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#### The Fundamental Theorem of Calculus

Three Different Quantities
The Whole as Sum of Partial Changes
The Indefinite Integral as Antiderivative
The FTC and the Chain Rule

#### 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
Revised Table of Integrals
Substitution for Definite Integrals

#### Area Between Curves

The Slice and Dice Principle
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
Volumes Rotating About the $y$-axis

Behind IBP
Examples
Going in Circles

#### Integrals of Trig Functions

Basic Trig Functions
Product of Sines and Cosines (1)
Product of Sines and Cosines (2)
Product of Secants and Tangents
Other Cases

#### Trig Substitutions

How it works
Examples
Completing the Square

#### Partial Fractions

Introduction
Linear Factors
Improper Rational Functions and Long Division
Summary

#### Strategies of Integration

Substitution
Integration by Parts
Trig Integrals
Trig Substitutions
Partial Fractions

#### Improper Integrals

Type I Integrals
Type II Integrals
Comparison Tests for Convergence

#### Differential Equations

Introduction
Separable Equations
Mixing and Dilution

#### Models of Growth

Exponential Growth and Decay
Logistic Growth

#### Infinite Sequences

Close is Good Enough (revisited)
Examples
Limit Laws for Sequences
Monotonic Convergence

#### Infinite Series

Introduction
Geometric Series
Limit Laws for Series
Telescoping Sums and the FTC

#### Integral Test

The Integral Test
When the Integral Diverges
When the Integral Converges

#### 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 Test
The Root Test
Examples

#### Strategies for testing Series

List of Major Convergence Tests
Examples

#### Power Series

Finding the Interval of Convergence
Other Power Series

#### 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

What are Taylor Polynomials?
How Accurate are Taylor Polynomials?
What can go Wrong?
Other Uses of Taylor Polynomials

#### Partial Derivatives

Definitions and Rules
The 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

One Variable at the Time
Fubini's Theorem
Notation and Order

#### Double Integrals over General Regions

Type I and Type II regions
Examples
Order of Integration
Area and Volume Revisited

### Order of Integration

Some regions can be viewed either as Type I or Type II. In that case we can set up an iterated integral in two ways. Depending on the integrand, one can be a lot easier than the other!

Sometimes you're given an impossible-looking iterated integral, and you can solve it by swapping (i.e., reversing) the order of integration. This means

 Realizing that the iterated integral is a double integral over some region $D$. Expressing $D$ as the other type of region (Type II if it was originally set up as Type I, and Type I if it was originally set up as Type II). Re-writing the integral over $D$ as an interated integral in a new way. If had originally been $dx \,dy$, it should now be $dy \,dx$, and vice-versa. Doing the new iterated integral.

An example is worked in detail in the video below.

 Example 1: Evaluate the integral $$I \ = \ \int \int_D\, x\,\sqrt{1+y^3}\, dA$$ when $D$ is the triangular region shown to the right enclosed by the $y$-axis and the lines $$y \ = \ \frac{1}{3}x\,, \qquad y \ = \ 2\,.$$ Bad Choice: Fix $x$ and integrate with respect $y$ along the red line. Then $$I \ = \ \int_0^6 \left(\int_{x/3}^2\, x\,\sqrt{1+y^3}\, dy\right)\,dx\,.$$ The trouble is that the inner integral involves requires evaluating the integral $$\int_{x/3}^2\, \sqrt{1+y^3}\, dy\,.$$ Nothing you've learned so far in calculus will work here! The other order of integration is needed. Good Choice: Fix $y$ and integrate with respect to $x$ along the black line. Then $$I \ = \ \int_0^2 \left(\int_{0}^{3y}\, x\,\sqrt{1+y^3}\, dy\right)dx\,.$$ Now the inner integral involves requires evaluating the integral $$\int_{0}^{3y}\, x\, dx\ = \ \Bigl[\,\frac{1}{2}x^2\, \Bigr]_0^{3y} \ = \ \frac{9}{2}y^2\,.$$ In this case, $$I \ = \ \frac{9}{2} \int_0^2\, y^2 \, \sqrt{1+y^3}\,dy\ = \ 3\int_1^3\, u^2\, du\ = \ 26\,,$$ using the substitution $u^2 = 1+y^3$.

Reversing the order of integration in a double integral always requires first looking carefully at a graph of the region of integration. Then it's a matter of algebra and inverse functions.

 Example 2: Reverse the order of integration in the iterated integral $$I \ = \ \int _0^2\left(\int_{x^2}^4\, f(x,\,y)\, dy\right)dx\,,$$ but make no attempt to evaluate either integral. Solution: The region of integration is the set $$D \ = \ \Bigl\{\,(x,\,y) : \, x^2 \le y \le 4\,, \ \ 0 \le x \le 2\,\Bigr\}$$ whose graph is shown below. The given repeated integral fixes $x$ and integrates with respect to $y$ along the vertical black line. To reverse the order of integration we need to fix $y$ and integrate with respect to $x$ along the red line. To set up the repeated integral we have to express $D$ in the form $$D \ = \ \Bigl\{\,(x,\,y) : \, \phi(y) \le x \le \psi(y)\,, \ \ c \le y \le d\,\Bigr\}$$ for suitably chosen $c,\, d$ and functions $\phi(y),\, \psi(y)$. Now by inverse functions, the parabola $y = x^2$ can be written as $x = \sqrt{y}$; this tells us how to find the right hand limit of integration $x = \psi(y)$. On the other hand, the graph above shows the left hand limit is $x = 0$. Thus $D$ can also be written as $$D \ = \ \Bigl\{\,(x,\,y) : \, 0 \le x \le \sqrt{y}\,, \ \ 0 \le y \le 4\,\Bigr\}\,.$$ Consequently, reversing the order of integration shows that $$I \ = \ \int _0^4\left(\int_0^{\sqrt{y}}\, f(x,\,y)\, dx\right)\,dy\,,$$ integrating now first with respect to $x$.

 Warning: The tricky part of swapping the order of integration is re-writing the limits of integration. This involves studying the region of integration. $$\int_a^b \int_{y=g(x)}^{h(x)}\, f(x,y) \,dy\, dx$$ does NOT become $$\int_{g(x)}^{h(x)} \,\int_a^b f(x,y) \,dx\, dy \hbox{ !!}$$