CMII_CH1 - CC: AD [ 2011-12 ] 243: College Mathematics II...

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Unformatted text preview: CC: AD [ 2011-12 ] 243: College Mathematics II — Notes — Chapter 1: Integration and its Applications [ 3 Lessons ] ª 1.1 Indefinite Integration ª 1.2 Definite Integration ª 1.3 Applications of Integration ª 1.4 Improper Integration Ch1: Integration and its Applications [ LN1-1 ] q 1.1 Indefinite Integration Function Differentiation Derivative f ( x ) Explicit / Logarithmic f ′ ( x ) f ( x , y ) = Implicit dy dx ¤ x = f ( t ) y = g ( t ) Parametric dy dx Integrand Integration Integral Indefinite ∫ f ( x ) dx f ( x ) Definite ∫ b a f ( x ) dx Improper ∫ ∞ a f ( x ) dx [ Q: ] Find y = F ( x ) such that dy dx = d dx F ( x ) = x . [ A: ] Since d dx [ x 2 ] = 2 x , we have d dx 1 2 x 2 = 1 2 d dx [ x 2 ] = 1 2 ( 2 x ) = x . In fact, d dx [ 1 2 x 2 + C ] = x , where C is any constant (independent of x ). Hence y = F ( x ) = 1 2 x 2 + C . Ø Primitive Function, Indefinite Integral, Integrand, Indefinite Integration For a given function f ( x ) , if F ( x ) is a function such that d dx [ F ( x )] = f ( x ) then F ( x ) is called the primitive function of f ( x ) . The indefinite integral of a function f ( x ) , denoted by ∫ f ( x ) dx , is defined to be a collection of all primitives of f ( x ) . If d dx [ F ( x )] = f ( x ) then ∫ f ( x ) dx = F ( x )+ C where f ( x ) is called the integrand , x is the variable of integration, C is the constant of integration or arbitrary constant, ∫ is the integral sign. The process of finding a function from its derivative, which reverses the operation of differentia- tion, is called Indefinite Integration / Anti-Differentiation . Remark ‹ The indefinite integral ∫ f ( x ) dx is only a notation and it does not mean that f ( x ) is multi- plied by dx . › An indefinite integral is NOT a function, but a collection of functions which differ by only an arbitrary constant C . AD: College Mathematics II [ 2011-12 ] Louis Lau 1 1.1 Indefinite Integration [ LN1-1 ] Example (a) ∫ 3 x 2 dx = x 3 + C ( ∵ d dx x 3 = 3 x 2 ) (b) ∫ x 2 dx = 1 3 x 3 + C ( ∵ d dx 1 3 x 3 = x 2 ) (c) ∫ e x dx = e x + C ( ∵ d dx [ e x ] = e x ) (d) ∫ cos x dx = sin x + C ( ∵ d dx [ sin x ] = cos x ) (e) ∫ sin x dx = − cos x + C ( ∵ d dx [ − cos x ] = sin x ) Example 1 Differentiate ln ( x 2 + 1 ) with respect to x . Hence, find ∫ x x 2 + 1 dx . Solution d dx ln ( x 2 + 1 ) = 1 x 2 + 1 d dx x 2 + 1 = 2 x x 2 + 1 . ∵ d dx 1 2 ln ( x 2 + 1 ) = x x 2 + 1 . ∴ ∫ x x 2 + 1 dx = 1 2 ln ( x 2 + 1 )+ C . Example 2 Differentiate x ln x with respect to x . Hence, find ∫ ln x dx . Solution d dx [ x ln x ] = ( 1 ) ln x + x ( 1 x ) = ln x + 1. ∵ d dx [ x ln x − x ] = ln x . ∴ ∫ ln x dx = x ln x − x + C ....
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This note was uploaded on 02/22/2012 for the course CHEM yscn0027 taught by Professor Drtong during the Fall '10 term at HKU.

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CMII_CH1 - CC: AD [ 2011-12 ] 243: College Mathematics II...

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