ...

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MTH 251
Konstantinos Beros
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