Skills Review for Green’s Theorem, Divergence and Curl, and Surface Integrals

Learning Outcomes

Apply basic derivative rules
Apply the chain rule together with the power and product rule
Find the general antiderivative of a given function

In the section about Green’s Theorem, we will learn how Green’s Theorem is an extension of the Fundamental Theorem of Calculus to two dimensions. Then, we will learn about divergence and curl, two important operations on vector fields. Lastly, we will explore how to integrate over a surface. Here we will review basic differentiation techniques and basic integration techniques.

Basic Derivative Rules

(See Module 6, Skills Review for Line Integrals and Conservative Vector Fields)

The Chain Rule

(See Module 6, Skills Review for Line Integrals and Conservative Vector Fields)

Apply Basic Integration Techniques

(also in Module 5, Skills Review for Double and Triple Integrals)

Indefinite Integrals

Definition

Given a function $f$ , the indefinite integral of $f$ , denoted

\int f (x) d x

$\displaystyle\int f(x) dx$ ,

is the most general antiderivative of $f$ . If $F$ is an antiderivative of $f$ , then

\int f (x) d x = F (x) + C

$\displaystyle\int f(x) dx=F(x)+C$

The expression $f(x)$ is called the integrand and the variable $x$ is the variable of integration.

Power Rule for Integrals

For $n \ne −1$ ,

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

$\displaystyle\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$

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.

Integration Formulas
Differentiation Formula	Indefinite Integral
$\frac{d}{dx}(k)=0$	$\displaystyle\int kdx=\displaystyle\int kx^0 dx=kx+C$
$\frac{d}{dx}(x^n)=nx^{n-1}$	$\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}+C$ for $n\ne −1$
$\frac{d}{dx}(\ln \|x\|)=\frac{1}{x}$	$\displaystyle\int \frac{1}{x}dx=\ln \|x\|+C$
$\frac{d}{dx}(e^x)=e^x$	$\displaystyle\int e^x dx=e^x+C$
$\frac{d}{dx}(\sin x)= \cos x$	$\displaystyle\int \cos x dx= \sin x+C$
$\frac{d}{dx}(\cos x)=− \sin x$	$\displaystyle\int \sin x dx=− \cos x+C$
$\frac{d}{dx}(\tan x)= \sec^2 x$	$\displaystyle\int \sec^2 x dx= \tan x+C$
$\frac{d}{dx}(\csc x)=−\csc x \cot x$	$\displaystyle\int \csc x \cot x dx=−\csc x+C$
$\frac{d}{dx}(\sec x)= \sec x \tan x$	$\displaystyle\int \sec x \tan x dx= \sec x+C$
$\frac{d}{dx}(\cot x)=−\csc^2 x$	$\displaystyle\int \csc^2 x dx=−\cot x+C$
$\frac{d}{dx}( \sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}$	$\displaystyle\int \frac{1}{\sqrt{1-x^2}} dx= \sin^{-1} x+C$
$\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}$	$\displaystyle\int \frac{1}{1+x^2} dx= \tan^{-1} x+C$
$\frac{d}{dx}(\sec^{-1} \|x\|)=\frac{1}{x\sqrt{x^2-1}}$	$\displaystyle\int \frac{1}{x\sqrt{x^2-1}} dx= \sec^{-1} \|x\|+C$

Properties of Indefinite Integrals

Let $F$ and $G$ be antiderivatives of $f$ and $g$ , respectively, and let $k$ be any real number.

Sums and Differences

\int (f (x) \pm g (x)) d x = F (x) \pm G (x) + C

$\displaystyle\int (f(x) \pm g(x)) dx=F(x) \pm G(x)+C$

Constant Multiples

\int k f (x) d x = k F (x) + C

$\displaystyle\int kf(x) dx=kF(x)+C$

Example: Evaluating Indefinite Integrals

Evaluate each of the following indefinite integrals:

$\displaystyle\int (5x^3-7x^2+3x+4) dx$
$\displaystyle\int \frac{x^2+4\sqrt[3]{x}}{x} dx$
$\displaystyle\int \frac{4}{1+x^2} dx$
$\displaystyle\int \tan x \cos x dx$

Show Solution

Using the properties of indefinite integrals, we can integrate each of the four terms in the integrand separately. We obtain
$\displaystyle\int (5x^3-7x^2+3x+4) dx=\displaystyle\int 5x^3 dx-\displaystyle\int 7x^2 dx+\displaystyle\int 3x dx+\displaystyle\int 4 dx$

From the Constant Multiples property of indefinite integrals, each coefficient can be written in front of the integral sign, which gives

$\displaystyle\int 5x^3 dx-\displaystyle\int 7x^2 dx+\displaystyle\int 3x dx+\displaystyle\int 4 dx=5\displaystyle\int x^3 dx-7\displaystyle\int x^2 dx+3\displaystyle\int x dx+4\displaystyle\int 1 dx$

Using the power rule for integrals, we conclude that

$\displaystyle\int (5x^3-7x^2+3x+4) dx=\frac{5}{4}x^4-\frac{7}{3}x^3+\frac{3}{2}x^2+4x+C$
Rewrite the integrand as
$\frac{x^2+4\sqrt[3]{x}}{x}=\frac{x^2}{x}+\frac{4\sqrt[3]{x}}{x}$

Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have

$\begin{array}{ll} \displaystyle\int (x+\frac{4}{x^{2/3}}) dx & =\displaystyle\int x dx+4\displaystyle\int x^{-2/3} dx \\ & =\frac{1}{2}x^2+4\frac{1}{(\frac{-2}{3})+1}x^{(-2/3)+1}+C \\ & =\frac{1}{2}x^2+12x^{1/3}+C \end{array}$
Using the properties of indefinite integrals, write the integral as
$4\displaystyle\int \frac{1}{1+x^2} dx$ .

Then, use the fact that $\tan^{-1} (x)$ is an antiderivative of $\frac{1}{1+x^2}$ to conclude that

$\displaystyle\int \frac{4}{1+x^2} dx=4 \tan^{-1} (x)+C$
Rewrite the integrand as
$\tan x \cos x=\frac{ \sin x}{ \cos x} \cos x= \sin x$ .

Therefore,

$\displaystyle\int \tan x \cos x dx=\displaystyle\int \sin x dx=− \cos x+C$

Try It

Evaluate $\displaystyle\int (4x^3-5x^2+x-7) dx$

Hint

Show Solution

$x^4-\frac{5}{3}x^3+\frac{1}{2}x^2-7x+C$

Vector Calculus: Prerequisite Material