REVERSE CHAIN RULE CALCULUS 7. Dr Adrian Jannetta MIMA CMath FRAS INU0115/515 (MATHS 2) Reverse Chain Rule 1/12 Adrian Jannetta

Size: px
Start display at page:

Download "REVERSE CHAIN RULE CALCULUS 7. Dr Adrian Jannetta MIMA CMath FRAS INU0115/515 (MATHS 2) Reverse Chain Rule 1/12 Adrian Jannetta"

Transcription

1 REVERSE CHAIN RULE CALCULUS 7 INU05/55 (MATHS 2) Dr Adrian Jannetta MIMA CMath FRAS Reverse Chain Rule /2 Adrian Jannetta

2 Reversing the chain rule In differentiation the chain rule is used to get the derivative of a composite function. For example d dx (e2x )=2e 2x Integration is the reverse of differentiation. Therefore, we could reverse this statement and write: 2e 2x dx= e 2x (Ignore the constant for a moment). Dividing both sides by 2 we find e 2x dx= e2x 2 () Similarly, consider d dx [sin(3x+2)]=3cos(3x+2). Reversing this process: 3cos(3x+2)dx=sin(3x+2) Therefore: cos(3x+2)dx= sin(3x+2) 3 Can you a pattern for directly integrating the expressions given in () and (2)? Reverse Chain Rule 2/2 Adrian Jannetta (2)

3 Reversing the chain rule Here s another example of reversing the chain rule. d dx (2x 5)3 =(3)(2)(2x 5) 2 In words this process is Subtract from the power. Multiply by the old power and multiply by the derivative of the inner function. Since integration is the reverse of differentiation: (3)(2)(2x 5) 2 dx=(2x 5) 3 (Again, let s ignore the constant C for a moment). Dividing both sides by the factor(3)(2) we get (2x 5) 2 dx= (2x 5)3 (3)(2) Notice that the (3) is the power and the (2) is the derivative of the inner function(2x 5) (2x 5) 2 dx= (2x 5)3 6 The first term on the RHS is equivalent in words to Add to the power. Divide by the new power and divide by the derivative of the inner function. It is the reverse of applying the chain rule. Reverse Chain Rule 3/2 Adrian Jannetta

4 Reverse Chain Rule The examples just seen illustrate a simple procedure for integrating composite functions. It is the reverse of how we d carry out the chain rule in differentiation. In words, the general procedure doing reverse chain rule is Integrate the outer function. Divide by the derivative of the inner function Here some examples. We ll include the constant from now on! sin0x dx= cos0x + C 0 e 4x dx= e 4x 4 + C Reverse Chain Rule 4/2 Adrian Jannetta

5 Reverse chain rule Use the reverse chain rule to integrate First, write the integral in a more suitable form (3 5x) 4 dx (3 5x) 4 dx Now integrate. Use the rule for integrating powers (in this case, like x 4 ); Add to the power. Divide by the new power and divide by the derivative of the inner function. (3 5x) 4 dx= (3 5x) 3 ( 3)( 5) + C=(3 5x) 3 + C 5 It can be written without the negative power like this: (3 5x) dx= 4 5(3 5x) 3+ C Reverse Chain Rule 5/2 Adrian Jannetta

6 When can reverse chain rule be used? The reverse chain rule can only be used when the inner function is linear, which means it has the form mx + c (where m and c are constants). We cannot use the reverse chain rule when the inner function is NOT LINEAR! For example, we could not use it to find function +x 2 is not linear. Therefore (+x 2 ) 5 dx (+x2 ) 6 + C 6(2x) (+x 2 ) 5 dx because the inner (You can check this by finding the derivative of the RHS - it will not be equal to the LHS!) Reverse Chain Rule 6/2 Adrian Jannetta

7 Test yourself Which of the following can be integrated with the reverse chain rule? ( 3x) 4 dx 2e x dx x 2 dx 3x dx sinx dx cos(3x π) dx +2x dx 2x sinx dx Answers... Yes 3 No 5 No 7 Yes 2 Yes 4 Yes 6 Yes 8 No Reverse Chain Rule 7/2 Adrian Jannetta

8 Further examples Rational power Find 2x+5dx with the reverse chain rule. First rewrite without the square-root: (2x+5) 2 dx = (2x+5) 3 2 ( 3 2 )(2) + C = (2x+5) C 3 (2x+5) 3 = + C 3 You can always check the answer by differentiating it! You should end up with the original integrand. Reverse Chain Rule 8/2 Adrian Jannetta

9 A definite integral Evaluate 0 (2x+) 3 dx Using the reverse chain rule we see that (2x+) 3 dx = Looking forward... 0 = 8 = 8 (2x+) 4 (4)(2) 0 (2x+) = 8 (8 ) = 0 The reverse chain rule is a useful shortcut for simple situations. A related technique, called integration by substitution, will be demonstrated in the next lecture. Reverse Chain Rule 9/2 Adrian Jannetta 0

10 Integration of f (x)/f(x) Here is a result that we saw during our study of differentiation: d dx ln f(x) = f (x) f(x) It s a consequence of differentiating logarithm functions using the chain rule. Naturally, we can reverse this to produce an integration result: f (x) f(x) dx=ln f(x) +C (3) This is a result we should learn to recognise when it appears! Reverse Chain Rule 0/2 Adrian Jannetta

11 Integration of f (x)/f(x) 2x Find x 2 9 dx. This integral matches the form of equation (3) exactly. Therefore: Integration of f (x)/f(x) 4sin x Find +cos x dx. 2x x 2 9 dx= ln x C This integral almost fits the form of equation (3). We can rewrite it to make it the same: 4sin x sin x +cosx dx= 4 +cos x dx We took out a factor of 4 so that the numerator is now the derivative of the denominator. Therefore: 4sin x dx= 4ln +cos x +C +cosx Reverse Chain Rule /2 Adrian Jannetta

12 Test yourself Use of reverse chain rule Use the reverse chain rule to find these integrals: ( 3x) 4 dx 5 π 9 0 sin 3x dx e x dx 5+2x dx 2cos( 3 x π) dx x dx cosx sin x dx Answers... 5 ( 3x)5 + C 2 2e x + C 3 2 ln 5+2x +C 4 6sin( 3 x π)+c ln sin x + C Reverse Chain Rule 2/2 Adrian Jannetta

Introduction Derivation General formula List of series Convergence Applications Test SERIES 4 INU0114/514 (MATHS 1)

Introduction Derivation General formula List of series Convergence Applications Test SERIES 4 INU0114/514 (MATHS 1) MACLAURIN SERIES SERIES 4 INU0114/514 (MATHS 1) Dr Adrian Jannetta MIMA CMath FRAS Maclaurin Series 1/ 21 Adrian Jannetta Recap: Binomial Series Recall that some functions can be rewritten as a power series

More information

Introduction Derivation General formula Example 1 List of series Convergence Applications Test SERIES 4 INU0114/514 (MATHS 1)

Introduction Derivation General formula Example 1 List of series Convergence Applications Test SERIES 4 INU0114/514 (MATHS 1) MACLAURIN SERIES SERIES 4 INU0114/514 (MATHS 1) Dr Adrian Jannetta MIMA CMath FRAS Maclaurin Series 1/ 19 Adrian Jannetta Background In this presentation you will be introduced to the concept of a power

More information

PRODUCT & QUOTIENT RULES CALCULUS 2. Dr Adrian Jannetta MIMA CMath FRAS INU0115/515 (MATHS 2) Product & quotient rules 1/13 Adrian Jannetta

PRODUCT & QUOTIENT RULES CALCULUS 2. Dr Adrian Jannetta MIMA CMath FRAS INU0115/515 (MATHS 2) Product & quotient rules 1/13 Adrian Jannetta PRODUCT & QUOTIENT RULES CALCULUS 2 INU0115/515 (MATHS 2) Dr Adrian Jannetta MIMA CMath FRAS Proct & quotient rules 1/13 Adrian Jannetta Objectives In this presentation we ll continue learning how to differentiate

More information

BASIC ALGEBRA ALGEBRA 1. Dr Adrian Jannetta MIMA CMath FRAS INU0114/514 (MATHS 1) Basic algebra 1/ 17 Adrian Jannetta

BASIC ALGEBRA ALGEBRA 1. Dr Adrian Jannetta MIMA CMath FRAS INU0114/514 (MATHS 1) Basic algebra 1/ 17 Adrian Jannetta BASIC ALGEBRA ALGEBRA 1 INU0114/514 (MATHS 1) Dr Adrian Jannetta MIMA CMath FRAS Basic algebra 1/ 17 Adrian Jannetta Overview In this presentation we will review some basic definitions and skills required

More information

Section 3.6 The chain rule 1 Lecture. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

Section 3.6 The chain rule 1 Lecture. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I Section 3.6 The chain rule 1 Lecture College of Science MATHS 101: Calculus I (University of Bahrain) Logarithmic Differentiation 1 / 1 Motivation Goal: We want to derive rules to find the derivative of

More information

Section 3.6 The chain rule 1 Lecture. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

Section 3.6 The chain rule 1 Lecture. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I Section 3.6 The chain rule 1 Lecture College of Science MATHS 101: Calculus I (University of Bahrain) Logarithmic Differentiation 1 / 23 Motivation Goal: We want to derive rules to find the derivative

More information

PARTIAL DIFFERENTIATION

PARTIAL DIFFERENTIATION PARTIAL DIFFERENTIATION CALCULUS 13 INU0115/515 (MATHS 2) Dr Adrian Jannetta MIMA CMath FRAS Partial Differentiation 1/ 15 Adrian Jannetta Functions of many variables The functions and derivatives which

More information

Math 106 Fall 2014 Exam 2.1 October 31, ln(x) x 3 dx = 1. 2 x 2 ln(x) + = 1 2 x 2 ln(x) + 1. = 1 2 x 2 ln(x) 1 4 x 2 + C

Math 106 Fall 2014 Exam 2.1 October 31, ln(x) x 3 dx = 1. 2 x 2 ln(x) + = 1 2 x 2 ln(x) + 1. = 1 2 x 2 ln(x) 1 4 x 2 + C Math 6 Fall 4 Exam. October 3, 4. The following questions have to do with the integral (a) Evaluate dx. Use integration by parts (x 3 dx = ) ( dx = ) x3 x dx = x x () dx = x + x x dx = x + x 3 dx dx =

More information

Section 5.5 More Integration Formula (The Substitution Method) 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

Section 5.5 More Integration Formula (The Substitution Method) 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I Section 5.5 More Integration Formula (The Substitution Method) 2 Lectures College of Science MATHS : Calculus I (University of Bahrain) Integrals / 7 The Substitution Method Idea: To replace a relatively

More information

DETERMINANTS, MINORS AND COFACTORS ALGEBRA 6. Dr Adrian Jannetta MIMA CMath FRAS INU0114/514 (MATHS 1)

DETERMINANTS, MINORS AND COFACTORS ALGEBRA 6. Dr Adrian Jannetta MIMA CMath FRAS INU0114/514 (MATHS 1) DETERMINANTS, MINORS AND COFACTORS ALGEBRA 6 INU0114/514 (MATHS 1) Dr Adrian Jannetta MIMA CMath FRAS Determinants, minors and cofactors 1/ 15 Adrian Jannetta Determinants Every square matrix has a number

More information

Change of Variables: Indefinite Integrals

Change of Variables: Indefinite Integrals Change of Variables: Indefinite Integrals Mathematics 11: Lecture 39 Dan Sloughter Furman University November 29, 2007 Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November

More information

IMPLICIT DIFFERENTIATION

IMPLICIT DIFFERENTIATION IMPLICIT DIFFERENTIATION CALCULUS 3 INU0115/515 (MATHS 2) Dr Arian Jannetta MIMA CMath FRAS Implicit Differentiation 1/ 11 Arian Jannetta Explicit an implicit functions Explicit functions An explicit function

More information

SMALL INCREMENTS CALCULUS 13. Dr Adrian Jannetta MIMA CMath FRAS INU0115/515 (MATHS 2) Small increments 1/15 Adrian Jannetta

SMALL INCREMENTS CALCULUS 13. Dr Adrian Jannetta MIMA CMath FRAS INU0115/515 (MATHS 2) Small increments 1/15 Adrian Jannetta SMALL INCREMENTS CALCULUS 13 INU0115/515 (MATHS 2) Dr Adrian Jannetta MIMA CMath FRAS Small increments 1/15 Adrian Jannetta Objectives In this presentation we re going to look at one of the applications

More information

COMPLEX NUMBERS ALGEBRA 7. Dr Adrian Jannetta MIMA CMath FRAS INU0114/514 (MATHS 1) Complex Numbers 1/ 22 Adrian Jannetta

COMPLEX NUMBERS ALGEBRA 7. Dr Adrian Jannetta MIMA CMath FRAS INU0114/514 (MATHS 1) Complex Numbers 1/ 22 Adrian Jannetta COMPLEX NUMBERS ALGEBRA 7 INU0114/514 (MATHS 1) Dr Adrian Jannetta MIMA CMath FRAS Complex Numbers 1/ 22 Adrian Jannetta Objectives This presentation will cover the following: Introduction to complex numbers.

More information

Introduction Growthequations Decay equations Forming differential equations Case studies Shifted equations Test INU0115/515 (MATHS 2)

Introduction Growthequations Decay equations Forming differential equations Case studies Shifted equations Test INU0115/515 (MATHS 2) GROWTH AND DECAY CALCULUS 12 INU0115/515 (MATHS 2) Dr Adrian Jannetta MIMA CMath FRAS Growth and Decay 1/ 24 Adrian Jannetta Introduction Some of the simplest systems that can be modelled by differential

More information

AREA UNDER A CURVE CALCULUS 8. Dr Adrian Jannetta MIMA CMath FRAS INU0115/515 (MATHS 2) Area under a curve 1/15 Adrian Jannetta

AREA UNDER A CURVE CALCULUS 8. Dr Adrian Jannetta MIMA CMath FRAS INU0115/515 (MATHS 2) Area under a curve 1/15 Adrian Jannetta AREA UNDER A CURVE CALCULUS 8 INU0115/515 (MATHS 2) Dr Adrian Jannetta MIMA CMath FRAS Area under a curve 1/15 Adrian Jannetta The area beneath a curve =f() The total area beneath the curve is approimatel

More information

RATIONAL FUNCTIONS AND

RATIONAL FUNCTIONS AND RATIONAL FUNCTIONS AND GRAPHS ALGEBRA 5 INU0114/514 (MATHS 1) Dr Adrian Jannetta MIMA CMath FRAS Rational functions and graphs 1/ 20 Adrian Jannetta Objectives In this lecture (and next seminar) we will

More information

How might we evaluate this? Suppose that, by some good luck, we knew that. x 2 5. x 2 dx 5

How might we evaluate this? Suppose that, by some good luck, we knew that. x 2 5. x 2 dx 5 8.4 1 8.4 Partial Fractions Consider the following integral. 13 2x (1) x 2 x 2 dx How might we evaluate this? Suppose that, by some good luck, we knew that 13 2x (2) x 2 x 2 = 3 x 2 5 x + 1 We could then

More information

Chapter 5: Integrals

Chapter 5: Integrals Chapter 5: Integrals Section 5.3 The Fundamental Theorem of Calculus Sec. 5.3: The Fundamental Theorem of Calculus Fundamental Theorem of Calculus: Sec. 5.3: The Fundamental Theorem of Calculus Fundamental

More information

Practice Problems: Integration by Parts

Practice Problems: Integration by Parts Practice Problems: Integration by Parts Answers. (a) Neither term will get simpler through differentiation, so let s try some choice for u and dv, and see how it works out (we can always go back and try

More information

Integration by Substitution

Integration by Substitution November 22, 2013 Introduction 7x 2 cos(3x 3 )dx =? 2xe x2 +5 dx =? Chain rule The chain rule: d dx (f (g(x))) = f (g(x)) g (x). Use the chain rule to find f (x) and then write the corresponding anti-differentiation

More information

Calculus I Announcements

Calculus I Announcements Slide 1 Calculus I Announcements Read sections 4.2,4.3,4.4,4.1 and 5.3 Do the homework from sections 4.2,4.3,4.4,4.1 and 5.3 Exam 3 is Thursday, November 12th See inside for a possible exam question. Slide

More information

Math 106: Review for Exam II - SOLUTIONS

Math 106: Review for Exam II - SOLUTIONS Math 6: Review for Exam II - SOLUTIONS INTEGRATION TIPS Substitution: usually let u a function that s inside another function, especially if du (possibly off by a multiplying constant) is also present

More information

Section 4.8 Anti Derivative and Indefinite Integrals 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

Section 4.8 Anti Derivative and Indefinite Integrals 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I Section 4.8 Anti Derivative and Indefinite Integrals 2 Lectures College of Science MATHS 101: Calculus I (University of Bahrain) 1 / 28 Indefinite Integral Given a function f, if F is a function such that

More information

INTEGRATION: THE FUNDAMENTAL THEOREM OF CALCULUS MR. VELAZQUEZ AP CALCULUS

INTEGRATION: THE FUNDAMENTAL THEOREM OF CALCULUS MR. VELAZQUEZ AP CALCULUS INTEGRATION: THE FUNDAMENTAL THEOREM OF CALCULUS MR. VELAZQUEZ AP CALCULUS RECALL: ANTIDERIVATIVES When we last spoke of integration, we examined a physics problem where we saw that the area under the

More information

Math 106: Review for Exam II - SOLUTIONS

Math 106: Review for Exam II - SOLUTIONS Math 6: Review for Exam II - SOLUTIONS INTEGRATION TIPS Substitution: usually let u a function that s inside another function, especially if du (possibly off by a multiplying constant) is also present

More information

Dr. Abdulla Eid. Section 3.8 Derivative of the inverse function and logarithms 3 Lecture. Dr. Abdulla Eid. MATHS 101: Calculus I. College of Science

Dr. Abdulla Eid. Section 3.8 Derivative of the inverse function and logarithms 3 Lecture. Dr. Abdulla Eid. MATHS 101: Calculus I. College of Science Section 3.8 Derivative of the inverse function and logarithms 3 Lecture College of Science MATHS 101: Calculus I (University of Bahrain) Logarithmic Differentiation 1 / 19 Topics 1 Inverse Functions (1

More information

Applied Calculus I. Lecture 29

Applied Calculus I. Lecture 29 Applied Calculus I Lecture 29 Integrals of trigonometric functions We shall continue learning substitutions by considering integrals involving trigonometric functions. Integrals of trigonometric functions

More information

Section 5.6. Integration By Parts. MATH 126 (Section 5.6) Integration By Parts The University of Kansas 1 / 10

Section 5.6. Integration By Parts. MATH 126 (Section 5.6) Integration By Parts The University of Kansas 1 / 10 Section 5.6 Integration By Parts MATH 126 (Section 5.6) Integration By Parts The University of Kansas 1 / 10 Integration By Parts Manipulating the Product Rule d dx (f (x) g(x)) = f (x) g (x) + f (x) g(x)

More information

If y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy. du du. If y = f (u) then y = f (u) u

If y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy. du du. If y = f (u) then y = f (u) u Section 3 4B Lecture The Chain Rule If y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy du du dx or If y = f (u) then y = f (u) u The Chain Rule

More information

Lecture 7 - Separable Equations

Lecture 7 - Separable Equations Lecture 7 - Separable Equations Separable equations is a very special type of differential equations where you can separate the terms involving only y on one side of the equation and terms involving only

More information

Math 250 Skills Assessment Test

Math 250 Skills Assessment Test Math 5 Skills Assessment Test Page Math 5 Skills Assessment Test The purpose of this test is purely diagnostic (before beginning your review, it will be helpful to assess both strengths and weaknesses).

More information

Updated: January 16, 2016 Calculus II 7.4. Math 230. Calculus II. Brian Veitch Fall 2015 Northern Illinois University

Updated: January 16, 2016 Calculus II 7.4. Math 230. Calculus II. Brian Veitch Fall 2015 Northern Illinois University Math 30 Calculus II Brian Veitch Fall 015 Northern Illinois University Integration of Rational Functions by Partial Fractions From algebra, we learned how to find common denominators so we can do something

More information

Core Mathematics 3 Differentiation

Core Mathematics 3 Differentiation http://kumarmaths.weebly.com/ Core Mathematics Differentiation C differentiation Page Differentiation C Specifications. By the end of this unit you should be able to : Use chain rule to find the derivative

More information

The goal of today is to determine what u-substitution to use for trigonometric integrals. The most common substitutions are the following:

The goal of today is to determine what u-substitution to use for trigonometric integrals. The most common substitutions are the following: Trigonometric Integrals The goal of today is to determine what u-substitution to use for trigonometric integrals. The most common substitutions are the following: Substitution u sinx u cosx u tanx u secx

More information

Math 180, Final Exam, Fall 2012 Problem 1 Solution

Math 180, Final Exam, Fall 2012 Problem 1 Solution Math 80, Final Exam, Fall 0 Problem Solution. Find the derivatives of the following functions: (a) ln(ln(x)) (b) x 6 + sin(x) e x (c) tan(x ) + cot(x ) (a) We evaluate the derivative using the Chain Rule.

More information

11.5. The Chain Rule. Introduction. Prerequisites. Learning Outcomes

11.5. The Chain Rule. Introduction. Prerequisites. Learning Outcomes The Chain Rule 11.5 Introduction In this Section we will see how to obtain the derivative of a composite function (often referred to as a function of a function ). To do this we use the chain rule. This

More information

Grade: The remainder of this page has been left blank for your workings. VERSION D. Midterm D: Page 1 of 12

Grade: The remainder of this page has been left blank for your workings. VERSION D. Midterm D: Page 1 of 12 First Name: Student-No: Last Name: Section: Grade: The remainder of this page has been left blank for your workings. Midterm D: Page of 2 Indefinite Integrals. 9 marks Each part is worth marks. Please

More information

Solutions to Exam 1, Math Solution. Because f(x) is one-to-one, we know the inverse function exists. Recall that (f 1 ) (a) =

Solutions to Exam 1, Math Solution. Because f(x) is one-to-one, we know the inverse function exists. Recall that (f 1 ) (a) = Solutions to Exam, Math 56 The function f(x) e x + x 3 + x is one-to-one (there is no need to check this) What is (f ) ( + e )? Solution Because f(x) is one-to-one, we know the inverse function exists

More information

Integration by Substitution

Integration by Substitution Integration by Substitution Dr. Philippe B. Laval Kennesaw State University Abstract This handout contains material on a very important integration method called integration by substitution. Substitution

More information

Differentiation by taking logarithms

Differentiation by taking logarithms Differentiation by taking logarithms In this unit we look at how we can use logarithms to simplify certain functions before we differentiate them. In order to master the techniques explained here it is

More information

MATH 1231 MATHEMATICS 1B Calculus Section 4.4: Taylor & Power series.

MATH 1231 MATHEMATICS 1B Calculus Section 4.4: Taylor & Power series. MATH 1231 MATHEMATICS 1B 2010. For use in Dr Chris Tisdell s lectures. Calculus Section 4.4: Taylor & Power series. 1. What is a Taylor series? 2. Convergence of Taylor series 3. Common Maclaurin series

More information

Mathematics 1 Lecture Notes Chapter 1 Algebra Review

Mathematics 1 Lecture Notes Chapter 1 Algebra Review Mathematics 1 Lecture Notes Chapter 1 Algebra Review c Trinity College 1 A note to the students from the lecturer: This course will be moving rather quickly, and it will be in your own best interests to

More information

Math 230 Mock Final Exam Detailed Solution

Math 230 Mock Final Exam Detailed Solution Name: Math 30 Mock Final Exam Detailed Solution Disclaimer: This mock exam is for practice purposes only. No graphing calulators TI-89 is allowed on this test. Be sure that all of your work is shown and

More information

MATH 250 TOPIC 13 INTEGRATION. 13B. Constant, Sum, and Difference Rules

MATH 250 TOPIC 13 INTEGRATION. 13B. Constant, Sum, and Difference Rules Math 5 Integration Topic 3 Page MATH 5 TOPIC 3 INTEGRATION 3A. Integration of Common Functions Practice Problems 3B. Constant, Sum, and Difference Rules Practice Problems 3C. Substitution Practice Problems

More information

Study 5.5, # 1 5, 9, 13 27, 35, 39, 49 59, 63, 69, 71, 81. Class Notes: Prof. G. Battaly, Westchester Community College, NY Homework.

Study 5.5, # 1 5, 9, 13 27, 35, 39, 49 59, 63, 69, 71, 81. Class Notes: Prof. G. Battaly, Westchester Community College, NY Homework. Goals: 1. Recognize an integrand that is the derivative of a composite function. 2. Generalize the Basic Integration Rules to include composite functions. 3. Use substitution to simplify the process of

More information

QUADRATIC GRAPHS ALGEBRA 2. Dr Adrian Jannetta MIMA CMath FRAS INU0114/514 (MATHS 1) Quadratic Graphs 1/ 16 Adrian Jannetta

QUADRATIC GRAPHS ALGEBRA 2. Dr Adrian Jannetta MIMA CMath FRAS INU0114/514 (MATHS 1) Quadratic Graphs 1/ 16 Adrian Jannetta QUADRATIC GRAPHS ALGEBRA 2 INU0114/514 (MATHS 1) Dr Adrian Jannetta MIMA CMath FRAS Quadratic Graphs 1/ 16 Adrian Jannetta Objectives Be able to sketch the graph of a quadratic function Recognise the shape

More information

1 Lecture 39: The substitution rule.

1 Lecture 39: The substitution rule. Lecture 39: The substitution rule. Recall the chain rule and restate as the substitution rule. u-substitution, bookkeeping for integrals. Definite integrals, changing limits. Symmetry-integrating even

More information

Access to Science, Engineering and Agriculture: Mathematics 2 MATH00040 Chapter 4 Solutions

Access to Science, Engineering and Agriculture: Mathematics 2 MATH00040 Chapter 4 Solutions Access to Science, Engineering and Agriculture: Mathematics MATH4 Chapter 4 Solutions In all these solutions, c will represent an arbitrary constant.. (a) Since f(x) 5 is a constant, 5dx 5x] 5. (b) Since

More information

Exploring Substitution

Exploring Substitution I. Introduction Exploring Substitution Math Fall 08 Lab We use the Fundamental Theorem of Calculus, Part to evaluate a definite integral. If f is continuous on [a, b] b and F is any antiderivative of f

More information

Math 480 The Vector Space of Differentiable Functions

Math 480 The Vector Space of Differentiable Functions Math 480 The Vector Space of Differentiable Functions The vector space of differentiable functions. Let C (R) denote the set of all infinitely differentiable functions f : R R. Then C (R) is a vector space,

More information

Math 226 Calculus Spring 2016 Exam 2V1

Math 226 Calculus Spring 2016 Exam 2V1 Math 6 Calculus Spring 6 Exam V () (35 Points) Evaluate the following integrals. (a) (7 Points) tan 5 (x) sec 3 (x) dx (b) (8 Points) cos 4 (x) dx Math 6 Calculus Spring 6 Exam V () (Continued) Evaluate

More information

Chapter 5: Integrals

Chapter 5: Integrals Chapter 5: Integrals Section 5.5 The Substitution Rule (u-substitution) Sec. 5.5: The Substitution Rule We know how to find the derivative of any combination of functions Sum rule Difference rule Constant

More information

Graphs of Antiderivatives, Substitution Integrals

Graphs of Antiderivatives, Substitution Integrals Unit #10 : Graphs of Antiderivatives, Substitution Integrals Goals: Relationship between the graph of f(x) and its anti-derivative F (x) The guess-and-check method for anti-differentiation. The substitution

More information

8.3 Partial Fraction Decomposition

8.3 Partial Fraction Decomposition 8.3 partial fraction decomposition 575 8.3 Partial Fraction Decomposition Rational functions (polynomials divided by polynomials) and their integrals play important roles in mathematics and applications,

More information

MATH 1231 MATHEMATICS 1B Calculus Section 1: - Integration.

MATH 1231 MATHEMATICS 1B Calculus Section 1: - Integration. MATH 1231 MATHEMATICS 1B 2007. For use in Dr Chris Tisdell s lectures: Tues 11 + Thur 10 in KBT Calculus Section 1: - Integration. 1. Motivation 2. What you should already know 3. Useful integrals 4. Integrals

More information

The Chain Rule. The Chain Rule. dy dy du dx du dx. For y = f (u) and u = g (x)

The Chain Rule. The Chain Rule. dy dy du dx du dx. For y = f (u) and u = g (x) AP Calculus Mrs. Jo Brooks The Chain Rule To find the derivative of more complicated functions, we use something called the chain rule. It can be confusing unless you keep yourself organized as you go

More information

MATH 1207 R02 MIDTERM EXAM 2 SOLUTION

MATH 1207 R02 MIDTERM EXAM 2 SOLUTION MATH 7 R MIDTERM EXAM SOLUTION FALL 6 - MOON Name: Write your answer neatly and show steps. Except calculators, any electronic devices including laptops and cell phones are not allowed. () (5 pts) Find

More information

Example. Evaluate. 3x 2 4 x dx.

Example. Evaluate. 3x 2 4 x dx. 3x 2 4 x 3 + 4 dx. Solution: We need a new technique to integrate this function. Notice that if we let u x 3 + 4, and we compute the differential du of u, we get: du 3x 2 dx Going back to our integral,

More information

MAT 132 Midterm 1 Spring 2017

MAT 132 Midterm 1 Spring 2017 MAT Midterm Spring 7 Name: ID: Problem 5 6 7 8 Total ( pts) ( pts) ( pts) ( pts) ( pts) ( pts) (5 pts) (5 pts) ( pts) Score Instructions: () Fill in your name and Stony Brook ID number at the top of this

More information

f(g(x)) g (x) dx = f(u) du.

f(g(x)) g (x) dx = f(u) du. 1. Techniques of Integration Section 8-IT 1.1. Basic integration formulas. Integration is more difficult than derivation. The derivative of every rational function or trigonometric function is another

More information

A.P. Calculus BC Test Two Section One Multiple-Choice Calculators Allowed Time 40 minutes Number of Questions 15

A.P. Calculus BC Test Two Section One Multiple-Choice Calculators Allowed Time 40 minutes Number of Questions 15 A.P. Calculus BC Test Two Section One Multiple-Choice Calculators Allowed Time 40 minutes Number of Questions 15 The scoring for this section is determined by the formula [C (0.25 I)] 1.8 where C is the

More information

The Relation between the Integral and the Derivative Graphs. Unit #10 : Graphs of Antiderivatives, Substitution Integrals

The Relation between the Integral and the Derivative Graphs. Unit #10 : Graphs of Antiderivatives, Substitution Integrals Graphs of Antiderivatives - Unit #0 : Graphs of Antiderivatives, Substitution Integrals Goals: Relationship between the graph of f(x) and its anti-derivative F (x) The guess-and-check method for anti-differentiation.

More information

Infinite series, improper integrals, and Taylor series

Infinite series, improper integrals, and Taylor series Chapter Infinite series, improper integrals, and Taylor series. Determine which of the following sequences converge or diverge (a) {e n } (b) {2 n } (c) {ne 2n } (d) { 2 n } (e) {n } (f) {ln(n)} 2.2 Which

More information

If y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy. du du. If y = f (u) then y = f (u) u

If y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy. du du. If y = f (u) then y = f (u) u Section 3 4B The Chain Rule If y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy du du dx or If y = f (u) then f (u) u The Chain Rule with the Power

More information

Substitutions and by Parts, Area Between Curves. Goals: The Method of Substitution Areas Integration by Parts

Substitutions and by Parts, Area Between Curves. Goals: The Method of Substitution Areas Integration by Parts Week #7: Substitutions and by Parts, Area Between Curves Goals: The Method of Substitution Areas Integration by Parts 1 Week 7 The Indefinite Integral The Fundamental Theorem of Calculus, b a f(x) dx =

More information

Math 261 Calculus I. Test 1 Study Guide. Name. Decide whether the limit exists. If it exists, find its value. 1) lim x 1. f(x) 2) lim x -1/2 f(x)

Math 261 Calculus I. Test 1 Study Guide. Name. Decide whether the limit exists. If it exists, find its value. 1) lim x 1. f(x) 2) lim x -1/2 f(x) Math 261 Calculus I Test 1 Study Guide Name Decide whether the it exists. If it exists, find its value. 1) x 1 f(x) 2) x -1/2 f(x) Complete the table and use the result to find the indicated it. 3) If

More information

Grade: The remainder of this page has been left blank for your workings. VERSION E. Midterm E: Page 1 of 12

Grade: The remainder of this page has been left blank for your workings. VERSION E. Midterm E: Page 1 of 12 First Name: Student-No: Last Name: Section: Grade: The remainder of this page has been left blank for your workings. Midterm E: Page of Indefinite Integrals. 9 marks Each part is worth 3 marks. Please

More information

Assignment # 8, Math 370, Fall 2018 SOLUTIONS:

Assignment # 8, Math 370, Fall 2018 SOLUTIONS: Assignment # 8, Math 370, Fall 018 SOLUTIONS: Problem 1: Solve the equations (a) y y = 3x + x 4, (i) y(0) = 1, y (0) = 1, y (0) = 1. Characteristic equation: α 3 α = 0 so α 1, = 0 and α 3 =. y c = C 1

More information

Math 111 lecture for Friday, Week 10

Math 111 lecture for Friday, Week 10 Math lecture for Friday, Week Finding antiderivatives mean reversing the operation of taking derivatives. Today we ll consider reversing the chain rule and the product rule. Substitution technique. Recall

More information

f(x) g(x) = [f (x)g(x) dx + f(x)g (x)dx

f(x) g(x) = [f (x)g(x) dx + f(x)g (x)dx Chapter 7 is concerned with all the integrals that can t be evaluated with simple antidifferentiation. Chart of Integrals on Page 463 7.1 Integration by Parts Like with the Chain Rule substitutions with

More information

Math 180, Exam 2, Practice Fall 2009 Problem 1 Solution. f(x) = arcsin(2x + 1) = sin 1 (3x + 1), lnx

Math 180, Exam 2, Practice Fall 2009 Problem 1 Solution. f(x) = arcsin(2x + 1) = sin 1 (3x + 1), lnx Math 80, Exam, Practice Fall 009 Problem Solution. Differentiate the functions: (do not simplify) f(x) = x ln(x + ), f(x) = xe x f(x) = arcsin(x + ) = sin (3x + ), f(x) = e3x lnx Solution: For the first

More information

OCR A2 Level Mathematics Core Mathematics Scheme of Work

OCR A2 Level Mathematics Core Mathematics Scheme of Work OCR A Level Mathematics Core Mathematics Scheme of Work Examination in June of Year 13 The Solomen press worksheets are an excellent resource and incorporated into the SOW NUMERICAL METHODS (6 ) (Solomen

More information

Methods of Integration

Methods of Integration Methods of Integration Professor D. Olles January 8, 04 Substitution The derivative of a composition of functions can be found using the chain rule form d dx [f (g(x))] f (g(x)) g (x) Rewriting the derivative

More information

Prelim 1 Solutions V2 Math 1120

Prelim 1 Solutions V2 Math 1120 Feb., Prelim Solutions V Math Please show your reasoning and all your work. This is a 9 minute exam. Calculators are not needed or permitted. Good luck! Problem ) ( Points) Calculate the following: x a)

More information

p324 Section 5.2: The Natural Logarithmic Function: Integration

p324 Section 5.2: The Natural Logarithmic Function: Integration p324 Section 5.2: The Natural Logarithmic Function: Integration Theorem 5.5: Log Rule for Integration Let u be a differentiable function of x 1. 2. Example 1: Using the Log Rule for Integration ** Note:

More information

Integrated Calculus II Exam 1 Solutions 2/6/4

Integrated Calculus II Exam 1 Solutions 2/6/4 Integrated Calculus II Exam Solutions /6/ Question Determine the following integrals: te t dt. We integrate by parts: u = t, du = dt, dv = e t dt, v = dv = e t dt = e t, te t dt = udv = uv vdu = te t (

More information

Math 112 Rahman. Week Taylor Series Suppose the function f has the following power series:

Math 112 Rahman. Week Taylor Series Suppose the function f has the following power series: Math Rahman Week 0.8-0.0 Taylor Series Suppose the function f has the following power series: fx) c 0 + c x a) + c x a) + c 3 x a) 3 + c n x a) n. ) Can we figure out what the coefficients are? Yes, yes

More information

University Calculus I. Worksheet # 8 Mar b. sin tan e. sin 2 sin 1 5. b. tan. c. sec sin 1 ( x )) cos 1 ( x )) f. csc. c.

University Calculus I. Worksheet # 8 Mar b. sin tan e. sin 2 sin 1 5. b. tan. c. sec sin 1 ( x )) cos 1 ( x )) f. csc. c. MATH 6 WINTER 06 University Calculus I Worksheet # 8 Mar. 06-0 The topic covered by this worksheet is: Derivative of Inverse Functions and the Inverse Trigonometric functions. SamplesolutionstoallproblemswillbeavailableonDL,

More information

DRAFT - Math 102 Lecture Note - Dr. Said Algarni

DRAFT - Math 102 Lecture Note - Dr. Said Algarni Math02 - Term72 - Guides and Exercises - DRAFT 7 Techniques of Integration A summery for the most important integrals that we have learned so far: 7. Integration by Parts The Product Rule states that if

More information

Math 229 Mock Final Exam Solution

Math 229 Mock Final Exam Solution Name: Math 229 Mock Final Exam Solution Disclaimer: This mock exam is for practice purposes only. No graphing calulators TI-89 is allowed on this test. Be sure that all of your work is shown and that it

More information

Chapter 7: Techniques of Integration

Chapter 7: Techniques of Integration Chapter 7: Techniques of Integration MATH 206-01: Calculus II Department of Mathematics University of Louisville last corrected September 14, 2013 1 / 43 Chapter 7: Techniques of Integration 7.1. Integration

More information

Integration 1/10. Integration. Student Guidance Centre Learning Development Service

Integration 1/10. Integration. Student Guidance Centre Learning Development Service Integration / Integration Student Guidance Centre Learning Development Service lds@qub.ac.uk Integration / Contents Introduction. Indefinite Integration....................... Definite Integration.......................

More information

Math 181, Exam 2, Study Guide 2 Problem 1 Solution. 1 + dx. 1 + (cos x)2 dx. 1 + cos2 xdx. = π ( 1 + cos π 2

Math 181, Exam 2, Study Guide 2 Problem 1 Solution. 1 + dx. 1 + (cos x)2 dx. 1 + cos2 xdx. = π ( 1 + cos π 2 Math 8, Exam, Study Guide Problem Solution. Use the trapezoid rule with n to estimate the arc-length of the curve y sin x between x and x π. Solution: The arclength is: L b a π π + ( ) dy + (cos x) + cos

More information

Chapter 7 Rational Expressions, Equations, and Functions

Chapter 7 Rational Expressions, Equations, and Functions Chapter 7 Rational Expressions, Equations, and Functions Section 7.1: Simplifying, Multiplying, and Dividing Rational Expressions and Functions Section 7.2: Adding and Subtracting Rational Expressions

More information

WeBWorK, Problems 2 and 3

WeBWorK, Problems 2 and 3 WeBWorK, Problems 2 and 3 7 dx 2. Evaluate x ln(6x) This can be done using integration by parts or substitution. (Most can not). However, it is much more easily done using substitution. This can be written

More information

To calculate the x=moment, we multiply the integrand above by x to get = )

To calculate the x=moment, we multiply the integrand above by x to get = ) . ( points) The region shown below is the area between the curves y = 3x and y = x. Find the center of mass of this region. 8 6 3 We must calculate the area, x-moment, and y-moment to find the center of

More information

MATH 116, LECTURE 13, 14 & 15: Derivatives

MATH 116, LECTURE 13, 14 & 15: Derivatives MATH 116, LECTURE 13, 14 & 15: Derivatives 1 Formal Definition of the Derivative We have seen plenty of limits so far, but very few applications. In particular, we have seen very few functions for which

More information

S56 (5.3) Further Calculus.notebook March 24, 2016

S56 (5.3) Further Calculus.notebook March 24, 2016 Daily Practice 16.3.2016 Today we will be learning how to differentiate using the Chain Rule. Homework Solutions Video online - please mark 2009 P2 Polynomials HW Online due 22.3.16 We use the Chain Rule

More information

Math Final Exam Review

Math Final Exam Review Math - Final Exam Review. Find dx x + 6x +. Name: Solution: We complete the square to see if this function has a nice form. Note we have: x + 6x + (x + + dx x + 6x + dx (x + + Note that this looks a lot

More information

Final Exam Review Exercise Set A, Math 1551, Fall 2017

Final Exam Review Exercise Set A, Math 1551, Fall 2017 Final Exam Review Exercise Set A, Math 1551, Fall 2017 This review set gives a list of topics that we explored throughout this course, as well as a few practice problems at the end of the document. A complete

More information

Unit #10 : Graphs of Antiderivatives; Substitution Integrals

Unit #10 : Graphs of Antiderivatives; Substitution Integrals Unit #10 : Graphs of Antiderivatives; Substitution Integrals Goals: Relationship between the graph of f(x) and its anti-derivative F(x) The guess-and-check method for anti-differentiation. The substitution

More information

8.7 MacLaurin Polynomials

8.7 MacLaurin Polynomials 8.7 maclaurin polynomials 67 8.7 MacLaurin Polynomials In this chapter you have learned to find antiderivatives of a wide variety of elementary functions, but many more such functions fail to have an antiderivative

More information

Review for the Final Exam

Review for the Final Exam Math 171 Review for the Final Exam 1 Find the limits (4 points each) (a) lim 4x 2 3; x x (b) lim ( x 2 x x 1 )x ; (c) lim( 1 1 ); x 1 ln x x 1 sin (x 2) (d) lim x 2 x 2 4 Solutions (a) The limit lim 4x

More information

Section 10.7 Taylor series

Section 10.7 Taylor series Section 10.7 Taylor series 1. Common Maclaurin series 2. s and approximations with Taylor polynomials 3. Multiplication and division of power series Math 126 Enhanced 10.7 Taylor Series The University

More information

( ) is called the dependent variable because its

( ) is called the dependent variable because its page 1 of 16 CLASS NOTES: 3 8 thru 4 3 and 11 7 Functions, Exponents and Polynomials 3 8: Function Notation A function is a correspondence between two sets, the domain (x) and the range (y). An example

More information

Review of Integration Techniques

Review of Integration Techniques A P P E N D I X D Brief Review of Integration Techniques u-substitution The basic idea underlying u-substitution is to perform a simple substitution that converts the intergral into a recognizable form

More information

MATH 1A - FINAL EXAM DELUXE - SOLUTIONS. x x x x x 2. = lim = 1 =0. 2) Then ln(y) = x 2 ln(x) 3) ln(x)

MATH 1A - FINAL EXAM DELUXE - SOLUTIONS. x x x x x 2. = lim = 1 =0. 2) Then ln(y) = x 2 ln(x) 3) ln(x) MATH A - FINAL EXAM DELUXE - SOLUTIONS PEYAM RYAN TABRIZIAN. ( points, 5 points each) Find the following limits (a) lim x x2 + x ( ) x lim x2 + x x2 + x 2 + + x x x x2 + + x x 2 + x 2 x x2 + + x x x2 +

More information

Calculus II/III Summer Packet

Calculus II/III Summer Packet Calculus II/III Summer Packet First of all, have a great summer! Enjoy your time away from school. Come back fired up and ready to learn. I know that I will be ready to have a great year of calculus with

More information

t 2 + 2t dt = (t + 1) dt + 1 = arctan t x + 6 x(x 3)(x + 2) = A x +

t 2 + 2t dt = (t + 1) dt + 1 = arctan t x + 6 x(x 3)(x + 2) = A x + MATH 06 0 Practice Exam #. (0 points) Evaluate the following integrals: (a) (0 points). t +t+7 This is an irreducible quadratic; its denominator can thus be rephrased via completion of the square as a

More information