In the sections that discuss double and triple integrals, we will learn how to integrate functions with more than one independent variable over various types of regions in various coordinate systems. Here we will review basic integration techniques, evaluating trigonometric functions, evaluating definite integrals using the Second Fundamental Theorem of Calculus, manipulating equations, and how to find limits at infinity.

Indefinite Integrals

Evaluating indefinite integrals for some other functions is also a straightforward calculation. The following table lists the indefinite integrals for several common functions. A more complete list appears in Appendix B: Table of Derivatives.

Evaluate Trigonometric Functions

(also in Module 1, Skills Review for Polar Coordinates)
The unit circle tells us the value of cosine and sine at any of the given angle measures seen below. The first coordinate in each ordered pair is the value of cosine at the given angle measure, while the second coordinate in each ordered pair is the value of sine at the given angle measure. You will learn that all trigonometric functions can be written in terms of sine and cosine. Thus, if you can evaluate sine and cosine at various angle values, you can also evaluate the other trigonometric functions at various angle values. Take time to learn the $\left(x,y\right)$ coordinates of all of the major angles in the first quadrant of the unit circle.

Remember, every angle in quadrant two, three, or four has a reference angle that lies in quadrant one. The quadrant of the original angle only affects the sign (positive or negative) of a trigonometric function’s value at a given angle.

Below are the values of all six trigonometric functions evaluated at common angle measures from Quadrant I.

Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem

(also in Module 3, Skills Review for Arc Length and Curvature)

The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.

Manipulate Equations

It will sometimes be necessary to change the bounds of an integral which will even require manipulating the given equation.

Manipulate Rational Equations

Equations that contain rational expressions are called rational equations. For example, $\frac{2y+1}{4}=\frac{x}{3}$ is a rational equation (of two variables).

One of the most straightforward ways to solve a rational equation for the indicated variable is to eliminate denominators with the common denominator and then use properties of equality to isolate the indicated variable.

Solve for $y$ in the equation $\frac{1}{2}y-3=2-\frac{3}{4}x$ by clearing the fractions in the equation first.

Multiply both sides of the equation by $4$, the common denominator of the fractional coefficients.

$\begin{array}{c}\frac{1}{2}y-3=2-\frac{3}{4}x\\ 4\left(\frac{1}{2}y-3\right)=4\left(2-\frac{3}{4}x\right)\\\text{}\,\,\,\,4\left(\frac{1}{2}y\right)-4\left(3\right)=4\left(2\right)+4\left(-\frac{3}{4}x\right)\\2y-12=8-3x\\\underline{+12}\,\,\,\,\,\,\underline{+12}\\ 2y=20-3x\\ \frac{2y}{2}=\frac{20-3x}{2} \\ y=10-\frac{3}{2}x\end{array}$

In the next example, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they do not share any common factors.

Manipulate Radical Equations

Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as

Radical equations are manipulated by eliminating each radical, one at a time until you have solved for the indicated variable.

Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, ${16}^{\frac{1}{2}}$ is another way of writing $\sqrt{16}$ and ${8}^{\frac{2}{3}}$ is another way of writing $\left(\sqrt[3]{8}\right)^2$.

We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, $\frac{2}{3}\left(\frac{3}{2}\right)=1$.

Take Limits at Infinity

Recall that $\underset{x \to a}{\lim}f(x)=L$ means $f(x)$ becomes arbitrarily close to $L$ as long as $x$ is sufficiently close to $a$. We can extend this idea to limits at infinity. For example, consider the function $f(x)=2+\frac{1}{x}$. As can be seen graphically below, as the values of $x$ get larger, the values of $f(x)$ approach 2. We say the limit as $x$ approaches $\infty$ of $f(x)$ is 2 and write $\underset{x\to \infty }{\lim}f(x)=2$. Similarly, for $x<0$, as the values $|x|$ get larger, the values of $f(x)$ approaches 2. We say the limit as $x$ approaches $−\infty$ of $f(x)$ is 2 and write $\underset{x\to a}{\lim}f(x)=2$.

The function approaches the asymptote $y=2$ as $x$ approaches $\pm \infty$.

More generally, for any function $f$, we say the limit as $x \to \infty$ of $f(x)$ is $L$ if $f(x)$ becomes arbitrarily close to $L$ as long as $x$ is sufficiently large. In that case, we write $\underset{x\to \infty}{\lim}f(x)=L$. Similarly, we say the limit as $x\to −\infty$ of $f(x)$ is $L$ if $f(x)$ becomes arbitrarily close to $L$ as long as $x<0$ and $|x|$ is sufficiently large. In that case, we write $\underset{x\to −\infty }{\lim}f(x)=L$. We now look at the definition of a function having a limit at infinity.

Definition

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(Informal) If the values of $f(x)$ become arbitrarily close to $L$ as $x$ becomes sufficiently large, we say the function $f$ has a limit at infinity and write

$\underset{x\to \infty }{\lim}f(x)=L$

 

If the values of $f(x)$ becomes arbitrarily close to $L$ for $x<0$ as $|x|$ becomes sufficiently large, we say that the function $f$ has a limit at negative infinity and write

$\underset{x\to -\infty }{\lim}f(x)=L$

 

If the values $f(x)$ are getting arbitrarily close to some finite value $L$ as $x\to \infty$ or $x\to −\infty$, the graph of $f$ approaches the line $y=L$. In that case, the line $y=L$ is a horizontal asymptote of $f$ (Figure 2). For example, for the function $f(x)=\frac{1}{x}$, since $\underset{x\to \infty }{\lim}f(x)=0$, the line $y=0$ is a horizontal asymptote of $f(x)=\frac{1}{x}$.

Definition

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If $\underset{x\to \infty }{\lim}f(x)=L$ or $\underset{x \to −\infty}{\lim}f(x)=L$, we say the line $y=L$ is a horizontal asymptote of $f$.

Figure 2. (a) As $x\to \infty$, the values of $f$ are getting arbitrarily close to $L$. The line $y=L$ is a horizontal asymptote of $f$. (b) As $x\to −\infty$, the values of $f$ are getting arbitrarily close to $M$. The line $y=M$ is a horizontal asymptote of $f$.

A function cannot cross a vertical asymptote because the graph must approach infinity (or negative infinity) from at least one direction as $x$ approaches the vertical asymptote. However, a function may cross a horizontal asymptote. In fact, a function may cross a horizontal asymptote an unlimited number of times. For example, the function $f(x)=\frac{ \cos x}{x}+1$ shown in Figure 3 intersects the horizontal asymptote $y=1$ an infinite number of times as it oscillates around the asymptote with ever-decreasing amplitude.

Figure 3. The graph of $f(x)=\cos x/x+1$ crosses its horizontal asymptote $y=1$ an infinite number of times.

Example: Computing Limits at Infinity

For each of the following functions $f$, evaluate $\underset{x\to \infty }{\lim}f(x)$ and $\underset{x\to −\infty }{\lim}f(x)$.

1. $f(x)=5-\frac{2}{x^2}$
2. $f(x)=\dfrac{\sin x}{x}$

Show Solution

1. Using the algebraic limit laws, we have $\underset{x\to \infty }{\lim}(5-\frac{2}{x^2})=\underset{x\to \infty }{\lim}5-2(\underset{x\to \infty }{\lim}\frac{1}{x})(\underset{x\to \infty }{\lim}\frac{1}{x})=5-2 \cdot 0=5$. Similarly, $\underset{x\to -\infty }{\lim}f(x)=5$.
2. nce $-1\le \sin x\le 1$ for all $x$, we have
   $\frac{-1}{x}\le \frac{\sin x}{x}\le \frac{1}{x}$

   for all $x \ne 0$. Also, since

   $\underset{x\to \infty }{\lim}\frac{-1}{x}=0=\underset{x\to \infty }{\lim}\frac{1}{x}$,

   we can apply the squeeze theorem to conclude that

   $\underset{x\to \infty }{\lim}\frac{\sin x}{x}=0$

   Similarly,

   $\underset{x\to −\infty}{\lim}\frac{\sin x}{x}=0$

Try It

Evaluate $\underset{x\to −\infty}{\lim}\left(3+\frac{4}{x}\right)$ and $\underset{x\to \infty }{\lim}\left(3+\frac{4}{x}\right)$.

Hint

$\underset{x\to \pm \infty }{\lim}1/x=0$

Show Solution

Both limits are 3.

Infinite Limits at Infinity

Sometimes the values of a function $f$ become arbitrarily large as $x\to \infty$ (or as $x\to −\infty )$. In this case, we write $\underset{x\to \infty }{\lim}f(x)=\infty$ (or $\underset{x\to −\infty }{\lim}f(x)=\infty )$. On the other hand, if the values of $f$ are negative but become arbitrarily large in magnitude as $x\to \infty$ (or as $x\to −\infty )$, we write $\underset{x\to \infty }{\lim}f(x)=−\infty$ (or $\underset{x\to −\infty }{\lim}f(x)=−\infty )$.

For example, consider the function $f(x)=x^3$. The $\underset{x\to \infty }{\lim}x^3=\infty$. On the other hand, as $x\to −\infty$, the values of $f(x)=x^3$ are negative but become arbitrarily large in magnitude. Consequently, $\underset{x\to −\infty }{\lim}x^3=−\infty$.

Values of a power function as $x\to \pm \infty$

$x$	10	20	50	100	1000
$x^3$	1000	8000	125,000	1,000,000	1,000,000,000
$x$	-10	-20	-50	-100	-1000
$x^3$	-1000	-8000	-125,000	-1,000,000	-1,000,000,000

For this function, the functional values approach infinity as $x\to \pm \infty$.

Definition

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(Informal) We say a function $f$ has an infinite limit at infinity and write

$\underset{x\to \infty }{\lim}f(x)=\infty$

 

if $f(x)$ becomes arbitrarily large for $x$ sufficiently large. We say a function has a negative infinite limit at infinity and write

$\underset{x\to \infty }{\lim}f(x)=−\infty$

 

if $f(x)<0$ and $|f(x)|$ becomes arbitrarily large for $x$ sufficiently large. Similarly, we can define infinite limits as $x\to −\infty$.

Try It

Find $\underset{x\to \infty }{\lim}3x^2$.

Show Solution

$\infty$

Try It