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 [latex]f[/latex], the indefinite integral of [latex]f[/latex], denoted
is the most general antiderivative of [latex]f[/latex]. If [latex]F[/latex] is an antiderivative of [latex]f[/latex], then
The expression [latex]f(x)[/latex] is called the integrand and the variable [latex]x[/latex] is the variable of integration.
Power Rule for Integrals
For [latex]n \ne −1[/latex],
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.
Differentiation Formula | Indefinite Integral |
---|---|
[latex]\frac{d}{dx}(k)=0[/latex] | [latex]\displaystyle\int kdx=\displaystyle\int kx^0 dx=kx+C[/latex] |
[latex]\frac{d}{dx}(x^n)=nx^{n-1}[/latex] | [latex]\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}+C[/latex] for [latex]n\ne −1[/latex] |
[latex]\frac{d}{dx}(\ln |x|)=\frac{1}{x}[/latex] | [latex]\displaystyle\int \frac{1}{x}dx=\ln |x|+C[/latex] |
[latex]\frac{d}{dx}(e^x)=e^x[/latex] | [latex]\displaystyle\int e^x dx=e^x+C[/latex] |
[latex]\frac{d}{dx}(\sin x)= \cos x[/latex] | [latex]\displaystyle\int \cos x dx= \sin x+C[/latex] |
[latex]\frac{d}{dx}(\cos x)=− \sin x[/latex] | [latex]\displaystyle\int \sin x dx=− \cos x+C[/latex] |
[latex]\frac{d}{dx}(\tan x)= \sec^2 x[/latex] | [latex]\displaystyle\int \sec^2 x dx= \tan x+C[/latex] |
[latex]\frac{d}{dx}(\csc x)=−\csc x \cot x[/latex] | [latex]\displaystyle\int \csc x \cot x dx=−\csc x+C[/latex] |
[latex]\frac{d}{dx}(\sec x)= \sec x \tan x[/latex] | [latex]\displaystyle\int \sec x \tan x dx= \sec x+C[/latex] |
[latex]\frac{d}{dx}(\cot x)=−\csc^2 x[/latex] | [latex]\displaystyle\int \csc^2 x dx=−\cot x+C[/latex] |
[latex]\frac{d}{dx}( \sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}[/latex] | [latex]\displaystyle\int \frac{1}{\sqrt{1-x^2}} dx= \sin^{-1} x+C[/latex] |
[latex]\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}[/latex] | [latex]\displaystyle\int \frac{1}{1+x^2} dx= \tan^{-1} x+C[/latex] |
[latex]\frac{d}{dx}(\sec^{-1} |x|)=\frac{1}{x\sqrt{x^2-1}}[/latex] | [latex]\displaystyle\int \frac{1}{x\sqrt{x^2-1}} dx= \sec^{-1} |x|+C[/latex] |
Properties of Indefinite Integrals
Let [latex]F[/latex] and [latex]G[/latex] be antiderivatives of [latex]f[/latex] and [latex]g[/latex], respectively, and let [latex]k[/latex] be any real number.
Sums and Differences
Constant Multiples
Example: Evaluating Indefinite Integrals
Evaluate each of the following indefinite integrals:
- [latex]\displaystyle\int (5x^3-7x^2+3x+4) dx[/latex]
- [latex]\displaystyle\int \frac{x^2+4\sqrt[3]{x}}{x} dx[/latex]
- [latex]\displaystyle\int \frac{4}{1+x^2} dx[/latex]
- [latex]\displaystyle\int \tan x \cos x dx[/latex]
Try It
Evaluate [latex]\displaystyle\int (4x^3-5x^2+x-7) dx[/latex]
Try It
Candela Citations
- Calculus Volume 1. Provided by: Lumen Learning. Located at: https://courses.lumenlearning.com/calculus1/. License: CC BY: Attribution
- Calculus Volume 2. Provided by: Lumen Learning. Located at: https://courses.lumenlearning.com/calculus2/. License: CC BY: Attribution