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The Fundamental Theorem of Calculus

Three Different Concepts
The Fundamental Theorem of Calculus (Part 2)
The Fundamental Theorem of Calculus (Part 1)
More FTC 1

The Indefinite Integral and the Net Change

Indefinite Integrals and Anti-derivatives
A Table of Common Anti-derivatives
The Net Change Theorem
The NCT and Public Policy

Substitution

Substitution for Indefinite Integrals
Examples to Try
Revised Table of Integrals
Substitution for Definite Integrals
Examples

Area Between Curves

Computation Using Integration
To Compute a Bulk Quantity
The Area Between Two Curves
Horizontal Slicing
Summary

Volumes

Slicing and Dicing Solids
Solids of Revolution 1: Disks
Solids of Revolution 2: Washers
More Practice

Integration by Parts

Integration by Parts
Examples
Integration by Parts with a definite integral
Going in Circles
Tricks of the Trade

Integrals of Trig Functions

Antiderivatives of Basic Trigonometric Functions
Product of Sines and Cosines (mixed even and odd powers or only odd powers)
Product of Sines and Cosines (only even powers)
Product of Secants and Tangents
Other Cases

Trig Substitutions

How Trig Substitution Works
Summary of trig substitution options
Examples
Completing the Square

Partial Fractions

Introduction
Linear Factors
Irreducible Quadratic Factors
Improper Rational Functions and Long Division
Summary

Strategies of Integration

Substitution
Integration by Parts
Trig Integrals
Trig Substitutions
Partial Fractions

Improper Integrals

Type 1 - Improper Integrals with Infinite Intervals of Integration
Type 2 - Improper Integrals with Discontinuous Integrands
Comparison Tests for Convergence

Differential Equations

Introduction
Separable Equations
Mixing and Dilution

Models of Growth

Exponential Growth and Decay
Logistic Growth

Infinite Sequences

Approximate Versus Exact Answers
Examples of Infinite Sequences
Limit Laws for Sequences
Theorems for and Examples of Computing Limits of Sequences
Monotonic Covergence

Infinite Series

Introduction
Geometric Series
Limit Laws for Series
Test for Divergence and Other Theorems
Telescoping Sums and the FTC

Integral Test

Road Map
The Integral Test
Estimates of Value of the Series

Comparison Tests

The Basic Comparison Test
The Limit Comparison Test

Convergence of Series with Negative Terms

Introduction, Alternating Series,and the AS Test
Absolute Convergence
Rearrangements

The Ratio and Root Tests

The Ratio Test
The Root Test
Examples

Strategies for testing Series

Strategy to Test Series and a Review of Tests
Examples, Part 1
Examples, Part 2

Power Series

Radius and Interval of Convergence
Finding the Interval of Convergence
Power Series Centered at $x=a$

Representing Functions as Power Series

Functions as Power Series
Derivatives and Integrals of Power Series
Applications and Examples

Taylor and Maclaurin Series

The Formula for Taylor Series
Taylor Series for Common Functions
Adding, Multiplying, and Dividing Power Series
Miscellaneous Useful Facts

Applications of Taylor Polynomials

Taylor Polynomials
When Functions Are Equal to Their Taylor Series
When a Function Does Not Equal Its Taylor Series
Other Uses of Taylor Polynomials

Partial Derivatives

Visualizing Functions in 3 Dimensions
Definitions and Examples
An Example from DNA
Geometry of Partial Derivatives
Higher Order Derivatives
Differentials and Taylor Expansions

Multiple Integrals

Background
What is a Double Integral?
Volumes as Double Integrals

Iterated Integrals over Rectangles

How To Compute Iterated Integrals
Examples of Iterated Integrals
Cavalieri's Principle
Fubini's Theorem
Summary and an Important Example

Double Integrals over General Regions

Type I and Type II regions
Examples 1-4
Examples 5-7
Order of Integration


Derivatives and Integrals of Power Series

As long as we are strictly inside the interval of convergence, we can take derivatives and integrals of power series allowing us to get new series. 

(Let $R$ be the radius of convergence of a series.  $x$ is strictly inside the interval of convergence of the series when $-R<x<R$, so $x$ is not equal to either of the two possible endpoints $R$ or $-R$ of the interval.) 

To see how this works with a series centered at the origin, first consider that for any constant $c_n$, $\frac d{dx}\left(c_nx^n\right)=nc_nx^{n-1}$.  Similarly, $\int c_nx^n\,dx=c_n\frac{x^{n+1}}{n+1} + C$.  Now consider the power series $\displaystyle\sum_{n=0}^\infty c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5+\cdots$.  When $x$ is strictly inside the interval of convergence of this series, we can differentiate and integrate term by term.  A beautiful fact is that the radius of convergence will not change when you differentiate or antidifferentiate.

To differentiate, we simply differentiate each term (not worrying that we have infinitely many terms) and then put the terms back into summation notation.  Notice that in the derivative series we must change our index to begin at $n=1$. (Why?)
$\begin{eqnarray}
\frac{d}{dx}\left(\sum_{n=0}^\infty c_n \,x^n\right)&=&\frac{d}{dx}(c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5+\cdots)\\
&=&0+c_1+2c_2x+3c_3x^2+4c_4x^3+5c_5x^4+\cdots\\
&=&
\sum_{n=1}^\infty n\,c_n \,x^{n-1}
\end{eqnarray}$

Similarly, while we get an infinite integrand, we don't worry and just antidifferentiate each term, and then add a constant.  We do not need to change our index starting point here.  (Why not?)

$\begin{eqnarray}
\int\left(\sum_{n=0}^\infty c_n \,x^n\right)\,dx&=&\int(c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5+\cdots)\,dx\\
&=&c_0x+c_1\frac{x^2}{2}+c_2\frac{x^3}{3}+c_3\frac{x^4}{4}+c_4\frac{x^5}{5}+c_5\frac{x^6}{6}\cdots\\
&=&
\left(\sum_{n=0}^\infty \,c_n \,\frac{x^{n+1}}{n+1}\right)+C
\end{eqnarray}$

Succinctly, we get the following for power series centered at the origin:

Let $\displaystyle \sum_{n=0}^\infty c_n \,x^n$ have radius of convergence $R$.  As long as $x$ is strictly inside the interval of convergence of the series, i.e. $-R<x<R$,
$$\frac{d}{dx}\left(\sum_{n=0}^\infty c_n \,x^n\right)=\sum_{n=1}^\infty n\,c_n \,x^{n-1}$$
$$ \int\left(\sum_{n=0}^\infty c_n \,x^n\right)\, dx = \left(\sum_{n=0}^\infty c_n \frac{x^{n+1}}{n+1}\right)+C$$ and the new series have the same $R$ as the original series.

The same holds for power series centered at $a$.

Let $\displaystyle \sum_{n=0}^\infty c_n \,(x-a)^n$ have radius of convergence $R$.  As long as $x$ is strictly inside the interval of convergence of the series, i.e. $a-R<x<a+R$,
$$\frac{d}{dx}\left(\sum_{n=0}^\infty c_n \,(x-a)^n\right)=\sum_{n=1}^\infty n\,c_n \,(x-a)^{n-1}$$
$$ \int\left(\sum_{n=0}^\infty c_n \,(x-a)^n\right)\, dx = \left(\sum_{n=0}^\infty c_n \frac{(x-a)^{n+1}}{n+1}\right)+C$$and the new series have the same $R$ as the original series


This video will discuss the derivatives and antiderivatives of power series, and explain that they have the same radius of convergence as the original series.  Convergence at the endpoints does not carry through to the derivatives and antiderivatives, where convergence at the endpoints may be different.  The new series must be tested at the endpoints to determine their convergence.