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 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


Slicing and Dicing Solids
Solids of Revolution 1: Disks
Solids of Revolution 2: Washers
Volumes Rotating About the $y$-axis

Integration by Parts

Behind IBP
Going in Circles
Tricks of the Trade

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
Completing the Square

Partial Fractions

Linear Factors
Quadratic Factors
Improper Rational Functions and Long Division

Strategies of Integration

Integration by Parts
Trig Integrals
Trig Substitutions
Partial Fractions

Improper Integrals

Type I Integrals
Type II Integrals
Comparison Tests for Convergence

Differential Equations

Separable Equations
Mixing and Dilution

Models of Growth

Exponential Growth and Decay
Logistic Growth

Infinite Sequences

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

Infinite Series

Geometric Series
Limit Laws for Series
Telescoping Sums and the FTC

Integral Test

Road Map
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

Alternating Series and the AS Test
Absolute Convergence

The Ratio and Root Tests

The Ratio Test
The Root Test

Strategies for testing Series

List of Major Convergence Tests

Power Series

Radius and Interval of Convergence
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

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
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

  1. Realizing that the iterated integral is a double integral over some region $D$.

  2. 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).

  3. 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.

  4. 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 $$\int_{g(x)}^{h(x)} \,\int_a^b f(x,y) \,dx\, dy \hbox{ !!}$$