Chapter7(0211)

Chapter7(0211) - Ch7/MATH0211/YMC/2009-10 1 Chapter 7...

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Unformatted text preview: Ch7/MATH0211/YMC/2009-10 1 Chapter 7. Integration 7.1. Indefinite Integrals We have studied how to differentiate a given function f ( x ) in the last two chapters. Starting with this section, we would like to know what function we diffferentiate to get f ( x ) , or equivalently, we would like to find a function F whose derivative is the given function f . If such a function F exists, it is called the antiderivative of f : Definition 7.1 An antiderivative of the function f is a function F such that F ( x ) = f ( x ) For example, x 2 is the antiderivative of 2 x as d dx x 2 = 2 x . However, it is not the only antiderivative of 2 x : d dx ( x 2 + 1) = 2 x and d dx ( x 2- 3) = 2 x. In fact, any function of the form x 2 + C , for any constant C , gives an antiderivative of 2 x . It follows that 2 x has infinitely many antiderivatives, and any two of them differ only by a constant. We can refer to x 2 + C the most general antiderivative of 2 x and denote it by R 2 xdx , that is, Z 2 xdx = x 2 + C. This is the called an indefinite integral of 2 x . The symbol R is called the integral sign , the function 2 x is called the integrand and C is the constant of integration . The dx is part of the integral notation and indicates the variable involved. If F is any antiderivative of f , then Z f ( x ) dx = F ( x ) + C where C is a constant Here are some important properties of indefinite integrals: Properties of the Indefinite Integration 1. Z k dx = kx + C where k is constant. 2. Z kf ( x ) dx = k Z f ( x ) dx where k is constant. 3. Z ( f ( x ) ± g ( x )) dx = Z f ( x ) dx ± Z g ( x )) dx . Ch7/MATH0211/YMC/2009-10 2 To actually compute indefinite integrals we begin with some of the basic integration formulas: Basic Integration Formulas 1. Z x n dx = x n +1 n + 1 + C for n 6 =- 1 . 2. Z 1 x dx = Z dx x = ln | x | + C . 3. Z e x dx = e x + C Example 7.1 Find the following indefinite integrals: ( a ) Z (3 + 5 x- x 4 ) dx ( b ) Z 1 √ t dt ( c ) Z x 3- 1 x + 4 e x dx Solution . (a) Z (3 + 5 x- x 4 ) dx = Z 3 dx + Z 5 xdx- Z x 4 dx = (3 x + C 1 ) + 5 x 2 2 + C 2- x 5 5 + C 3 = 3 x + 5 2 x 2 + 1 5 x 5 + ( C 1 + C 2- C 3 ) For convenience, we will replace the constant C 1 + C 2- C 3 by a single constant C . We then have Z (3 + 5 x- x 4 ) dx = 3 x + 5 2 x 2 + 1 5 x 5 + C....
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This note was uploaded on 01/02/2012 for the course MATH 0211 taught by Professor Chan during the Spring '10 term at HKU.

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Chapter7(0211) - Ch7/MATH0211/YMC/2009-10 1 Chapter 7...

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