At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
Integration by substitution is the chain rule in reverse.
NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200
Slides: On the Chi Square and Higher-Order Chi Distances for Approximating f-...Frank Nielsen
Slides for the paper:
On the Chi Square and Higher-Order Chi Distances for Approximating f-Divergences
published in IEEE SPL:
http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=6654274
Lesson 26: The Fundamental Theorem of Calculus (slides)Matthew Leingang
The document discusses the Fundamental Theorem of Calculus, which has two parts. The first part states that if a function f is continuous on an interval, then the derivative of the integral of f is equal to f. This is proven using Riemann sums. The second part relates the integral of a function f to the integral of its derivative F'. Examples are provided to illustrate how the area under a curve relates to these concepts.
The document discusses curve sketching of functions by analyzing their derivatives. It provides:
1) A checklist for graphing a function which involves finding where the function is positive/negative/zero, its monotonicity from the first derivative, and concavity from the second derivative.
2) An example of graphing the cubic function f(x) = 2x^3 - 3x^2 - 12x through analyzing its derivatives.
3) Explanations of the increasing/decreasing test and concavity test to determine monotonicity and concavity from a function's derivatives.
This document discusses numerical integration techniques including the trapezoidal rule, Simpson's 1/3 rule, Simpson's 3/8 rule, and Gaussian integration formulas. It provides the formulas for calculating integration numerically using these methods and notes that accuracy increases with smaller interval widths h. Errors are estimated to be order h^2 for trapezoidal rule and order h^4 for Simpson's rules.
This document contains lecture notes on continuity from a Calculus I class at New York University. It begins with announcements about office hours and homework grades. It then reviews the definition of a limit and introduces the definition of continuity as a function having a limit equal to its value at a point. Examples are provided to demonstrate showing a function is continuous. The document states that polynomials, rational functions, and trigonometric functions are continuous based on their definitions and limit properties. It concludes by explaining the continuity of inverse trigonometric functions.
The document discusses numerical methods for solving nonlinear equations, including root finding and systems of nonlinear equations. It covers the basics of nonlinear solvers like bisection, Newton's method, and fixed-point iteration. For one-dimensional root finding, it analyzes the convergence properties and order of convergence for these methods. It then extends the discussion to systems of nonlinear equations and shows how Newton's method can be applied by taking derivatives to form the Jacobian matrix.
1) The document discusses how statistical learning techniques from other disciplines can inform econometric modeling and central bank policymaking.
2) It covers topics like high-dimensional data analysis, nonparametric regression, causal inference challenges, and model selection methods.
3) The key message is that econometrics can benefit from adopting techniques from fields like machine learning and statistics to develop more flexible, data-driven models.
This document provides an introduction to multiattribute decision making and decision theories. It discusses several key aspects of multiattribute choice models, including:
1) The number and nature of attributes that are used to differentiate decision alternatives.
2) The structure of the feasible set of alternatives.
3) The basis of evaluation, such as preference relations or criterion functions.
4) Independence and separability assumptions that are required to obtain additive representations of preferences.
The document outlines some classic evaluation theories under certainty that do not involve probabilities, and discusses the concept of separability, which reduces complexity by allowing decentralized preferences across attribute groups.
Lesson 27: Integration by Substitution (Section 041 slides)Matthew Leingang
The document contains notes from a Calculus I class at New York University on December 13, 2010. It discusses using the substitution method for indefinite and definite integrals. Examples are provided to demonstrate how to use substitutions to evaluate integrals involving trigonometric, exponential, and polynomial functions. The key steps are to make a substitution for the variable in terms of a new variable, determine the differential of the substitution, and substitute into the integral to transform it into an integral involving only the new variable.
This document summarizes several methods for estimating copula densities from sample data in a nonparametric way, including using kernel density estimation with different types of kernels and variable transformations. It describes the standard kernel estimate, issues with it near boundaries, a mirror kernel estimate, using beta kernels, a probit transformation of variables, and improved probit transformation estimators that use local polynomial approximations. The goal is to find estimators that are consistent along the boundaries of the copula support and improve inference about the copula density.
Image sciences, image processing, image restoration, photo manipulation. Image and videos representation. Digital versus analog imagery. Quantization and sampling. Sources and models of noises in digital CCD imagery: photon, thermal and readout noises. Sources and models of blurs. Convolutions and point spread functions. Overview of other standard models, problems and tasks: salt-and-pepper and impulse noises, half toning, inpainting, super-resolution, compressed sensing, high dynamic range imagery, demosaicing. Short introduction to other types of imagery: SAR, Sonar, ultrasound, CT and MRI. Linear and ill-posed restoration problems.
This document summarizes Arthur Charpentier's presentation on data science and actuarial science using R. It discusses S3 and S4 classes in R for defining custom object classes. It also covers matrices, vectors, numbers and memory management in R. The presentation references advanced R techniques and packages for insurance and data mining applications.
This document presents a new coupled fixed point theorem for mappings having the mixed monotone property in partially ordered metric spaces. Specifically:
1) The theorem establishes the existence of a coupled fixed point for a mapping F that satisfies a contraction-type condition and has the mixed monotone property in a partially ordered, complete metric space.
2) It is shown that the coupled fixed point can be unique under additional conditions involving midpoint lower or upper bound properties.
3) An estimate is provided for the convergence rate as the iterates of the mapping F converge to the coupled fixed point.
This document discusses moment-generating functions and their properties and applications in probability theory. It defines the moment-generating function for both discrete and continuous random variables. Some key properties of moment-generating functions are that each probability distribution has a unique moment-generating function, and they can be used to find the distribution of sums of random variables. The document also describes several common probability distributions and derives their corresponding moment-generating functions.
JEE Mathematics/ Lakshmikanta Satapathy/ Sequences and Series QA part 3/ JEE question on the sum of a special sequence solved with the related concepts
JEE Physics/ Lakshmikanta Satapathy/ Gravitation QA part 3/ JEE question on finding the escape velocity from the surface of a planet from velocity of projection and height gained by a bullet
JEE Physics/ Lakshmikanta Satapathy/ Direct Current QA part 9/ Question on Power consumed by an external resistance connected across a parallel combination of two identical cells
JEE Physics/ Lakshmikanta Satapathy/ Alternating Current Theory part 1/ Mean value of Alternating Current and its relation with Peak value discussed with the related concepts
JEE Physics/ Lakshmikanta Satapathy/ Direct current theory part 5/ Theory of EMF Terminal Potential Difference and Internal Resistance of a cell with complete explanation of the related concepts
A man was walking towards a vertical pillar at a uniform speed. At point A, the angle of elevation of the top of the pillar was 30 degrees. After 10 minutes at point B, the angle was 60 degrees. Using trigonometry relationships in the triangles formed, the distance traveled from A to B was calculated. Given the uniform speed, the time taken to reach the pillar from B was determined to be 5 minutes.
JEE Mathematics/ Lakshmikanta Satapathy/ Sequences and Series QA part 2/ JEE Question on Arithmetic Mean and Geometric Mean solved with the related concepts
JEE Physics/ Lakshmikanta Satapathy/ Table of Units and Dimensional Formulae with the related Key equations necessary to determine the units and dimensions of Fundamental and Derived Physical quantities
JEE Mathematics/ Lakshmikanta Satapathy/ 3D Geometry QA 10/ JEE question on equation of plane passing through the line of intersection of two planes and at a given distance from a given point
JEE Physics/ Lakshmikanta Satapathy/ Laws of Motion QA part 2/ Question on Equilibrium of forces solved by resolution of forces into rectangular components
JEE Mathematics/ Lakshmikanta Satapathy/ Set Theory part 2/ Theory of Union Intersection and Difference of two sets Complement of a Set, Formulae necessary for solving practical problems involving Cardinality of Sets discussed in details
JEE Physics/ Lakshmikanta Satapathy/ Laws of Motion QA part 5/ JEE Question on motion of a balloon and the effect of change in mass on its acceleration solved with the related concepts
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
An antiderivative of a function is a function whose derivative is the given function. The problem of antidifferentiation is interesting, complicated, and useful, especially when discussing motion.
This is the slideshow version from class.
This document is from a Calculus I class at New York University and covers antiderivatives. It begins with announcements about an upcoming quiz. The objectives are to find antiderivatives of simple functions, remember that a function whose derivative is zero must be constant, and solve rectilinear motion problems. It then outlines finding antiderivatives through tabulation, graphically, and with rectilinear motion examples. The document provides examples of finding antiderivatives of power functions by using the power rule in reverse.
This document is from a Calculus I class at New York University and covers antiderivatives. It begins with announcements about an upcoming quiz. The objectives are to find antiderivatives of simple functions, remember that a function whose derivative is zero must be constant, and solve rectilinear motion problems. It then outlines finding antiderivatives through tabulation, graphically, and with rectilinear motion examples. Examples are provided of finding the antiderivative of power functions like x^3 through identifying the power rule relationship between a function and its derivative.
An antiderivative of a function is a function whose derivative is the given function. The problem of antidifferentiation is interesting, complicated, and useful, especially when discussing motion.
An antiderivative of a function is a function whose derivative is the given function. The problem of antidifferentiation is interesting, complicated, and useful, especially when discussing motion.
An antiderivative of a function is a function whose derivative is the given function. The problem of antidifferentiation is interesting, complicated, and useful, especially when discussing motion.
This is the handout version to take notes on.
Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 041 ...Matthew Leingang
This document is a lecture on derivatives of exponential and logarithmic functions from a Calculus I class at New York University. It covers the objectives and outline, which include finding derivatives of exponential functions with any base, logarithmic functions with any base, and using logarithmic differentiation. It provides proofs and examples of finding derivatives, such as the derivative of the natural exponential function being itself and the derivative of the natural logarithm function.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
This document is a section from a Calculus I course at New York University dated September 20, 2010. It discusses continuity of functions, beginning with definitions and examples of determining continuity. Key points covered include the definition of continuity as a function having a limit equal to its value, and theorems stating polynomials, rational functions, and combinations of continuous functions are also continuous. Trigonometric functions like sin, cos, tan, and cot are shown to be continuous on their domains. The document provides examples, explanations, and questions to illustrate the concept of continuity.
- The document is a lecture on calculus from an NYU course. It discusses using derivatives to determine the monotonicity and concavity of functions.
- There will be a quiz this week covering sections 3.3, 3.4, 3.5, and 3.7. Homework is due November 24.
- The lecture covers using the first derivative to determine if a function is increasing or decreasing over an interval, and using the second derivative to determine if a graph is concave up or down.
Lesson 15: Exponential Growth and Decay (Section 041 slides)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
This document is from a Calculus I class at New York University and covers continuity. It provides announcements about office hours and homework deadlines. It then discusses the objectives of understanding the definition of continuity and applying it to piecewise functions. Examples are provided to demonstrate how to show a function is continuous at a point by evaluating the limit as x approaches the point and showing it equals the function value. The students are asked to determine at which other points the example function is continuous.
This document is from a Calculus I class at New York University and covers continuity. It provides announcements about office hours and homework deadlines. It then discusses the objectives of understanding the definition of continuity and applying it to piecewise functions. Examples are provided to demonstrate how to show a function is continuous at a point by evaluating the limit as x approaches the point and showing it equals the function value. The students are asked to determine at which other points the example function is continuous.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few more good examples.
This document discusses partial derivatives of functions with multiple variables. It defines partial derivatives as derivatives of a function where all but one variable is held constant. For a function z=f(x,y), the partial derivatives with respect to x and y are defined. Higher order partial derivatives and partial derivatives of functions with more than two variables are also introduced. Examples are provided to demonstrate calculating first and second order partial derivatives.
This document contains lecture notes on evaluating definite integrals. It introduces the definition of the definite integral as a limit of Riemann sums, and properties of integrals such as additivity and comparison properties. It also states the Second Fundamental Theorem of Calculus, which relates definite integrals to indefinite integrals via the derivative of the integrand function. Examples are provided to illustrate how to use these properties and theorems to evaluate definite integrals.
Similar to Lesson 23: Antiderivatives (Section 041 slides) (20)
This document provides guidance on developing effective lesson plans for calculus instructors. It recommends starting by defining specific learning objectives and assessments. Examples should be chosen carefully to illustrate concepts and engage students at a variety of levels. The lesson plan should include an introductory problem, definitions, theorems, examples, and group work. Timing for each section should be estimated. After teaching, the lesson can be improved by analyzing what was effective and what needs adjustment for the next time. Advanced preparation is key to looking prepared and ensuring students learn.
Streamlining assessment, feedback, and archival with auto-multiple-choiceMatthew Leingang
Auto-multiple-choice (AMC) is an open-source optical mark recognition software package built with Perl, LaTeX, XML, and sqlite. I use it for all my in-class quizzes and exams. Unique papers are created for each student, fixed-response items are scored automatically, and free-response problems, after manual scoring, have marks recorded in the same process. In the first part of the talk I will discuss AMC’s many features and why I feel it’s ideal for a mathematics course. My contributions to the AMC workflow include some scripts designed to automate the process of returning scored papers
back to students electronically. AMC provides an email gateway, but I have written programs to return graded papers via the DAV protocol to student’s dropboxes on our (Sakai) learning management systems. I will also show how graded papers can be archived, with appropriate metadata tags, into an Evernote notebook.
This document discusses electronic grading of paper assessments using PDF forms. Key points include:
- Various tools for creating fillable PDF forms using LaTeX packages or desktop software.
- Methods for stamping completed forms onto scanned documents including using pdftk or overlaying in TikZ.
- Options for grading on tablets or desktops including GoodReader, PDFExpert, Adobe Acrobat.
- Extracting data from completed forms can be done in Adobe Acrobat or via command line with pdftk.
Lesson 26: The Fundamental Theorem of Calculus (slides)Matthew Leingang
g(x) represents the area under the curve of f(t) between 0 and x.
.
x
What can you say about g? 2 4 6 8 10f
The First Fundamental Theorem of Calculus
Theorem (First Fundamental Theorem of Calculus)
Let f be a con nuous func on on [a, b]. Define the func on F on [a, b] by
∫ x
F(x) = f(t) dt
a
Then F is con nuous on [a, b] and differentiable on (a, b) and for all x in (a, b),
F′(x
Lesson 27: Integration by Substitution (handout)Matthew Leingang
This document contains lecture notes on integration by substitution from a Calculus I class. It introduces the technique of substitution for both indefinite and definite integrals. For indefinite integrals, the substitution rule is presented, along with examples of using substitutions to evaluate integrals involving polynomials, trigonometric, exponential, and other functions. For definite integrals, the substitution rule is extended and examples are worked through both with and without first finding the indefinite integral. The document emphasizes that substitution often simplifies integrals and makes them easier to evaluate.
Lesson 26: The Fundamental Theorem of Calculus (handout)Matthew Leingang
1) The document discusses lecture notes on Section 5.4: The Fundamental Theorem of Calculus from a Calculus I course. 2) It covers stating and explaining the Fundamental Theorems of Calculus and using the first fundamental theorem to find derivatives of functions defined by integrals. 3) The lecture outlines the first fundamental theorem, which relates differentiation and integration, and gives examples of applying it.
This document contains notes from a calculus class lecture on evaluating definite integrals. It discusses using the evaluation theorem to evaluate definite integrals, writing derivatives as indefinite integrals, and interpreting definite integrals as the net change of a function over an interval. The document also contains examples of evaluating definite integrals, properties of integrals, and an outline of the key topics covered.
This document contains lecture notes from a Calculus I class covering Section 5.3 on evaluating definite integrals. The notes discuss using the Evaluation Theorem to calculate definite integrals, writing derivatives as indefinite integrals, and interpreting definite integrals as the net change of a function over an interval. Examples are provided to demonstrate evaluating definite integrals using the midpoint rule approximation. Properties of integrals such as additivity and the relationship between definite and indefinite integrals are also outlined.
Lesson 24: Areas and Distances, The Definite Integral (handout)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
Lesson 24: Areas and Distances, The Definite Integral (slides)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
This document contains lecture notes from a Calculus I class discussing optimization problems. It begins with announcements about upcoming exams and courses the professor is teaching. It then presents an example problem about finding the rectangle of a fixed perimeter with the maximum area. The solution uses calculus techniques like taking the derivative to find the critical points and determine that the optimal rectangle is a square. The notes discuss strategies for solving optimization problems and summarize the key steps to take.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
The document contains lecture notes on curve sketching from a Calculus I class. It discusses using the first and second derivative tests to determine properties of a function like monotonicity, concavity, maxima, minima, and points of inflection in order to sketch the graph of the function. It then provides an example of using these tests to sketch the graph of the cubic function f(x) = 2x^3 - 3x^2 - 12x.
Lesson 20: Derivatives and the Shapes of Curves (slides)Matthew Leingang
This document contains lecture notes on derivatives and the shapes of curves from a Calculus I class taught by Professor Matthew Leingang at New York University. The notes cover using derivatives to determine the intervals where a function is increasing or decreasing, classifying critical points as maxima or minima, using the second derivative to determine concavity, and applying the first and second derivative tests. Examples are provided to illustrate finding intervals of monotonicity for various functions.
Lesson 20: Derivatives and the Shapes of Curves (handout)Matthew Leingang
This document contains lecture notes on calculus from a Calculus I course. It covers determining the monotonicity of functions using the first derivative test. Key points include using the sign of the derivative to determine if a function is increasing or decreasing over an interval, and using the first derivative test to classify critical points as local maxima, minima, or neither. Examples are provided to demonstrate finding intervals of monotonicity for various functions and applying the first derivative test.
The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we can derive tests for the monotonicity (increase or decrease) and concavity of a function.
There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.
There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.
Quality Patents: Patents That Stand the Test of TimeAurora Consulting
Is your patent a vanity piece of paper for your office wall? Or is it a reliable, defendable, assertable, property right? The difference is often quality.
Is your patent simply a transactional cost and a large pile of legal bills for your startup? Or is it a leverageable asset worthy of attracting precious investment dollars, worth its cost in multiples of valuation? The difference is often quality.
Is your patent application only good enough to get through the examination process? Or has it been crafted to stand the tests of time and varied audiences if you later need to assert that document against an infringer, find yourself litigating with it in an Article 3 Court at the hands of a judge and jury, God forbid, end up having to defend its validity at the PTAB, or even needing to use it to block pirated imports at the International Trade Commission? The difference is often quality.
Quality will be our focus for a good chunk of the remainder of this season. What goes into a quality patent, and where possible, how do you get it without breaking the bank?
** Episode Overview **
In this first episode of our quality series, Kristen Hansen and the panel discuss:
⦿ What do we mean when we say patent quality?
⦿ Why is patent quality important?
⦿ How to balance quality and budget
⦿ The importance of searching, continuations, and draftsperson domain expertise
⦿ Very practical tips, tricks, examples, and Kristen’s Musts for drafting quality applications
https://www.aurorapatents.com/patently-strategic-podcast.html
How to Avoid Learning the Linux-Kernel Memory ModelScyllaDB
The Linux-kernel memory model (LKMM) is a powerful tool for developing highly concurrent Linux-kernel code, but it also has a steep learning curve. Wouldn't it be great to get most of LKMM's benefits without the learning curve?
This talk will describe how to do exactly that by using the standard Linux-kernel APIs (locking, reference counting, RCU) along with a simple rules of thumb, thus gaining most of LKMM's power with less learning. And the full LKMM is always there when you need it!
Transcript: Details of description part II: Describing images in practice - T...BookNet Canada
This presentation explores the practical application of image description techniques. Familiar guidelines will be demonstrated in practice, and descriptions will be developed “live”! If you have learned a lot about the theory of image description techniques but want to feel more confident putting them into practice, this is the presentation for you. There will be useful, actionable information for everyone, whether you are working with authors, colleagues, alone, or leveraging AI as a collaborator.
Link to presentation recording and slides: https://bnctechforum.ca/sessions/details-of-description-part-ii-describing-images-in-practice/
Presented by BookNet Canada on June 25, 2024, with support from the Department of Canadian Heritage.
Coordinate Systems in FME 101 - Webinar SlidesSafe Software
If you’ve ever had to analyze a map or GPS data, chances are you’ve encountered and even worked with coordinate systems. As historical data continually updates through GPS, understanding coordinate systems is increasingly crucial. However, not everyone knows why they exist or how to effectively use them for data-driven insights.
During this webinar, you’ll learn exactly what coordinate systems are and how you can use FME to maintain and transform your data’s coordinate systems in an easy-to-digest way, accurately representing the geographical space that it exists within. During this webinar, you will have the chance to:
- Enhance Your Understanding: Gain a clear overview of what coordinate systems are and their value
- Learn Practical Applications: Why we need datams and projections, plus units between coordinate systems
- Maximize with FME: Understand how FME handles coordinate systems, including a brief summary of the 3 main reprojectors
- Custom Coordinate Systems: Learn how to work with FME and coordinate systems beyond what is natively supported
- Look Ahead: Gain insights into where FME is headed with coordinate systems in the future
Don’t miss the opportunity to improve the value you receive from your coordinate system data, ultimately allowing you to streamline your data analysis and maximize your time. See you there!
An invited talk given by Mark Billinghurst on Research Directions for Cross Reality Interfaces. This was given on July 2nd 2024 as part of the 2024 Summer School on Cross Reality in Hagenberg, Austria (July 1st - 7th)
Paradigm Shifts in User Modeling: A Journey from Historical Foundations to Em...Erasmo Purificato
Slide of the tutorial entitled "Paradigm Shifts in User Modeling: A Journey from Historical Foundations to Emerging Trends" held at UMAP'24: 32nd ACM Conference on User Modeling, Adaptation and Personalization (July 1, 2024 | Cagliari, Italy)
Kief Morris rethinks the infrastructure code delivery lifecycle, advocating for a shift towards composable infrastructure systems. We should shift to designing around deployable components rather than code modules, use more useful levels of abstraction, and drive design and deployment from applications rather than bottom-up, monolithic architecture and delivery.
Sustainability requires ingenuity and stewardship. Did you know Pigging Solutions pigging systems help you achieve your sustainable manufacturing goals AND provide rapid return on investment.
How? Our systems recover over 99% of product in transfer piping. Recovering trapped product from transfer lines that would otherwise become flush-waste, means you can increase batch yields and eliminate flush waste. From raw materials to finished product, if you can pump it, we can pig it.
How RPA Help in the Transportation and Logistics Industry.pptxSynapseIndia
Revolutionize your transportation processes with our cutting-edge RPA software. Automate repetitive tasks, reduce costs, and enhance efficiency in the logistics sector with our advanced solutions.
MYIR Product Brochure - A Global Provider of Embedded SOMs & SolutionsLinda Zhang
This brochure gives introduction of MYIR Electronics company and MYIR's products and services.
MYIR Electronics Limited (MYIR for short), established in 2011, is a global provider of embedded System-On-Modules (SOMs) and
comprehensive solutions based on various architectures such as ARM, FPGA, RISC-V, and AI. We cater to customers' needs for large-scale production, offering customized design, industry-specific application solutions, and one-stop OEM services.
MYIR, recognized as a national high-tech enterprise, is also listed among the "Specialized
and Special new" Enterprises in Shenzhen, China. Our core belief is that "Our success stems from our customers' success" and embraces the philosophy
of "Make Your Idea Real, then My Idea Realizing!"
2. . . . . . .
Announcements
Quiz 5 in recitation this
week on §§4.1–4.4
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 2 / 35
3. . . . . . .
Objectives
Given a ”simple“
elementary function, find a
function whose derivative
is that function.
Remember that a function
whose derivative is zero
along an interval must be
zero along that interval.
Solve problems involving
rectilinear motion.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 3 / 35
4. . . . . . .
Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Antiderivatives of piecewise functions
Finding Antiderivatives Graphically
Rectilinear motion
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 4 / 35
5. . . . . . .
What is an antiderivative?
Definition
Let f be a function. An antiderivative for f is a function F such that
F′
= f.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 5 / 35
6. . . . . . .
Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
7. . . . . . .
Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
Solution
???
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
8. . . . . . .
Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
Solution
???
Example
is F(x) = x ln x − x an antiderivative for f(x) = ln x?
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
9. . . . . . .
Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
Solution
???
Example
is F(x) = x ln x − x an antiderivative for f(x) = ln x?
Solution
d
dx
(x ln x − x)
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
10. . . . . . .
Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
Solution
???
Example
is F(x) = x ln x − x an antiderivative for f(x) = ln x?
Solution
d
dx
(x ln x − x) = 1 · ln x + x ·
1
x
− 1
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
11. . . . . . .
Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
Solution
???
Example
is F(x) = x ln x − x an antiderivative for f(x) = ln x?
Solution
d
dx
(x ln x − x) = 1 · ln x + x ·
1
x
− 1 = ln x
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
12. . . . . . .
Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
Solution
???
Example
is F(x) = x ln x − x an antiderivative for f(x) = ln x?
Solution
d
dx
(x ln x − x) = 1 · ln x + x ·
1
x
− 1 = ln x
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
13. . . . . . .
Why the MVT is the MITC
Most Important Theorem In Calculus!
Theorem
Let f′
= 0 on an interval (a, b). Then f is constant on (a, b).
Proof.
Pick any points x and y in (a, b) with x y. Then f is continuous on
[x, y] and differentiable on (x, y). By MVT there exists a point z in (x, y)
such that
f(y) − f(x)
y − x
= f′
(z) =⇒ f(y) = f(x) + f′
(z)(y − x)
But f′
(z) = 0, so f(y) = f(x). Since this is true for all x and y in (a, b),
then f is constant.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 7 / 35
14. . . . . . .
When two functions have the same derivative
Theorem
Suppose f and g are two differentiable functions on (a, b) with f′
= g′
.
Then f and g differ by a constant. That is, there exists a constant C
such that f(x) = g(x) + C.
Proof.
Let h(x) = f(x) − g(x)
Then h′
(x) = f′
(x) − g′
(x) = 0 on (a, b)
So h(x) = C, a constant
This means f(x) − g(x) = C on (a, b)
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 8 / 35
15. . . . . . .
Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Antiderivatives of piecewise functions
Finding Antiderivatives Graphically
Rectilinear motion
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 9 / 35
16. . . . . . .
Antiderivatives of power functions
Recall that the derivative of a
power function is a power
function.
Fact (The Power Rule)
If f(x) = xr
, then f′
(x) = rxr−1
.
..
x
.
y
.
f(x) = x2
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 35
17. . . . . . .
Antiderivatives of power functions
Recall that the derivative of a
power function is a power
function.
Fact (The Power Rule)
If f(x) = xr
, then f′
(x) = rxr−1
.
..
x
.
y
.
f(x) = x2
.
f′
(x) = 2x
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 35
18. . . . . . .
Antiderivatives of power functions
Recall that the derivative of a
power function is a power
function.
Fact (The Power Rule)
If f(x) = xr
, then f′
(x) = rxr−1
.
..
x
.
y
.
f(x) = x2
.
f′
(x) = 2x
.
F(x) = ?
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 35
19. . . . . . .
Antiderivatives of power functions
Recall that the derivative of a
power function is a power
function.
Fact (The Power Rule)
If f(x) = xr
, then f′
(x) = rxr−1
.
So in looking for antiderivatives
of power functions, try power
functions!
..
x
.
y
.
f(x) = x2
.
f′
(x) = 2x
.
F(x) = ?
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 35
20. . . . . . .
Example
Find an antiderivative for the function f(x) = x3
.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
21. . . . . . .
Example
Find an antiderivative for the function f(x) = x3
.
Solution
Try a power function F(x) = axr
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
22. . . . . . .
Example
Find an antiderivative for the function f(x) = x3
.
Solution
Try a power function F(x) = axr
Then F′
(x) = arxr−1
, so we want arxr−1
= x3
.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
23. . . . . . .
Example
Find an antiderivative for the function f(x) = x3
.
Solution
Try a power function F(x) = axr
Then F′
(x) = arxr−1
, so we want arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
24. . . . . . .
Example
Find an antiderivative for the function f(x) = x3
.
Solution
Try a power function F(x) = axr
Then F′
(x) = arxr−1
, so we want arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =
1
4
.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
25. . . . . . .
Example
Find an antiderivative for the function f(x) = x3
.
Solution
Try a power function F(x) = axr
Then F′
(x) = arxr−1
, so we want arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =
1
4
.
So F(x) =
1
4
x4
is an antiderivative.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
26. . . . . . .
Example
Find an antiderivative for the function f(x) = x3
.
Solution
Try a power function F(x) = axr
Then F′
(x) = arxr−1
, so we want arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =
1
4
.
So F(x) =
1
4
x4
is an antiderivative.
Check:
d
dx
(
1
4
x4
)
= 4 ·
1
4
x4−1
= x3
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
27. . . . . . .
Example
Find an antiderivative for the function f(x) = x3
.
Solution
Try a power function F(x) = axr
Then F′
(x) = arxr−1
, so we want arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =
1
4
.
So F(x) =
1
4
x4
is an antiderivative.
Check:
d
dx
(
1
4
x4
)
= 4 ·
1
4
x4−1
= x3
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
28. . . . . . .
Example
Find an antiderivative for the function f(x) = x3
.
Solution
Try a power function F(x) = axr
Then F′
(x) = arxr−1
, so we want arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =
1
4
.
So F(x) =
1
4
x4
is an antiderivative.
Check:
d
dx
(
1
4
x4
)
= 4 ·
1
4
x4−1
= x3
Any others?
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
29. . . . . . .
Example
Find an antiderivative for the function f(x) = x3
.
Solution
Try a power function F(x) = axr
Then F′
(x) = arxr−1
, so we want arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =
1
4
.
So F(x) =
1
4
x4
is an antiderivative.
Check:
d
dx
(
1
4
x4
)
= 4 ·
1
4
x4−1
= x3
Any others? Yes, F(x) =
1
4
x4
+ C is the most general form.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
30. . . . . . .
Extrapolating to general power functions
Fact (The Power Rule for antiderivatives)
If f(x) = xr
, then
F(x) =
1
r + 1
xr+1
is an antiderivative for f…
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 12 / 35
31. . . . . . .
Extrapolating to general power functions
Fact (The Power Rule for antiderivatives)
If f(x) = xr
, then
F(x) =
1
r + 1
xr+1
is an antiderivative for f as long as r ̸= −1.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 12 / 35
32. . . . . . .
Extrapolating to general power functions
Fact (The Power Rule for antiderivatives)
If f(x) = xr
, then
F(x) =
1
r + 1
xr+1
is an antiderivative for f as long as r ̸= −1.
Fact
If f(x) = x−1
=
1
x
, then
F(x) = ln |x| + C
is an antiderivative for f.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 12 / 35
33. . . . . . .
What's with the absolute value?
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
34. . . . . . .
What's with the absolute value?
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x|
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
35. . . . . . .
What's with the absolute value?
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x)
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
36. . . . . . .
What's with the absolute value?
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
37. . . . . . .
What's with the absolute value?
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
38. . . . . . .
What's with the absolute value?
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
If x 0,
d
dx
ln |x|
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
39. . . . . . .
What's with the absolute value?
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
If x 0,
d
dx
ln |x| =
d
dx
ln(−x)
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
40. . . . . . .
What's with the absolute value?
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
If x 0,
d
dx
ln |x| =
d
dx
ln(−x) =
1
−x
· (−1)
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
41. . . . . . .
What's with the absolute value?
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
If x 0,
d
dx
ln |x| =
d
dx
ln(−x) =
1
−x
· (−1) =
1
x
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
42. . . . . . .
What's with the absolute value?
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
If x 0,
d
dx
ln |x| =
d
dx
ln(−x) =
1
−x
· (−1) =
1
x
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
43. . . . . . .
What's with the absolute value?
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
If x 0,
d
dx
ln |x| =
d
dx
ln(−x) =
1
−x
· (−1) =
1
x
We prefer the antiderivative with the larger domain.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
44. . . . . . .
Graph of ln |x|
.. x.
y
. f(x) = 1/x
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 14 / 35
45. . . . . . .
Graph of ln |x|
.. x.
y
. f(x) = 1/x.
F(x) = ln(x)
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 14 / 35
46. . . . . . .
Graph of ln |x|
.. x.
y
. f(x) = 1/x.
F(x) = ln |x|
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 14 / 35
47. . . . . . .
Combinations of antiderivatives
Fact (Sum and Constant Multiple Rule for Antiderivatives)
If F is an antiderivative of f and G is an antiderivative of g, then
F + G is an antiderivative of f + g.
If F is an antiderivative of f and c is a constant, then cF is an
antiderivative of cf.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 15 / 35
48. . . . . . .
Combinations of antiderivatives
Fact (Sum and Constant Multiple Rule for Antiderivatives)
If F is an antiderivative of f and G is an antiderivative of g, then
F + G is an antiderivative of f + g.
If F is an antiderivative of f and c is a constant, then cF is an
antiderivative of cf.
Proof.
These follow from the sum and constant multiple rule for derivatives:
If F′
= f and G′
= g, then
(F + G)′
= F′
+ G′
= f + g
Or, if F′
= f,
(cF)′
= cF′
= cf
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 15 / 35
49. . . . . . .
Antiderivatives of Polynomials
..
Example
Find an antiderivative for f(x) = 16x + 5.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 35
50. . . . . . .
Antiderivatives of Polynomials
..
Example
Find an antiderivative for f(x) = 16x + 5.
Solution
The expression
1
2
x2
is an antiderivative for x, and x is an antiderivative for 1.
So
F(x) = 16 ·
(
1
2
x2
)
+ 5 · x + C = 8x2
+ 5x + C
is the antiderivative of f.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 35
51. . . . . . .
Antiderivatives of Polynomials
..
Example
Find an antiderivative for f(x) = 16x + 5.
Solution
The expression
1
2
x2
is an antiderivative for x, and x is an antiderivative for 1.
So
F(x) = 16 ·
(
1
2
x2
)
+ 5 · x + C = 8x2
+ 5x + C
is the antiderivative of f.
Question
Do we need two C’s or just one?
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 35
52. . . . . . .
Antiderivatives of Polynomials
..
Example
Find an antiderivative for f(x) = 16x + 5.
Solution
The expression
1
2
x2
is an antiderivative for x, and x is an antiderivative for 1.
So
F(x) = 16 ·
(
1
2
x2
)
+ 5 · x + C = 8x2
+ 5x + C
is the antiderivative of f.
Question
Do we need two C’s or just one?
Answer
Just one. A combination of two arbitrary constants is still an arbitrary constant.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 35
54. . . . . . .
Exponential Functions
Fact
If f(x) = ax
, f′
(x) = (ln a)ax
.
Accordingly,
Fact
If f(x) = ax
, then F(x) =
1
ln a
ax
+ C is the antiderivative of f.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 17 / 35
55. . . . . . .
Exponential Functions
Fact
If f(x) = ax
, f′
(x) = (ln a)ax
.
Accordingly,
Fact
If f(x) = ax
, then F(x) =
1
ln a
ax
+ C is the antiderivative of f.
Proof.
Check it yourself.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 17 / 35
56. . . . . . .
Exponential Functions
Fact
If f(x) = ax
, f′
(x) = (ln a)ax
.
Accordingly,
Fact
If f(x) = ax
, then F(x) =
1
ln a
ax
+ C is the antiderivative of f.
Proof.
Check it yourself.
In particular,
Fact
If f(x) = ex
, then F(x) = ex
+ C is the antiderivative of f.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 17 / 35
57. . . . . . .
Logarithmic functions?
Remember we found
F(x) = x ln x − x
is an antiderivative of f(x) = ln x.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 18 / 35
58. . . . . . .
Logarithmic functions?
Remember we found
F(x) = x ln x − x
is an antiderivative of f(x) = ln x.
This is not obvious. See Calc II for the full story.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 18 / 35
59. . . . . . .
Logarithmic functions?
Remember we found
F(x) = x ln x − x
is an antiderivative of f(x) = ln x.
This is not obvious. See Calc II for the full story.
However, using the fact that loga x =
ln x
ln a
, we get:
Fact
If f(x) = loga(x)
F(x) =
1
ln a
(x ln x − x) + C = x loga x −
1
ln a
x + C
is the antiderivative of f(x).
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 18 / 35
60. . . . . . .
Trigonometric functions
Fact
d
dx
sin x = cos x
d
dx
cos x = − sin x
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 19 / 35
61. . . . . . .
Trigonometric functions
Fact
d
dx
sin x = cos x
d
dx
cos x = − sin x
So to turn these around,
Fact
The function F(x) = − cos x + C is the antiderivative of f(x) = sin x.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 19 / 35
62. . . . . . .
Trigonometric functions
Fact
d
dx
sin x = cos x
d
dx
cos x = − sin x
So to turn these around,
Fact
The function F(x) = − cos x + C is the antiderivative of f(x) = sin x.
The function F(x) = sin x + C is the antiderivative of f(x) = cos x.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 19 / 35
63. . . . . . .
More Trig
Example
Find an antiderivative of f(x) = tan x.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
64. . . . . . .
More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
65. . . . . . .
More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
Answer
F(x) = ln(sec x).
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
66. . . . . . .
More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
Answer
F(x) = ln(sec x).
Check
d
dx
=
1
sec x
·
d
dx
sec x
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
67. . . . . . .
More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
Answer
F(x) = ln(sec x).
Check
d
dx
=
1
sec x
·
d
dx
sec x =
1
sec x
· sec x tan x
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
68. . . . . . .
More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
Answer
F(x) = ln(sec x).
Check
d
dx
=
1
sec x
·
d
dx
sec x =
1
sec x
· sec x tan x = tan x
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
69. . . . . . .
More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
Answer
F(x) = ln(sec x).
Check
d
dx
=
1
sec x
·
d
dx
sec x =
1
sec x
· sec x tan x = tan x
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
70. . . . . . .
More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
Answer
F(x) = ln(sec x).
Check
d
dx
=
1
sec x
·
d
dx
sec x =
1
sec x
· sec x tan x = tan x
More about this later.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
71. . . . . . .
Antiderivatives of piecewise functions
Example
Let f(x) =
{
x if 0 ≤ x ≤ 1;
1 − x2
if 1 x.
Find the antiderivative of f with
F(0) = 1.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 21 / 35
72. . . . . . .
Antiderivatives of piecewise functions
Example
Let f(x) =
{
x if 0 ≤ x ≤ 1;
1 − x2
if 1 x.
Find the antiderivative of f with
F(0) = 1.
Solution
We can antidifferentiate each piece:
F(x) =
1
2
x2
+ C1 if 0 ≤ x ≤ 1;
x −
1
3
x3
+ C2 if 1 x.
The constants need to be chosen so that F(0) = 1 and F is continuous
(at 1).
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 21 / 35
73. . . . . . .
F(x) =
1
2
x2
+ C1 if 0 ≤ x ≤ 1;
x −
1
3
x3
+ C2 if 1 x.
Note
F(0) =
1
2
02
+ C1 = C1 =⇒ C1 = 1
This means lim
x→1−
F(x) =
1
2
12
+ 1 =
3
2
. Now
lim
x→1+
F(x) = 1 −
1
3
+ C2 =
2
3
+ C2
So for F to be continuous we need
3
2
=
2
3
+ C2 =⇒ C2 =
5
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 22 / 35
74. . . . . . .
Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Antiderivatives of piecewise functions
Finding Antiderivatives Graphically
Rectilinear motion
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 23 / 35
75. . . . . . .
Finding Antiderivatives Graphically
Problem
Below is the graph of a function f. Draw the graph of an antiderivative
for f.
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
y = f(x)
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 24 / 35
76. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
77. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
78. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
79. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
80. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
81. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
82. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
83. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
84. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
85. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
86. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
87. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
88. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
89. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
90. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
91. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
92. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
93. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
94. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
95. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
96. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
97. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
98. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
99. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
100. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
101. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
102. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
..
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
103. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
...
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
104. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
....
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
105. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
.....
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
106. . . . . . .
Using f to make a sign chart for F
Assuming F′
= f, we can make a sign chart for f and f′
to find the
intervals of monotonicity and concavity for F:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
?
.
?
.
?
.
?
.
?
.
?
The only question left is: What are the function values?
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
107. . . . . . .
Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
Solution
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
108. . . . . . .
Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
Solution
We start with F(1) = 0. ..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
109. . . . . . .
Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
Solution
We start with F(1) = 0.
Using the sign chart, we
draw arcs with the
specified monotonicity and
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
110. . . . . . .
Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
Solution
We start with F(1) = 0.
Using the sign chart, we
draw arcs with the
specified monotonicity and
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
..
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
111. . . . . . .
Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
Solution
We start with F(1) = 0.
Using the sign chart, we
draw arcs with the
specified monotonicity and
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
..
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
112. . . . . . .
Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
Solution
We start with F(1) = 0.
Using the sign chart, we
draw arcs with the
specified monotonicity and
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
...
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
113. . . . . . .
Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
Solution
We start with F(1) = 0.
Using the sign chart, we
draw arcs with the
specified monotonicity and
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
...
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
114. . . . . . .
Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
Solution
We start with F(1) = 0.
Using the sign chart, we
draw arcs with the
specified monotonicity and
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
....
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
115. . . . . . .
Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
Solution
We start with F(1) = 0.
Using the sign chart, we
draw arcs with the
specified monotonicity and
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
....
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
116. . . . . . .
Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
Solution
We start with F(1) = 0.
Using the sign chart, we
draw arcs with the
specified monotonicity and
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
.....
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
117. . . . . . .
Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
Solution
We start with F(1) = 0.
Using the sign chart, we
draw arcs with the
specified monotonicity and
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
......
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
118. . . . . . .
Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
Solution
We start with F(1) = 0.
Using the sign chart, we
draw arcs with the
specified monotonicity and
concavity
It’s harder to tell if/when F
crosses the axis; more
about that later.
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
......
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
119. . . . . . .
Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Antiderivatives of piecewise functions
Finding Antiderivatives Graphically
Rectilinear motion
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 27 / 35
120. . . . . . .
Say what?
“Rectilinear motion” just means motion along a line.
Often we are given information about the velocity or acceleration
of a moving particle and we want to know the equations of motion.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 28 / 35
121. . . . . . .
Application: Dead Reckoning
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 29 / 35
122. . . . . . .
Application: Dead Reckoning
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 29 / 35
123. . . . . . .
Problem
Suppose a particle of mass m is acted upon by a constant force F.
Find the position function s(t), the velocity function v(t), and the
acceleration function a(t).
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 30 / 35
124. . . . . . .
Problem
Suppose a particle of mass m is acted upon by a constant force F.
Find the position function s(t), the velocity function v(t), and the
acceleration function a(t).
Solution
By Newton’s Second Law (F = ma) a constant force induces a
constant acceleration. So a(t) = a =
F
m
.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 30 / 35
125. . . . . . .
Problem
Suppose a particle of mass m is acted upon by a constant force F.
Find the position function s(t), the velocity function v(t), and the
acceleration function a(t).
Solution
By Newton’s Second Law (F = ma) a constant force induces a
constant acceleration. So a(t) = a =
F
m
.
Since v′
(t) = a(t), v(t) must be an antiderivative of the constant
function a. So
v(t) = at + C = at + v0
where v0 is the initial velocity.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 30 / 35
126. . . . . . .
Problem
Suppose a particle of mass m is acted upon by a constant force F.
Find the position function s(t), the velocity function v(t), and the
acceleration function a(t).
Solution
By Newton’s Second Law (F = ma) a constant force induces a
constant acceleration. So a(t) = a =
F
m
.
Since v′
(t) = a(t), v(t) must be an antiderivative of the constant
function a. So
v(t) = at + C = at + v0
where v0 is the initial velocity.
Since s′
(t) = v(t), s(t) must be an antiderivative of v(t), meaning
s(t) =
1
2
at2
+ v0t + C =
1
2
at2
+ v0t + s0
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 30 / 35
127. . . . . . .
An earlier Hatsumon
Example
Drop a ball off the roof of the Silver Center. What is its velocity when it
hits the ground?
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 31 / 35
128. . . . . . .
An earlier Hatsumon
Example
Drop a ball off the roof of the Silver Center. What is its velocity when it
hits the ground?
Solution
Assume s0 = 100 m, and v0 = 0. Approximate a = g ≈ −10. Then
s(t) = 100 − 5t2
So s(t) = 0 when t =
√
20 = 2
√
5. Then
v(t) = −10t,
so the velocity at impact is v(2
√
5) = −20
√
5 m/s.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 31 / 35
129. . . . . . .
Finding initial velocity from stopping distance
Example
The skid marks made by an automobile indicate that its brakes were
fully applied for a distance of 160 ft before it came to a stop. Suppose
that the car in question has a constant deceleration of 20 ft/s2 under the
conditions of the skid. How fast was the car traveling when its brakes
were first applied?
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 32 / 35
130. . . . . . .
Finding initial velocity from stopping distance
Example
The skid marks made by an automobile indicate that its brakes were
fully applied for a distance of 160 ft before it came to a stop. Suppose
that the car in question has a constant deceleration of 20 ft/s2 under the
conditions of the skid. How fast was the car traveling when its brakes
were first applied?
Solution (Setup)
While braking, the car has acceleration a(t) = −20
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 32 / 35
131. . . . . . .
Finding initial velocity from stopping distance
Example
The skid marks made by an automobile indicate that its brakes were
fully applied for a distance of 160 ft before it came to a stop. Suppose
that the car in question has a constant deceleration of 20 ft/s2 under the
conditions of the skid. How fast was the car traveling when its brakes
were first applied?
Solution (Setup)
While braking, the car has acceleration a(t) = −20
Measure time 0 and position 0 when the car starts braking. So
s(0) = 0.
The car stops at time some t1, when v(t1) = 0.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 32 / 35
132. . . . . . .
Finding initial velocity from stopping distance
Example
The skid marks made by an automobile indicate that its brakes were
fully applied for a distance of 160 ft before it came to a stop. Suppose
that the car in question has a constant deceleration of 20 ft/s2 under the
conditions of the skid. How fast was the car traveling when its brakes
were first applied?
Solution (Setup)
While braking, the car has acceleration a(t) = −20
Measure time 0 and position 0 when the car starts braking. So
s(0) = 0.
The car stops at time some t1, when v(t1) = 0.
We know that when s(t1) = 160.
We want to know v(0), or v0.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 32 / 35
133. . . . . . .
Implementing the Solution
In general,
s(t) = s0 + v0t +
1
2
at2
Since s0 = 0 and a = −20, we have
s(t) = v0t − 10t2
v(t) = v0 − 20t
for all t.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 33 / 35
134. . . . . . .
Implementing the Solution
In general,
s(t) = s0 + v0t +
1
2
at2
Since s0 = 0 and a = −20, we have
s(t) = v0t − 10t2
v(t) = v0 − 20t
for all t. Plugging in t = t1,
160 = v0t1 − 10t2
1
0 = v0 − 20t1
We need to solve these two equations.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 33 / 35
135. . . . . . .
Solving
We have
v0t1 − 10t2
1 = 160 v0 − 20t1 = 0
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 34 / 35
136. . . . . . .
Solving
We have
v0t1 − 10t2
1 = 160 v0 − 20t1 = 0
The second gives t1 = v0/20, so substitute into the first:
v0 ·
v0
20
− 10
( v0
20
)2
= 160
or
v2
0
20
−
10v2
0
400
= 160
2v2
0 − v2
0 = 160 · 40 = 6400
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 34 / 35
137. . . . . . .
Solving
We have
v0t1 − 10t2
1 = 160 v0 − 20t1 = 0
The second gives t1 = v0/20, so substitute into the first:
v0 ·
v0
20
− 10
( v0
20
)2
= 160
or
v2
0
20
−
10v2
0
400
= 160
2v2
0 − v2
0 = 160 · 40 = 6400
So v0 = 80 ft/s ≈ 55 mi/hr
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 34 / 35
138. . . . . . .
Summary
Antiderivatives are a useful
concept, especially in
motion
We can graph an
antiderivative from the
graph of a function
We can compute
antiderivatives, but not
always
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.......
F
f(x) = e−x2
f′
(x) = ???
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 35 / 35