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About *Calculus Volume 2*

*Calculus* is designed for the typical two- or three-semester general calculus course, incorporating innovative features to enhance student learning. The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them. Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency. Volume 2 covers integration, differential equations, sequences and series, and parametric equations and polar coordinates.

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Integration

Introduction

Approximating Areas

Sigma (Summation) Notation

As mentioned, we will use shapes of known area to approximate the area of an irregular region bounded by curves. This process often requires adding up long strings of numbers. To make it easier to write down these lengthy sums, we look at some new notation here, called sigma notation (also known as summation notation). The Greek capital letter $\Sigma ,$ sigma, is used to express long sums of values in a compact form. For example, if we want to add all the integers from 1 to 20 without sigma notation, we have to write

Approximating Area

Now that we have the necessary notation, we return to the problem at hand: approximating the area under a curve. Let $f\left(x\right)$ be a continuous, nonnegative function defined on the closed interval $\left[a,b\right].$ We want to approximate the area *A* bounded by $f\left(x\right)$ above, the *x*-axis below, the line $x=a$ on the left, and the line $x=b$ on the right ([link]).

Forming Riemann Sums

So far we have been using rectangles to approximate the area under a curve. The heights of these rectangles have been determined by evaluating the function at either the right or left endpoints of the subinterval $\left[{x}_{i-1},{x}_{i}\right].$ In reality, there is no reason to restrict evaluation of the function to one of these two points only. We could evaluate the function at any point ${x}_{i}^{\*}$ in the subinterval $\left[{x}_{i-1},{x}_{i}\right],$ and use $f\left({x}_{i}^{\*}\right)$ as the height of our rectangle. This gives us an estimate for the area of the form

The Definite Integral

Definition and Notation

The definite integral generalizes the concept of the area under a curve. We lift the requirements that $f\left(x\right)$ be continuous and nonnegative, and define the definite integral as follows.

Evaluating Definite Integrals

Evaluating definite integrals this way can be quite tedious because of the complexity of the calculations. Later in this chapter we develop techniques for evaluating definite integrals *without* taking limits of Riemann sums. However, for now, we can rely on the fact that definite integrals represent the area under the curve, and we can evaluate definite integrals by using geometric formulas to calculate that area. We do this to confirm that definite integrals do, indeed, represent areas, so we can then discuss what to do in the case of a curve of a function dropping below the *x*-axis.

Area and the Definite Integral

When we defined the definite integral, we lifted the requirement that $f\left(x\right)$ be nonnegative. But how do we interpret “the area under the curve” when $f\left(x\right)$ is negative?

Properties of the Definite Integral

The properties of indefinite integrals apply to definite integrals as well. Definite integrals also have properties that relate to the limits of integration. These properties, along with the rules of integration that we examine later in this chapter, help us manipulate expressions to evaluate definite integrals.

Average Value of a Function

We often need to find the average of a set of numbers, such as an average test grade. Suppose you received the following test scores in your algebra class: 89, 90, 56, 78, 100, and 69. Your semester grade is your average of test scores and you want to know what grade to expect. We can find the average by adding all the scores and dividing by the number of scores. In this case, there are six test scores. Thus,

The Fundamental Theorem of Calculus

The Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at some point in that interval. The theorem guarantees that if $f\left(x\right)$ is continuous, a point *c* exists in an interval $\left[a,b\right]$ such that the value of the function at *c* is equal to the average value of $f\left(x\right)$ over $\left[a,b\right].$ We state this theorem mathematically with the help of the formula for the average value of a function that we presented at the end of the preceding section.

Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives

As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, Part 1, is stated here. Part 1 establishes the relationship between differentiation and integration.

Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem

The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. Using calculus, astronomers could finally determine distances in space and map planetary orbits. Everyday financial problems such as calculating marginal costs or predicting total profit could now be handled with simplicity and accuracy. Engineers could calculate the bending strength of materials or the three-dimensional motion of objects. Our view of the world was forever changed with calculus.

Integration Formulas and the Net Change Theorem

Basic Integration Formulas

Recall the integration formulas given in the table in Antiderivatives and the rule on properties of definite integrals. Let’s look at a few examples of how to apply these rules.

The Net Change Theorem

The net change theorem considers the integral of a *rate of change*. It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.

Applying the Net Change Theorem

The net change theorem can be applied to the flow and consumption of fluids, as shown in [link].

Integrating Even and Odd Functions

We saw in Functions and Graphs that an even function is a function in which $f\left(\text{−}x\right)=f\left(x\right)$ for all *x* in the domain—that is, the graph of the curve is unchanged when *x* is replaced with −*x*. The graphs of even functions are symmetric about the *y*-axis. An odd function is one in which $f\left(\text{−}x\right)=\text{−}f\left(x\right)$ for all *x* in the domain, and the graph of the function is symmetric about the origin.

Substitution

Proof

Let *f*, *g*, *u*, and *F* be as specified in the theorem. Then

Substitution for Definite Integrals

Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.

Integrals Involving Exponential and Logarithmic Functions

Integrals of Exponential Functions

The exponential function is perhaps the most efficient function in terms of the operations of calculus. The exponential function, $y={e}^{x},$ is its own derivative and its own integral.

Integrals Involving Logarithmic Functions

Integrating functions of the form $f\left(x\right)={x}^{-1}$ result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as $f\left(x\right)=\text{ln}{\rule{0.1em}{0ex}}x$ and $f\left(x\right)={\text{log}}_{a}x,$ are also included in the rule.

Integrals Resulting in Inverse Trigonometric Functions

Integrals that Result in Inverse Sine Functions

Let us begin this last section of the chapter with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.

Integrals Resulting in Other Inverse Trigonometric Functions

There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use. The only difference is whether the integrand is positive or negative. Rather than memorizing three more formulas, if the integrand is negative, simply factor out −1 and evaluate the integral using one of the formulas already provided. To close this section, we examine one more formula: the integral resulting in the inverse tangent function.

Applications of Integration

Introduction

Areas between Curves

Area of a Region between Two Curves

Let $f\left(x\right)$ and $g\left(x\right)$ be continuous functions over an interval $\left[a,b\right]$ such that $f\left(x\right)\ge g\left(x\right)$ on $\left[a,b\right].$ We want to find the area between the graphs of the functions, as shown in the following figure.

Areas of Compound Regions

So far, we have required $f\left(x\right)\ge g\left(x\right)$ over the entire interval of interest, but what if we want to look at regions bounded by the graphs of functions that cross one another? In that case, we modify the process we just developed by using the absolute value function.

Regions Defined with Respect to *y*

In [link], we had to evaluate two separate integrals to calculate the area of the region. However, there is another approach that requires only one integral. What if we treat the curves as functions of $y,$ instead of as functions of $x?$ Review [link]. Note that the left graph, shown in red, is represented by the function $y=f\left(x\right)={x}^{2}.$ We could just as easily solve this for $x$ and represent the curve by the function $x=v\left(y\right)=\sqrt{y}.$ (Note that $x=\text{−}\sqrt{y}$ is also a valid representation of the function $y=f\left(x\right)={x}^{2}$ as a function of $y.$ However, based on the graph, it is clear we are interested in the positive square root.) Similarly, the right graph is represented by the function $y=g\left(x\right)=2-x,$ but could just as easily be represented by the function $x=u\left(y\right)=2-y.$ When the graphs are represented as functions of $y,$ we see the region is bounded on the left by the graph of one function and on the right by the graph of the other function. Therefore, if we integrate with respect to $y,$ we need to evaluate one integral only. Let’s develop a formula for this type of integration.

Determining Volumes by Slicing

Volume and the Slicing Method

Just as area is the numerical measure of a two-dimensional region, volume is the numerical measure of a three-dimensional solid. Most of us have computed volumes of solids by using basic geometric formulas. The volume of a rectangular solid, for example, can be computed by multiplying length, width, and height: $V=lwh.$ The formulas for the volume of a sphere $\left(V=\frac{4}{3}\pi {r}^{3}\right),$ a cone $\left(V=\frac{1}{3}\pi {r}^{2}h\right),$ and a pyramid $\left(V=\frac{1}{3}Ah\right)$ have also been introduced. Although some of these formulas were derived using geometry alone, all these formulas can be obtained by using integration.

Solids of Revolution

If a region in a plane is revolved around a line in that plane, the resulting solid is called a solid of revolution, as shown in the following figure.

The Disk Method

When we use the slicing method with solids of revolution, it is often called the disk method because, for solids of revolution, the slices used to over approximate the volume of the solid are disks. To see this, consider the solid of revolution generated by revolving the region between the graph of the function $f\left(x\right)={\left(x-1\right)}^{2}+1$ and the $x\text{-axis}$ over the interval $\left[-1,3\right]$ around the $x\text{-axis}\text{.}$ The graph of the function and a representative disk are shown in [link](a) and (b). The region of revolution and the resulting solid are shown in [link](c) and (d).

The Washer Method

Some solids of revolution have cavities in the middle; they are not solid all the way to the axis of revolution. Sometimes, this is just a result of the way the region of revolution is shaped with respect to the axis of revolution. In other cases, cavities arise when the region of revolution is defined as the region between the graphs of two functions. A third way this can happen is when an axis of revolution other than the $x\text{-axis}$ or $y\text{-axis}$ is selected.

Volumes of Revolution: Cylindrical Shells

The Method of Cylindrical Shells

Again, we are working with a solid of revolution. As before, we define a region $R,$ bounded above by the graph of a function $y=f\left(x\right),$ below by the $x\text{-axis,}$ and on the left and right by the lines $x=a$ and $x=b,$ respectively, as shown in [link](a). We then revolve this region around the *y*-axis, as shown in [link](b). Note that this is different from what we have done before. Previously, regions defined in terms of functions of $x$ were revolved around the $x\text{-axis}$ or a line parallel to it.

Which Method Should We Use?

We have studied several methods for finding the volume of a solid of revolution, but how do we know which method to use? It often comes down to a choice of which integral is easiest to evaluate. [link] describes the different approaches for solids of revolution around the $x\text{-axis}.$ It’s up to you to develop the analogous table for solids of revolution around the $y\text{-axis}.$

Arc Length of a Curve and Surface Area

Arc Length of the Curve *y* = *f*(*x*)

In previous applications of integration, we required the function $f\left(x\right)$ to be integrable, or at most continuous. However, for calculating arc length we have a more stringent requirement for $f\left(x\right).$ Here, we require $f\left(x\right)$ to be differentiable, and furthermore we require its derivative, ${f}^{\prime }\left(x\right),$ to be continuous. Functions like this, which have continuous derivatives, are called *smooth*. (This property comes up again in later chapters.)

Arc Length of the Curve *x* = *g*(*y*)

We have just seen how to approximate the length of a curve with line segments. If we want to find the arc length of the graph of a function of $y,$ we can repeat the same process, except we partition the $y\text{-axis}$ instead of the $x\text{-axis}.$ [link] shows a representative line segment.

Area of a Surface of Revolution

The concepts we used to find the arc length of a curve can be extended to find the surface area of a surface of revolution. Surface area is the total area of the outer layer of an object. For objects such as cubes or bricks, the surface area of the object is the sum of the areas of all of its faces. For curved surfaces, the situation is a little more complex. Let $f\left(x\right)$ be a nonnegative smooth function over the interval $\left[a,b\right].$ We wish to find the surface area of the surface of revolution created by revolving the graph of $y=f\left(x\right)$ around the $x\text{-axis}$ as shown in the following figure.

Physical Applications

Mass and Density

We can use integration to develop a formula for calculating mass based on a density function. First we consider a thin rod or wire. Orient the rod so it aligns with the $x\text{-axis,}$ with the left end of the rod at $x=a$ and the right end of the rod at $x=b$ ([link]). Note that although we depict the rod with some thickness in the figures, for mathematical purposes we assume the rod is thin enough to be treated as a one-dimensional object.

Work Done by a Force

We now consider work. In physics, work is related to force, which is often intuitively defined as a push or pull on an object. When a force moves an object, we say the force does work on the object. In other words, work can be thought of as the amount of energy it takes to move an object. According to physics, when we have a constant force, work can be expressed as the product of force and distance.

Work Done in Pumping

Consider the work done to pump water (or some other liquid) out of a tank. Pumping problems are a little more complicated than spring problems because many of the calculations depend on the shape and size of the tank. In addition, instead of being concerned about the work done to move a single mass, we are looking at the work done to move a volume of water, and it takes more work to move the water from the bottom of the tank than it does to move the water from the top of the tank.

Hydrostatic Force and Pressure

In this last section, we look at the force and pressure exerted on an object submerged in a liquid. In the English system, force is measured in pounds. In the metric system, it is measured in newtons. Pressure is force per unit area, so in the English system we have pounds per square foot (or, perhaps more commonly, pounds per square inch, denoted psi). In the metric system we have newtons per square meter, also called *pascals*.

Moments and Centers of Mass

Center of Mass and Moments

Let’s begin by looking at the center of mass in a one-dimensional context. Consider a long, thin wire or rod of negligible mass resting on a fulcrum, as shown in [link](a). Now suppose we place objects having masses ${m}_{1}$ and ${m}_{2}$ at distances ${d}_{1}$ and ${d}_{2}$ from the fulcrum, respectively, as shown in [link](b).

Center of Mass of Thin Plates

So far we have looked at systems of point masses on a line and in a plane. Now, instead of having the mass of a system concentrated at discrete points, we want to look at systems in which the mass of the system is distributed continuously across a thin sheet of material. For our purposes, we assume the sheet is thin enough that it can be treated as if it is two-dimensional. Such a sheet is called a lamina. Next we develop techniques to find the center of mass of a lamina. In this section, we also assume the density of the lamina is constant.

The Symmetry Principle

We stated the symmetry principle earlier, when we were looking at the centroid of a rectangle. The symmetry principle can be a great help when finding centroids of regions that are symmetric. Consider the following example.

Theorem of Pappus

This section ends with a discussion of the theorem of Pappus for volume, which allows us to find the volume of particular kinds of solids by using the centroid. (There is also a theorem of Pappus for surface area, but it is much less useful than the theorem for volume.)

Integrals, Exponential Functions, and Logarithms

The Natural Logarithm as an Integral

Recall the power rule for integrals:

Properties of the Natural Logarithm

Because of the way we defined the natural logarithm, the following differentiation formula falls out immediately as a result of to the Fundamental Theorem of Calculus.

Defining the Number *e*

Now that we have the natural logarithm defined, we can use that function to define the number $e.$

The Exponential Function

We now turn our attention to the function ${e}^{x}.$ Note that the natural logarithm is one-to-one and therefore has an inverse function. For now, we denote this inverse function by $\text{exp}{\rule{0.2em}{0ex}}x.$ Then,

Properties of the Exponential Function

Since the exponential function was defined in terms of an inverse function, and not in terms of a power of $e,$ we must verify that the usual laws of exponents hold for the function ${e}^{x}.$

General Logarithmic and Exponential Functions

We close this section by looking at exponential functions and logarithms with bases other than $e.$ Exponential functions are functions of the form $f\left(x\right)={a}^{x}.$ Note that unless $a=e,$ we still do not have a mathematically rigorous definition of these functions for irrational exponents. Let’s rectify that here by defining the function $f\left(x\right)={a}^{x}$ in terms of the exponential function ${e}^{x}.$ We then examine logarithms with bases other than $e$ as inverse functions of exponential functions.

Exponential Growth and Decay

Exponential Growth Model

Many systems exhibit exponential growth. These systems follow a model of the form $y={y}_{0}{e}^{kt},$ where ${y}_{0}$ represents the initial state of the system and $k$ is a positive constant, called the *growth constant*. Notice that in an exponential growth model, we have

Exponential Decay Model

Exponential functions can also be used to model populations that shrink (from disease, for example), or chemical compounds that break down over time. We say that such systems exhibit exponential decay, rather than exponential growth. The model is nearly the same, except there is a negative sign in the exponent. Thus, for some positive constant $k,$ we have $y={y}_{0}{e}^{\text{−}kt}.$

Calculus of the Hyperbolic Functions

Derivatives and Integrals of the Hyperbolic Functions

Recall that the hyperbolic sine and hyperbolic cosine are defined as

Calculus of Inverse Hyperbolic Functions

Looking at the graphs of the hyperbolic functions, we see that with appropriate range restrictions, they all have inverses. Most of the necessary range restrictions can be discerned by close examination of the graphs. The domains and ranges of the inverse hyperbolic functions are summarized in the following table.

Applications

One physical application of hyperbolic functions involves hanging cables. If a cable of uniform density is suspended between two supports without any load other than its own weight, the cable forms a curve called a catenary. High-voltage power lines, chains hanging between two posts, and strands of a spider’s web all form catenaries. The following figure shows chains hanging from a row of posts.

Techniques of Integration

Introduction

Integration by Parts

Abstract:
* Recognize when to use integration by parts.
* Use the integration-by-parts formula to solve integration problems.
* Use the integration-by-parts formula for definite integrals.

The Integration-by-Parts Formula

If, $h\left(x\right)=f\left(x\right)g\left(x\right),$ then by using the product rule, we obtain ${h}^{\prime }\left(x\right)={f}^{\prime }\left(x\right)g\left(x\right)+{g}^{\prime }\left(x\right)f\left(x\right).$ Although at first it may seem counterproductive, let’s now integrate both sides of this equation: $\int {h}^{\prime }\left(x\right)dx=\int \left(g\left(x\right){f}^{\prime }\left(x\right)+f\left(x\right){g}^{\prime }\left(x\right)\right)dx.$

Integration by Parts for Definite Integrals

Now that we have used integration by parts successfully to evaluate indefinite integrals, we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.

Trigonometric Integrals

Abstract:
* Solve integration problems involving products and powers of $\text{sin}x$ and $\text{cos}x.$
* Solve integration problems involving products and powers of $\text{tan}x$ and $\text{sec}x.$
* Use reduction formulas to solve trigonometric integrals.

Integrating Products and Powers of sin*x* and cos*x*

A key idea behind the strategy used to integrate combinations of products and powers of $\text{sin}{\rule{0.1em}{0ex}}x$ and $\text{cos}{\rule{0.1em}{0ex}}x$ involves rewriting these expressions as sums and differences of integrals of the form $\int {\text{sin}}^{j}x{\rule{0.1em}{0ex}}\text{cos}{\rule{0.1em}{0ex}}x{\rule{0.2em}{0ex}}dx$ or $\int {\text{cos}}^{j}x{\rule{0.1em}{0ex}}\text{sin}{\rule{0.1em}{0ex}}x{\rule{0.2em}{0ex}}dx.$ After rewriting these integrals, we evaluate them using *u*-substitution. Before describing the general process in detail, let’s take a look at the following examples.

Integrating Products and Powers of tan*x* and sec*x*

Before discussing the integration of products and powers of $\text{tan}{\rule{0.1em}{0ex}}x$ and $\text{sec}{\rule{0.1em}{0ex}}x,$ it is useful to recall the integrals involving $\text{tan}{\rule{0.1em}{0ex}}x$ and $\text{sec}{\rule{0.1em}{0ex}}x$ we have already learned:

Reduction Formulas

Evaluating $\int {\text{sec}}^{n}x{\rule{0.2em}{0ex}}dx$ for values of $n$ where $n$ is odd requires integration by parts. In addition, we must also know the value of $\int {\text{sec}}^{n-2}x{\rule{0.2em}{0ex}}dx$ to evaluate $\int {\text{sec}}^{n}x{\rule{0.2em}{0ex}}dx.$ The evaluation of $\int {\text{tan}}^{n}x{\rule{0.2em}{0ex}}dx$ also requires being able to integrate $\int {\text{tan}}^{n-2}x{\rule{0.2em}{0ex}}dx.$ To make the process easier, we can derive and apply the following power reduction formulas. These rules allow us to replace the integral of a power of $\text{sec}{\rule{0.1em}{0ex}}x$ or $\text{tan}{\rule{0.1em}{0ex}}x$ with the integral of a lower power of $\text{sec}{\rule{0.1em}{0ex}}x$ or $\text{tan}{\rule{0.1em}{0ex}}x.$

Trigonometric Substitution

Abstract: * Solve integration problems involving the square root of a sum or difference of two squares.

Integrals Involving $\sqrt{{a}^{2}-{x}^{2}}$

Before developing a general strategy for integrals containing $\sqrt{{a}^{2}-{x}^{2}},$ consider the integral $\int \sqrt{9-{x}^{2}}dx.$ This integral cannot be evaluated using any of the techniques we have discussed so far. However, if we make the substitution $x=3{\rule{0.1em}{0ex}}\text{sin}{\rule{0.1em}{0ex}}\theta ,$ we have $dx=3{\rule{0.1em}{0ex}}\text{cos}{\rule{0.1em}{0ex}}\theta d\theta .$ After substituting into the integral, we have

Integrating Expressions Involving $\sqrt{{a}^{2}+{x}^{2}}$

For integrals containing $\sqrt{{a}^{2}+{x}^{2},}$ let’s first consider the domain of this expression. Since $\sqrt{{a}^{2}+{x}^{2}}$ is defined for all real values of $x,$ we restrict our choice to those trigonometric functions that have a range of all real numbers. Thus, our choice is restricted to selecting either $x=a{\rule{0.1em}{0ex}}\text{tan}{\rule{0.1em}{0ex}}\theta$ or $x=a{\rule{0.1em}{0ex}}\text{cot}{\rule{0.1em}{0ex}}\theta .$ Either of these substitutions would actually work, but the standard substitution is $x=a{\rule{0.1em}{0ex}}\text{tan}{\rule{0.1em}{0ex}}\theta$ or, equivalently, $\text{tan}{\rule{0.1em}{0ex}}\theta =x\text{/}a.$ With this substitution, we make the assumption that $\text{−}\left(\pi \text{/}2\right)<\theta <\pi \text{/}2,$ so that we also have $\theta ={\text{tan}}^{-1}\left(x\text{/}a\right).$ The procedure for using this substitution is outlined in the following problem-solving strategy.

Integrating Expressions Involving $\sqrt{{x}^{2}-{a}^{2}}$

The domain of the expression $\sqrt{{x}^{2}-{a}^{2}}$ is $\left(\text{−}\infty ,\text{−}a\right]\cup \left[a,\text{+}\infty \right).$ Thus, either $x\le \text{−}a$ or $x\ge a.$ Hence, $\frac{x}{a}\le -1$ or $\frac{x}{a}\ge 1.$ Since these intervals correspond to the range of $\text{sec}{\rule{0.1em}{0ex}}\theta$ on the set $\left[0,\frac{\pi }{2}\right)\cup \left(\frac{\pi }{2},\pi \right],$ it makes sense to use the substitution $\text{sec}{\rule{0.1em}{0ex}}\theta =\frac{x}{a}$ or, equivalently, $x=a{\rule{0.1em}{0ex}}\text{sec}{\rule{0.1em}{0ex}}\theta ,$ where $0\le \theta <\frac{\pi }{2}$ or $\frac{\pi }{2}<\theta \le \pi .$ The corresponding substitution for $dx$ is $dx=a{\rule{0.1em}{0ex}}\text{sec}{\rule{0.1em}{0ex}}\theta {\rule{0.1em}{0ex}}\text{tan}{\rule{0.1em}{0ex}}\theta d\theta .$ The procedure for using this substitution is outlined in the following problem-solving strategy.

Partial Fractions

Abstract:
* Integrate a rational function using the method of partial fractions.
* Recognize simple linear factors in a rational function.
* Recognize repeated linear factors in a rational function.
* Recognize quadratic factors in a rational function.

Nonrepeated Linear Factors

If $Q\left(x\right)$ can be factored as $\left({a}_{1}x+{b}_{1}\right)\left({a}_{2}x+{b}_{2}\right)\text{…}\left({a}_{n}x+{b}_{n}\right),$ where each linear factor is distinct, then it is possible to find constants ${A}_{1},{A}_{2}\text{,…}{\rule{0.2em}{0ex}}{A}_{n}$ satisfying

Repeated Linear Factors

For some applications, we need to integrate rational expressions that have denominators with repeated linear factors—that is, rational functions with at least one factor of the form ${\left(ax+b\right)}^{n},$ where $n$ is a positive integer greater than or equal to $2.$ If the denominator contains the repeated linear factor ${\left(ax+b\right)}^{n},$ then the decomposition must contain

The General Method

Now that we are beginning to get the idea of how the technique of partial fraction decomposition works, let’s outline the basic method in the following problem-solving strategy.

Simple Quadratic Factors

Now let’s look at integrating a rational expression in which the denominator contains an irreducible quadratic factor. Recall that the quadratic $a{x}^{2}+bx+c$ is irreducible if $a{x}^{2}+bx+c=0$ has no real zeros—that is, if ${b}^{2}-4ac<0.$

Other Strategies for Integration

Abstract:
* Use a table of integrals to solve integration problems.
* Use a computer algebra system (CAS) to solve integration problems.

Tables of Integrals

Integration tables, if used in the right manner, can be a handy way either to evaluate or check an integral quickly. Keep in mind that when using a table to check an answer, it is possible for two completely correct solutions to look very different. For example, in Trigonometric Substitution, we found that, by using the substitution $x=\text{tan}{\rule{0.1em}{0ex}}\theta ,$ we can arrive at

Computer Algebra Systems

If available, a CAS is a faster alternative to a table for solving an integration problem. Many such systems are widely available and are, in general, quite easy to use.

Numerical Integration

Abstract:
* Approximate the value of a definite integral by using the midpoint and trapezoidal rules.
* Determine the absolute and relative error in using a numerical integration technique.
* Estimate the absolute and relative error using an error-bound formula.
* Recognize when the midpoint and trapezoidal rules over- or underestimate the true value of an integral.
* Use Simpson’s rule to approximate the value of a definite integral to a given accuracy.

The Midpoint Rule

Earlier in this text we defined the definite integral of a function over an interval as the limit of Riemann sums. In general, any Riemann sum of a function $f\left(x\right)$ over an interval $\left[a,b\right]$ may be viewed as an estimate of ${\int }_{a}^{b}f\left(x\right)dx.$ Recall that a Riemann sum of a function $f\left(x\right)$ over an interval $\left[a,b\right]$ is obtained by selecting a partition

The Trapezoidal Rule

We can also approximate the value of a definite integral by using trapezoids rather than rectangles. In [link], the area beneath the curve is approximated by trapezoids rather than by rectangles.

Absolute and Relative Error

An important aspect of using these numerical approximation rules consists of calculating the error in using them for estimating the value of a definite integral. We first need to define absolute error and relative error.

Simpson’s Rule

With the midpoint rule, we estimated areas of regions under curves by using rectangles. In a sense, we approximated the curve with piecewise constant functions. With the trapezoidal rule, we approximated the curve by using piecewise linear functions. What if we were, instead, to approximate a curve using piecewise quadratic functions? With Simpson’s rule, we do just this. We partition the interval into an even number of subintervals, each of equal width. Over the first pair of subintervals we approximate ${\int }_{{x}_{0}}^{{x}_{2}}f\left(x\right)dx$ with ${\int }_{{x}_{0}}^{{x}_{2}}p\left(x\right)dx,$ where $p\left(x\right)=A{x}^{2}+Bx+C$ is the quadratic function passing through $\left({x}_{0},f\left({x}_{0}\right)\right),$ $\left({x}_{1},f\left({x}_{1}\right)\right),$ and $\left({x}_{2},f\left({x}_{2}\right)\right)$ ([link]). Over the next pair of subintervals we approximate ${\int }_{{x}_{2}}^{{x}_{4}}f\left(x\right)dx$ with the integral of another quadratic function passing through $\left({x}_{2},f\left({x}_{2}\right)\right),$ $\left({x}_{3},f\left({x}_{3}\right)\right),$ and $\left({x}_{4},f\left({x}_{4}\right)\right).$ This process is continued with each successive pair of subintervals.

Improper Integrals

Abstract:
* Evaluate an integral over an infinite interval.
* Evaluate an integral over a closed interval with an infinite discontinuity within the interval.
* Use the comparison theorem to determine whether a definite integral is convergent.

Integrating over an Infinite Interval

How should we go about defining an integral of the type ${\int }_{a}^{+\infty }f\left(x\right)dx?$ We can integrate ${\int }_{a}^{t}f\left(x\right)dx$ for any value of $t,$ so it is reasonable to look at the behavior of this integral as we substitute larger values of $t.$ [link] shows that ${\int }_{a}^{t}f\left(x\right)dx$ may be interpreted as area for various values of $t.$ In other words, we may define an improper integral as a limit, taken as one of the limits of integration increases or decreases without bound.

Integrating a Discontinuous Integrand

Now let’s examine integrals of functions containing an infinite discontinuity in the interval over which the integration occurs. Consider an integral of the form ${\int }_{a}^{b}f\left(x\right)dx,$ where $f\left(x\right)$ is continuous over $\left[a,b\right)$ and discontinuous at $b.$ Since the function $f\left(x\right)$ is continuous over $\left[a,t\right]$ for all values of $t$ satisfying $a the integral ${\int }_{a}^{t}f\left(x\right)dx$ is defined for all such values of $t.$ Thus, it makes sense to consider the values of ${\int }_{a}^{t}f\left(x\right)dx$ as $t$ approaches $b$ for $a That is, we define ${\int }_{a}^{b}f\left(x\right)dx=\underset{t\to {b}^{-}}{\text{lim}}{\int }_{a}^{t}f\left(x\right)dx,$ provided this limit exists. [link] illustrates ${\int }_{a}^{t}f\left(x\right)dx$ as areas of regions for values of $t$ approaching $b.$

A Comparison Theorem

It is not always easy or even possible to evaluate an improper integral directly; however, by comparing it with another carefully chosen integral, it may be possible to determine its convergence or divergence. To see this, consider two continuous functions $f\left(x\right)$ and $g\left(x\right)$ satisfying $0\le f\left(x\right)\le g\left(x\right)$ for $x\ge a$ ([link]). In this case, we may view integrals of these functions over intervals of the form $\left[a,t\right]$ as areas, so we have the relationship

Introduction to Differential Equations

Introduction

Basics of Differential Equations

Abstract:
* Identify the order of a differential equation.
* Explain what is meant by a solution to a differential equation.
* Distinguish between the general solution and a particular solution of a differential equation.
* Identify an initial-value problem.
* Identify whether a given function is a solution to a differential equation or an initial-value problem.

General Differential Equations

Consider the equation ${y}^{\prime }=3{x}^{2},$ which is an example of a differential equation because it includes a derivative. There is a relationship between the variables $x$ and $y\text{:}{\rule{0.2em}{0ex}}y$ is an unknown function of $x.$ Furthermore, the left-hand side of the equation is the derivative of $y.$ Therefore we can interpret this equation as follows: Start with some function $y=f\left(x\right)$ and take its derivative. The answer must be equal to $3{x}^{2}.$ What function has a derivative that is equal to $3{x}^{2}?$ One such function is $y={x}^{3},$ so this function is considered a solution to a differential equation.

General and Particular Solutions

We already noted that the differential equation ${y}^{\prime }=2x$ has at least two solutions: $y={x}^{2}$ and $y={x}^{2}+4.$ The only difference between these two solutions is the last term, which is a constant. What if the last term is a different constant? Will this expression still be a solution to the differential equation? In fact, any function of the form $y={x}^{2}+C,$ where $C$ represents any constant, is a solution as well. The reason is that the derivative of ${x}^{2}+C$ is $2x,$ regardless of the value of $C.$ It can be shown that any solution of this differential equation must be of the form $y={x}^{2}+C.$ This is an example of a general solution to a differential equation. A graph of some of these solutions is given in [link]. (*Note*: in this graph we used even integer values for $C$ ranging between $-4$ and $4.$ In fact, there is no restriction on the value of $C;$ it can be an integer or not.)

Initial-Value Problems

Usually a given differential equation has an infinite number of solutions, so it is natural to ask which one we want to use. To choose one solution, more information is needed. Some specific information that can be useful is an initial value, which is an ordered pair that is used to find a particular solution.

Direction Fields and Numerical Methods

Abstract:
* Draw the direction field for a given first-order differential equation.
* Use a direction field to draw a solution curve of a first-order differential equation.
* Use Euler’s Method to approximate the solution to a first-order differential equation.

Creating Direction Fields

Direction fields (also called slope fields) are useful for investigating first-order differential equations. In particular, we consider a first-order differential equation of the form

Using Direction Fields

We can use a direction field to predict the behavior of solutions to a differential equation without knowing the actual solution. For example, the direction field in [link] serves as a guide to the behavior of solutions to the differential equation $y\prime =3x+2y-4.$

Euler’s Method

Consider the initial-value problem

Separable Equations

Abstract:
* Use separation of variables to solve a differential equation.
* Solve applications using separation of variables.

Separation of Variables

We start with a definition and some examples.

Applications of Separation of Variables

Many interesting problems can be described by separable equations. We illustrate two types of problems: solution concentrations and Newton’s law of cooling.

The Logistic Equation

Abstract:
* Describe the concept of environmental carrying capacity in the logistic model of population growth.
* Draw a direction field for a logistic equation and interpret the solution curves.
* Solve a logistic equation and interpret the results.

Population Growth and Carrying Capacity

To model population growth using a differential equation, we first need to introduce some variables and relevant terms. The variable $t.$ will represent time. The units of time can be hours, days, weeks, months, or even years. Any given problem must specify the units used in that particular problem. The variable $P$ will represent population. Since the population varies over time, it is understood to be a function of time. Therefore we use the notation $P\left(t\right)$ for the population as a function of time. If $P\left(t\right)$ is a differentiable function, then the first derivative $\frac{dP}{dt}$ represents the instantaneous rate of change of the population as a function of time.

Solving the Logistic Differential Equation

The logistic differential equation is an autonomous differential equation, so we can use separation of variables to find the general solution, as we just did in [link].

First-order Linear Equations

Abstract:
* Write a first-order linear differential equation in standard form.
* Find an integrating factor and use it to solve a first-order linear differential equation.
* Solve applied problems involving first-order linear differential equations.

Standard Form

Consider the differential equation

Integrating Factors

We now develop a solution technique for any first-order linear differential equation. We start with the standard form of a first-order linear differential equation:

Applications of First-order Linear Differential Equations

We look at two different applications of first-order linear differential equations. The first involves air resistance as it relates to objects that are rising or falling; the second involves an electrical circuit. Other applications are numerous, but most are solved in a similar fashion.

Sequences and Series

Introduction

Sequences

Abstract:
* Find the formula for the general term of a sequence.
* Calculate the limit of a sequence if it exists.
* Determine the convergence or divergence of a given sequence.

Terminology of Sequences

To work with this new topic, we need some new terms and definitions. First, an infinite sequence is an ordered list of numbers of the form

Limit of a Sequence

A fundamental question that arises regarding infinite sequences is the behavior of the terms as $n$ gets larger. Since a sequence is a function defined on the positive integers, it makes sense to discuss the limit of the terms as $n\to \infty .$ For example, consider the following four sequences and their different behaviors as $n\to \infty$ (see [link]):

Bounded Sequences

We now turn our attention to one of the most important theorems involving sequences: the Monotone Convergence Theorem. Before stating the theorem, we need to introduce some terminology and motivation. We begin by defining what it means for a sequence to be bounded.

Infinite Series

Abstract:
* Explain the meaning of the sum of an infinite series.
* Calculate the sum of a geometric series.
* Evaluate a telescoping series.

Sums and Series

An infinite series is a sum of infinitely many terms and is written in the form

Telescoping Series

Consider the series $\sum _{n=1}^{\infty }\frac{1}{n\left(n+1\right)}.$ We discussed this series in [link], showing that the series converges by writing out the first several partial sums ${S}_{1},{S}_{2}\text{,…},{S}_{6}$ and noticing that they are all of the form ${S}_{k}=\frac{k}{k+1}.$ Here we use a different technique to show that this series converges. By using partial fractions, we can write

The Divergence and Integral Tests

Abstract:
* Use the divergence test to determine whether a series converges or diverges.
* Use the integral test to determine the convergence of a series.
* Estimate the value of a series by finding bounds on its remainder term.

Divergence Test

A series $\sum _{n=1}^{\infty }{a}_{n}$ being convergent is equivalent to the convergence of the sequence of partial sums (${S}_{k}$) as $k\to \infty .$

Integral Test

In the previous section, we proved that the harmonic series diverges by looking at the sequence of partial sums $\left\{{S}_{k}\right\}$ and showing that ${S}_{{2}^{k}}>1+k\text{/}2$ for all positive integers $k.$ In this section we use a different technique to prove the divergence of the harmonic series. This technique is important because it is used to prove the divergence or convergence of many other series. This test, called the integral test, compares an infinite sum to an improper integral. It is important to note that this test can only be applied when we are considering a series whose terms are all positive.

The *p*-Series

The harmonic series $\sum _{n=1}^{\infty }1\text{/}n$ and the series $\sum _{n=1}^{\infty }1\text{/}{n}^{2}$ are both examples of a type of series called a *p*-series.

Estimating the Value of a Series

Suppose we know that a series $\sum _{n=1}^{\infty }{a}_{n}$ converges and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum $\sum _{n=1}^{N}{a}_{n}$ where $N$ is any positive integer. The question we address here is, for a convergent series $\sum _{n=1}^{\infty }{a}_{n},$ how good is the approximation $\sum _{n=1}^{N}{a}_{n}\text{?}$ More specifically, if we let

Comparison Tests

Abstract:
* Use the comparison test to test a series for convergence.
* Use the limit comparison test to determine convergence of a series.

Comparison Test

In the preceding two sections, we discussed two large classes of series: geometric series and *p*-series. We know exactly when these series converge and when they diverge. Here we show how to use the convergence or divergence of these series to prove convergence or divergence for other series, using a method called the comparison test.

Limit Comparison Test

The comparison test works nicely if we can find a comparable series satisfying the hypothesis of the test. However, sometimes finding an appropriate series can be difficult. Consider the series

Alternating Series

Abstract:
* Use the alternating series test to test an alternating series for convergence.
* Estimate the sum of an alternating series.
* Explain the meaning of absolute convergence and conditional convergence.

The Alternating Series Test

A series whose terms alternate between positive and negative values is an alternating series. For example, the series

Remainder of an Alternating Series

It is difficult to explicitly calculate the sum of most alternating series, so typically the sum is approximated by using a partial sum. When doing so, we are interested in the amount of error in our approximation. Consider an alternating series

Absolute and Conditional Convergence

Consider a series $\sum _{n=1}^{\infty }{a}_{n}$ and the related series $\sum _{n=1}^{\infty }|{a}_{n}|.$ Here we discuss possibilities for the relationship between the convergence of these two series. For example, consider the alternating harmonic series $\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\text{/}n.$ The series whose terms are the absolute value of these terms is the harmonic series, since $\sum _{n=1}^{\infty }|{\left(-1\right)}^{n+1}\text{/}n|=\sum _{n=1}^{\infty }1\text{/}n.$ Since the alternating harmonic series converges, but the harmonic series diverges, we say the alternating harmonic series exhibits conditional convergence.

Ratio and Root Tests

Abstract:
* Use the ratio test to determine absolute convergence of a series.
* Use the root test to determine absolute convergence of a series.
* Describe a strategy for testing the convergence of a given series.

Ratio Test

Consider a series $\sum _{n=1}^{\infty }{a}_{n}.$ From our earlier discussion and examples, we know that $\underset{n\to \infty }{\text{lim}}{a}_{n}=0$ is not a sufficient condition for the series to converge. Not only do we need ${a}_{n}\to 0,$ but we need ${a}_{n}\to 0$ quickly enough. For example, consider the series $\sum _{n=1}^{\infty }1\text{/}n$ and the series $\sum _{n=1}^{\infty }1\text{/}{n}^{2}.$ We know that $1\text{/}n\to 0$ and $1\text{/}{n}^{2}\to 0.$ However, only the series $\sum _{n=1}^{\infty }1\text{/}{n}^{2}$ converges. The series $\sum _{n=1}^{\infty }1\text{/}n$ diverges because the terms in the sequence $\left\{1\text{/}n\right\}$ do not approach zero fast enough as $n\to \infty .$ Here we introduce the ratio test, which provides a way of measuring how fast the terms of a series approach zero.

Choosing a Convergence Test

At this point, we have a long list of convergence tests. However, not all tests can be used for all series. When given a series, we must determine which test is the best to use. Here is a strategy for finding the best test to apply.

Power Series

Introduction

Power Series and Functions

Abstract:
* Identify a power series and provide examples of them.
* Determine the radius of convergence and interval of convergence of a power series.
* Use a power series to represent a function.

Form of a Power Series

A series of the form

Convergence of a Power Series

Since the terms in a power series involve a variable *x*, the series may converge for certain values of *x* and diverge for other values of *x*. For a power series centered at $x=a,$ the value of the series at $x=a$ is given by ${c}_{0}.$ Therefore, a power series always converges at its center. Some power series converge only at that value of *x*. Most power series, however, converge for more than one value of *x*. In that case, the power series either converges for all real numbers *x* or converges for all *x* in a finite interval. For example, the geometric series $\sum _{n=0}^{\infty }{x}^{n}$ converges for all *x* in the interval $\left(-1,1\right),$ but diverges for all *x* outside that interval. We now summarize these three possibilities for a general power series.

Representing Functions as Power Series

Being able to represent a function by an “infinite polynomial” is a powerful tool. Polynomial functions are the easiest functions to analyze, since they only involve the basic arithmetic operations of addition, subtraction, multiplication, and division. If we can represent a complicated function by an infinite polynomial, we can use the polynomial representation to differentiate or integrate it. In addition, we can use a truncated version of the polynomial expression to approximate values of the function. So, the question is, when can we represent a function by a power series?

Properties of Power Series

Abstract:
* Combine power series by addition or subtraction.
* Create a new power series by multiplication by a power of the variable or a constant, or by substitution.
* Multiply two power series together.
* Differentiate and integrate power series term-by-term.

Combining Power Series

If we have two power series with the same interval of convergence, we can add or subtract the two series to create a new power series, also with the same interval of convergence. Similarly, we can multiply a power series by a power of *x* or evaluate a power series at ${x}^{m}$ for a positive integer *m* to create a new power series. Being able to do this allows us to find power series representations for certain functions by using power series representations of other functions. For example, since we know the power series representation for $f\left(x\right)=\frac{1}{1-x},$ we can find power series representations for related functions, such as

Multiplication of Power Series

We can also create new power series by multiplying power series. Being able to multiply two power series provides another way of finding power series representations for functions.

Differentiating and Integrating Power Series

Consider a power series $\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\text{⋯}$ that converges on some interval *I*, and let $f$ be the function defined by this series. Here we address two questions about $f.$

Taylor and Maclaurin Series

Abstract:
* Describe the procedure for finding a Taylor polynomial of a given order for a function.
* Explain the meaning and significance of Taylor’s theorem with remainder.
* Estimate the remainder for a Taylor series approximation of a given function.

Overview of Taylor/Maclaurin Series

Consider a function $f$ that has a power series representation at $x=a.$ Then the series has the form

Taylor Polynomials

The *n*th partial sum of the Taylor series for a function $f$ at $a$ is known as the *n*th Taylor polynomial. For example, the 0th, 1st, 2nd, and 3rd partial sums of the Taylor series are given by

Taylor’s Theorem with Remainder

Recall that the *n*th Taylor polynomial for a function $f$ at *a* is the *n*th partial sum of the Taylor series for $f$ at *a*. Therefore, to determine if the Taylor series converges, we need to determine whether the sequence of Taylor polynomials $\left\{{p}_{n}\right\}$ converges. However, not only do we want to know if the sequence of Taylor polynomials converges, we want to know if it converges to $f.$ To answer this question, we define the remainder ${R}_{n}\left(x\right)$ as

Representing Functions with Taylor and Maclaurin Series

We now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.

Working with Taylor Series

Abstract:
* Write the terms of the binomial series.
* Recognize the Taylor series expansions of common functions.
* Recognize and apply techniques to find the Taylor series for a function.
* Use Taylor series to solve differential equations.
* Use Taylor series to evaluate nonelementary integrals.

The Binomial Series

Our first goal in this section is to determine the Maclaurin series for the function $f\left(x\right)={\left(1+x\right)}^{r}$ for all real numbers $r.$ The Maclaurin series for this function is known as the binomial series. We begin by considering the simplest case: $r$ is a nonnegative integer. We recall that, for $r=0,1,2,3,4,f\left(x\right)={\left(1+x\right)}^{r}$ can be written as

Common Functions Expressed as Taylor Series

At this point, we have derived Maclaurin series for exponential, trigonometric, and logarithmic functions, as well as functions of the form $f\left(x\right)={\left(1+x\right)}^{r}.$ In [link], we summarize the results of these series. We remark that the convergence of the Maclaurin series for $f\left(x\right)=\text{ln}{\rule{0.1em}{0ex}}\left(1+x\right)$ at the endpoint $x=1$ and the Maclaurin series for $f\left(x\right)={\text{tan}}^{-1}x$ at the endpoints $x=1$ and $x=-1$ relies on a more advanced theorem than we present here. (Refer to Abel’s theorem for a discussion of this more technical point.)

Solving Differential Equations with Power Series

Consider the differential equation

Evaluating Nonelementary Integrals

Solving differential equations is one common application of power series. We now turn to a second application. We show how power series can be used to evaluate integrals involving functions whose antiderivatives cannot be expressed using elementary functions.

Parametric Equations and Polar Coordinates

Introduction

Parametric Equations

Abstract:
* Plot a curve described by parametric equations.
* Convert the parametric equations of a curve into the form $y=f\left(x\right).$
* Recognize the parametric equations of basic curves, such as a line and a circle.
* Recognize the parametric equations of a cycloid.

Parametric Equations and Their Graphs

Consider the orbit of Earth around the Sun. Our year lasts approximately 365.25 days, but for this discussion we will use 365 days. On January 1 of each year, the physical location of Earth with respect to the Sun is nearly the same, except for leap years, when the lag introduced by the extra $\frac{1}{4}$ day of orbiting time is built into the calendar. We call January 1 “day 1” of the year. Then, for example, day 31 is January 31, day 59 is February 28, and so on.

Eliminating the Parameter

To better understand the graph of a curve represented parametrically, it is useful to rewrite the two equations as a single equation relating the variables *x* and *y.* Then we can apply any previous knowledge of equations of curves in the plane to identify the curve. For example, the equations describing the plane curve in [link]b. are

Cycloids and Other Parametric Curves

Imagine going on a bicycle ride through the country. The tires stay in contact with the road and rotate in a predictable pattern. Now suppose a very determined ant is tired after a long day and wants to get home. So he hangs onto the side of the tire and gets a free ride. The path that this ant travels down a straight road is called a cycloid ([link]). A cycloid generated by a circle (or bicycle wheel) of radius *a* is given by the parametric equations

Calculus of Parametric Curves

Abstract:
* Determine derivatives and equations of tangents for parametric curves.
* Find the area under a parametric curve.
* Use the equation for arc length of a parametric curve.
* Apply the formula for surface area to a volume generated by a parametric curve.

Derivatives of Parametric Equations

We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations

Second-Order Derivatives

Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function $y=f\left(x\right)$ is defined to be the derivative of the first derivative; that is,

Integrals Involving Parametric Equations

Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations $x\left(t\right)=t-\text{sin}{\rule{0.2em}{0ex}}t,{\rule{1em}{0ex}}y\left(t\right)=1-\text{cos}{\rule{0.2em}{0ex}}t.$ Suppose we want to find the area of the shaded region in the following graph.

Arc Length of a Parametric Curve

In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point *A* to point *B* along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.

Surface Area Generated by a Parametric Curve

Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function $y=f\left(x\right)$ from $x=a$ to $x=b,$ revolved around the *x*-axis:

Polar Coordinates

Abstract:
* Locate points in a plane by using polar coordinates.
* Convert points between rectangular and polar coordinates.
* Sketch polar curves from given equations.
* Convert equations between rectangular and polar coordinates.
* Identify symmetry in polar curves and equations.

Defining Polar Coordinates

To find the coordinates of a point in the polar coordinate system, consider [link]. The point $P$ has Cartesian coordinates $\left(x,y\right).$ The line segment connecting the origin to the point $P$ measures the distance from the origin to $P$ and has length $r.$ The angle between the positive $x$-axis and the line segment has measure $\theta .$ This observation suggests a natural correspondence between the coordinate pair $\left(x,y\right)$ and the values $r$ and $\theta .$ This correspondence is the basis of the polar coordinate system. Note that every point in the Cartesian plane has two values (hence the term *ordered pair*) associated with it. In the polar coordinate system, each point also has two values associated with it: $r$ and $\theta .$

Polar Curves

Now that we know how to plot points in the polar coordinate system, we can discuss how to plot curves. In the rectangular coordinate system, we can graph a function $y=f\left(x\right)$ and create a curve in the Cartesian plane. In a similar fashion, we can graph a curve that is generated by a function $r=f\left(\theta \right).$

Symmetry in Polar Coordinates

When studying symmetry of functions in rectangular coordinates (i.e., in the form $y=f\left(x\right)\right),$ we talk about symmetry with respect to the *y*-axis and symmetry with respect to the origin. In particular, if $f\left(\text{−}x\right)=f\left(x\right)$ for all $x$ in the domain of $f,$ then $f$ is an even function and its graph is symmetric with respect to the *y*-axis. If $f\left(\text{−}x\right)=\text{−}f\left(x\right)$ for all $x$ in the domain of $f,$ then $f$ is an odd function and its graph is symmetric with respect to the origin. By determining which types of symmetry a graph exhibits, we can learn more about the shape and appearance of the graph. Symmetry can also reveal other properties of the function that generates the graph. Symmetry in polar curves works in a similar fashion.

Area and Arc Length in Polar Coordinates

Abstract:
* Apply the formula for area of a region in polar coordinates.
* Determine the arc length of a polar curve.

Areas of Regions Bounded by Polar Curves

We have studied the formulas for area under a curve defined in rectangular coordinates and parametrically defined curves. Now we turn our attention to deriving a formula for the area of a region bounded by a polar curve. Recall that the proof of the Fundamental Theorem of Calculus used the concept of a Riemann sum to approximate the area under a curve by using rectangles. For polar curves we use the Riemann sum again, but the rectangles are replaced by sectors of a circle.

Arc Length in Polar Curves

Here we derive a formula for the arc length of a curve defined in polar coordinates.

Conic Sections

Abstract:
* Identify the equation of a parabola in standard form with given focus and directrix.
* Identify the equation of an ellipse in standard form with given foci.
* Identify the equation of a hyperbola in standard form with given foci.
* Recognize a parabola, ellipse, or hyperbola from its eccentricity value.
* Write the polar equation of a conic section with eccentricity $e$.
* Identify when a general equation of degree two is a parabola, ellipse, or hyperbola.

Parabolas

A parabola is generated when a plane intersects a cone parallel to the generating line. In this case, the plane intersects only one of the nappes. A parabola can also be defined in terms of distances.

Ellipses

An ellipse can also be defined in terms of distances. In the case of an ellipse, there are two foci (plural of focus), and two directrices (plural of directrix). We look at the directrices in more detail later in this section.

Hyperbolas

A hyperbola can also be defined in terms of distances. In the case of a hyperbola, there are two foci and two directrices. Hyperbolas also have two asymptotes.

Eccentricity and Directrix

An alternative way to describe a conic section involves the directrices, the foci, and a new property called eccentricity. We will see that the value of the eccentricity of a conic section can uniquely define that conic.

Polar Equations of Conic Sections

Sometimes it is useful to write or identify the equation of a conic section in polar form. To do this, we need the concept of the focal parameter. The focal parameter of a conic section *p* is defined as the distance from a focus to the nearest directrix. The following table gives the focal parameters for the different types of conics, where *a* is the length of the semi-major axis (i.e., half the length of the major axis), *c* is the distance from the origin to the focus, and *e* is the eccentricity. In the case of a parabola, *a* represents the distance from the vertex to the focus.

General Equations of Degree Two

A general equation of degree two can be written in the form

Table of Integrals

Basic Integrals

1. $\int {u}^{n}{\rule{0.2em}{0ex}}du=\frac{{u}^{n+1}}{n+1}+C,n\ne \text{−}1$

Trigonometric Integrals

18. $\int {\text{sin}}^{2}u{\rule{0.2em}{0ex}}du=\frac{1}{2}u-\frac{1}{4}\text{sin}{\rule{0.2em}{0ex}}2u+C$

Exponential and Logarithmic Integrals

42. $\int u{e}^{au}{\rule{0.2em}{0ex}}du=\frac{1}{{a}^{2}}\left(au-1\right){e}^{au}+C$

Hyperbolic Integrals

49. $\int \text{sinh}{\rule{0.2em}{0ex}}u{\rule{0.2em}{0ex}}du=\text{cosh}{\rule{0.2em}{0ex}}u+C$

Inverse Trigonometric Integrals

59. $\int {\text{sin}}^{-1}u{\rule{0.2em}{0ex}}du=u{\rule{0.2em}{0ex}}{\text{sin}}^{-1}u+\sqrt{1-{u}^{2}}+C$

Integrals Involving *a*2 + *u*2, *a* > 0

68. $\int \sqrt{{a}^{2}+{u}^{2}}{\rule{0.2em}{0ex}}du=\frac{u}{2}\sqrt{{a}^{2}+{u}^{2}}+\frac{{a}^{2}}{2}\text{ln}{\rule{0.1em}{0ex}}\left(u+\sqrt{{a}^{2}+{u}^{2}}\right)+C$

Integrals Involving *u*2 − *a*2, *a* > 0

77. $\int \sqrt{{u}^{2}-{a}^{2}}{\rule{0.2em}{0ex}}du=\frac{u}{2}\sqrt{{u}^{2}-{a}^{2}}-\frac{{a}^{2}}{2}\text{ln}{\rule{0.1em}{0ex}}|u+\sqrt{{u}^{2}-{a}^{2}}|+C$

Integrals Involving *a*2 − *u*2, *a* > 0

85. $\int \sqrt{{a}^{2}-{u}^{2}}{\rule{0.2em}{0ex}}du=\frac{u}{2}\sqrt{{a}^{2}-{u}^{2}}+\frac{{a}^{2}}{2}{\text{sin}}^{-1}\frac{u}{a}+C$

Integrals Involving 2*au* − *u*2, *a* > 0

94. $\int \sqrt{2au-{u}^{2}}{\rule{0.2em}{0ex}}du=\frac{u-a}{2}\sqrt{2au-{u}^{2}}+\frac{{a}^{2}}{2}{\text{cos}}^{-1}\left(\frac{a-u}{a}\right)+C$

Integrals Involving *a* + *bu*, *a* ≠ 0

98. $\int \frac{u{\rule{0.2em}{0ex}}du}{a+bu}=\frac{1}{{b}^{2}}\left(a+bu-a\text{ln}{\rule{0.1em}{0ex}}|a+bu|\right)+C$

Table of Derivatives

General Formulas

1. $\frac{d}{dx}\left(c\right)=0$

Trigonometric Functions

9. $\frac{d}{dx}\left(\text{sin}{\rule{0.1em}{0ex}}x\right)=\text{cos}{\rule{0.1em}{0ex}}x$

Inverse Trigonometric Functions

15. $\frac{d}{dx}\left({\text{sin}}^{-1}x\right)=\frac{1}{\sqrt{1-{x}^{2}}}$

Exponential and Logarithmic Functions

21. $\frac{d}{dx}\left({e}^{x}\right)={e}^{x}$

Hyperbolic Functions

25. $\frac{d}{dx}\left(\text{sinh}{\rule{0.1em}{0ex}}x\right)=\text{cosh}{\rule{0.1em}{0ex}}x$

Inverse Hyperbolic Functions

31. $\frac{d}{dx}\left({\text{sinh}}^{-1}x\right)=\frac{1}{\sqrt{{x}^{2}+1}}$

Review of Pre-Calculus

Formulas from Geometry

$A=\text{area},$ $V=\text{Volume},{\rule{0.2em}{0ex}}\text{and}$ $S=\text{lateral surface area}$

Formulas from Algebra

Formulas from Trigonometry