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

Area Between Curves

Computation Using Integration
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
More Practice

Integration by Parts

Integration by Parts
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
Completing the Square

Partial Fractions

Linear Factors
Irreducible Quadratic Factors
Improper Rational Functions and Long Division

Strategies of Integration

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

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

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

The Ratio and Root Tests

The Ratio Test
The Root Test

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

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

Examples 5-7

In the previous Examples 3 and 4, the way we specified $D$ and $R$ suggested how to write the integral as a iterated integral.  Sometimes conditions are best interpreted graphically before deciding on whether to evaluate as Type I or Type II.

Example 5:  Evaluate the integral $\displaystyle I \ = \iint_D\, (3x +4y)\, dA$,
where $D$ is the bounded region enclosed by $y = x$ and $y=x^2$.

DO:  Without looking, graph the region $D$.  Set up a Type I integral.

Solution 5:  $D$ is enclosed by the straight line $y = x$ and the parabola $y = x^2$ as shown here. To determine the limits of integration we first need to find the points of intersection of $y = x$ and $y = x^2$. These occur when $x^2 = x$.  This means $x(x-1)=0$, so $x = 0,\, 1$.
Treating $D$ as a Type I region, we fix $x$ between $x=0$ and $x=1$, and integrate with respect to $y$ along the black vertical line, getting the iterated integral $\displaystyle I \ = \ \int_0^1\left(\int_{x^2}^{x}\, (3x + 4y)\, dy\right)\,dx$ 
:  Evaluate $I$.


In Example 5 it was simpler to fix $x$ and integrate first with respect to $y$ because the bounding curves were already given to us in terms of $x$, as $$y \ = \ f_1(x)\ = \ x^2\,,\qquad y \ = \ f_2(x)\ = \ x\,.$$ Had they been given as $x \,=\, g_1(y)$ and $x\,=\, g_2(y)\,$ it might have been easier to think of $D$ as a Type II region. We would fix $y$ and first integrate with respect to $x$.  Sometimes the geometry of the region of integration makes it obvious whether we have Type I or Type II, as in the following examples (which we will not evaluate).

Example 6:  The region $D$ that is shown here is Type I but is not Type II:  Fixing $x$ and integrating first with respect to $y$ along the vertical black line makes good sense because then we have the same curve $f_1(x)$ along the bottom and $f_2(x)$ along the top: $$ D \ = \ \Bigl\{\,(x,\,y) : f_1(x) \le y \le f_2(x),\ \ a \le x \le b\,\Bigl\}$$ for suitable choices of $a,\, b$ and functions $f_1(x),\, f_2(x)$ giving us: $$ \iint_D\, f(x,\,y)\, dA = \int_a^b \left(\int_{f_1(x)}^{f_2(x)}\, f(x,\,y)\, dy\right) dx\,.$$But if we had chosen to fix $y$, then the integral with respect to $x$ would sometimes split into two parts, as is shown with the red horizontal lines.  This would make evaluate of this integral more complicated -- for one thing, there would be more than one integral.


Example 7:  Similarly, the region $D$ shown here is Type II but not Type I.  Fixing $y$ and integrating first with respect to $x$ along the horizontal black line makes good sense because then $$ D \ = \ \Bigl\{\,(x,\,y) : g_1(y) \le x \le g_2(y),\ \ c \le y \le d\,\Bigl\}$$ for suitable choices of $c,\, d$ and functions $g_1(y),\, g_2(y)$.  In this case $$ \iint_D\, f(x,\,y)\, dxdy = \int_c^d \left(\int_{g_1(y)}^{g_2(y)}\, f(x,\,y)\, dx\right) dy\,.$$But if we had chosen to fix $x$, then the integral with respect to $y$ would sometimes splits into two parts as shown by the red vertical lines.  Again, this would make the integral(s) more complicated.