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

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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
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Ch1: Integration and its Applications [ LN1-1 ] q 1.1 Indefinite Integration Function Differentiation Derivative f ( x ) Explicit / Logarithmic f ( x ) f ( x , y ) = 0 Implicit d y d x ¤ x = f ( t ) y = g ( t ) Parametric d y d x Integrand Integration Integral Indefinite f ( x ) d x f ( x ) Definite b a f ( x ) d x Improper a f ( x ) d x [ Q: ] Find y = F ( x ) such that d y d x = d d x F ( x ) = x . [ A: ] Since d d x [ x 2 ] = 2 x , we have d d x 1 2 x 2 = 1 2 d d x [ x 2 ] = 1 2 ( 2 x ) = x . In fact, d d x [ 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 d x [ 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 ) d x , is defined to be a collection of all primitives of f ( x ) . If d d x [ F ( x )] = f ( x ) then f ( x ) d x = 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 ) d x is only a notation and it does not mean that f ( x ) is multi- plied by d x . 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
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1.1 Indefinite Integration [ LN1-1 ] Example (a) 3 x 2 d x = x 3 + C ( d d x x 3 = 3 x 2 ) (b) x 2 d x = 1 3 x 3 + C ( d d x 1 3 x 3 = x 2 ) (c) e x d x = e x + C ( d d x [ e x ] = e x ) (d) cos x d x = sin x + C ( d d x [ sin x ] = cos x ) (e) sin x d x = cos x + C ( d d x [ cos x ] = sin x ) Example 1 Differentiate ln ( x 2 + 1 ) with respect to x . Hence, find x x 2 + 1 d x . Solution d d x ln ( x 2 + 1 ) = 1 x 2 + 1 d d x x 2 + 1 = 2 x x 2 + 1 . d d x 1 2 ln ( x 2 + 1 ) = x x 2 + 1 . x x 2 + 1 d x = 1 2 ln ( x 2 + 1 ) + C . Example 2 Differentiate x ln x with respect to x . Hence, find ln x d x . Solution d d x [ x ln x ] = ( 1 ) ln x + x ( 1 x ) = ln x + 1. d d x [ x ln x x ] = ln x . ln x d x = x ln x x + C . AD: College Mathematics II [ 2011-12 ] Louis Lau 2
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1.1 Indefinite Integration [ LN1-1 ] v Standard Indefinite Integrals f ( x ) f ( x ) d x f ( x ) f ( x ) d x x n , n ̸ = 1 1 n + 1 x n + 1 + C ( Ax + B ) n , n ̸ = 1 1 A ( n + 1 ) ( Ax + B ) n + 1 + C x 1 = 1 x ln | x | + C ( Ax + B ) 1 = 1 Ax + B 1 A ln | Ax + B | + C e x e x + C e Ax + B 1 A e Ax + B + C a x 1 ln a a x + C a Ax + B 1 A ln a a Ax + B + C sin x cos x + C sin ( Ax + B ) 1 A cos ( Ax + B ) + C cos x sin x + C cos ( Ax + B ) 1 A sin ( Ax + B ) + C 1 p a 2 x 2 sin 1 x a + C 1 p a 2 ( Ax + B ) 2 1 A sin 1 Ax + B a + C 1 a 2 + x 2 1 a tan 1 x a + C 1 a 2 + ( Ax + B ) 2 1 aA tan 1 Ax + B a + C sec 2 x tan x + C sec 2 ( Ax + B ) 1 A tan ( Ax + B ) + C csc 2 x cot x + C csc 2 ( Ax + B ) 1 A cot ( Ax + B ) + C sec x tan x sec x + C sec ( Ax + B ) tan ( Ax + B ) 1 A sec ( Ax + B ) + C csc x
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