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Unformatted text preview: Unit 2: The Antiderivative Definition 2.1. F ( x ) is said to be an antiderivative of f ( x ) on an interval if F ( x ) = f ( x ) or equivalently d dx ( F ( x )) = f ( x ) for every value of x on the interval. Example 1 . Find an antiderivative of x 3 Solution: The exponent on x is 3, so in searching for an antiderivative (thinking of power rule in reverse) it makes sense to begin our search with x 4 . Testing this hypothesis we see that d dx ( x 4 ) = 4 x 3 We require x 3 which we observe to be 1 4 (4 x 3 ). With this in mind we see that d dx ( 1 4 ( x 4 )) = x 3 Definition 2.2. Notice that we defined an antiderivative as opposed to the antiderivative of a function. The reason for this can be demonstrated quite easily as follows: First we see for example that d dx ( 1 4 ( x 4 )) = x 3 1 and that d dx ( 1 4 ( x 4 ) + 3) = x 3 + 0 = x 3 and that in general d dx ( 1 4 ( x 4 ) + C ) = x 3 , where C is any arbitrary constant For this reason we now define the general antiderivative of a function f ( x ) to be the sum of an antiderivative of f ( x ) and an arbitrary constant C. Example 2 . Find the general antiderivative, F ( x ), of f ( x ) = x 4 Solution: After some thought (see example 1) we see that 1 5 ( x 5 ) is an antiderivative of x 4 . In light of definition 2 we then have F ( x ) = 1 5 ( x 5 ) + C is the general antiderivative of x 4 . We now introduce some general rules and notation: The general antiderivative of f ( x ) will be denoted Z f ( x ) dx where the sybol R is called the integral sign, f ( x ) dx is called the integrand , and the presence of dx signifies that x is the variable of integration , i.e. we are integrating with respect to x. We will now refer to the operation of finding an antiderivative as integration . Example 3 . ( a ) Z x 3 dx = 1 4 x 4 + C . 2 ( b ) Z x 4 dx = 1 5 x 5 + C . Rules of Integration R x n dx = 1 n +1 ( x n +1 ) + C for n not equal to 1 . This is known as the Power Rule for integration. The special case when n = 1 is described by R x 1 dx = R 1 x dx = ln  x  + C . The use of these two rules can be easily demonstrated. Example 4 . Solve the following integral Z x 17 dx Solution: Since the power on x is not 1 we may apply the power rule with n = 17 to obtain Z x 1 7 dx = 1 18 x 1 8 + C Example 5 . Solve the following integral Z √ x 3 dx 3 Solution: Once again we invoke the power rule, this time with n = 3 2 ....
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This note was uploaded on 04/15/2010 for the course MATH 20C taught by Professor Lit during the Spring '10 term at UCLA.
 Spring '10
 lit
 Calculus, Derivative

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